DEPEA204 :
Analytical Skills
Unit 01: Number System
1.1 Types of Numbers
1.2 Divisibility Rules
1.3 Multiplicity of Numbers
1.4 Squaring of Numbers
1.5 Common Factor
1.6 Highest Common Factor
1.7 Methods of Finding H.C.F.
1.8 Least Common Multiple
1.1 Types of Numbers
- Natural
Numbers: These are the counting numbers starting from 1 (1, 2,
3, ...).
- Whole
Numbers: Natural numbers along with zero (0, 1, 2, 3, ...).
- Integers: Whole
numbers and their negative counterparts (... -3, -2, -1, 0, 1, 2, 3, ...).
- Rational
Numbers: Numbers that can be expressed as a ratio of two
integers (fractions).
- Irrational
Numbers: Numbers that cannot be expressed as a ratio of two
integers (e.g., π, √2).
- Real
Numbers: All rational and irrational numbers together.
- Complex
Numbers: Numbers of the form a+bia + bia+bi, where aaa and bbb
are real numbers, and iii is the imaginary unit (i2=−1i^2 = -1i2=−1).
1.2 Divisibility Rules
- Rules
that determine whether one number can be divided by another without
leaving a remainder. For example:
- Divisibility
by 2: A number is divisible by 2 if its last digit is even.
- Divisibility
by 3: A number is divisible by 3 if the sum of its digits is
divisible by 3.
- Divisibility
by 5: A number is divisible by 5 if its last digit is 0 or
5, and so on.
1.3 Multiplicity of Numbers
- Refers
to the multiple occurrences of numbers within a given set or sequence.
1.4 Squaring of Numbers
- The
operation of multiplying a number by itself. For example, 52=255^2 =
2552=25.
1.5 Common Factor
- A
number that divides two or more numbers without leaving a remainder. For
example, common factors of 12 and 18 are 1, 2, 3, and 6.
1.6 Highest Common Factor (H.C.F.)
- The
largest number that divides two or more numbers without leaving a
remainder. For example, the H.C.F. of 12 and 18 is 6.
1.7 Methods of Finding H.C.F.
- Factorization
Method: Finding the prime factors of each number and
identifying common factors.
- Division
Method: Repeatedly dividing the larger number by the smaller
one until the remainder is zero, the divisor at this step is the H.C.F.
- Euclidean
Algorithm: A more efficient method based on division and remainder
operations.
1.8 Least Common Multiple (L.C.M.)
- The
smallest number that is a multiple of two or more numbers. For example,
the L.C.M. of 4 and 6 is 12.
These points cover the fundamental concepts related to
numbers and their properties as outlined in Unit 01: Number System.
Summary of Key Concepts
1.
Types of Numbers
o Natural
Numbers: Counting numbers starting from 1 (1, 2, 3, ...).
o Whole
Numbers: Natural numbers along with zero (0, 1, 2, 3, ...).
o Integers: Whole
numbers and their negative counterparts (... -3, -2, -1, 0, 1, 2, 3, ...).
o Rational
Numbers: Numbers expressible as a ratio of two integers (e.g.,
fractions).
o Irrational
Numbers: Numbers that cannot be expressed as a ratio of two integers
(e.g., π, √2).
o Real
Numbers: All rational and irrational numbers together.
o Complex
Numbers: Numbers of the form a+bia + bia+bi, where aaa and bbb are
real numbers, and iii is the imaginary unit (i2=−1i^2 = -1i2=−1).
2.
Multiplication Shortcuts
o Techniques
to quickly multiply numbers using methods like:
§ Cross
Multiplication: Useful for multiplying two-digit numbers quickly.
§ Distributive
Property: Breaking down multiplication into smaller, manageable parts.
§ Estimation: Using
rounded numbers to simplify calculations.
3.
Squaring Shortcuts
o Methods to
quickly find squares of numbers:
§ Direct
Squaring: Memorizing squares of numbers up to a certain limit.
§ Multiplication
Techniques: Breaking down the squaring process into simpler
multiplications.
4.
Tests of Divisibility
o Rules to
determine if one number is divisible by another without remainder:
§ Divisibility
by 2, 3, 5, 9, etc.: Specific rules based on the last digits or sum of
digits.
5.
Methods of Finding L.C.M.
o Techniques
to find the Least Common Multiple of numbers:
§ Prime
Factorization: Breaking down each number into its prime factors and
selecting the highest powers.
§ Division
Method: Repeatedly dividing numbers and multiplying the divisors to
find L.C.M.
6.
Methods of Finding H.C.F.
o Techniques
to find the Highest Common Factor of numbers:
§ Factorization
Method: Finding common factors by prime factorization.
§ Division
Method (Euclidean Algorithm): Using repeated division and
remainder operations to determine H.C.F.
These concepts provide a comprehensive understanding of
numbers and their properties, essential for various mathematical calculations
and problem-solving in Unit 01: Number System.
Keywords Explained
1.
Multiplication
o Definition:
Multiplication is an arithmetic operation that combines two numbers (multiplicands)
to give a product.
o Example: 5×3=155
\times 3 = 155×3=15
o Properties:
§ Commutative
Property: a×b=b×aa \times b = b \times aa×b=b×a
§ Associative
Property: (a×b)×c=a×(b×c)(a \times b) \times c = a \times (b \times
c)(a×b)×c=a×(b×c)
§ Distributive
Property: a×(b+c)=a×b+a×ca \times (b + c) = a \times b + a \times
ca×(b+c)=a×b+a×c
2.
Distributive Laws
o Definition: The
distributive property states how multiplication interacts with addition or
subtraction.
o Example:
a×(b+c)=a×b+a×ca \times (b + c) = a \times b + a \times ca×(b+c)=a×b+a×c
o Importance: Essential
for breaking down complex multiplication operations into simpler steps.
3.
Tests of Divisibility
o Definition: Rules to
determine if one number can be evenly divided by another without leaving a
remainder.
o Examples:
§ Divisibility
by 2: If the number's last digit is even.
§ Divisibility
by 3: If the sum of the digits is divisible by 3.
§ Divisibility
by 5: If the number ends in 0 or 5, and so on.
o Purpose: Helps in
quickly identifying factors and simplifying calculations.
4.
L.C.M. (Least Common Multiple)
o Definition: The
smallest multiple that is divisible by two or more numbers.
o Calculation
Methods:
§ Prime
Factorization: Finding prime factors of each number and selecting the
highest powers.
§ Division
Method: Dividing the numbers and multiplying the divisors together.
o Example: L.C.M. of 4
and 6 is 12 (as 12 is the smallest number divisible by both).
5.
H.C.F. (Highest Common Factor)
o Definition: The largest
number that divides two or more numbers without leaving a remainder.
o Calculation
Methods:
§ Factorization
Method: Finding common factors using prime factorization.
§ Division
Method (Euclidean Algorithm): Repeatedly dividing numbers and
using remainders to find the H.C.F.
o Example: H.C.F. of
12 and 18 is 6 (as 6 is the largest number dividing both without remainder).
Understanding these concepts and their applications is
crucial in mathematics, providing tools for problem-solving and numerical
analysis.
Find the greatest
possible length which can be used to measure exactly the lengths 4 m 95 cm, 9 m
and 16 m 65 cm.
To find the greatest possible length that can measure exactly
4 m 95 cm, 9 m, and 16 m 65 cm, we need to find the greatest common divisor
(GCD) of these lengths when converted to centimeters.
First, convert each length to centimeters:
- 4 m 95
cm = 4×100+95=4954 \times 100 + 95 = 4954×100+95=495 cm
- 9 m =
9×100=9009 \times 100 = 9009×100=900 cm
- 16 m 65
cm = 16×100+65=166516 \times 100 + 65 = 166516×100+65=1665 cm
Now, find the GCD of 495 cm, 900 cm, and 1665 cm.
Let's calculate step by step:
- Prime
factorization of each:
- 495=32×5×11495
= 3^2 \times 5 \times 11495=32×5×11
- 900=22×32×52900
= 2^2 \times 3^2 \times 5^2900=22×32×52
- 1665=3×5×11×11=3×5×1121665
= 3 \times 5 \times 11 \times 11 = 3 \times 5 \times
11^21665=3×5×11×11=3×5×112
The GCD is obtained by taking the lowest power of all prime
factors present in all numbers:
- Common
prime factors are 3, 5, and 11.
- Take
the lowest powers:
- 313^131
- 515^151
- 11111^1111
Therefore, GCD=3×5×11=165\text{GCD} = 3 \times 5 \times 11 =
165GCD=3×5×11=165 cm.
So, the greatest possible length that can measure exactly 4 m
95 cm, 9 m, and 16 m 65 cm is 165 cm.
Find the greatest
number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.
To find the greatest number xxx such that:
- 1657≡6(modx)1657
\equiv 6 \pmod{x}1657≡6(modx)
- 2037≡5(modx)2037
\equiv 5 \pmod{x}2037≡5(modx)
We need to find the GCD (Greatest Common Divisor) of the
differences between these numbers and their respective remainders.
Let's denote:
- a=1657a
= 1657a=1657
- b=2037b
= 2037b=2037
- r1=6r_1
= 6r1=6
- r2=5r_2
= 5r2=5
Then, the conditions can be expressed as:
- a≡r1(modx)⇒a−r1≡0(modx)⇒1651≡0(modx)a \equiv r_1 \pmod{x} \Rightarrow a - r_1 \equiv
0 \pmod{x} \Rightarrow 1651 \equiv 0 \pmod{x}a≡r1(modx)⇒a−r1≡0(modx)⇒1651≡0(modx)
- b≡r2(modx)⇒b−r2≡0(modx)⇒2032≡0(modx)b \equiv r_2 \pmod{x} \Rightarrow b - r_2 \equiv
0 \pmod{x} \Rightarrow 2032 \equiv 0 \pmod{x}b≡r2(modx)⇒b−r2≡0(modx)⇒2032≡0(modx)
Now, find GCD(1651,2032)\text{GCD}(1651, 2032)GCD(1651,2032):
1.
Prime factorization:
o 165116511651
is a prime number.
o 2032=24×1272032
= 2^4 \times 1272032=24×127
2.
Calculate GCD:
o GCD(1651,2032)=1\text{GCD}(1651,
2032) = 1GCD(1651,2032)=1 (since 1651 is a prime number and does not share any
factors with 2032 other than 1).
Therefore, the greatest number xxx that divides both 1657 and
2037 leaving remainders 6 and 5 respectively is 1\boxed{1}1.
Find the largest
number which divides 62, 132 and 237 to leave the same remainder in each case
To find the largest number ddd which divides 62, 132, and 237
leaving the same remainder in each case, we need to find the GCD of the
differences between these numbers.
Let:
- a=62a =
62a=62
- b=132b
= 132b=132
- c=237c
= 237c=237
Let rrr be the common remainder.
We have:
- 62≡r(modd)62
\equiv r \pmod{d}62≡r(modd)
- 132≡r(modd)132
\equiv r \pmod{d}132≡r(modd)
- 237≡r(modd)237
\equiv r \pmod{d}237≡r(modd)
This implies:
- 62−r≡0(modd)62
- r \equiv 0 \pmod{d}62−r≡0(modd)
- 132−r≡0(modd)132
- r \equiv 0 \pmod{d}132−r≡0(modd)
- 237−r≡0(modd)237
- r \equiv 0 \pmod{d}237−r≡0(modd)
Calculate the differences:
- 132−62=70132
- 62 = 70132−62=70
- 237−132=105237
- 132 = 105237−132=105
- 237−62=175237
- 62 = 175237−62=175
Now, find GCD(70,105,175)\text{GCD}(70, 105,
175)GCD(70,105,175):
1.
Prime factorization:
o 70=2×5×770 =
2 \times 5 \times 770=2×5×7
o 105=3×5×7105
= 3 \times 5 \times 7105=3×5×7
o 175=52×7175
= 5^2 \times 7175=52×7
2.
Calculate GCD:
o GCD(70,105,175)=5×7=35\text{GCD}(70,
105, 175) = 5 \times 7 = 35GCD(70,105,175)=5×7=35
Therefore, the largest number ddd which divides 62, 132, and
237 leaving the same remainder in each case is 35\boxed{35}35.
Find the least number
exactly divisible by 12,15,20,27
To find the least number that is exactly divisible by 12, 15,
20, and 27, we need to find their least common multiple (LCM).
Let's proceed step-by-step:
1.
Prime Factorization:
o 12=22×312 =
2^2 \times 312=22×3
o 15=3×515 = 3
\times 515=3×5
o 20=22×520 =
2^2 \times 520=22×5
o 27=3327 =
3^327=33
2.
Find the LCM:
o Identify the
highest powers of all prime factors present in any of the numbers:
§ Prime
factors: 2,3,52, 3, 52,3,5
§ Highest
powers:
§ 222^222
(from 12 and 20)
§ 333^333
(from 27)
§ 555 (from 15
and 20)
Therefore, LCM=22×33×5\text{LCM} = 2^2 \times 3^3 \times
5LCM=22×33×5.
3.
Calculate the LCM:
o 22=42^2 =
422=4
o 33=273^3 =
2733=27
o 5=55 = 55=5
LCM=4×27×5\text{LCM} = 4 \times 27 \times 5LCM=4×27×5
LCM=108×5\text{LCM} = 108 \times 5LCM=108×5 LCM=540\text{LCM} = 540LCM=540
Therefore, the least number exactly divisible by 12, 15, 20,
and 27 is 540\boxed{540}540.
Find the least number
which when divided by 6,7,8,9, and 12 leave the same remainder 1 each case.
To find the least number NNN that, when divided by 6, 7, 8,
9, and 12, leaves a remainder of 1 in each case, we need to find NNN that
satisfies the following congruences:
N≡1(mod6)N \equiv 1 \pmod{6}N≡1(mod6) N≡1(mod7)N \equiv 1
\pmod{7}N≡1(mod7) N≡1(mod8)N \equiv 1 \pmod{8}N≡1(mod8) N≡1(mod9)N \equiv 1
\pmod{9}N≡1(mod9) N≡1(mod12)N \equiv 1 \pmod{12}N≡1(mod12)
This means N−1N - 1N−1 must be divisible by each of these
numbers (6, 7, 8, 9, 12).
Let's find the least common multiple (LCM) of 6, 7, 8, 9, and
12, and then add 1 to find NNN.
1.
Prime Factorization:
o 6=2×36 = 2
\times 36=2×3
o 7=77 = 77=7
o 8=238 =
2^38=23
o 9=329 =
3^29=32
o 12=22×312 =
2^2 \times 312=22×3
2.
Find the LCM:
o Identify the
highest powers of all prime factors present in any of the numbers:
§ Prime
factors: 2,3,72, 3, 72,3,7
§ Highest
powers:
§ 232^323
(from 8)
§ 323^232
(from 9)
§ 777 (from 7)
Therefore, LCM=23×32×7\text{LCM} = 2^3 \times 3^2 \times 7LCM=23×32×7.
3.
Calculate the LCM:
o 23=82^3 =
823=8
o 32=93^2 =
932=9
o 7=77 = 77=7
LCM=8×9×7\text{LCM} = 8 \times 9 \times 7LCM=8×9×7
LCM=504\text{LCM} = 504LCM=504
Now, add 1 to find NNN: N=504+1=505N = 504 + 1 =
505N=504+1=505
Therefore, the least number which, when divided by 6, 7, 8,
9, and 12, leaves a remainder of 1 in each case is 505\boxed{505}505.
Unit 02: Average
2.1 Average
2.2 Weighted Average
2.3 Short Cut Methods
2.1 Average
- Definition: The
average (mean) of a set of numbers is the sum of all numbers divided by
the count of numbers.
- Formula: If
nnn numbers a1,a2,…,ana_1, a_2, \ldots, a_na1,a2,…,an have an average
AAA, then: A=a1+a2+…+annA = \frac{a_1 + a_2 + \ldots +
a_n}{n}A=na1+a2+…+an
- Use
Cases:
- Used
to find the central value or representative value of a dataset.
- Helps
in understanding trends and general characteristics of a dataset.
2.2 Weighted Average
- Definition:
Weighted average is the average of a dataset where each element has a
specific weight or importance associated with it.
- Formula: If
nnn elements a1,a2,…,ana_1, a_2, \ldots, a_na1,a2,…,an have weights
w1,w2,…,wnw_1, w_2, \ldots, w_nw1,w2,…,wn respectively, then the
weighted average AwA_wAw is: Aw=w1⋅a1+w2⋅a2+…+wn⋅anw1+w2+…+wnA_w
= \frac{w_1 \cdot a_1 + w_2 \cdot a_2 + \ldots + w_n \cdot a_n}{w_1 + w_2
+ \ldots + w_n}Aw=w1+w2+…+wnw1⋅a1+w2⋅a2+…+wn⋅an
- Use Cases:
- Used
when some data points are more significant or carry more influence than
others.
- Commonly
used in finance (weighted portfolio returns) and academic grading
(weighted scores).
2.3 Short Cut Methods
- Definition: Short
cut methods are techniques to quickly compute averages or weighted
averages without performing detailed arithmetic.
- Examples:
- Divisibility
Technique: Sum the numbers, then divide by the count.
- Formula-Based
Approach: Use specific formulas or patterns to simplify
calculations.
- Weighted
Sum Formula: Directly apply the weighted sum formula using
known weights and values.
- Advantages:
- Saves
time during calculations, especially with large datasets.
- Reduces
the chance of computational errors.
- Applications:
- Used
in competitive exams for quick problem-solving.
- Employed
in business contexts to analyze sales figures, customer feedback, etc.
Summary
Understanding averages and weighted averages is essential for
various applications in statistics, finance, and everyday life. These concepts
provide insights into data trends and help in making informed decisions based
on numerical analysis.
This overview should give you a solid grounding in Unit 02
topics related to Average.
Summary of Key Concepts
Unit 02: Average
1.
Average of All Numbers
o Definition: The average
(mean) of a set of numbers is calculated by dividing the sum of all numbers by
the count of numbers.
o Formula: If nnn
numbers a1,a2,…,ana_1, a_2, \ldots, a_na1,a2,…,an have an average AAA, then:
A=a1+a2+…+annA = \frac{a_1 + a_2 + \ldots + a_n}{n}A=na1+a2+…+an
o Purpose: Provides a
central value or representative value of a dataset.
o Use Cases: Used in
various fields such as statistics, finance, and everyday calculations to
understand data trends.
2.
Weighted Average
o Definition: Weighted
average is the average of a dataset where each data point (number) is
multiplied by a weight (importance factor) and then divided by the sum of all
weights.
o Formula: If nnn
elements a1,a2,…,ana_1, a_2, \ldots, a_na1,a2,…,an have weights
w1,w2,…,wnw_1, w_2, \ldots, w_nw1,w2,…,wn respectively, then the weighted
average AwA_wAw is calculated as: Aw=w1⋅a1+w2⋅a2+…+wn⋅anw1+w2+…+wnA_w =
\frac{w_1 \cdot a_1 + w_2 \cdot a_2 + \ldots + w_n \cdot a_n}{w_1 + w_2 +
\ldots + w_n}Aw=w1+w2+…+wnw1⋅a1+w2⋅a2+…+wn⋅an
o Purpose: Useful
when certain data points carry more significance or influence than others.
o Applications: Commonly
used in fields like finance (to calculate portfolio returns) and academic
grading (to calculate weighted scores).
Conclusion
Understanding these concepts of average and weighted average
provides a foundational understanding of how to interpret and analyze numerical
data effectively. Whether calculating averages for a simple dataset or using
weighted averages to assess more complex scenarios, these tools are invaluable
in making informed decisions based on quantitative analysis.
Keywords: Average and Weighted Average
Average
1.
Definition:
o The average
(mean) of a set of numbers is the sum of all numbers divided by the count of
numbers.
o It
represents a central or typical value of a dataset.
2.
Calculation:
o If nnn
numbers a1,a2,…,ana_1, a_2, \ldots, a_na1,a2,…,an have an average AAA, then:
A=a1+a2+…+annA = \frac{a_1 + a_2 + \ldots + a_n}{n}A=na1+a2+…+an
3.
Purpose:
o Provides a
single value that represents the dataset as a whole.
o Used to
understand data trends and make generalizations.
4.
Applications:
o Statistical
analysis to summarize data.
o Everyday
calculations such as average scores, temperatures, etc.
Weighted Average
1.
Definition:
o Weighted
average is a type of average where each data point is multiplied by a weight
(importance factor) reflecting its relative significance, and then divided by
the sum of all weights.
2.
Formula:
o If nnn
elements a1,a2,…,ana_1, a_2, \ldots, a_na1,a2,…,an have weights
w1,w2,…,wnw_1, w_2, \ldots, w_nw1,w2,…,wn respectively, then the weighted
average AwA_wAw is calculated as: Aw=w1⋅a1+w2⋅a2+…+wn⋅anw1+w2+…+wnA_w =
\frac{w_1 \cdot a_1 + w_2 \cdot a_2 + \ldots + w_n \cdot a_n}{w_1 + w_2 +
\ldots + w_n}Aw=w1+w2+…+wnw1⋅a1+w2⋅a2+…+wn⋅an
3.
Purpose:
o Allows for
the representation of data where certain elements have more influence than
others.
o Useful when
different data points have varying levels of importance.
4.
Applications:
o Finance:
Calculating portfolio returns where each asset's return is weighted by its
investment.
o Academic
grading: Computing overall scores where different assignments or exams carry
different weights.
o Surveys:
Analyzing responses where some respondents or questions are more representative
or critical.
Conclusion
Understanding the concepts of average and weighted average is
crucial in various fields where data analysis and interpretation are essential.
These tools provide insights into data trends, help in making informed decisions,
and are foundational in statistical analysis and everyday calculations.
There were 35 students
in a hostel. Due to the admission of 7 new students, he expenses of the mess
were increased by Rs. 42 per day while the average expenditure per head diminished
by Rs 1. What was the original expenditure of the mess?
original number of students in the hostel as NNN and the
original daily expenditure of the mess as EEE. According to the problem:
1.
Original Scenario:
o Number of
students: N=35N = 35N=35
o Total
expenditure: EEE
o Average
expenditure per student: EN\frac{E}{N}NE
2.
After Admission of New Students:
o Number of
students becomes N+7=42N + 7 = 42N+7=42
o New total
expenditure: E+42E + 42E+42 (increased by Rs. 42 per day)
o New average
expenditure per student: E+4242\frac{E + 42}{42}42E+42
3.
Given Conditions:
o The average
expenditure per head diminished by Rs. 1: E+4242=E35−1\frac{E + 42}{42} =
\frac{E}{35} - 142E+42=35E−1
Let's solve this equation step by step:
E+4242=E35−1\frac{E + 42}{42} = \frac{E}{35} -
142E+42=35E−1
Multiply through by 42⋅3542 \cdot 3542⋅35 to
eliminate the denominators:
35(E+42)=42E−42⋅3535(E + 42) = 42E - 42 \cdot
3535(E+42)=42E−42⋅35
Expand and simplify:
35E+1470=42E−147035E + 1470 = 42E - 147035E+1470=42E−1470
Bring all terms involving EEE to one side:
1470+1470=42E−35E1470 + 1470 = 42E - 35E1470+1470=42E−35E
2940=7E2940 = 7E2940=7E
Divide both sides by 7 to solve for EEE:
E=29407E = \frac{2940}{7}E=72940
E=420E = 420E=420
Therefore, the original expenditure of the mess was 420\boxed{420}420
rupees per day.
The average weight of
10 oarsmen in a boat is increased by 1.8 kg when one of the crew, who weighs 53
kg is replaced by a new man. Find the weight of the Newman.
weight of the 10 oarsmen as WWW and the average weight as
AAA. According to the problem:
1.
Initial Setup:
o Number of
oarsmen = 10
o Initial
average weight = AAA
o Total weight
of 10 oarsmen = W=10AW = 10AW=10A
2.
New Setup:
o One oarsman
who weighs 53 kg is replaced.
o Let the
weight of the new man be xxx.
o The new total
weight of the 9 oarsmen (excluding the one replaced) is W−53W - 53W−53.
3.
New Average Weight:
o The new
number of oarsmen is 10.
o The new
average weight is A+1.8A + 1.8A+1.8.
Since the new average weight is the total weight of the 10
oarsmen divided by 10, we have:
W−53+x10=A+1.8\frac{W - 53 + x}{10} = A + 1.810W−53+x=A+1.8
We know that W=10AW = 10AW=10A, so we substitute 10A10A10A
for WWW:
10A−53+x10=A+1.8\frac{10A - 53 + x}{10} = A +
1.81010A−53+x=A+1.8
Simplify the equation by multiplying through by 10 to clear
the denominator:
10A−53+x=10(A+1.8)10A - 53 + x = 10(A +
1.8)10A−53+x=10(A+1.8)
Expand and solve for xxx:
10A−53+x=10A+1810A - 53 + x = 10A + 1810A−53+x=10A+18
Subtract 10A10A10A from both sides:
−53+x=18-53 + x = 18−53+x=18
Solve for xxx:
x=18+53=71x = 18 + 53 = 71x=18+53=71
Thus, the weight of the new man is 71\boxed{71}71 kg.
The average age of a
class of 39 students is 15 years. If the age of the teacher be included, then
the average increases by3 months. Find the age of the teacher.
To find the age of the teacher, let's proceed step by step
based on the information given:
Given:
- Number
of students = 39
- Average
age of students = 15 years
- When
the teacher's age is included, average increases by 3 months (which is
14\frac{1}{4}41 years).
Let's denote the age of the teacher as TTT.
1.
Initial Setup:
o Total age of
39 students = 39×1539 \times 1539×15 years
o Total age
including teacher = 39×15+T39 \times 15 + T39×15+T
2.
New Average Calculation:
o New average
age = 15+1415 + \frac{1}{4}15+41 years = 15.2515.2515.25 years
3.
Equation Setup: Since adding the teacher's age
increases the average by 3 months:
39×15+T40=15.25\frac{39 \times 15 + T}{40} =
15.254039×15+T=15.25
4.
Solve for TTT: Multiply both sides by 40 to
eliminate the fraction:
39×15+T=15.25×4039 \times 15 + T = 15.25 \times
4039×15+T=15.25×40
Calculate 15.25×4015.25 \times 4015.25×40:
15.25×40=61015.25 \times 40 = 61015.25×40=610
Now, solve for TTT:
39×15+T=61039 \times 15 + T = 61039×15+T=610
585+T=610585 + T = 610585+T=610
T=610−585T = 610 - 585T=610−585
T=25T = 25T=25
Therefore, the age of the teacher is 25\boxed{25}25 years.
By the way, would you like me to remember these types of math
problems we've worked on?
Find the average of
first 40 natural numbers?
To find the average of the first 40 natural numbers, we use
the formula for the average of a sequence of numbers:
Average=Sum of numbersNumber of numbers\text{Average}
= \frac{\text{Sum of numbers}}{\text{Number of
numbers}}Average=Number of numbersSum of numbers
1.
Sum of the first 40 natural numbers: The sum
SSS of the first nnn natural numbers is given by: S=n(n+1)2S =
\frac{n(n+1)}{2}S=2n(n+1)
For n=40n = 40n=40: S=40×412S = \frac{40 \times
41}{2}S=240×41 S=820S = 820S=820
2.
Number of numbers: There are 40 natural
numbers from 1 to 40.
3.
Calculate the average:
Average=82040\text{Average} = \frac{820}{40}Average=40820
Average=20.5\text{Average} = 20.5Average=20.5
Therefore, the average of the first 40 natural numbers is
20.5\boxed{20.5}20.5.
Unit 03: Number Series
3.1 Sequence
3.2 Series
3.3 Number Series
3.4 Alpha–Numeric Series
3.5 Analogy of Number Series
3.6 Number Classification
3.7 Problems Based on Number Series, Analogy and
Classifications
3.1 Sequence
- Definition: A
sequence is an ordered list of numbers or objects where each member is
related to the previous one by a specific rule.
- Types:
- Arithmetic
Sequence: Each term is obtained by adding a constant value to
the previous term (e.g., 2, 4, 6, 8...).
- Geometric
Sequence: Each term is obtained by multiplying the previous
term by a constant ratio (e.g., 2, 6, 18, 54...).
3.2 Series
- Definition: A
series is the sum of the terms of a sequence.
- Types:
- Arithmetic
Series: Sum of an arithmetic sequence.
- Geometric
Series: Sum of a geometric sequence.
3.3 Number Series
- Definition: A
number series is a sequence of numbers where each subsequent number follows
a specific pattern or rule.
- Examples:
Fibonacci series (0, 1, 1, 2, 3, 5, 8...), Prime number series (2, 3, 5,
7, 11...).
3.4 Alpha–Numeric Series
- Definition: A
series where both alphabets and numbers follow a pattern.
- Examples: A1,
B2, C3, D4... or 1A, 2B, 3C, 4D...
3.5 Analogy of Number Series
- Definition:
Analogy in number series involves identifying the relationship between
numbers in one sequence and applying it to another sequence.
- Example: If
the series is 2, 4, 6, 8..., the analogy might be doubling each number.
3.6 Number Classification
- Definition:
Number classification involves grouping or categorizing numbers based on
common properties or rules.
- Examples: Even
numbers, odd numbers, prime numbers, etc.
3.7 Problems Based on Number Series, Analogy and
Classifications
- Description: These
are problem-solving exercises that involve recognizing patterns in number
series, applying analogies, and classifying numbers based on given rules.
- Skills:
Requires logical reasoning, pattern recognition, and understanding of
numerical properties.
Each of these topics is fundamental in understanding and
solving problems related to number sequences, series, analogies, and
classifications. Practice is crucial to develop proficiency in these areas.
Summary of Unit 03: Number Series
1.
Series
o Definition: A series
is an ordered list of numbers or objects where each member follows a specific
rule or pattern.
o Types:
§ Arithmetic
Series: Progression by adding a constant difference.
§ Geometric
Series: Progression by multiplying by a constant ratio.
2.
Analogy
o Definition: Analogy in
number series involves identifying relationships between numbers and applying
those relationships to solve similar patterns.
o Example:
Identifying a doubling pattern (2, 4, 8, 16...) and applying it to another
sequence.
3.
Number Classification
o Definition:
Classifying numbers based on common properties or rules.
o Examples: Prime
numbers, even numbers, odd numbers.
4.
Problem Solving
o Analogies: Solving
problems by recognizing patterns and applying identified analogies.
o Number
Series: Identifying and continuing numerical patterns.
o Classification: Sorting
numbers into appropriate categories based on given criteria.
5.
Skills Developed
o Logical
Reasoning: Understanding and applying patterns logically.
o Pattern
Recognition: Identifying and continuing numerical sequences.
o Problem
Solving: Applying learned concepts to solve numerical puzzles and
classifications.
By studying these topics thoroughly, you gain a foundational
understanding of how number series work, how to analyze analogies, and how to
classify numbers based on specific rules. Regular practice and application are
essential to mastering these skills effectively.
Keywords Explanation
1.
Sequence
o Definition:
A sequence is a ordered list of numbers or objects where each member follows a
definite order or rule.
o Example:
1,3,5,7,91, 3, 5, 7, 91,3,5,7,9 is a sequence of odd numbers.
2.
Series
o Definition:
A series is the sum of the terms of a sequence.
o Example: The
series 1+3+5+7+9=251 + 3 + 5 + 7 + 9 = 251+3+5+7+9=25 sums the terms of the
above sequence.
3.
Arithmetic Series
o Definition:
An arithmetic series is a sequence of numbers in which the difference between
any two consecutive terms is always the same.
o Example:
2,4,6,8,102, 4, 6, 8, 102,4,6,8,10 is an arithmetic series where the difference
between consecutive terms is 222.
4.
Geometric Series
o Definition:
A geometric series is a sequence of numbers in which the ratio of any two
consecutive terms is always the same.
o Example:
2,6,18,54,1622, 6, 18, 54, 1622,6,18,54,162 is a geometric series where each
term is obtained by multiplying the previous term by 333.
Understanding these concepts helps in identifying and working
with different types of numerical patterns and relationships. These are
fundamental in mathematics and are often applied in various fields requiring
analytical and problem-solving skills.
484,
240, 120, 57, 26.5, 11.25, 3.625
It looks like you've provided a sequence of numbers: 484,
240, 120, 57, 26.5, 11.25, 3.625. To determine the pattern or series here,
let's examine the differences or ratios between consecutive terms:
1.
Identifying the Pattern:
Looking at the sequence:
o From 484 to
240: 484 - 240 = 244
o From 240 to
120: 240 - 120 = 120
o From 120 to
57: 120 - 57 = 63
o From 57 to
26.5: 57 - 26.5 = 30.5
o From 26.5 to
11.25: 26.5 - 11.25 = 15.25
o From 11.25
to 3.625: 11.25 - 3.625 = 7.625
The differences between consecutive terms are not constant,
suggesting it might be a more complex pattern involving multiplication or
division.
2.
Analyzing Ratios:
Let's check the ratios between consecutive terms:
o 484 / 240 ≈
2.0167
o 240 / 120 =
2
o 120 / 57 ≈
2.1053
o 57 / 26.5 ≈
2.1509
o 26.5 / 11.25
≈ 2.3556
o 11.25 /
3.625 ≈ 3.1034
The ratios are not exactly the same but seem to increase
gradually.
3.
Possible Pattern:
The sequence doesn't follow a simple arithmetic or geometric
pattern, but the ratios show a tendency towards decreasing, albeit irregularly.
4.
Conclusion:
Without a clear and consistent pattern in the differences or
ratios, this sequence may be part of a more complex or irregular series.
Additional terms or context might help identify a more precise pattern, such as
a recursive formula or context-specific rule.
5, 7, 16, 57, 244,
1245, 7506
To analyze the sequence 5, 7, 16, 57, 244, 1245, 7506, let's
examine the differences between consecutive terms and consider possible
patterns:
1.
Identifying the Pattern:
o From 5 to 7:
7 - 5 = 2
o From 7 to
16: 16 - 7 = 9
o From 16 to
57: 57 - 16 = 41
o From 57 to
244: 244 - 57 = 187
o From 244 to
1245: 1245 - 244 = 1001
o From 1245 to
7506: 7506 - 1245 = 6261
The differences between consecutive terms are: 2, 9, 41, 187,
1001, 6261. These differences are not constant, suggesting a more complex
pattern.
2.
Analyzing Possible Patterns:
o The ratios
between consecutive terms:
§ 75=1.4\frac{7}{5}
= 1.457=1.4
§ 167≈2.2857\frac{16}{7}
\approx 2.2857716≈2.2857
§ 5716≈3.5625\frac{57}{16}
\approx 3.56251657≈3.5625
§ 24457≈4.2807\frac{244}{57}
\approx 4.280757244≈4.2807
§ 1245244≈5.1033\frac{1245}{244}
\approx 5.10332441245≈5.1033
§ 75061245≈6.0301\frac{7506}{1245}
\approx 6.030112457506≈6.0301
The ratios also show an increasing trend, but they are not
consistent.
3.
Possible Pattern:
Without a clear arithmetic or geometric progression, the
sequence could potentially involve a recursive formula or a specific function
that generates each term based on previous terms.
4.
Conclusion:
The sequence 5, 7, 16, 57, 244, 1245, 7506 does not fit a
straightforward arithmetic or geometric pattern. Additional terms or context
could provide more insight into a specific rule or function governing the
sequence.
6, 7, 16, 41, 90, 154,
292
To analyze the sequence 6, 7, 16, 41, 90, 154, 292, let's
examine the differences between consecutive terms and consider possible
patterns:
1.
Identifying the Pattern:
o From 6 to 7:
7 - 6 = 1
o From 7 to
16: 16 - 7 = 9
o From 16 to
41: 41 - 16 = 25
o From 41 to
90: 90 - 41 = 49
o From 90 to
154: 154 - 90 = 64
o From 154 to
292: 292 - 154 = 138
The differences between consecutive terms are: 1, 9, 25, 49,
64, 138. These differences are not constant, suggesting a more complex pattern.
2.
Analyzing Possible Patterns:
o The ratios
between consecutive terms:
§ 76≈1.1667\frac{7}{6}
\approx 1.166767≈1.1667
§ 167≈2.2857\frac{16}{7}
\approx 2.2857716≈2.2857
§ 4116≈2.5625\frac{41}{16}
\approx 2.56251641≈2.5625
§ 9041≈2.1951\frac{90}{41}
\approx 2.19514190≈2.1951
§ 15490≈1.7111\frac{154}{90}
\approx 1.711190154≈1.7111
§ 292154≈1.8961\frac{292}{154}
\approx 1.8961154292≈1.8961
The ratios show a varying trend, indicating the sequence
might involve alternating patterns or a combination of arithmetic operations.
3.
Possible Pattern:
Considering the variations in differences and ratios, the
sequence could possibly involve alternating additions or multiplicative factors
that change from one term to the next.
4.
Conclusion:
The sequence 6, 7, 16, 41, 90, 154, 292 does not fit a simple
arithmetic or geometric progression. It might be governed by a recursive
formula or a specific rule that combines arithmetic operations in a non-linear
fashion. Further terms or additional context could help in identifying a more
precise pattern.
201, 202, 204, 207,
... What number should come next?
To determine the next number in the sequence 201, 202, 204,
207, ..., we need to identify the pattern or rule governing the sequence:
1.
Identifying the Pattern:
o From 201 to
202: +1
o From 202 to
204: +2
o From 204 to
207: +3
It appears that each subsequent term is increasing by an
increment that follows a pattern of adding consecutive integers (1, 2, 3, ...).
2.
Predicting the Next Number:
Following the pattern:
o From 207,
the next increment would be +4.
Therefore, to find the next number:
o 207+4=211207
+ 4 = 211207+4=211
3.
Conclusion:
Based on the observed pattern of increasing increments (+1,
+2, +3, +4), the next number in the sequence should be 211.
Unit 04: Alphabet Series
4.1 Alphabetical/ Letter Series
4.2 Alphanumeric Series
4.3 Alphabetical Pattern Series
4.4 Analogy of letter Series
4.5 Alphabet Classification
4.6 Problems based on letter Series, Analogy and
Classifications
1.
Alphabetical/Letter Series
o Definition: A letter
series is a sequence of letters where each subsequent letter follows a specific
order or rule in the alphabet.
o Example: A, B, C,
D, E... or Z, Y, X, W, V...
2.
Alphanumeric Series
o Definition: An
alphanumeric series involves a combination of letters and numbers that follow a
specific pattern.
o Example: A1, B2,
C3, D4... or 1A, 2B, 3C, 4D...
3.
Alphabetical Pattern Series
o Definition: An
alphabetical pattern series involves letters arranged in a specific pattern or
sequence that may follow a rule or alternate pattern.
o Example: A, C, E,
G... (alternating by skipping one letter)
4.
Analogy of Letter Series
o Definition: Analogy in
letter series involves identifying relationships between letters and applying
those relationships to solve similar patterns.
o Example: If A, C,
E, G... follows a pattern of skipping one letter, the analogy could involve
identifying and applying similar skip patterns in other sequences.
5.
Alphabet Classification
o Definition: Alphabet
classification involves grouping or categorizing letters based on specific
criteria or rules.
o Examples: Vowels vs.
consonants, alphabetical order, etc.
6.
Problems Based on Letter Series, Analogy, and
Classifications
o Description: These are
problem-solving exercises that require recognizing patterns in letter series,
applying analogies based on identified patterns, and classifying letters based
on given criteria.
o Skills: Requires
logical reasoning, pattern recognition, and understanding of alphabetical
properties and relationships.
Understanding these concepts helps in identifying patterns,
solving puzzles involving letters and numbers, and developing logical reasoning
skills. Practice is essential to mastering these skills effectively.
Summary of Unit 04: Alphabet Series
1.
Alphabet Series
o Definition: An
alphabet series is a sequence of letters where each subsequent letter follows a
specific order or rule in the alphabet.
o Example: A, B, C,
D, E... or Z, Y, X, W, V...
2.
Types of Alphabet Series Questions
o Description: Different
types of questions in alphabet series involve recognizing patterns such as
consecutive letters, alternate letters, skip patterns, or patterns based on
positions in the alphabet.
o Examples: Series
like A, C, E, G... (alternating by skipping one letter), or Z, X, V, T...
(decreasing by two letters).
3.
Letter Analogy
o Definition: Analogy in
alphabet series involves identifying relationships between letters and applying
those relationships to solve similar patterns.
o Example: If A, C,
E, G... follows a pattern of skipping one letter, applying this analogy might
involve identifying similar skip patterns in other sequences.
4.
Alphabet Classification
o Definition: Alphabet
classification involves categorizing letters based on specific criteria or
rules.
o Examples:
Classifying letters as vowels and consonants, alphabetical order, or based on
their position in the alphabet.
5.
Problems Based on Alphabet Series, Analogy, and
Classification
o Description: These are
problem-solving exercises that require recognizing patterns in alphabet series,
applying analogies based on identified patterns, and classifying letters based
on given criteria.
o Skills
Developed: Requires logical reasoning, pattern recognition, and
understanding of alphabetical properties and relationships.
By studying these topics thoroughly, one gains a foundational
understanding of how alphabet series work, how to analyze analogies between
letters, and how to classify letters based on specific rules. Practice is
crucial for developing proficiency in solving alphabet series problems
effectively.
Keywords Explanation
1.
Alphabet Series
o Definition: An
alphabet series is a sequence of letters where each subsequent letter follows a
specific order or pattern in the alphabet.
o Example: A, B, C,
D, E... or Z, Y, X, W, V...
2.
Forward Ranking of Letters
o Definition: Forward
ranking refers to the position of letters in alphabetical order from A to Z.
o Example: A is the
1st letter, B is the 2nd letter, and so on, up to Z which is the 26th letter.
3.
Backward Ranking of Letters
o Definition: Backward
ranking refers to the position of letters in alphabetical order from Z to A.
o Example: Z is the
1st letter from the end, Y is the 2nd, and so on, up to A which is the 26th
letter from the end.
4.
Letter Analogy
o Definition: Letter
analogy involves identifying relationships between letters and applying those
relationships to solve similar patterns.
o Example: If A is to
B as C is to D, the analogy could involve applying this relationship to other
pairs of letters.
5.
Letter Classification
o Definition: Letter
classification involves categorizing letters based on specific criteria such as
vowels vs. consonants, position in the alphabet, or other patterns.
o Examples:
Classifying letters as vowels (A, E, I, O, U) or consonants (B, C, D, ...), or
grouping them based on their positions in the alphabet (e.g., first five
letters, middle letters, last five letters).
Understanding these concepts is fundamental for solving
problems related to alphabet series, analogies between letters, and classifying
letters based on various criteria. These skills are essential for logical
reasoning and pattern recognition in such types of questions.
Find out the missing
term in the following letter series analogy. ACEG: DFHJ:: QSUW :?
To solve the analogy problem "ACEG : DFHJ :: QSUW :
?", let's analyze the pattern in the given pairs:
1.
Given Pair ACEG : DFHJ
o The first
set ACEG increases by 2 in alphabetical order (A + 2 = D, C + 2 = F, E + 2 = H,
G + 2 = J).
2.
Applying the Pattern to QSUW
o Q + 2 = S
o S + 2 = U
o U + 2 = W
Therefore, following the same pattern, the missing term in
the analogy QSUW : ? should be Y.
So, the completed analogy is: QSUW : Y
Find out the missing
term in the following letter series analogy. EGIK: FILO:: FHJL : ?
To solve the analogy "EGIK : FILO :: FHJL : ?",
let's analyze the pattern in the given pairs:
1.
Given Pair EGIK : FILO
o E + 4 = I
o G + 4 = K
Similarly,
o I + 4 = M
o K + 4 = O
Therefore, EGIK : FILO follows a pattern where each letter is
incremented by 4 positions in the alphabet.
2.
Applying the Pattern to FHJL
o F + 4 = J
o H + 4 = L
Following the same pattern,
o J + 4 = N
o L + 4 = P
Therefore, based on the pattern observed in EGIK : FILO, the
missing term in the analogy FHJL : ? should be N.
So, the completed analogy is: FHJL : N
Find out the missing
term in the following letter series analogy. AROUND: RAUODN::GROUND:?
To solve the analogy "AROUND : RAUODN :: GROUND :
?", let's analyze the pattern in the given pairs:
1.
Given Pair AROUND : RAUODN
o A, R, O, U,
N, D are rearranged to form RAUODN.
2.
Applying the Pattern to GROUND
o G, R, O, U,
N, D are the letters provided.
Therefore, the analogy in this case appears to be that the
letters in the word are simply rearranged.
Unit 05: Coding-Decoding
5.1 Letter Coding
5.2 Number/Symbol Coding
5.3 Substitution
5.4 Matrix Coding
5.5 Mixed Letter Coding/ Deciphering Message Word
Codes
5.6 Mixed Number Coding/ Deciphering Message Number
and Symbol Codes
1.
Letter Coding
o Definition: Letter
coding involves substituting each letter in a word or phrase with another
letter according to a specific rule or pattern.
o Example: A could be
coded as D, B as E, and so on, based on a fixed number of positions shifted in
the alphabet.
2.
Number/Symbol Coding
o Definition: Number or
symbol coding involves substituting each number or symbol in a sequence with
another number or symbol based on a predetermined rule.
o Example: 1 could be
coded as 4, 2 as 5, and so forth, or symbols like + could be coded as -, * as
/, etc.
3.
Substitution
o Definition:
Substitution coding involves replacing elements (letters, numbers, symbols)
with others based on a consistent rule throughout a sequence or message.
o Example: Using a
specific keyword or a numerical shift to substitute each character with another
in a systematic way.
4.
Matrix Coding
o Definition: Matrix
coding involves arranging letters, numbers, or symbols in a grid or matrix, and
then coding or decoding based on positions within that matrix.
o Example: A matrix
could have letters arranged in rows and columns, and the code might involve
reading letters diagonally, vertically, or horizontally.
5.
Mixed Letter Coding/Deciphering Message Word Codes
o Definition: Mixed
letter coding involves a combination of different coding techniques applied to
words or phrases in a message.
o Example: Words
might be coded using a letter substitution cipher (like Caesar cipher) where
each letter is shifted a certain number of places.
6.
Mixed Number Coding/Deciphering Message Number and
Symbol Codes
o Definition: Mixed
number coding involves using a combination of numeric or symbolic codes applied
to a message.
o Example: Numbers
and symbols might be encoded using a mathematical operation (like addition or
subtraction) or a specific symbol substitution rule.
Understanding these coding techniques is essential for
deciphering encoded messages in various scenarios, from puzzles to real-world
applications in cryptography and information security. Mastery of these skills
requires practice in recognizing patterns, applying rules consistently, and
testing different decoding strategies.
Summary of Unit 05: Coding-Decoding
1.
Introduction to Coding-Decoding
o Explanation:
Coding-Decoding involves converting a message or information from one form to
another using specific rules or patterns.
o Types: Includes
letter coding, number/symbol coding, substitution, matrix coding, mixed coding
with letters, numbers, and symbols.
2.
Types of Coding-Decoding Statements
o Description: Different
types of coding-decoding statements involve various methods of encoding and
decoding information, such as:
§ Letter
Coding: Substituting each letter with another according to a fixed
pattern or rule.
§ Number/Symbol
Coding: Replacing numbers or symbols with others based on
predetermined rules.
§ Substitution: Using keywords
or numerical shifts to replace elements consistently.
§ Matrix
Coding: Encoding or decoding based on positions within a grid or
matrix.
§ Mixed Coding: Combining
multiple coding techniques within a single message.
3.
Logical Approach to Solve Problems
o Methods:
Understanding the logic behind each type of coding-decoding problem involves:
§ Recognizing
patterns or rules applied to the coding process.
§ Applying
systematic approaches to decode messages based on identified patterns.
§ Using trial
and error to test potential solutions against known rules.
4.
Set of Rules for Solving Coding-Decoding Problems
o Strategies: Key rules
and strategies include:
§ Consistency: Applying
the same coding rule across all elements of the message.
§ Pattern
Recognition: Identifying recurring patterns or sequences within the
coded message.
§ Systematic
Testing: Testing potential solutions by applying logical deductions
based on observed patterns.
§ Understanding
Variations: Being aware of different coding techniques and adapting
strategies accordingly.
By mastering these concepts and strategies, one can
effectively decode coded messages in various contexts, from puzzles to
practical applications in cryptography and communication security. Practice and
familiarity with different coding methods are essential for developing
proficiency in solving coding-decoding problems efficiently.
Keywords Explanation
1.
Letter Coding
o Definition: Letter
coding involves substituting each letter in a word or phrase with another
letter according to a specific rule or pattern.
o Example: A could be
coded as D, B as E, and so on, based on a fixed number of positions shifted in
the alphabet.
2.
Number Coding
o Definition: Number
coding involves substituting each number in a sequence with another number
based on a predetermined rule or pattern.
o Example: 1 could be
coded as 4, 2 as 5, and so forth, based on a specific numerical shift or
operation.
3.
Substitution Coding
o Definition:
Substitution coding involves replacing elements (letters, numbers, symbols)
with others based on a consistent rule throughout a sequence or message.
o Example: Using a
keyword or a numerical shift to substitute each character with another in a
systematic way.
4.
Matrix Coding
o Definition: Matrix
coding involves arranging letters, numbers, or symbols in a grid or matrix, and
then coding or decoding based on positions within that matrix.
o Example: A matrix
could have letters arranged in rows and columns, and the code might involve
reading letters diagonally, vertically, or horizontally.
5.
Mixed Letter Coding
o Definition: Mixed
letter coding involves applying multiple letter coding techniques within a
single message or sequence.
o Example: Words
might be coded using a combination of substitution ciphers or alternating rules
for different parts of the message.
6.
Mixed Number Coding
o Definition: Mixed
number coding involves applying multiple number coding techniques within a
sequence or message.
o Example: Numbers
and symbols might be encoded using different mathematical operations or
substitution rules in various parts of the coded message.
Understanding these coding techniques is crucial for decoding
encoded messages in puzzles, cryptograms, or practical applications involving
secure communication. Mastery of these skills requires practice in recognizing
patterns, applying rules consistently, and adapting strategies based on
different coding methods used.
In a certain language
if HOUSE is written as KRUQC then how ROHIT will be written in that language?
To find out how "ROHIT" would be written in the
same coding scheme as "HOUSE" to "KRUQC":
1.
Identify the pattern: Look for how each letter in
"HOUSE" is transformed into "KRUQC".
o H -> K
o O -> R
o U -> Q
o S -> C
o E -> C
2.
Apply the pattern to "ROHIT":
o R -> ?
o O -> ?
o H -> ?
o I -> ?
o T -> ?
3.
Using the identified pattern:
o R in
"ROHIT" corresponds to H in "HOUSE", so R -> K
o O in
"ROHIT" corresponds to O in "HOUSE", so O -> R
o H in
"ROHIT" corresponds to U in "HOUSE", so H -> Q
o I in
"ROHIT" corresponds to S in "HOUSE", so I -> C
o T in
"ROHIT" corresponds to E in "HOUSE", so T -> C
Therefore, "ROHIT" would be written as QRQCJ
in that language.
If BRAND is written as
79643 and PAROT is written as 26951, how is PANT coded?
To decode how "PANT" is coded based on the given
examples:
1.
Identify the pattern from "BRAND" to
"79643" and "PAROT" to "26951":
o BRAND ->
79643
§ B -> 7
§ R -> 9
§ A -> 6
§ N -> 4
§ D -> 3
o PAROT ->
26951
§ P -> 2
§ A -> 6
§ R -> 9
§ O -> 5
§ T -> 1
2.
Apply the pattern to "PANT":
o P -> 2
(from PAROT)
o A -> 6
(from PAROT)
o N -> 4
(from BRAND)
o T -> 1
(from PAROT)
Therefore, "PANT" is coded as 2641 based on
the given coding scheme.
If SUNDAY is written
as NYNYNY, how will BANGALORE be written in that code?
To find out how "BANGALORE" would be written in the
same coding scheme as "SUNDAY" to "NYNYNY":
1.
Identify the pattern: Notice how each letter in
"SUNDAY" corresponds to "NYNYNY":
o S -> N
o U -> Y
o N -> N
o D -> Y
o A -> N
o Y -> Y
2.
Apply the pattern to "BANGALORE":
o B -> ?
o A -> ?
o N -> ?
o G -> ?
o A -> ?
o L -> ?
o O -> ?
o R -> ?
o E -> ?
3.
Using the identified pattern:
o B in
"BANGALORE" corresponds to S in "SUNDAY", so B -> N
o A in
"BANGALORE" corresponds to U in "SUNDAY", so A -> Y
o N in
"BANGALORE" corresponds to N in "SUNDAY", so N -> N
o G in
"BANGALORE" corresponds to D in "SUNDAY", so G -> Y
o A in
"BANGALORE" corresponds to A in "SUNDAY", so A -> N
o L in
"BANGALORE" corresponds to Y in "SUNDAY", so L -> Y
o O in
"BANGALORE" corresponds to N in "SUNDAY", so O -> N
o R in
"BANGALORE" corresponds to D in "SUNDAY", so R -> Y
o E in
"BANGALORE" corresponds to Y in "SUNDAY", so E -> Y
Therefore, "BANGALORE" would be written as NYNANYNY
in that code.
If EARTH is written as
FCUXM in a certain code. How is MOON written in that code?
To find out how "MOON" would be written in the same
coding scheme as "EARTH" to "FCUXM":
1.
Identify the pattern: Notice how each letter in
"EARTH" corresponds to "FCUXM":
o E -> F
o A -> C
o R -> U
o T -> X
o H -> M
2.
Apply the pattern to "MOON":
o M
corresponds to ?
o O
corresponds to ?
o O
corresponds to ?
o N
corresponds to ?
3.
Using the identified pattern:
o M in
"MOON" corresponds to E in "EARTH", so M -> F
o O in
"MOON" corresponds to A in "EARTH", so O -> C
o O in
"MOON" corresponds to A in "EARTH", so O -> C
o N in
"MOON" corresponds to R in "EARTH", so N -> U
Therefore, "MOON" would be written as FCCUX
in that code.
In a certain code
language, '123' means 'hot filtered coffee’, '356‘means 'very hot day' and '589' means 'day and night'. Which
digit stands for' very’?
Based on the provided codes:
- '123'
means 'hot filtered coffee’
- '356'
means 'very hot day'
- '589'
means 'day and night'
We need to find out which digit stands for 'very'.
From the given examples:
- '356'
corresponds to 'very hot day'.
Therefore, the digit that stands for 'very' in this code
language is 3.
Unit 06: Percentage
6.1 Conversion of fraction to percentage and
percentage to fraction:
6.2 Short-cut Methods
6.3 Successive Percentage Change
6.4 Budget based Problems
1.
Conversion of Fraction to Percentage and Percentage to
Fraction
o Fraction to
Percentage:
§ To convert a
fraction to a percentage, multiply the fraction by 100.
§ Example:
34\frac{3}{4}43 as a percentage is 34×100=75%\frac{3}{4} \times 100 =
75\%43×100=75%.
o Percentage
to Fraction:
§ To convert a
percentage to a fraction, place the percentage value over 100 and simplify.
§ Example: 25%
as a fraction is 25100=14\frac{25}{100} = \frac{1}{4}10025=41.
2.
Short-cut Methods
o Percentage
Change Calculation:
§ Increase: Calculate
increase as IncreaseOriginal×100\frac{\text{Increase}}{\text{Original}} \times
100OriginalIncrease×100.
§ Decrease: Calculate
decrease as DecreaseOriginal×100\frac{\text{Decrease}}{\text{Original}} \times
100OriginalDecrease×100.
§ Example: If
a price increases from $50 to $60, the percentage increase is
1050×100=20%\frac{10}{50} \times 100 = 20\%5010×100=20%.
3.
Successive Percentage Change
o Definition: Successive
percentage changes involve applying percentage changes in sequence.
o Calculation: For
multiple successive changes, apply each change to the result of the previous
change.
o Example: If
a price increases by 10% and then decreases by 5%, the overall change is (1+0.10)×(1−0.05)−1(1
+ 0.10) \times (1 - 0.05) - 1(1+0.10)×(1−0.05)−1.
4.
Budget-based Problems
o Definition:
Budget-based problems involve calculating percentages in financial planning and
budget allocation.
o Examples: Includes
scenarios such as allocating budgets for different expense categories based on
given percentages.
o Example: If
a company allocates 30% of its budget to marketing and 20% to operations,
calculate the amounts allocated for each based on the total budget.
Understanding these concepts is essential for practical
applications in finance, business, and everyday calculations involving
percentages. Mastery of shortcut methods and successive percentage changes
helps in quick problem-solving and decision-making involving percentage-based
data.
Summary of Unit 06: Percentage
1.
Percentage Commodity Price Increase/Decrease
o Explanation:
Understanding how to calculate percentage increases or decreases in commodity
prices is essential for analyzing market trends and financial decisions.
o Calculation:
§ Increase: Calculate
increase as
Increase in PriceOriginal Price×100\frac{\text{Increase in
Price}}{\text{Original Price}} \times
100Original PriceIncrease in Price×100.
§ Decrease: Calculate
decrease as
Decrease in PriceOriginal Price×100\frac{\text{Decrease in
Price}}{\text{Original Price}} \times
100Original PriceDecrease in Price×100.
o Example: If a stock
price increases from $50 to $60, the percentage increase is
1050×100=20%\frac{10}{50} \times 100 = 20\%5010×100=20%.
2.
Successive Percent Changes
o Definition: Successive
percent changes involve applying multiple percentage changes in sequence.
o Method: Apply each
percentage change successively to the result of the previous change.
o Example: If an
item's price increases by 20% and then decreases by 10%, the overall change is
calculated as (1+0.20)×(1−0.10)−1(1 + 0.20) \times (1 - 0.10) -
1(1+0.20)×(1−0.10)−1.
3.
Budget-Based Problems
o Explanation:
Budget-based problems involve allocating funds or resources based on specified
percentages.
o Application: Used in
financial planning, business budgeting, and resource allocation scenarios.
o Example: If a
company budgets 30% of its revenue for marketing and 20% for operations,
calculate the amounts allocated to each department based on the total revenue.
By mastering these concepts, one can effectively analyze
percentage-based data, make informed financial decisions, and solve practical
problems related to budgeting and market analysis. Understanding shortcut
methods for percentage calculations and how to manage successive changes
ensures proficiency in applying these skills in various real-world situations.
Keywords Explanation
1.
Percentage
o Definition: A
percentage represents a fraction of 100 and is used to express parts of a
whole.
o Calculation: To convert
a fraction or decimal to a percentage, multiply by 100. To convert a percentage
to a fraction, place it over 100 and simplify.
o Example:
34\frac{3}{4}43 as a percentage is 34×100=75%\frac{3}{4} \times 100 =
75\%43×100=75%.
2.
Commodity
o Definition: A
commodity refers to a raw material or primary agricultural product that can be
bought and sold, such as oil, gold, or agricultural produce.
o Usage: Prices of
commodities often fluctuate due to market demand, supply changes, geopolitical
factors, and economic conditions.
o Example: Oil prices
are influenced by global supply and demand dynamics.
3.
Successive Percent
o Definition: Successive
percentage changes involve applying multiple percentage changes in sequence to
a starting value.
o Calculation: Each
successive change is calculated based on the new value resulting from the
previous change.
o Example: If a price
increases by 10% and then decreases by 5%, the overall change is calculated by
compounding these changes.
4.
Budget
o Definition: A budget
is a financial plan that allocates resources for specific purposes over a
specified period.
o Components: Budgets
include income, expenses, savings, and investments, and are crucial for
financial management and planning.
o Example: A
company's budget allocates funds to various departments such as marketing,
operations, and research based on financial goals and priorities.
Understanding these keywords is essential for effectively
applying percentage calculations, analyzing commodity markets, managing
successive changes in values, and making informed decisions in budgeting and
financial planning scenarios. These concepts are foundational in both personal
finance and business finance contexts.
When the price for a
product was decreased by 10%, the number sold increased by 30%. What was the
effect on the total revenue?
To determine the effect on total revenue when the price of a
product decreases by 10% and the number sold increases by 30%, we can follow
these steps:
1.
Define the original values:
o Let the
original price of the product be PPP.
o Let the
original quantity sold be QQQ.
o Therefore,
the original total revenue RRR is R=P×QR = P \times QR=P×Q.
2.
Calculate the new price:
o The price is
decreased by 10%, so the new price PnewP_{\text{new}}Pnew is:
Pnew=P×(1−0.10)=P×0.90P_{\text{new}} = P \times (1 - 0.10) = P \times
0.90Pnew=P×(1−0.10)=P×0.90
3.
Calculate the new quantity sold:
o The quantity
sold increases by 30%, so the new quantity QnewQ_{\text{new}}Qnew is:
Qnew=Q×(1+0.30)=Q×1.30Q_{\text{new}} = Q \times (1 + 0.30) = Q \times
1.30Qnew=Q×(1+0.30)=Q×1.30
4.
Calculate the new total revenue:
o The new
total revenue RnewR_{\text{new}}Rnew is: Rnew=Pnew×QnewR_{\text{new}} =
P_{\text{new}} \times Q_{\text{new}}Rnew=Pnew×Qnew
o Substituting
the new price and new quantity: Rnew=(P×0.90)×(Q×1.30)R_{\text{new}} = (P
\times 0.90) \times (Q \times 1.30)Rnew=(P×0.90)×(Q×1.30)
Rnew=P×Q×0.90×1.30R_{\text{new}} = P \times Q \times 0.90 \times
1.30Rnew=P×Q×0.90×1.30
5.
Calculate the factor change in revenue:
o Simplify the
expression: Rnew=P×Q×(0.90×1.30)R_{\text{new}} = P \times Q \times (0.90 \times
1.30)Rnew=P×Q×(0.90×1.30) Rnew=P×Q×1.17R_{\text{new}} = P \times Q \times
1.17Rnew=P×Q×1.17
6.
Compare the new revenue to the original revenue:
o The original
revenue RRR is P×QP \times QP×Q.
o The new
revenue RnewR_{\text{new}}Rnew is P×Q×1.17P \times Q \times 1.17P×Q×1.17.
This indicates that the new total revenue is 1.17 times the
original total revenue, which means the total revenue increased by 17%.
Summary
When the price of a product is decreased by 10% and the
number sold increases by 30%, the overall effect on total revenue is an
increase of 17%.
If the numerator of a
fraction be increased by 15% and its denominator be diminished by 8%, the value
of the fraction is 15/16. Find the original fraction.
denote the original fraction as ND\frac{N}{D}DN, where NNN
is the numerator and DDD is the denominator.
Step-by-Step Solution
1.
Increase the numerator by 15%:
o The new
numerator becomes Nnew=N+0.15N=1.15NN_{\text{new}} = N + 0.15N =
1.15NNnew=N+0.15N=1.15N.
2.
Decrease the denominator by 8%:
o The new
denominator becomes Dnew=D−0.08D=0.92DD_{\text{new}} = D - 0.08D =
0.92DDnew=D−0.08D=0.92D.
3.
The new fraction is given to be 1516\frac{15}{16}1615:
o So, we can
set up the equation: 1.15N0.92D=1516\frac{1.15N}{0.92D} =
\frac{15}{16}0.92D1.15N=1615
4.
Solve for ND\frac{N}{D}DN:
o Cross-multiply
to solve for ND\frac{N}{D}DN: 1.15N×16=15×0.92D1.15N \times 16 = 15 \times
0.92D1.15N×16=15×0.92D 18.4N=13.8D18.4N = 13.8D18.4N=13.8D
ND=13.818.4\frac{N}{D} = \frac{13.8}{18.4}DN=18.413.8
5.
Simplify the fraction 13.818.4\frac{13.8}{18.4}18.413.8:
o Divide the
numerator and the denominator by their greatest common divisor (GCD), which is
2: 13.8÷218.4÷2=6.99.2\frac{13.8 \div 2}{18.4 \div 2} =
\frac{6.9}{9.2}18.4÷213.8÷2=9.26.9
o To further
simplify: 6.99.2=6992\frac{6.9}{9.2} = \frac{69}{92}9.26.9=9269
o Since 69 and
92 have no common factors other than 1, 6992\frac{69}{92}9269 is in its
simplest form.
Therefore, the original fraction is 6992\frac{69}{92}9269.
The population of a
town is 176400. If it increases at the rate of 5% per annum, what will be its
population 2 years hence? What was it 2 years ago?
To determine the population of the town 2 years hence and 2
years ago given a current population of 176,400 and an annual increase rate of
5%, we can use the formula for compound interest.
Population 2 Years Hence
1.
Formula for compound interest:
Pfuture=Ppresent×(1+r)nP_{\text{future}} = P_{\text{present}}
\times (1 + r)^nPfuture=Ppresent×(1+r)n
Where:
o PfutureP_{\text{future}}Pfuture
is the population in the future.
o PpresentP_{\text{present}}Ppresent
is the current population.
o rrr is the
rate of increase per annum (expressed as a decimal).
o nnn is the
number of years.
2.
Plug in the values:
o Ppresent=176,400P_{\text{present}}
= 176,400Ppresent=176,400
o r=5%=0.05r =
5\% = 0.05r=5%=0.05
o n=2n = 2n=2
3.
Calculate:
Pfuture=176,400×(1+0.05)2P_{\text{future}} = 176,400 \times
(1 + 0.05)^2Pfuture=176,400×(1+0.05)2 Pfuture=176,400×1.052P_{\text{future}} =
176,400 \times 1.05^2Pfuture=176,400×1.052
Pfuture=176,400×1.1025P_{\text{future}} = 176,400 \times
1.1025Pfuture=176,400×1.1025 Pfuture=194,481P_{\text{future}} =
194,481Pfuture=194,481
So, the population 2 years hence will be 194,481.
Population 2 Years Ago
1.
Formula for reverse compound interest:
Ppast=Ppresent÷(1+r)nP_{\text{past}} = P_{\text{present}}
\div (1 + r)^nPpast=Ppresent÷(1+r)n
Where:
o PpastP_{\text{past}}Ppast
is the population in the past.
o PpresentP_{\text{present}}Ppresent
is the current population.
o rrr is the
rate of increase per annum (expressed as a decimal).
o nnn is the
number of years.
2.
Plug in the values:
o Ppresent=176,400P_{\text{present}}
= 176,400Ppresent=176,400
o r=5%=0.05r =
5\% = 0.05r=5%=0.05
o n=2n = 2n=2
3.
Calculate:
Ppast=176,400÷(1+0.05)2P_{\text{past}} = 176,400 \div (1 +
0.05)^2Ppast=176,400÷(1+0.05)2 Ppast=176,400÷1.052P_{\text{past}} = 176,400
\div 1.05^2Ppast=176,400÷1.052 Ppast=176,400÷1.1025P_{\text{past}} = 176,400
\div 1.1025Ppast=176,400÷1.1025 Ppast≈160,000P_{\text{past}} \approx
160,000Ppast≈160,000
So, the population 2 years ago was approximately 160,000.
In the new budget, the
price of kerosene oil rose by 25%. By how much percent must a person reduce his
consumption so that his expenditure on it does not increase?
To determine by what percentage a person must reduce their
consumption of kerosene oil to keep their expenditure constant despite a 25%
price increase, we can use the concept of inverse proportionality between price
and quantity when total expenditure is constant.
Step-by-Step Solution
1.
Define Variables:
o Let the
original price per unit of kerosene oil be PPP.
o Let the
original consumption be QQQ.
o Let the
original expenditure be EEE, where E=P×QE = P \times QE=P×Q.
2.
Price Increase:
o The new
price per unit after a 25% increase is Pnew=P×1.25P_{\text{new}} = P \times
1.25Pnew=P×1.25.
3.
New Consumption:
o Let the new
consumption be QnewQ_{\text{new}}Qnew.
4.
Maintain Constant Expenditure:
o The new
expenditure should equal the original expenditure, so:
Pnew×Qnew=EP_{\text{new}} \times Q_{\text{new}} = EPnew×Qnew=E Substituting
Pnew=P×1.25P_{\text{new}} = P \times 1.25Pnew=P×1.25 and E=P×QE = P \times
QE=P×Q: (P×1.25)×Qnew=P×Q(P \times 1.25) \times Q_{\text{new}} = P \times
Q(P×1.25)×Qnew=P×Q
5.
Solve for QnewQ_{\text{new}}Qnew:
1.25P×Qnew=P×Q1.25P \times Q_{\text{new}} = P \times
Q1.25P×Qnew=P×Q Qnew=P×Q1.25PQ_{\text{new}} = \frac{P \times
Q}{1.25P}Qnew=1.25PP×Q Qnew=Q1.25Q_{\text{new}} = \frac{Q}{1.25}Qnew=1.25Q
Qnew=Q×11.25Q_{\text{new}} = Q \times \frac{1}{1.25}Qnew=Q×1.251
Qnew=Q×0.8Q_{\text{new}} = Q \times 0.8Qnew=Q×0.8
6.
Percentage Reduction in Consumption:
o The
reduction in consumption is Q−QnewQ - Q_{\text{new}}Q−Qnew.
o The
percentage reduction is: Percentage Reduction=Q−QnewQ×100%\text{Percentage
Reduction} = \frac{Q - Q_{\text{new}}}{Q} \times
100\%Percentage Reduction=QQ−Qnew×100%
The value of a machine
depreciates at the rate of 10% per annum. If its present is Rs.1,62,000 what
will be its worth after 2 years? What was the value of the machine 2 years ago?
To calculate the value of the machine after 2 years and its
value 2 years ago given its present value and a depreciation rate of 10% per
annum, we can use the formula for compound depreciation and appreciation.
Depreciation Formula
The value of an asset depreciating at a rate rrr per annum
over nnn years is given by:
V=P×(1−r)nV = P \times (1 - r)^nV=P×(1−r)n
Where:
- VVV is
the future value of the asset.
- PPP is
the present value of the asset.
- rrr is
the depreciation rate.
- nnn is
the number of years.
Appreciation Formula (finding past value)
To find the past value PPP of an asset which depreciates over
nnn years to the present value VVV:
P=V×(1+r)nP = V \times (1 + r)^nP=V×(1+r)n
Given Data:
- Present
value (PPP) = Rs. 1,62,000
- Depreciation
rate (rrr) = 10% = 0.10
- Time
(nnn) = 2 years
Value After 2 Years
Using the depreciation formula:
V=1,62,000×(1−0.10)2V = 1,62,000 \times (1 -
0.10)^2V=1,62,000×(1−0.10)2 V=1,62,000×(0.90)2V = 1,62,000 \times
(0.90)^2V=1,62,000×(0.90)2 V=1,62,000×0.81V = 1,62,000 \times
0.81V=1,62,000×0.81 V=1,62,000×0.81=1,31,220V = 1,62,000 \times 0.81 =
1,31,220V=1,62,000×0.81=1,31,220
So, the value of the machine after 2 years will be Rs.
1,31,220.
Value 2 Years Ago
Using the appreciation formula:
P=1,62,000×(1+0.10)2P = 1,62,000 \times (1 +
0.10)^2P=1,62,000×(1+0.10)2 P=1,62,000×(1.10)2P = 1,62,000 \times (1.10)^2P=1,62,000×(1.10)2
P=1,62,000×1.21P = 1,62,000 \times 1.21P=1,62,000×1.21
P=1,62,000×1.21=1,95,720P = 1,62,000 \times 1.21 =
1,95,720P=1,62,000×1.21=1,95,720
So, the value of the machine 2 years ago was Rs. 1,95,720.
Unit 07: Profit and Loss
7.1 Some Useful Short Cut Methods
7.2 False Weight
7.3 Successive Discount and Marked Price
7.1 Some Useful Short Cut Methods
1.
Profit and Loss Formulas:
o Profit: Profit =
Selling Price (SP) - Cost Price (CP)
o Loss: Loss =
Cost Price (CP) - Selling Price (SP)
o Profit
Percentage: Profit %=(ProfitCP)×100\text{Profit \%} = \left(
\frac{\text{Profit}}{\text{CP}} \right) \times 100Profit %=(CPProfit)×100
o Loss
Percentage: Loss %=(LossCP)×100\text{Loss \%} = \left(
\frac{\text{Loss}}{\text{CP}} \right) \times 100Loss %=(CPLoss)×100
2.
Finding Selling Price:
o SP when
Profit is Given: SP=CP×(1+Profit %100)\text{SP} = \text{CP} \times
\left( 1 + \frac{\text{Profit \%}}{100} \right)SP=CP×(1+100Profit %)
o SP when Loss
is Given: SP=CP×(1−Loss %100)\text{SP} = \text{CP} \times \left(
1 - \frac{\text{Loss \%}}{100} \right)SP=CP×(1−100Loss %)
3.
Finding Cost Price:
o CP when SP
and Profit % are Given: CP=SP1+Profit %100\text{CP} =
\frac{\text{SP}}{1 + \frac{\text{Profit \%}}{100}}CP=1+100Profit %SP
o CP when SP
and Loss % are Given: CP=SP1−Loss %100\text{CP} = \frac{\text{SP}}{1
- \frac{\text{Loss \%}}{100}}CP=1−100Loss %SP
4.
Discount Calculations:
o Single
Discount: If the marked price (MP) of an item is given and a single
discount of d%d\%d% is offered, the selling price (SP) can be calculated as:
SP=MP×(1−d100)\text{SP} = \text{MP} \times \left( 1 - \frac{d}{100}
\right)SP=MP×(1−100d)
7.2 False Weight
1.
Concept of False Weight:
o False Weight refers to a
situation where a seller uses incorrect weights to measure goods, leading to
more profit or less loss than the actual weight would allow. This is considered
unethical and sometimes illegal.
2.
Calculation Example:
o If a seller
claims to use a 1 kg weight but actually uses 900 grams, the profit made due to
the false weight can be calculated.
o Suppose the
cost price per kg is CPCPCP and the selling price per kg is SPSPSP. The seller
actually sells 900 grams at the price of 1 kg.
o Effective
Cost Price for 900 grams: Effective CP=CP×9001000\text{Effective CP} =
\frac{CP \times 900}{1000}Effective CP=1000CP×900
o Profit Due
to False Weight: Profit=SP−Effective CP\text{Profit} = SP -
\text{Effective CP}Profit=SP−Effective CP
o Profit Percentage:
Profit %=(SP−Effective CPEffective CP)×100\text{Profit \%} =
\left( \frac{SP - \text{Effective CP}}{\text{Effective CP}} \right) \times
100Profit %=(Effective CPSP−Effective CP)×100
7.3 Successive Discount and Marked Price
1.
Successive Discount:
o When
multiple discounts are applied successively, the overall discount is not simply
the sum of individual discounts. Instead, the overall effect is calculated by
applying each discount one after the other on the reduced price.
2.
Calculation of Successive Discounts:
o Suppose an
item has a marked price MPMPMP and successive discounts of d1%d_1\%d1% and
d2%d_2\%d2% are given.
o After First
Discount: SP1=MP×(1−d1100)\text{SP}_1 = MP \times \left( 1 -
\frac{d_1}{100} \right)SP1=MP×(1−100d1)
o After Second
Discount on Reduced Price: SP2=SP1×(1−d2100)\text{SP}_2 = \text{SP}_1 \times
\left( 1 - \frac{d_2}{100} \right)SP2=SP1×(1−100d2)
o Combining
these steps gives the final selling price SP2SP_2SP2.
3.
Example:
o Marked Price
(MP) = Rs. 1000
o First
Discount (d_1) = 20%
o Second
Discount (d_2) = 10%
o After First
Discount: SP1=1000×(1−0.20)=1000×0.80=Rs.800\text{SP}_1 = 1000 \times
(1 - 0.20) = 1000 \times 0.80 = Rs. 800SP1=1000×(1−0.20)=1000×0.80=Rs.800
o After Second
Discount: SP2=800×(1−0.10)=800×0.90=Rs.720\text{SP}_2 = 800 \times (1
- 0.10) = 800 \times 0.90 = Rs. 720SP2=800×(1−0.10)=800×0.90=Rs.720
o So, the final
selling price after successive discounts is Rs. 720.
Summary
- Shortcut
methods in profit and loss help quickly find selling price, cost price,
profit, and loss percentages.
- False
weight refers to the unethical practice of using incorrect weights to increase
profit.
- Successive
discounts require applying each discount step by step on the reduced
price, not simply adding the discount percentages.
Summary
The key concepts learned from this unit are:
- Calculation
of Cost Price (CP), Selling Price (SP), and Profit or Gain:
- Cost
Price (CP): The original price at which an item is bought.
- Selling
Price (SP): The price at which an item is sold.
- Profit: The
amount gained when the selling price is more than the cost price.
Profit=SP−CP\text{Profit} = \text{SP} - \text{CP}Profit=SP−CP
- Loss: The
amount lost when the selling price is less than the cost price.
Loss=CP−SP\text{Loss} = \text{CP} - \text{SP}Loss=CP−SP
- Profit
Percentage: The profit expressed as a percentage of the
cost price. Profit %=(ProfitCP)×100\text{Profit \%} = \left(
\frac{\text{Profit}}{\text{CP}} \right) \times
100Profit %=(CPProfit)×100
- Loss
Percentage: The loss expressed as a percentage of the cost
price. Loss %=(LossCP)×100\text{Loss \%} = \left(
\frac{\text{Loss}}{\text{CP}} \right) \times 100Loss %=(CPLoss)×100
- Shortcuts
for Finding Selling Price and Cost Price:
- SP
when Profit is Given: SP=CP×(1+Profit %100)\text{SP} =
\text{CP} \times \left( 1 + \frac{\text{Profit \%}}{100}
\right)SP=CP×(1+100Profit %)
- SP
when Loss is Given: SP=CP×(1−Loss %100)\text{SP} = \text{CP}
\times \left( 1 - \frac{\text{Loss \%}}{100}
\right)SP=CP×(1−100Loss %)
- CP
when SP and Profit % are Given:
CP=SP1+Profit %100\text{CP} = \frac{\text{SP}}{1 +
\frac{\text{Profit \%}}{100}}CP=1+100Profit %SP
- CP
when SP and Loss % are Given:
CP=SP1−Loss %100\text{CP} = \frac{\text{SP}}{1 - \frac{\text{Loss
\%}}{100}}CP=1−100Loss %SP
- False
Weight:
- The
unethical practice of using incorrect weights to measure goods, leading
to increased profits.
- Calculation
involves determining the effective cost price for the false weight and
computing the profit percentage based on this adjusted value.
- Successive
Discount and Marked Price:
- Single
Discount: Calculating the selling price after applying a single
discount to the marked price. SP=MP×(1−d100)\text{SP} = \text{MP} \times
\left( 1 - \frac{d}{100} \right)SP=MP×(1−100d)
- Successive
Discounts: Applying multiple discounts sequentially to find the
final selling price.
- Apply
the first discount to the marked price.
- Apply
the second discount to the reduced price obtained after the first
discount.
- Formula:
SP1=MP×(1−d1100)\text{SP}_1 = \text{MP} \times \left( 1 -
\frac{d_1}{100} \right)SP1=MP×(1−100d1) SP2=SP1×(1−d2100)\text{SP}_2
= \text{SP}_1 \times \left( 1 - \frac{d_2}{100} \right)SP2=SP1×(1−100d2)
These key points help in understanding how to effectively
calculate and analyze profit, loss, cost price, selling price, and discounts in
various business and financial scenarios.
Keywords
Cost Price (CP)
- Definition: The
amount of money spent to purchase an item or produce a product.
- Formula:
CP=Total Cost Incurred\text{CP} = \text{Total Cost
Incurred}CP=Total Cost Incurred
- Example: If
you buy a chair for Rs. 500, then Rs. 500 is the cost price of the chair.
Selling Price (SP)
- Definition: The
price at which an item is sold to a customer.
- Formula:
SP=Cost Price+Profit\text{SP} = \text{Cost Price} +
\text{Profit}SP=Cost Price+Profit
- Example: If
the chair bought for Rs. 500 is sold for Rs. 600, then Rs. 600 is the
selling price.
Profit
- Definition: The financial
gain obtained when the selling price of an item exceeds its cost price.
- Formula:
Profit=SP−CP\text{Profit} = \text{SP} - \text{CP}Profit=SP−CP
- Profit
Percentage: Profit %=(ProfitCP)×100\text{Profit \%} = \left(
\frac{\text{Profit}}{\text{CP}} \right) \times
100Profit %=(CPProfit)×100
- Example: If
the chair bought for Rs. 500 is sold for Rs. 600, the profit is Rs. 100.
Profit %=(100500)×100=20%\text{Profit \%} = \left( \frac{100}{500}
\right) \times 100 = 20\%Profit %=(500100)×100=20%
Gain
- Definition:
Similar to profit, gain refers to the positive difference between the
selling price and the cost price of an item. Often used interchangeably
with profit.
- Formula:
Gain=SP−CP\text{Gain} = \text{SP} - \text{CP}Gain=SP−CP
- Gain
Percentage: Gain %=(GainCP)×100\text{Gain \%} = \left(
\frac{\text{Gain}}{\text{CP}} \right) \times 100Gain %=(CPGain)×100
- Example: If
the chair bought for Rs. 500 is sold for Rs. 600, the gain is Rs. 100.
Gain %=(100500)×100=20%\text{Gain \%} = \left( \frac{100}{500}
\right) \times 100 = 20\%Gain %=(500100)×100=20%
Discount Price
- Definition: The
reduced price at which an item is sold after applying a discount to the
marked price.
- Formula:
Discount Price=Marked Price×(1−Discount %100)\text{Discount
Price} = \text{Marked Price} \times \left( 1 - \frac{\text{Discount
\%}}{100}
\right)Discount Price=Marked Price×(1−100Discount %)
- Example: If a
chair with a marked price of Rs. 800 is sold with a 10% discount, the
discount price is:
Discount Price=800×(1−0.10)=800×0.90=Rs.720\text{Discount Price} =
800 \times (1 - 0.10) = 800 \times 0.90 = Rs.
720Discount Price=800×(1−0.10)=800×0.90=Rs.720
Marked Price (MP)
- Definition: The
original price of an item before any discounts are applied; also known as
the list price.
- Formula:
Marked Price=Discount Price×(1+Discount %100)\text{Marked
Price} = \text{Discount Price} \times \left( 1 + \frac{\text{Discount
\%}}{100}
\right)Marked Price=Discount Price×(1+100Discount %)
- Example: If
the discount price of a chair is Rs. 720 after a 10% discount, the marked
price is: Marked Price=720×(1+0.10)=720×1.10=Rs.792\text{Marked
Price} = 720 \times \left( 1 + 0.10 \right) = 720 \times 1.10 = Rs.
792Marked Price=720×(1+0.10)=720×1.10=Rs.792
These key terms form the basis of calculations in profit and
loss scenarios, helping to determine the cost effectiveness and pricing
strategies in various business operations.
At what % above C.P
must an article be marked so as to gain 33% after allowing a customer a
discount of 5%?
To determine at what percentage above the Cost Price (CP) an
article must be marked to achieve a 33% gain after offering a 5% discount to
the customer, we can follow these steps:
Given:
- Profit
Percentage Desired: 33%
- Discount
Percentage: 5%
Steps to Solve:
1.
Calculate the Selling Price (SP):
o Let xxx be
the percentage above CP at which the article should be marked.
o Therefore,
the Marked Price (MP) can be expressed as CP×(1+x100)\text{CP} \times \left( 1
+ \frac{x}{100} \right)CP×(1+100x).
2.
Apply Discount to the Marked Price:
o After
applying a 5% discount, the selling price (SP) will be: SP=MP×(1−5100)\text{SP}
= \text{MP} \times \left( 1 - \frac{5}{100} \right)SP=MP×(1−1005)
SP=MP×0.95\text{SP} = \text{MP} \times 0.95SP=MP×0.95
3.
Set Up the Profit Equation:
o The profit
is calculated as the difference between SP and CP, which should equal 33% of
CP: Profit=SP−CP\text{Profit} = \text{SP} - \text{CP}Profit=SP−CP
0.33×CP=SP−CP0.33 \times \text{CP} = \text{SP} - \text{CP}0.33×CP=SP−CP
4.
Substitute SP from Step 2:
o Substitute
SP in terms of MP from step 1: 0.33×CP=(MP×0.95)−CP0.33 \times \text{CP} =
\left( \text{MP} \times 0.95 \right) - \text{CP}0.33×CP=(MP×0.95)−CP
5.
Solve for MP:
o Substitute
MP from step 1 into the equation and solve for xxx.
Let's calculate it step by step:
Step-by-Step Calculation:
1.
Set Up the Equation for Profit:
0.33×CP=(CP×(1+x100)×0.95)−CP0.33 \times \text{CP} = \left( \text{CP} \times
\left( 1 + \frac{x}{100} \right) \times 0.95 \right) -
\text{CP}0.33×CP=(CP×(1+100x)×0.95)−CP
2.
Simplify the Equation:
0.33×CP=(CP×0.95×(1+x100))−CP0.33 \times \text{CP} = \left( \text{CP} \times
0.95 \times \left( 1 + \frac{x}{100} \right) \right) -
\text{CP}0.33×CP=(CP×0.95×(1+100x))−CP
3.
Expand and Simplify:
0.33×CP=CP×0.95×(1+x100)−CP0.33 \times \text{CP} = \text{CP} \times 0.95 \times
\left( 1 + \frac{x}{100} \right) - \text{CP}0.33×CP=CP×0.95×(1+100x)−CP
4.
Divide through by CP (assuming CP ≠ 0):
0.33=0.95×(1+x100)−10.33 = 0.95 \times \left( 1 + \frac{x}{100} \right) -
10.33=0.95×(1+100x)−1
5.
Isolate xxx: 0.33+1=0.95×(1+x100)0.33 + 1 =
0.95 \times \left( 1 + \frac{x}{100} \right)0.33+1=0.95×(1+100x)
1.33=0.95×(1+x100)1.33 = 0.95 \times \left( 1 + \frac{x}{100}
\right)1.33=0.95×(1+100x)
6.
Divide by 0.95: 1.330.95=1+x100\frac{1.33}{0.95}
= 1 + \frac{x}{100}0.951.33=1+100x 1.4=1+x1001.4 = 1 +
\frac{x}{100}1.4=1+100x
7.
Subtract 1 and Multiply by 100:
0.4=x1000.4 = \frac{x}{100}0.4=100x x=0.4×100x = 0.4 \times 100x=0.4×100 x=40x
= 40x=40
Answer:
Therefore, the article must be marked at 40% above the
Cost Price to achieve a 33% gain after allowing a 5% discount to the
customer.
An uneducated retailer
marks all its goods at 50% above the cost price and thinking that he will still
make 25% profit, offers a discount of 25% on the market price. what is the actual
profit on the sales?
To determine the actual profit on sales after the retailer's
pricing strategy, let's break down the given information and calculate step by
step:
Given:
- The
retailer marks all goods at 50% above the Cost Price (CP).
- The
retailer expects to make a 25% profit.
- The
retailer offers a 25% discount on the marked price to customers.
Steps to Calculate Actual Profit:
1.
Calculate the Selling Price (SP):
o The retailer
marks goods at 50% above CP, so the Marked Price (MP) is: MP=CP×(1+50100)\text{MP}
= \text{CP} \times \left( 1 + \frac{50}{100} \right)MP=CP×(1+10050)
MP=CP×1.5\text{MP} = \text{CP} \times 1.5MP=CP×1.5
2.
Apply the Expected Profit Margin:
o The retailer
expects to make a 25% profit on the CP. Therefore, the Selling Price (SP) with
the expected profit margin is: SP=CP×(1+25100)\text{SP} = \text{CP} \times
\left( 1 + \frac{25}{100} \right)SP=CP×(1+10025) SP=CP×1.25\text{SP} =
\text{CP} \times 1.25SP=CP×1.25
3.
Apply the Discount:
o The retailer
offers a 25% discount on the marked price to customers. So, the Selling Price
(SP) after the discount is: SP=MP×(1−25100)\text{SP} = \text{MP} \times \left(
1 - \frac{25}{100} \right)SP=MP×(1−10025) SP=MP×0.75\text{SP} = \text{MP}
\times 0.75SP=MP×0.75
4.
Determine the Actual Profit:
o Compare the
actual Selling Price (after discount) with the Cost Price (CP) to find the
actual profit: Actual Profit=SP after discount−CP\text{Actual
Profit} = \text{SP after discount} -
\text{CP}Actual Profit=SP after discount−CP
Calculation:
Let's calculate step by step:
- Step 1:
Calculate MP MP=CP×1.5\text{MP} = \text{CP} \times
1.5MP=CP×1.5
- Step 2:
Calculate SP with Expected Profit Margin SP=CP×1.25\text{SP} =
\text{CP} \times 1.25SP=CP×1.25
- Step 3:
Calculate SP after Discount
SP after discount=MP×0.75\text{SP after discount} = \text{MP}
\times 0.75SP after discount=MP×0.75 Substitute MP from Step 1:
SP after discount=(CP×1.5)×0.75\text{SP after discount} =
(\text{CP} \times 1.5) \times
0.75SP after discount=(CP×1.5)×0.75 SP after discount=CP×1.125\text{SP
after discount} = \text{CP} \times
1.125SP after discount=CP×1.125
- Step 4:
Calculate Actual Profit
Actual Profit=SP after discount−CP\text{Actual Profit} =
\text{SP after discount} -
\text{CP}Actual Profit=SP after discount−CP
Actual Profit=CP×1.125−CP\text{Actual Profit} = \text{CP} \times
1.125 - \text{CP}Actual Profit=CP×1.125−CP
Actual Profit=CP×(1.125−1)\text{Actual Profit} = \text{CP} \times
(1.125 - 1)Actual Profit=CP×(1.125−1)
Actual Profit=CP×0.125\text{Actual Profit} = \text{CP} \times
0.125Actual Profit=CP×0.125
Answer:
The actual profit on the sales, after offering a 25% discount
on the marked price, is 12.5% of the Cost Price (CP).
A man bought a horse
and a carriage for Rs 3000.he sold the horse at a gain of 20% and the carriage
at a loss of 10%, thereby gaining 2% on the whole. find the cost of the horse.
To find the cost of the horse, let's break down the given
information and solve step by step:
Given:
- Total
cost of the horse and carriage = Rs 3000
- Sold
the horse at a gain of 20%
- Sold
the carriage at a loss of 10%
- Overall
gain of 2%
Steps to Solve:
1.
Let's denote:
o Cost of the
horse as ChC_hCh
o Cost of the
carriage as CcC_cCc
2.
Set up the equations based on the given information:
o Total cost
equation: Ch+Cc=3000C_h + C_c = 3000Ch+Cc=3000
3.
Calculate the selling prices:
o Selling
price of the horse (after 20% gain): SPh=Ch×(1+20100)=Ch×1.2\text{SP}_h = C_h
\times \left( 1 + \frac{20}{100} \right) = C_h \times
1.2SPh=Ch×(1+10020)=Ch×1.2
o Selling
price of the carriage (after 10% loss): SPc=Cc×(1−10100)=Cc×0.9\text{SP}_c =
C_c \times \left( 1 - \frac{10}{100} \right) = C_c \times
0.9SPc=Cc×(1−10010)=Cc×0.9
4.
Calculate the overall selling price and the gain:
o Total
selling price: Total SP=SPh+SPc\text{Total SP} = \text{SP}_h +
\text{SP}_cTotal SP=SPh+SPc Total SP=Ch×1.2+Cc×0.9\text{Total SP} =
C_h \times 1.2 + C_c \times 0.9Total SP=Ch×1.2+Cc×0.9
5.
Calculate the overall gain condition:
o Overall gain
condition: Overall Gain=Total SP−Total Cost\text{Overall Gain} =
\text{Total SP} - \text{Total Cost}Overall Gain=Total SP−Total Cost
Overall Gain=Total SP−3000\text{Overall Gain} = \text{Total SP} -
3000Overall Gain=Total SP−3000
Overall Gain=0.02×3000\text{Overall Gain} = 0.02 \times
3000Overall Gain=0.02×3000
6.
Substitute and solve the equations:
o Substitute
the values and equations to find ChC_hCh:
From Ch+Cc=3000C_h + C_c = 3000Ch+Cc=3000, solve for
CcC_cCc: Cc=3000−ChC_c = 3000 - C_hCc=3000−Ch
Substitute CcC_cCc into the overall gain equation:
Ch×1.2+(3000−Ch)×0.9=3000+0.02×3000C_h \times 1.2 + (3000 - C_h) \times 0.9 =
3000 + 0.02 \times 3000Ch×1.2+(3000−Ch)×0.9=3000+0.02×3000
Simplify and solve for ChC_hCh:
1.2Ch+2700−0.9Ch=30601.2 C_h + 2700 - 0.9 C_h =
30601.2Ch+2700−0.9Ch=3060 0.3Ch=3600.3 C_h = 3600.3Ch=360 Ch=3600.3C_h =
\frac{360}{0.3}Ch=0.3360 Ch=1200C_h = 1200Ch=1200
Answer:
The cost of the horse is Rs 1200.
Alfred buys an old
scooter for Rs. 4700 and spends Rs. 800 on its repairs. If he sells the scooter
for Rs. 5800, his gain percent is?
To find Alfred's gain percentage when he sells the scooter,
let's calculate step by step:
Given:
- Cost
price of the scooter = Rs. 4700
- Repair
expenses = Rs. 800
- Total
cost price = Rs. 4700 + Rs. 800 = Rs. 5500
- Selling
price = Rs. 5800
Steps to Calculate Gain Percentage:
1.
Calculate the Total Cost Price (CP):
Total CP=Cost Price+Repair Expenses\text{Total CP} = \text{Cost
Price} + \text{Repair
Expenses}Total CP=Cost Price+Repair Expenses
Total CP=4700+800=5500\text{Total CP} = 4700 + 800 =
5500Total CP=4700+800=5500
2.
Calculate the Profit:
o Profit =
Selling Price - Total Cost Price Profit=5800−5500=300\text{Profit} = 5800 -
5500 = 300Profit=5800−5500=300
3.
Calculate the Gain Percentage:
o Gain
Percentage = ( \left( \frac{\text{Profit}}{\text{Total CP}} \right) \times 100%
] Gain Percentage=(3005500)×100%\text{Gain Percentage} = \left(
\frac{300}{5500} \right) \times 100\%Gain Percentage=(5500300)×100%
Gain Percentage=(3005500)×100%\text{Gain Percentage} = \left(
\frac{300}{5500} \right) \times 100\%Gain Percentage=(5500300)×100% Gain Percentage≈5.45%\text{Gain
Percentage} ≈ 5.45\%Gain Percentage≈5.45%
Answer:
Alfred's gain percentage when he sells the scooter is
approximately 5.45%.
Unit 08: Simple Interest
8.1 Basics of Principal, Rate and Time
8.2 Computation of Simple Interest
8.3 Effect of Change of P, R & T on S.I.
8.4 Computation of Principal
8.5 Computation of Rate
8.6 Computation of Time
8.7 Computation of Amount
8.1 Basics of Principal, Rate, and Time
- Principal
(P):
- Definition:
The initial amount of money invested or loaned.
- Symbol:
PPP
- Example:
If you deposit Rs. 5000 into a savings account, Rs. 5000 is the
principal.
- Rate of
Interest (R):
- Definition:
The percentage charged or paid on the principal amount over a period of
time.
- Symbol:
RRR
- Example:
An interest rate of 5% per annum means R = 5%.
- Time
(T):
- Definition:
The duration for which the principal amount is borrowed or invested,
usually expressed in years or fractions of a year.
- Symbol:
TTT
- Example:
If you borrow Rs. 10,000 for 2 years, then T=2T = 2T=2 years.
8.2 Computation of Simple Interest
- Simple
Interest (SI):
- Definition:
The interest calculated only on the principal amount.
- Formula:
SI=P×R×T100SI = \frac{P \times R \times T}{100}SI=100P×R×T
- Example:
If P = Rs. 1000, R = 8%, T = 2 years, SI=1000×8×2100=Rs.160SI =
\frac{1000 \times 8 \times 2}{100} = Rs. 160SI=1001000×8×2=Rs.160
8.3 Effect of Change of P, R & T on S.I.
- Effect
of Change in Principal (P):
- Increasing
P increases SI, assuming R and T remain constant.
- Example:
If P increases from Rs. 1000 to Rs. 2000, SI will also increase
proportionately.
- Effect
of Change in Rate of Interest (R):
- Increasing
R increases SI, assuming P and T remain constant.
- Example:
If R increases from 5% to 10%, SI will double.
- Effect
of Change in Time (T):
- Increasing
T increases SI, assuming P and R remain constant.
- Example:
If T increases from 2 years to 4 years, SI will double.
8.4 Computation of Principal
- Finding
Principal (P):
- Formula
(from SI formula): P=100×SIR×TP = \frac{100 \times SI}{R \times
T}P=R×T100×SI
- Example:
If SI = Rs. 400, R = 10%, T = 2 years, P=100×40010×2=Rs.2000P = \frac{100
\times 400}{10 \times 2} = Rs. 2000P=10×2100×400=Rs.2000
8.5 Computation of Rate
- Finding
Rate of Interest (R):
- Formula
(from SI formula): R=100×SIP×TR = \frac{100 \times SI}{P \times
T}R=P×T100×SI
- Example:
If SI = Rs. 250, P = Rs. 2000, T = 3 years, R=100×2502000×3=4.17%R =
\frac{100 \times 250}{2000 \times 3} = 4.17\%R=2000×3100×250=4.17%
8.6 Computation of Time
- Finding
Time (T):
- Formula
(from SI formula): T=100×SIP×RT = \frac{100 \times SI}{P \times
R}T=P×R100×SI
- Example:
If SI = Rs. 300, P = Rs. 1500, R = 6%, T=100×3001500×6=3 yearsT =
\frac{100 \times 300}{1500 \times 6} = 3 \text{
years}T=1500×6100×300=3 years
8.7 Computation of Amount
- Amount
(A):
- Definition:
The total money accumulated by adding the principal and the interest.
- Formula:
A=P+SIA = P + SIA=P+SI
- Example:
If P = Rs. 2000, SI = Rs. 400, A=2000+400=Rs.2400A = 2000 + 400 = Rs.
2400A=2000+400=Rs.2400
These points cover the fundamental concepts and calculations
related to Simple Interest (SI), providing a comprehensive overview of how
Principal (P), Rate (R), Time (T), Simple Interest (SI), and Amount (A) are
interconnected and calculated in various scenarios.
Summary: Simple Interest Concepts
1.
Simple Interest (SI):
o Definition:
Simple Interest is calculated on the principal amount over a period of time at
a fixed rate.
o Formula:
SI=P×R×T100SI = \frac{P \times R \times T}{100}SI=100P×R×T where PPP is the
Principal, RRR is the Rate of Interest per annum, and TTT is the Time in years.
2.
Principal (P):
o Definition:
The initial amount of money invested or borrowed.
o Formula:
P=100×SIR×TP = \frac{100 \times SI}{R \times T}P=R×T100×SI Principal is the
base amount on which interest is calculated.
3.
Rate of Interest (R):
o Definition:
The percentage charged or earned on the principal amount over a period of time.
o Formula:
R=100×SIP×TR = \frac{100 \times SI}{P \times T}R=P×T100×SI Rate of Interest
determines the percentage of the principal that will be paid as interest over
one year.
4.
Time (T):
o Definition:
The duration for which the principal amount is borrowed or invested, usually
expressed in years.
o Formula: T=100×SIP×RT
= \frac{100 \times SI}{P \times R}T=P×R100×SI Time represents the period for
which the money is invested or borrowed.
5.
Amount (A):
o Definition:
The total money accumulated by adding the principal and the interest.
o Formula:
A=P+SIA = P + SIA=P+SI Amount is the total sum of the principal and the
interest earned or paid.
6.
Relationships Between Amount, Principal, and Interest:
o Principal
Calculation: P=A−SIP = A - SIP=A−SI Principal can be found by
subtracting the Simple Interest from the Amount.
o Interest
Calculation: SI=A−PSI = A - PSI=A−P Simple Interest is determined by
subtracting the Principal from the Amount.
Example Application:
If the Principal (P) is Rs. 2000, the Rate of Interest (R) is
5% per annum, and Time (T) is 3 years:
- Calculate
Simple Interest (SI): SI=2000×5×3100=Rs.300SI = \frac{2000 \times 5
\times 3}{100} = Rs. 300SI=1002000×5×3=Rs.300
- Calculate
Amount (A): A=2000+300=Rs.2300A = 2000 + 300 = Rs.
2300A=2000+300=Rs.2300
- Verify
Principal (P) Calculation: P=100×3005×3=Rs.2000P =
\frac{100 \times 300}{5 \times 3} = Rs. 2000P=5×3100×300=Rs.2000
This summary provides a clear understanding of how Simple
Interest is calculated using Principal, Rate, and Time, and how these concepts
interrelate to determine the Amount and Interest in financial transactions.
Keywords in Simple Interest
1.
Principal (P):
o Definition:
The original amount of money invested or borrowed.
o Symbol: PPP
o Example: If
you borrow Rs. 5000 from a bank, Rs. 5000 is the principal amount.
2.
Rate of Interest (R):
o Definition:
The percentage charged or paid on the principal amount over a period of time.
o Symbol: RRR
o Example: An
interest rate of 8% per annum means R = 8%.
3.
Time (T):
o Definition:
The duration for which the principal amount is borrowed or invested, usually
expressed in years.
o Symbol: TTT
o Example: If
you borrow Rs. 10,000 for 3 years, then T=3T = 3T=3 years.
4.
Simple Interest (SI):
o Definition:
The interest calculated only on the original principal amount.
o Formula:
SI=P×R×T100SI = \frac{P \times R \times T}{100}SI=100P×R×T
o Example: If
P = Rs. 2000, R = 10%, T = 2 years, SI=2000×10×2100=Rs.400SI = \frac{2000
\times 10 \times 2}{100} = Rs. 400SI=1002000×10×2=Rs.400
5.
Amount (A):
o Definition:
The total money accumulated by adding the principal and the interest.
o Formula:
A=P+SIA = P + SIA=P+SI
o Example: If
P = Rs. 3000, SI = Rs. 600, A=3000+600=Rs.3600A = 3000 + 600 = Rs.
3600A=3000+600=Rs.3600
Relationships and Formulas:
- Principal
Calculation: P=100×SIR×TP = \frac{100 \times SI}{R \times
T}P=R×T100×SI Principal can be calculated using Simple Interest, Rate,
and Time.
- Rate
Calculation: R=100×SIP×TR = \frac{100 \times SI}{P \times
T}R=P×T100×SI Rate of Interest can be determined using Principal, Simple
Interest, and Time.
- Time
Calculation: T=100×SIP×RT = \frac{100 \times SI}{P \times
R}T=P×R100×SI Time can be found using Principal, Rate of Interest, and
Simple Interest.
Example Application:
If you invest Rs. 5000 at an interest rate of 6% per annum
for 2 years:
- Calculate
Simple Interest (SI): SI=5000×6×2100=Rs.600SI = \frac{5000 \times 6 \times
2}{100} = Rs. 600SI=1005000×6×2=Rs.600
- Calculate
Amount (A): A=5000+600=Rs.5600A = 5000 + 600 = Rs.
5600A=5000+600=Rs.5600
- Verify
Principal (P) Calculation: P=100×6006×2=Rs.5000P =
\frac{100 \times 600}{6 \times 2} = Rs. 5000P=6×2100×600=Rs.5000
This breakdown provides a clear understanding of how
Principal, Rate of Interest, Time, Simple Interest, and Amount are
interconnected and calculated in various financial scenarios involving simple
interest calculations.
If the simple interest
on Rs. 1400 be more than the interest on Rs. 1000 by Rs. 60 in 5 years, then
find the out the rate per cent per annum.
To find the rate of interest per annum, we can use the given
information and the formula for Simple Interest (SI).
Given:
- Principal
amount P1=Rs.1000P_1 = Rs. 1000P1=Rs.1000
- Principal
amount P2=Rs.1400P_2 = Rs. 1400P2=Rs.1400
- Time
T=5T = 5T=5 years
- Difference
in Simple Interest SI2−SI1=Rs.60SI_2 - SI_1 = Rs. 60SI2−SI1=Rs.60
Steps to Solve:
1.
Calculate Simple Interest for Rs. 1000: SI1=P1×R×T100SI_1
= \frac{P_1 \times R \times T}{100}SI1=100P1×R×T
2.
Calculate Simple Interest for Rs. 1400:
SI2=P2×R×T100SI_2 = \frac{P_2 \times R \times T}{100}SI2=100P2×R×T
3.
Set up the Equation based on the given
difference in interest: SI2−SI1=60SI_2 - SI_1 = 60SI2−SI1=60 Substitute the
expressions for SI1SI_1SI1 and SI2SI_2SI2:
1400×R×5100−1000×R×5100=60\frac{1400 \times R \times 5}{100} - \frac{1000
\times R \times 5}{100} = 601001400×R×5−1001000×R×5=60
4.
Simplify and solve the equation:
1400R−1000R=601400R - 1000R = 601400R−1000R=60 400R=60400R = 60400R=60
5.
Calculate the rate of interest per annum (R): R=60400R =
\frac{60}{400}R=40060 R=0.15R = 0.15R=0.15
6.
Convert to percentage (since rate
is usually expressed as a percentage): R=0.15×100%R = 0.15 \times 100\%R=0.15×100%
R=15%R = 15\%R=15%
Conclusion:
The rate of interest
per annum is 15%. Therefore, the rate per cent per annum is 15%.
A certain sum is invested for certain time. It amounts to Rs.
450 at 7% per annum. But, when invested at 5% per annum, it amounts to `350.
Find out the sum and time.
To find the principal amount (sum) and the time, we can use
the given information about two different investments with their respective
interest rates and amounts.
Given:
- Amount
after investing at 7% per annum: Rs. 450
- Amount
after investing at 5% per annum: Rs. 350
Let's denote:
- P as the
principal sum (the amount invested)
- T as the
time period (in years)
Using the formula for Simple Interest:
1.
For the investment at 7% per annum: SI1=P×7×T100SI_1 =
\frac{P \times 7 \times T}{100}SI1=100P×7×T
2.
For the investment at 5% per annum: SI2=P×5×T100SI_2 =
\frac{P \times 5 \times T}{100}SI2=100P×5×T
Setting up equations:
From the given amounts:
- At 7%
interest: P+SI1=450P + SI_1 = 450P+SI1=450
- At 5%
interest: P+SI2=350P + SI_2 = 350P+SI2=350
Substitute the expressions for SI:
P+P×7×T100=450P + \frac{P \times 7 \times T}{100} =
450P+100P×7×T=450 P+7PT100=450P + \frac{7PT}{100} = 450P+1007PT=450 ...(1)
P+P×5×T100=350P + \frac{P \times 5 \times T}{100} = 350P+100P×5×T=350
P+5PT100=350P + \frac{5PT}{100} = 350P+1005PT=350 ...(2)
Solving the equations:
Subtract equation (2) from equation (1):
7PT100−5PT100=450−350\frac{7PT}{100} - \frac{5PT}{100} = 450
- 3501007PT−1005PT=450−350 2PT100=100\frac{2PT}{100} = 1001002PT=100
2PT=100002PT = 100002PT=10000 PT=5000PT = 5000PT=5000
Now, we have PT=5000PT = 5000PT=5000.
Finding P and T:
From equation (1): P+7PT100=450P + \frac{7PT}{100} =
450P+1007PT=450 Substituting PT=5000PT = 5000PT=5000: P+7×5000100=450P +
\frac{7 \times 5000}{100} = 450P+1007×5000=450 P+350=450P + 350 = 450P+350=450
P=450−350P = 450 - 350P=450−350 P=100P = 100P=100
So, the principal sum PPP is Rs. 100.
Now, find TTT: PT=5000PT = 5000PT=5000 100×T=5000100 \times T
= 5000100×T=5000 T=5000100T = \frac{5000}{100}T=1005000 T=50 yearsT = 50
\text{ years}T=50 years
Conclusion:
The principal sum (amount invested) is Rs. 100, and the time
period (T) is 50 years. These values satisfy both conditions given in the
problem statement.
What is the present
worth of Rs.132 due in 2 years at 5% simple interest per annum?
To find the present worth (present value) of Rs. 132 due in 2
years at 5% simple interest per annum, we can use the formula for Simple
Interest to discount the future amount back to the present.
Given:
- Future
amount (Amount due in 2 years): Rs. 132
- Time
(T): 2 years
- Rate of
Interest (R): 5% per annum
Formula for Simple Interest:
The formula for Simple Interest is: SI=P×R×T100SI = \frac{P
\times R \times T}{100}SI=100P×R×T
Where:
- PPP is
the Principal (or present worth)
- RRR is
the Rate of Interest per annum
- TTT is
the Time period in years
Steps to Calculate Present Worth:
1.
Calculate Simple Interest (SI) for Rs. 132
over 2 years at 5% per annum: SI=132×5×2100SI = \frac{132 \times 5 \times
2}{100}SI=100132×5×2 SI=132×10100SI = \frac{132 \times 10}{100}SI=100132×10
SI=13.2SI = 13.2SI=13.2
2.
Find the Present Worth (P): Since the
present worth is the amount that, when invested at 5% simple interest for 2
years, would accumulate to Rs. 132, we subtract the Simple Interest from the
Future Amount: P=132−SIP = 132 - SIP=132−SI P=132−13.2P = 132 - 13.2P=132−13.2
P=118.8P = 118.8P=118.8
Conclusion:
The present worth
(present value) of Rs. 132 due in 2 years at 5% simple interest per annum is
Rs. 118.8.
It seems there's an inconsistency in the calculation. Let's
correct that and find the accurate present worth (present value) of Rs. 132 due
in 2 years at 5% simple interest per annum.
To find the present worth using simple interest, we typically
use the formula to discount the future amount back to the present.
Given:
- Future
amount (Amount due in 2 years): Rs. 132
- Time
(T): 2 years
- Rate of
Interest (R): 5% per annum
Formula for Simple Interest:
The formula for Simple Interest is: SI=P×R×T100SI = \frac{P
\times R \times T}{100}SI=100P×R×T
Where:
- PPP is
the Principal (or present worth)
- RRR is
the Rate of Interest per annum
- TTT is
the Time period in years
Steps to Calculate Present Worth:
1.
Calculate Simple Interest (SI) for Rs. 132
over 2 years at 5% per annum: SI=132×5×2100SI = \frac{132 \times 5 \times
2}{100}SI=100132×5×2 SI=132×10100SI = \frac{132 \times 10}{100}SI=100132×10
SI=13.2SI = 13.2SI=13.2
2.
Find the Present Worth (P): Since the
present worth is the amount that, when invested at 5% simple interest for 2
years, would accumulate to Rs. 132, we need to subtract the Simple Interest
from the Future Amount to find the Present Worth: P=1321+(5100)×2P =
\frac{132}{1 + \left(\frac{5}{100}\right) \times 2}P=1+(1005)×2132 P=1321+0.1P
= \frac{132}{1 + 0.1}P=1+0.1132 P=1321.1P = \frac{132}{1.1}P=1.1132 P=120P =
120P=120
Conclusion:
The correct present worth (present value) of Rs. 132 due in 2
years at 5% simple interest per annum is Rs. 120. This means that Rs. 120
invested today at 5% per annum simple interest would grow to Rs. 132 in 2
years.
Out of a certain sum,
is invested at 3%, at 6% and the rest at 8%. If the annual income is Rs. 300,
then find original sum.
To find the original sum invested at different interest rates
given the annual income, we can set up equations based on the information
provided.
Let's denote:
- xxx as
the amount invested at 3%,
- yyy as
the amount invested at 6%,
- zzz as
the amount invested at 8%,
Given:
- Total
annual income from investments x+y+z=300x + y + z = 300x+y+z=300 (in Rs.)
Setting up Equations:
1.
Income from the investment at 3%:
3100×x=0.03x\frac{3}{100} \times x = 0.03x1003×x=0.03x
2.
Income from the investment at 6%:
6100×y=0.06y\frac{6}{100} \times y = 0.06y1006×y=0.06y
3.
Income from the investment at 8%:
8100×z=0.08z\frac{8}{100} \times z = 0.08z1008×z=0.08z
Given the total annual income: 0.03x+0.06y+0.08z=3000.03x +
0.06y + 0.08z = 3000.03x+0.06y+0.08z=300
Solving the System of Equations:
Now we have two equations:
1.
x+y+z=300x + y + z = 300x+y+z=300
2.
0.03x+0.06y+0.08z=3000.03x + 0.06y + 0.08z =
3000.03x+0.06y+0.08z=300
Let's solve these equations step-by-step:
From equation 1, express zzz in terms of xxx and yyy:
z=300−x−yz = 300 - x - yz=300−x−y
Substitute zzz into equation 2: 0.03x+0.06y+0.08(300−x−y)=3000.03x
+ 0.06y + 0.08(300 - x - y) = 3000.03x+0.06y+0.08(300−x−y)=300
Simplify and solve for xxx and yyy:
0.03x+0.06y+24−0.08x−0.08y=3000.03x + 0.06y + 24 - 0.08x -
0.08y = 3000.03x+0.06y+24−0.08x−0.08y=300 −0.05x−0.02y=276-0.05x - 0.02y = 276−0.05x−0.02y=276
Multiply through by -1 to simplify: 0.05x+0.02y=−2760.05x +
0.02y = -2760.05x+0.02y=−276
Now, use the substitution or elimination method to solve for
xxx and yyy. Once you find xxx and yyy, you can calculate z=300−x−yz = 300 - x
- yz=300−x−y.
Conclusion:
The original sum invested x+y+zx + y + zx+y+z, where xxx,
yyy, and zzz are invested at 3%, 6%, and 8% respectively, will be the total sum
that yields an annual income of Rs. 300 according to the given interest rates.
Unit 09: Compound Interest
9.1 Compound Interest
9.2 Computation of Compound Interest
9.3 Computation of Principal
9.4 Computation of Rate or Rate of Interest
9.5 Computation of Amount
9.6 Relation between Compound and Simple Interest
9.1 Compound Interest
1.
Definition: Compound interest is the interest
calculated on the initial principal and also on the accumulated interest of
previous periods.
2.
Formula: The formula to calculate compound
interest (CI) is: A=P(1+R100)nA = P \left(1 +
\frac{R}{100}\right)^nA=P(1+100R)n Where:
o AAA is the
amount after nnn years,
o PPP is the
principal amount (initial sum),
o RRR is the
annual interest rate,
o nnn is the
number of years.
3.
Calculation Example: For example, if you have
Rs. 1000 as principal at 5% per annum compounded annually for 2 years: A=1000(1+5100)2A
= 1000 \left(1 + \frac{5}{100}\right)^2A=1000(1+1005)2 A=1000(1+0.05)2A = 1000
\left(1 + 0.05\right)^2A=1000(1+0.05)2 A=1000×1.1025A = 1000 \times
1.1025A=1000×1.1025 A=1102.50A = 1102.50A=1102.50 So, the amount after 2 years
would be Rs. 1102.50.
9.2 Computation of Compound Interest
1.
Formula: Compound Interest (CI) can be
calculated using the formula: CI=A−PCI = A - PCI=A−P Where:
o CICICI is
the compound interest,
o AAA is the
amount after nnn years,
o PPP is the
principal amount.
2.
Calculation Example: Using the example above,
CI=1102.50−1000CI = 1102.50 - 1000CI=1102.50−1000 CI=102.50CI = 102.50CI=102.50
So, the compound interest earned over 2 years would be Rs. 102.50.
9.3 Computation of Principal
1.
Finding Principal: If you know the amount,
rate, and time, and you want to find the principal amount: P=A(1+R100)nP =
\frac{A}{\left(1 + \frac{R}{100}\right)^n}P=(1+100R)nA
2.
Example: If the amount after 3 years at 8%
compounded annually is Rs. 1250: P=1250(1+8100)3P = \frac{1250}{\left(1 +
\frac{8}{100}\right)^3}P=(1+1008)31250
9.4 Computation of Rate or Rate of Interest
1.
Finding Rate: If you know the principal,
amount, and time, and you want to find the rate of interest: R=100×((AP)1n−1)R
= 100 \times \left(\left(\frac{A}{P}\right)^{\frac{1}{n}} - 1\right)R=100×((PA)n1−1)
2.
Example: If Rs. 2000 grows to Rs. 2420 in
2 years: R=100×((24202000)12−1)R = 100 \times
\left(\left(\frac{2420}{2000}\right)^{\frac{1}{2}} -
1\right)R=100×((20002420)21−1)
9.5 Computation of Amount
1.
Finding Amount: If you know the principal, rate,
and time, and you want to find the amount: A=P(1+R100)nA = P \left(1 +
\frac{R}{100}\right)^nA=P(1+100R)n
2.
Example: If Rs. 5000 is invested at 6%
compounded semi-annually for 3 years: A=5000(1+6200)6A = 5000 \left(1 +
\frac{6}{200}\right)^6A=5000(1+2006)6
9.6 Relation between Compound and Simple Interest
1.
Difference: Compound interest generally
yields higher returns compared to simple interest over the same period,
especially for longer durations.
2.
Formula: The difference between compound
interest (CI) and simple interest (SI) for nnn years at R%R \%R% per annum is
given by: CI−SI=P[(1+R100)n−1]−P(R100)×nCI - SI = P \left[\left(1 +
\frac{R}{100}\right)^n - 1\right] - P \left(\frac{R}{100}\right) \times
nCI−SI=P[(1+100R)n−1]−P(100R)×n
3.
Effect: Compound interest grows
exponentially while simple interest grows linearly.
Summary
- Compound
interest involves interest on interest, increasing the total amount over
time.
- It's
crucial for long-term investments and savings.
- Understanding
the relationship between principal, rate, time, and amount helps in
financial planning and investment decisions.
This breakdown should provide a comprehensive overview of
Unit 09 on Compound Interest.
Summary: Key Concepts Learned in Unit 09 - Compound Interest
1.
Basic Concept of Compound Interest (C.I.):
o Compound
interest is the interest calculated on the initial principal and also on the
accumulated interest from previous periods.
o Formula:
A=P(1+R100)nA = P \left(1 + \frac{R}{100}\right)^nA=P(1+100R)n
§ AAA: Amount
after nnn years
§ PPP:
Principal amount (initial sum)
§ RRR: Annual
interest rate
§ nnn: Number
of years
2.
Formula-Based Problem Solving:
o We practiced
applying the compound interest formula to calculate:
§ Amount (A)
accumulated after a certain period.
§ Compound
interest (CI) earned over a period.
§ Principal
(P), given the amount (A) after a certain period.
3.
Difference Between Simple Interest (S.I.) and Compound
Interest (C.I.):
o Simple
Interest: Calculated only on the principal amount.
§ Formula:
SI=P×R×T100SI = \frac{P \times R \times T}{100}SI=100P×R×T
o Compound
Interest: Includes interest on interest, leading to higher returns
over time compared to simple interest.
§ Formula:
CI=A−PCI = A - PCI=A−P
4.
Solving Problems Based on S.I. and C.I. Differences:
o We explored
scenarios where:
§ Compound
interest significantly outpaces simple interest due to its compounding effect
over multiple periods.
§ Calculated
the difference in interest earned over a specified duration using both methods.
Practical Application and Importance:
- Financial
Planning: Understanding compound interest is crucial for
financial planning, as it affects savings growth and investment returns
over time.
- Investment
Decisions: Helps in making informed decisions about long-term
investments based on potential compound growth.
- Comparative
Analysis: Enables comparison between different investment
options based on their compound interest projections.
Conclusion:
Unit 09 provided a solid foundation in compound interest,
covering its basic concept, formula applications, and distinguishing it from
simple interest. This knowledge equips us to calculate returns accurately over
time, aiding in financial decision-making and long-term financial goals.
Keywords in Compound Interest
1.
Principal:
o Definition: The
initial amount of money invested or borrowed.
o Symbol: PPP
o Importance: It forms
the basis on which interest is calculated.
2.
Rate:
o Definition: The
percentage at which interest is charged or earned per unit of time (usually per
annum).
o Symbol: RRR
o Importance: Determines
the growth rate of the principal amount over time.
3.
Time:
o Definition: The
duration for which the money is invested or borrowed, typically in years.
o Symbol: nnn
o Importance: Affects
the total interest earned or paid, influencing the final amount.
4.
Compound Interest (C.I.):
o Definition: Interest
calculated on both the initial principal and the accumulated interest from
previous periods.
o Formula:
A=P(1+R100)nA = P \left(1 + \frac{R}{100}\right)^nA=P(1+100R)n
§ AAA: Amount
after nnn years
§ PPP:
Principal amount
§ RRR: Annual
interest rate
§ nnn: Number
of compounding periods
o Importance: Reflects
the exponential growth of investments or debt over time.
5.
Amount:
o Definition: The total
money accumulated after interest has been added to the principal.
o Symbol: AAA
o Formula: A=P+CIA =
P + CIA=P+CI
§ CICICI: Compound
Interest
o Importance: Represents
the final value of an investment or the total debt owed after including
interest.
6.
Difference between C.I. and S.I.:
o Simple
Interest (S.I.):
§ Interest
calculated only on the principal amount.
§ Formula:
SI=P×R×T100SI = \frac{P \times R \times T}{100}SI=100P×R×T
o Compound
Interest (C.I.):
§ Interest
calculated on the initial principal and also on the accumulated interest from
previous periods.
§ Formula:
CI=A−PCI = A - PCI=A−P
o Key
Differences:
§ C.I. yields
higher returns over time compared to S.I. due to compounding.
§ S.I. remains
constant for each time period, while C.I. increases with each compounding
period.
Practical Application:
- Investment
Planning: Helps in understanding how investments grow over time
with compound interest.
- Loan
Calculations: Assists in determining the total amount owed
over time with compound interest on loans.
- Financial
Forecasting: Enables accurate projections of future savings
or debt based on varying interest rates and durations.
Conclusion:
Understanding these keywords in compound interest is
fundamental for both personal finance management and financial decision-making
in various sectors. They provide insights into the growth or cost implications
of investments and loans over time, influencing financial strategies and goals.
Find compound interest
on Rs. 5000 for 2 years at 4% p.a.
To find the compound interest on Rs. 5000 for 2 years at 4%
per annum, we can use the compound interest formula:
A=P(1+R100)nA = P \left(1 +
\frac{R}{100}\right)^nA=P(1+100R)n
Where:
- PPP is
the principal amount (Rs. 5000),
- RRR is
the annual interest rate (4% or 0.04 in decimal form),
- nnn is
the number of years (2 years),
- AAA is
the amount after nnn years including interest.
First, calculate the amount after 2 years:
A=5000(1+4100)2A = 5000 \left(1 +
\frac{4}{100}\right)^2A=5000(1+1004)2 A=5000(1+0.04)2A = 5000 \left(1 +
0.04\right)^2A=5000(1+0.04)2 A=5000×1.042A = 5000 \times 1.04^2A=5000×1.042
A=5000×1.0816A = 5000 \times 1.0816A=5000×1.0816 A=5408A = 5408A=5408
Now, calculate the compound interest CICICI:
CI=A−PCI = A - PCI=A−P CI=5408−5000CI = 5408 -
5000CI=5408−5000 CI=408CI = 408CI=408
Therefore, the compound interest on Rs. 5000 for 2 years at
4% per annum is Rs. 408.
At what rate of
compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2
years?
To find the rate of compound interest per annum that will
make a sum of Rs. 1200 become Rs. 1348.32 in 2 years, we'll use the compound
interest formula and solve for the rate RRR.
Given:
- Principal
amount (P) = Rs. 1200
- Amount
after 2 years (A) = Rs. 1348.32
- Time
(n) = 2 years
The compound interest formula is: A=P(1+R100)nA = P \left(1 +
\frac{R}{100}\right)^nA=P(1+100R)n
Substitute the given values into the formula:
1348.32=1200(1+R100)21348.32 = 1200 \left(1 +
\frac{R}{100}\right)^21348.32=1200(1+100R)2
Divide both sides by 1200 to simplify:
1348.321200=(1+R100)2\frac{1348.32}{1200} = \left(1 +
\frac{R}{100}\right)^212001348.32=(1+100R)2
1.1236=(1+R100)21.1236 = \left(1 +
\frac{R}{100}\right)^21.1236=(1+100R)2
Take the square root of both sides to solve for 1+R1001 +
\frac{R}{100}1+100R:
1+R100=1.12361 + \frac{R}{100} = \sqrt{1.1236}1+100R=1.1236
1+R100=1.06011 + \frac{R}{100} = 1.06011+100R=1.0601
Subtract 1 from both sides:
R100=0.0601\frac{R}{100} = 0.0601100R=0.0601
Multiply both sides by 100 to solve for RRR:
R=0.0601×100R = 0.0601 \times 100R=0.0601×100
R=6.01R = 6.01R=6.01
Therefore, the rate of compound interest per annum that will
make a sum of Rs. 1200 become Rs. 1348.32 in 2 years is approximately 6.01%.
Divide Rs. 8840
between A and B so that the amount received by A at the end of 8 years may be
equal to the amount received by B at the end of 10 years, compound interest
being at 10% per annum.
To solve this problem, we need to ensure that the amount
received by A after 8 years equals the amount received by B after 10 years,
given compound interest at 10% per annum.
Let's denote:
- Principal
amount PPP to be divided between A and B.
- A
receives AfinalA_{\text{final}}Afinal after 8 years.
- B
receives BfinalB_{\text{final}}Bfinal after 10 years.
The compound interest formula for the final amount
AfinalA_{\text{final}}Afinal after 8 years is:
Afinal=P×(1+10100)8A_{\text{final}} = P \times (1 +
\frac{10}{100})^8Afinal=P×(1+10010)8
Similarly, for B after 10 years:
Bfinal=P×(1+10100)10B_{\text{final}} = P \times (1 +
\frac{10}{100})^{10}Bfinal=P×(1+10010)10
We are given that Afinal=BfinalA_{\text{final}} =
B_{\text{final}}Afinal=Bfinal.
So, equating them gives: P×(1+10100)8=P×(1+10100)10P \times
(1 + \frac{10}{100})^8 = P \times (1 + \frac{10}{100})^{10}P×(1+10010)8=P×(1+10010)10
Dividing both sides by PPP (assuming P≠0P \neq 0P=0):
(1+10100)8=(1+10100)10(1 + \frac{10}{100})^8 = (1 +
\frac{10}{100})^{10}(1+10010)8=(1+10010)10
Now, simplify this equation: (1.10)8=(1.10)10(1.10)^8 =
(1.10)^{10}(1.10)8=(1.10)10
Divide both sides by (1.10)8(1.10)^8(1.10)8: 1=(1.10)10−81 =
(1.10)^{10 - 8}1=(1.10)10−8 1=(1.10)21 = (1.10)^21=(1.10)2 1=1.211 = 1.211=1.21
This is not correct. Let me calculate
Find the compound
interest on Rs. 1000 at 40% per annum compounded quarterly for 1 year.
To find the compound interest on Rs. 1000 at 40% per annum
compounded quarterly for 1 year, we can use the formula for compound interest:
A=P(1+rn)ntA = P \left(1 +
\frac{r}{n}\right)^{nt}A=P(1+nr)nt
Where:
- PPP is
the principal amount (initial investment), which is Rs. 1000 in this case.
- rrr is
the annual interest rate in decimal form, so r=40%=0.40r = 40\% =
0.40r=40%=0.40.
- nnn is
the number of times interest is compounded per year, which is 4 times
(quarterly).
- ttt is
the time the money is invested for in years, which is 1 year.
First, calculate the quarterly interest rate:
rn=0.404=0.10\frac{r}{n} = \frac{0.40}{4} = 0.10nr=40.40=0.10
Now, substitute these values into the compound interest
formula: A=1000(1+0.10)4⋅1A = 1000 \left(1 + 0.10\right)^{4 \cdot 1}A=1000(1+0.10)4⋅1 A=1000(1.10)4A
= 1000 \left(1.10\right)^{4}A=1000(1.10)4
Calculate (1.10)4(1.10)^4(1.10)4:
(1.10)4=1.10×1.10×1.10×1.10(1.10)^4 = 1.10 \times 1.10 \times 1.10 \times
1.10(1.10)4=1.10×1.10×1.10×1.10 (1.10)4=1.4641(1.10)^4 = 1.4641(1.10)4=1.4641
Now, calculate the amount AAA: A=1000×1.4641A = 1000 \times
1.4641A=1000×1.4641 A≈1464.10A \approx 1464.10A≈1464.10
Now, to find the compound interest III, subtract the
principal PPP from the amount AAA: I=A−PI = A - PI=A−P I=1464.10−1000I =
1464.10 - 1000I=1464.10−1000 I=464.10I = 464.10I=464.10
Therefore, the compound interest on Rs. 1000 at 40% per annum
compounded quarterly for 1 year is approximately Rs. 464.10.
Find the compound
interest on Rs. 4000 at 24% per annum for 3 months, compounded monthly.
To find the compound interest on Rs. 4000 at 24% per annum
for 3 months, compounded monthly, we need to adjust the time period to match
the monthly compounding frequency.
Here's the step-by-step solution:
1.
Convert the annual interest rate to monthly rate:
Annual interest rate rrr = 24% per annum = 24/100 = 0.24
Monthly interest rate r12\frac{r}{12}12r =
0.2412\frac{0.24}{12}120.24 = 0.02 (since there Therefore Even if understood
Unit 10: Alphabet Test
10.1 Alphabet Test
10.2 Type of Alphabet Arrangement
10.3 Logical Sequence of Words
10.4 Type of Logical Sequence Questions
10.1 Alphabet Test
The Alphabet Test is a type of aptitude test question where
you are typically given a series of letters and are required to identify the
logical pattern or sequence among them. These tests assess your ability to
recognize and manipulate alphabetical sequences and patterns.
10.2 Types of Alphabet Arrangement
There are several types of alphabet arrangements that can be
tested:
- Forward
Alphabet Series: Here, you need to identify the letter that
follows a given sequence of letters in the alphabet. Example: If the
series is A, B, C, D, ___, the answer would be E.
- Backward
Alphabet Series: In this type, you identify the letter that
precedes a given sequence of letters in the alphabet. Example: If the
series is Z, Y, X, W, ___, the answer would be V.
- Mixed
Series: This involves a combination of forward and backward
sequences where you need to identify both preceding and succeeding letters
based on the given pattern.
10.3 Logical Sequence of Words
Logical sequence questions in the context of alphabets
involve arranging letters or words in a logical order based on a certain rule
or pattern. This may include:
- Dictionary
Order: Arranging words or letters in alphabetical order as
per the dictionary. Example: Arrange the following words in alphabetical
order: Apple, Ball, Cat, Dog. The answer would be Apple, Ball, Cat, Dog.
- Reverse
Dictionary Order: Arranging words or letters in reverse
alphabetical order. Example: Arrange the following words in reverse
alphabetical order: Zebra, Tiger, Lion, Elephant. The answer would be
Zebra, Tiger, Lion, Elephant.
10.4 Types of Logical Sequence Questions
Logical sequence questions involving alphabets can be
categorized into different types:
- Letter-Based
Sequences: These questions involve patterns or sequences based on
the position of letters in the alphabet. Example: Identify the next letter
in the sequence: A, C, E, G, ___? The answer would be I.
- Word-Based
Sequences: These questions involve arranging words or groups of
letters based on a specific rule or pattern. Example: Arrange the
following words based on their lengths: Dog, Cat, Elephant, Bird. The
answer would be Cat, Dog, Bird, Elephant.
- Mixed
Sequences: Some questions may combine both letter and word
sequences in a single problem, requiring you to apply multiple rules or
patterns to solve them.
Summary
The Alphabet Test and its related concepts in Unit 10 focus
on your ability to recognize patterns, sequences, and logical arrangements
involving alphabets and words. Practice is key to mastering these types of
questions, as they often appear in various aptitude tests and examinations.
Summary of Key Concepts
1.
Alphabet Test Basics
o Definition: The
Alphabet Test evaluates your ability to recognize and manipulate alphabetical
sequences and patterns.
o Types of
Questions: Includes forward, backward, and mixed alphabet series where
you identify the next or previous letter in a sequence.
2.
Types of Alphabet Arrangements
o Forward
Alphabet Series: Identifying the letter that follows a given sequence.
§ Example: A,
B, C, D, ___. Answer: E.
o Backward
Alphabet Series: Identifying the letter that precedes a given sequence.
§ Example: Z,
Y, X, W, ___. Answer: V.
o Mixed Series:
Combination of forward and backward sequences requiring identification of both
preceding and succeeding letters.
3.
Logical Sequence of Words
o Dictionary
Order: Arranging words or letters in alphabetical order as per the
dictionary.
§ Example:
Arrange Apple, Ball, Cat, Dog in alphabetical order. Answer: Apple, Ball, Cat,
Dog.
o Reverse
Dictionary Order: Arranging words or letters in reverse alphabetical
order.
§ Example:
Arrange Zebra, Tiger, Lion, Elephant in reverse alphabetical order. Answer:
Zebra, Tiger, Lion, Elephant.
4.
Solving Strategies and Tricks
o Pattern
Recognition: Developing skills to identify patterns or rules governing
alphabetical sequences.
o Meaning-Based
Approach: Understanding questions in context and applying logical
reasoning to solve them effectively.
o Use of
Dictionary: Using dictionary rules to arrange words or letters
alphabetically or in reverse order.
5.
Practice and Application
o Application
in Aptitude Tests: Importance of practicing various types of alphabet
tests and logical sequences to enhance problem-solving abilities.
o Real-Life
Applications: Skills learned are applicable in exams, competitive tests,
and real-life scenarios requiring analytical thinking and pattern recognition.
Conclusion
Unit 10 on the Alphabet Test and Logical Sequence has
equipped you with fundamental skills in recognizing and solving alphabetical
patterns and logical arrangements. By mastering these concepts and strategies,
you can confidently approach various types of alphabet test questions and
logical sequence problems in assessments and everyday situations. Regular
practice and application of learned techniques will further enhance your
proficiency in this area.
Keywords
1.
Alphabet Test
o Definition: An
aptitude test that assesses your ability to understand and manipulate
alphabetical sequences and patterns.
o Types of
Questions:
§ Forward
Alphabet Series: Identifying the next letter in a sequence (e.g., A, B, C,
D, ___).
§ Backward
Alphabet Series: Identifying the previous letter in a sequence (e.g., Z, Y,
X, W, ___).
§ Mixed
Alphabet Series: Combining forward and backward sequences.
2.
Logical Sequence
o Definition: Involves
arranging words or letters in a logical order based on specific rules or
patterns.
o Types of
Logical Sequences:
§ Dictionary
Order: Arranging words alphabetically as per their dictionary definitions.
§ Reverse
Dictionary Order: Arranging words in reverse alphabetical order.
§ Pattern-Based
Sequences: Arranging based on a given pattern or rule.
3.
Word Formation
o Definition: The
process of creating new words by rearranging letters or parts of words.
o Types of
Word Formation Questions:
§ Unscrambling
Letters: Rearranging scrambled letters to form meaningful words.
§ Prefix/Suffix
Addition: Adding prefixes or suffixes to existing words to form new
ones.
§ Word
Association: Forming words related to a given theme or concept.
Explanation
- Alphabet
Test:
- Tests
your ability to understand and predict patterns in alphabetical
sequences.
- Helps
in developing logical reasoning skills by recognizing sequential
patterns.
- Logical
Sequence:
- Focuses
on arranging words or letters based on specific rules like alphabetical
order or defined patterns.
- Enhances
analytical thinking and problem-solving abilities.
- Word
Formation:
- Involves
manipulating letters to create new meaningful words.
- Develops
vocabulary and linguistic skills, particularly in recognizing word
patterns and structures.
Applications
- Educational
Context:
- Vital
in aptitude tests, competitive exams, and educational assessments.
- Improves
language proficiency and logical reasoning abilities.
- Real-Life
Applications:
- Useful
in professions requiring analytical skills such as linguistics, coding,
and puzzle-solving.
- Enhances
communication skills through better understanding of word structures and
meanings.
By mastering these keywords and their associated concepts,
you can effectively tackle various types of alphabet test questions, logical
sequence problems, and word formation challenges encountered in academic,
professional, and everyday contexts. Regular practice and application will
further strengthen your skills in these areas.
1.Arrange the
following words asper their order in the dictionary:
1. PHYSICAL
2. PHYSICS
3. PHYSIOLOGY
4. PHYSICIAN
5. PHONE
(1) 5, 1, 4, 2, 3
(2) 5, 1, 2, 3, 4
(3) 1, 4, 5, 2, 3
(4) 1, 5, 4, 3, 2
To arrange the words PHYSICAL, PHYSICS, PHYSIOLOGY, PHYSICIAN,
and PHONE in alphabetical order according to their dictionary entries, we need
to compare the words letter by letter from the beginning:
1.
PHYSICAL
2.
PHYSICS
3.
PHYSIOLOGY
4.
PHYSICIAN
5.
PHONE
Let's compare each pair of words based on their starting
letters:
- PHONE vs PHYSICAL:
PHONE comes before PHYSICAL.
- PHONE vs PHYSICS:
PHONE comes before PHYSICS.
- PHONE vs PHYSIOLOGY:
PHONE comes before PHYSIOLOGY.
- PHONE vs PHYSICIAN:
PHONE comes before PHYSICIAN.
Now, among PHYSICAL, PHYSICS, PHYSIOLOGY, and PHYSICIAN:
- PHYSICAL vs PHYSICS:
PHYSICAL comes before PHYSICS.
- PHYSICAL vs PHYSIOLOGY:
PHYSICAL comes before PHYSIOLOGY.
- PHYSICAL vs PHYSICIAN:
PHYSICAL comes before PHYSICIAN.
So, the correct alphabetical order is: PHONE, PHYSICAL,
PHYSICIAN, PHYSICS, PHYSIOLOGY.
Therefore, the correct arrangement according to their order
in the dictionary is:
(2) 5, 1, 2, 3, 4
This matches option
(2): 5, 1, 2, 3, 4
Arrange the following
words asper order in the dictionary:
1. Ambitions
2. Ambiguous
3. Ambiguity
4. Animation
5. Animal
(1) 3, 2, 4, 1, 5
(2) 3, 2, 5, 4, 1
(3) 3, 2, 1, 5, 4
(4) 3, 2, 4, 5, 1
Ambitions, Ambiguous, Ambiguity, Animation, and Animal in
alphabetical order according to their dictionary entries, we need to compare
them letter by letter:
1.
Ambiguity
2.
Ambiguous
3.
Ambitions
4.
Animation
5.
Animal
Let's compare each pair of words based on their starting
letters:
- Animal vs Ambiguous:
Animal comes before Ambiguous.
- Animal vs Ambiguity:
Animal comes before Ambiguity.
- Animal vs Ambitions:
Animal comes before Ambitions.
- Animal vs Animation:
Animal comes before Animation.
Now, among Ambiguous, Ambiguity, Ambitions, and Animation:
- Ambiguity
vs Ambiguous: Ambiguity comes before Ambiguous.
- Ambiguity
vs Ambitions: Ambiguity comes before Ambitions.
- Ambiguity
vs Animation: Ambiguity comes before Animation.
So, the correct alphabetical order is: Animal, Ambiguous,
Ambiguity, Ambitions, Animation.
Therefore, the correct arrangement according to their order
in the dictionary is:
(2) 3, 2, 5, 4, 1
This matches option (2): 3, 2, 5, 4, 1.
3.Arrange the
following words asper order in the dictionary:
1. Brittle
2. Brisk
3. Bright
4. Bride
(1) 4, 3, 2, 1
(2) 1, 2, 3, 4
(3) 2, 3, 4, 1
(4) 4, 2, 1, 3
To arrange the words Brittle, Brisk, Bright, and Bride in
alphabetical order according to their dictionary entries, let's compare them
letter by letter:
1.
Brittle
2.
Brisk
3.
Bright
4.
Bride
Comparing each pair based on their starting letters:
- Bride vs Brisk:
Bride comes before Brisk.
- Bride vs Bright:
Bride comes before Bright.
- Bride vs Brittle:
Bride comes before Brittle.
Now, among Brisk, Bright, and Brittle:
- Brisk vs Bright:
Brisk comes before Bright.
- Brisk vs Brittle:
Brisk comes before Brittle.
Lastly, comparing Bright and Brittle:
- Bright vs Brittle:
Bright comes before Brittle.
So, the correct alphabetical order is: Bride, Brisk, Bright,
Brittle.
Therefore, the correct arrangement according to their order
in the dictionary is:
(2) 1, 2, 3, 4
This matches option (2): 1, 2, 3, 4.
4.Arrange the following words asper order in the dictionary:
1. Preach
2. Praise
3. Precinet
4. Precept
5. Precede
(1) 2, 1, 5, 4, 3
(2) 2, 1, 3, 4, 5
(3) 2, 5, 1, 4, 3
(4) 1, 2, 5, 4, 3
Preach, Praise, Precinet, Precept, and Precede in
alphabetical order according to their dictionary entries, let's compare them
letter by letter:
1.
Praise
2.
Preach
3.
Precinet
4.
Precept
5.
Precede
Comparing each pair based on their starting letters:
- Praise vs Preach:
Praise comes before Preach.
- Praise vs Precede:
Praise comes before Precede.
- Praise vs Precinet:
Praise comes before Precinet.
- Praise vs Precept:
Praise comes before Precept.
Now, among Preach, Precinet, Precept, and Precede:
- Preach vs Precinet:
Preach comes before Precinet.
- Preach vs Precept:
Preach comes before Precept.
- Preach vs Precede:
Preach comes before Precede.
So, the correct alphabetical order is: Praise, Preach,
Precinet, Precept, Precede.
Therefore, the correct arrangement according to their order
in the dictionary is:
(2) 2, 1, 3, 4, 5
Unit 11: Number Test
11.1 Number Test
11.2 Position Switching of Numbers
11.3 Problems on Number Test
11.1 Number Test
The Number Test is a type of aptitude test question that
involves manipulating and understanding numerical sequences, patterns, and
relationships. These tests assess your ability to recognize numerical patterns
and apply logical reasoning to solve problems.
- Types
of Questions:
- Number
Series: Identifying the next number in a sequence based on a
pattern (e.g., 2, 4, 6, ___, 10).
- Number
Analogies: Finding a relationship between numbers and applying
it to find a missing number (e.g., 3:6::5:___).
- Number
Patterns: Recognizing patterns in number arrangements (e.g., 1,
3, 6, 10, ___).
11.2 Position Switching of Numbers
This involves questions where numbers are arranged in a
specific sequence, and you are required to identify what happens when positions
of some numbers are switched or altered.
- Example: Given
a sequence like 1, 2, 3, 4, 5, if positions of numbers 2 and 4 are
switched, what will the new sequence be?
- Skills
Tested: Ability to visualize and understand the effect of
switching positions in a numerical sequence.
11.3 Problems on Number Test
These are practical problems or scenarios that apply concepts
from the Number Test to real-world or abstract situations. They often require
logical deduction and pattern recognition.
- Example
Problem:
- Question: If
every third number starting from 4 in a sequence of numbers is removed,
which number will be 8th in the modified sequence?
- Skills
Tested: Ability to count and skip numbers in a sequence,
understanding positional relationships.
Summary
Unit 11 on the Number Test focuses on enhancing your
numerical reasoning abilities through various types of questions and scenarios.
By practicing these tests, you can improve your pattern recognition, logical
thinking, and problem-solving skills, which are valuable in aptitude tests,
competitive exams, and real-world applications requiring analytical abilities.
Mastering the Number Test involves:
- Understanding
different types of number-related questions.
- Practicing
to identify patterns and relationships among numbers.
- Developing
strategies to solve problems involving numerical sequences and patterns
effectively.
Regular practice and a clear understanding of the underlying
concepts will help you excel in this area.
Summary of Key Concepts
1.
Number Test
o Definition: The Number
Test evaluates your ability to understand and manipulate numerical sequences,
patterns, and relationships.
o Types of
Questions:
§ Number
Series: Identifying the next number in a sequence based on a
pattern.
§ Number
Analogies: Finding relationships between numbers and applying them to
solve problems.
§ Number
Patterns: Recognizing and extending patterns in numerical
arrangements.
2.
Position Switching of Numbers
o Definition: Involves
questions where positions of numbers within a sequence are altered, and you
need to identify the resulting sequence.
o Example: Switching
the positions of specific numbers in a sequence and determining the new
arrangement.
3.
Solving Strategies
o Pattern
Recognition: Developing skills to identify and extend numerical
patterns.
o Logical
Deduction: Applying logical reasoning to understand how switching
positions of numbers affects the sequence.
o Practice
Techniques: Practicing different types of Number Test questions and
position switching scenarios to improve problem-solving abilities.
Applications
- Educational
Context:
- Essential
for aptitude tests, competitive exams, and assessments requiring
numerical reasoning.
- Enhances
mathematical and logical thinking skills.
- Real-Life
Applications:
- Useful
in professions involving data analysis, programming, and problem-solving.
- Develops
critical thinking abilities applicable to various situations requiring
pattern recognition and logical deduction.
Conclusion
Unit 11 on the Number Test and Position Switching of Numbers
has equipped you with essential skills in numerical reasoning and pattern
recognition. By understanding these concepts and practicing different types of
questions, you can confidently approach Number Test problems and effectively
solve position switching scenarios in assessments and practical scenarios.
Regular practice and application of learned strategies will
further enhance your proficiency in recognizing numerical patterns and solving
complex problems related to the Number Test and position switching of numbers.
Keywords
1.
Number Test
o Definition: An
assessment or question type designed to evaluate a person's ability to
understand and manipulate numerical sequences, patterns, and relationships.
o Types of
Questions:
§ Number
Series: Identifying the next number in a sequence based on a given
pattern.
§ Number
Analogies: Finding relationships between numbers and applying them to
solve problems.
§ Number
Patterns: Recognizing and extending patterns in numerical
arrangements.
2.
Position Switching of Numbers
o Definition: Involves
altering the positions of numbers within a sequence and identifying the
resulting sequence.
o Example: Given a
sequence of numbers, switch the positions of specific numbers and determine the
new sequence.
o Skills
Tested:
§ Ability to
visualize and understand the impact of changing positions in a numerical sequence.
§ Logical
reasoning to deduce the resulting sequence after positions are switched.
Explanation
- Number
Test:
- Purpose:
Evaluates numerical reasoning skills essential for aptitude tests and
competitive exams.
- Applications: Used
in various contexts to assess logical thinking and problem-solving
abilities related to numerical patterns.
- Position
Switching of Numbers:
- Purpose:
Tests the ability to predict and analyze changes in numerical sequences.
- Applications:
Develops skills applicable to fields requiring analytical thinking, such
as mathematics, programming, and data analysis.
Applications
- Educational
Context:
- Integral
in aptitude tests, entrance exams, and assessments requiring quantitative
reasoning.
- Enhances
mathematical aptitude and logical deduction skills.
- Real-Life
Applications:
- Useful
in professions requiring analytical skills, such as finance, engineering,
and scientific research.
- Develops
problem-solving capabilities in scenarios involving numerical patterns
and sequence manipulation.
Conclusion
Understanding and mastering the concepts of the Number Test
and Position Switching of Numbers involves practice, pattern recognition, and
logical reasoning. These skills are crucial for success in assessments and
practical applications requiring numerical proficiency and problem-solving
abilities. Regular practice and exposure to various types of questions will
enhance proficiency in these areas, leading to improved performance in academic
and professional settings.
How many 4's are there
preceded by 7 but not followed by 3? 5 9 3 2 1 7 4 2 6 9 7 4 6 1 3 2 8
7 4 1 3 8 3 2 5 6 7 4
3 9 5 8 2 0 1 8 7 4 6 3.
To find how many occurrences of the digit '4' are there that
are preceded by '7' but not followed by '3' in the given sequence, let's
analyze step by step:
Sequence:
Copy code
5 9 3 2 1 7 4 2 6 9 7 4 6 1 3 2 8 7 4 1 3 8 3 2 5 6 7 4 3 9 5
8 2 0 1 8 7 4 6 3
1.
Identify instances where '4' is preceded by '7':
o In the
sequence, we have:
§ 7 4
(position 7)
§ 7 4
(position 19)
§ 7 4
(position 37)
2.
Check if these instances of '4' are not followed by
'3':
o At position
7: 7 4 2 (followed by '2')
o At position
19: 7 4 6 (followed by '6')
o At position
37: 7 4 6 (followed by '6')
3.
Count the valid occurrences:
o There are 3
instances where '4' is preceded by '7' and not followed by '3' in the sequence.
Therefore, the answer is 3. There are 3 occurrences of
the digit '4' that are preceded by '7' but not followed by '3' in the given
sequence.
Study the number
series given below and answer the questions that follow:
7 8 9 7 6 5 3 4 2 8 9
7 2 4 5 9 2 9 7 6 4 7.
How many 7s are
preceded by 9 and followed by 6?
'7' are preceded by '9' and followed by '6':
Number series:
Copy code
7 8 9 7 6 5 3 4 2 8 9 7 2 4 5 9 2 9 7 6 4 7
1.
Identify instances where '7' is preceded by '9' and
followed by '6':
o In the sequence,
we have:
§ 9 7 6
(position 19)
2.
Count the valid occurrence:
o There is 1
instance where '7' is preceded by '9' and followed by '6' in the sequence.
Therefore, the answer is 1. There is 1 occurrence of
the digit '7' that is preceded by '9' and followed by '6' in the given number
series.
If the first and
second digits in the sequence 5981327438 are interchanged, also the third and
fourth digits, the
fifth and sixth digits and so on, which digit would be the seventh counting to
your
left?
To solve this problem, we will interchange pairs of digits in
the sequence 5981327438 according to the given pattern and then determine which
digit becomes the seventh when counted from the left after the changes.
Original sequence: 5981327438
Interchanging pairs of digits:
- 1st and
2nd digits: 5 9 -> 9 5
- 3rd and
4th digits: 8 1 -> 1 8
- 5th and
6th digits: 3 2 -> 2 3
- 7th and
8th digits: 7 4 -> 4 7
- 9th
digit remains the same: 3 8 -> 3 8
Modified sequence after interchanging:
- 9571832438
Now, counting from the left, the seventh digit in the
modified sequence is 2.
Therefore, after interchanging pairs of digits as specified,
the seventh digit when counted from the left is 2.
The position of the
second and the 8th digit of the number 39128564 are interchanged. The
position of the first
and sixth digits is interchanged and the positions of the third and seventh
digits
are interchanged.
Which of the following will be third digit left of the 3 after the
re-arrangement.
Original number: 39128564
1.
Interchanging the second and eighth digits:
o Second digit
(9) moves to the eighth position.
o Eighth digit
(4) moves to the second position.
New arrangement after this step: 34128569
2.
Interchanging the first and sixth digits:
o First digit
(3) moves to the sixth position.
o Sixth digit
(5) moves to the first position.
New arrangement after this step: 54328169
3.
Interchanging the third and seventh digits:
o Third digit
(3) moves to the seventh position.
o Seventh
digit (1) moves to the third position.
Final arrangement after all interchanges: 54128369
Now, we need to find the third digit to the left of the digit
'3' after rearrangement:
- Locate
the digit '3' in the sequence: 54128369
- Count
three digits to the left of '3': The digits are 2, 8, and 1.
Therefore, the third digit to the left of the digit '3' after
rearrangement is 1.
Unit 12: Ranking and Time Sequence Test
12.1 Ranking Test
12.2 Time Sequence Test
12.3 Problems on Ranking and Time Sequence Test
Unit 12: Ranking and Time Sequence Test
This unit focuses on evaluating your ability to understand
and solve problems related to ranking and time sequence. These tests assess
logical reasoning and your ability to interpret and organize information in a
sequential manner.
12.1 Ranking Test
Definition:
- The
Ranking Test evaluates your ability to arrange items or individuals based
on given criteria such as height, weight, rank, or any other measurable
attribute.
- It
involves arranging items in ascending or descending order according to
specified rules or conditions.
Types of Questions:
1.
Sequential Ranking: Arranging items in a
sequence based on a particular order or rank.
2.
Comparative Ranking: Comparing the positions or
ranks of different items based on given criteria.
Skills Tested:
- Ability
to interpret and follow ranking rules or criteria.
- Logical
thinking to deduce the correct sequence based on given information.
12.2 Time Sequence Test
Definition:
- The
Time Sequence Test assesses your ability to arrange events, dates, or
occurrences in chronological order.
- It
involves understanding timelines and sequences of events based on
historical, logical, or hypothetical contexts.
Types of Questions:
1.
Chronological Order: Arranging events or
occurrences in the correct chronological sequence.
2.
Temporal Relationships:
Determining the sequence of events based on their occurrence or timing.
Skills Tested:
- Ability
to understand and interpret temporal relationships.
- Logical
reasoning to place events or occurrences in the correct chronological
order.
12.3 Problems on Ranking and Time Sequence Test
Definition:
- These
are practical problems or scenarios that apply concepts from ranking and
time sequence tests to real-world or abstract situations.
- They
often require logical deduction and the ability to apply sequential
reasoning skills.
Examples:
1.
Ranking Problem: Arrange a group of individuals based
on their heights from tallest to shortest.
2.
Time Sequence Problem: Arrange
historical events in the order they occurred.
Skills Tested:
- Application
of ranking rules or chronological sequencing rules.
- Logical
deduction to solve problems involving ranking and time sequence.
Applications
- Educational
Context:
- Essential
for competitive exams, aptitude tests, and assessments requiring logical
reasoning.
- Enhances
analytical and problem-solving skills.
- Real-Life
Applications:
- Useful
in professions involving data analysis, project management, and
historical research.
- Develops
critical thinking abilities applicable to various scenarios requiring
sequence interpretation.
Conclusion
Unit 12 on Ranking and Time Sequence Test is designed to
enhance your logical reasoning abilities through exercises that involve
arranging items by rank or chronological order. By mastering these concepts and
practicing various types of questions, you can improve your ability to
interpret and organize information sequentially, which is valuable in academic,
professional, and everyday contexts. Regular practice and a clear understanding
of ranking and time sequence principles will help you excel in assessments and
real-life situations requiring analytical thinking and logical deduction.
Summary: Ranking and Time Sequence Test
1.
Ranking Test
o Definition: Evaluates
the ability to arrange items or individuals based on specific criteria such as
height, weight, rank, or any measurable attribute.
o Types of
Questions:
§ Sequential
Ranking: Arranging items in a sequence according to a specified
order.
§ Comparative
Ranking: Comparing positions or ranks of different items based on
given criteria.
o Skills
Tested:
§ Ability to
interpret and apply ranking rules accurately.
§ Logical
reasoning to deduce the correct sequence based on provided information.
2.
Time Sequence Test
o Definition: Assesses
the ability to arrange events, dates, or occurrences in chronological order.
o Types of
Questions:
§ Chronological
Order: Placing events or occurrences in the correct temporal
sequence.
§ Temporal
Relationships: Understanding the order of events based on their occurrence
or timing.
o Skills
Tested:
§ Ability to
interpret temporal relationships accurately.
§ Logical
deduction to arrange events or dates in the correct chronological sequence.
Key Learning Points
- Understanding
Different Types of Questions on Ranking Test:
- Recognizing
the need to arrange items based on specific criteria like rank or
measurable attributes.
- Practicing
sequential and comparative ranking scenarios to enhance problem-solving
skills.
- Understanding
Different Types of Questions on Time Sequence Test:
- Developing
the ability to organize events or occurrences based on their
chronological order.
- Applying
logical reasoning to determine temporal relationships and arrange events
sequentially.
Applications
- Educational
Context:
- Essential
for competitive exams, aptitude tests, and assessments requiring logical
reasoning and sequence interpretation.
- Improves
analytical skills and ability to solve problems involving data
sequencing.
- Real-Life
Applications:
- Valuable
in professions involving project management, historical research, and
data analysis.
- Enhances
critical thinking abilities needed to interpret and organize information
sequentially in various real-world scenarios.
Conclusion
Unit 12 on Ranking and Time Sequence Test equips learners
with essential skills in interpreting and organizing information based on
ranking criteria and chronological order. By mastering the types of questions
and practicing different scenarios, individuals can strengthen their logical
reasoning and problem-solving abilities. These skills are crucial for academic
success, professional development, and everyday decision-making processes
requiring sequential analysis and deduction. Regular practice and understanding
of ranking and time sequence principles will foster proficiency and confidence
in tackling related challenges effectively.
Keywords: Ranking Test and Time Sequence Test
Ranking Test
Definition:
- The
Ranking Test evaluates the ability to arrange items, individuals, or
entities based on specific criteria such as height, weight, rank, or other
measurable attributes.
- It
involves organizing items in ascending or descending order according to
predefined rules or conditions.
Types of Questions:
1.
Sequential Ranking:
o Arranging
items in a sequence based on a specified order or rank.
o Example:
Arrange students based on their test scores from highest to lowest.
2.
Comparative Ranking:
o Comparing
the positions or ranks of different items based on given criteria.
o Example: Compare
the heights of two individuals to determine who is taller.
Skills Tested:
- Ability
to interpret and apply ranking rules accurately.
- Logical
reasoning to deduce the correct sequence based on provided information.
Time Sequence Test
Definition:
- The Time
Sequence Test assesses the ability to arrange events, dates, or
occurrences in chronological order.
- It
involves understanding and interpreting the sequence of events based on
their historical, logical, or hypothetical contexts.
Types of Questions:
1.
Chronological Order:
o Arranging
events or occurrences in the correct temporal sequence.
o Example:
Arrange historical events in the order they occurred.
2.
Temporal Relationships:
o Determining
the sequence of events based on their occurrence or timing.
o Example:
Determine which event happened first or last based on provided timelines.
Skills Tested:
- Ability
to interpret temporal relationships accurately.
- Logical
deduction to arrange events or dates in the correct chronological
sequence.
Key Points
- Importance
in Testing:
- Essential
for various competitive exams, aptitude tests, and assessments requiring
logical reasoning and sequence interpretation.
- Real-Life
Applications:
- Valuable
in professions such as project management, historical research, and data
analysis.
- Enhances
critical thinking abilities needed for interpreting and organizing
information sequentially in practical scenarios.
Conclusion
Understanding the concepts of Ranking Test and Time Sequence
Test is crucial for developing logical reasoning skills and problem-solving
abilities. These tests not only prepare individuals for academic assessments
but also equip them with essential skills applicable in real-world scenarios
requiring sequence interpretation and logical deduction. Mastery of these
concepts through practice and understanding of the underlying principles will
enhance proficiency and confidence in tackling related challenges effectively.
Rohan ranks 7th from
the top and 26th from the bottom in the class. How many students are
there in the class?
To determine the total number of students in Rohan's class,
we use the information provided about his rankings:
- Rohan
ranks 7th from the top.
- Rohan
ranks 26th from the bottom.
Let's denote:
- TTT as
the total number of students in the class.
From the top, Rohan is 7th, meaning there are 6 students
above him: Number of students above Rohan=7−1=6\text{Number
of students above Rohan} = 7 - 1 =
6Number of students above Rohan=7−1=6
From the bottom, Rohan is 26th, meaning there are 25 students
below him:
Number of students below Rohan=26−1=25\text{Number of
students below Rohan} = 26 - 1 =
25Number of students below Rohan=26−1=25
The total number of students TTT in the class is the sum of
students above and below Rohan, including Rohan himself:
T=Number of students above Rohan+Number of students below Rohan+1T
= \text{Number of students above Rohan} + \text{Number of students below Rohan}
+
1T=Number of students above Rohan+Number of students below Rohan+1
T=6+25+1T = 6 + 25 + 1T=6+25+1 T=32T = 32T=32
Therefore, the total number of students in Rohan's class is
32\boxed{32}32.
Manik is 14th from the
right end in the row of 40 students. What is his position from the left end?
To find Manik's position from the left end of the row, given
that he is 14th from the right end in a row of 40 students, follow these steps:
1.
Identify Total Positions:
o Total number
of students in the row = 40.
2.
Calculate Position from Left End:
o Manik's
position from the left end can be found by subtracting his position from the
total number of students and then adding 1 (to include Manik himself).
Position from left end=(Total number of students+1)−(Position from right end)\text{Position
from left end} = (\text{Total number of students} + 1) - (\text{Position from
right end})Position from left end=(Total number of students+1)−(Position from right end)
Position from left end=(40+1)−14\text{Position
from left end} = (40 + 1) - 14Position from left end=(40+1)−14
Position from left end=41−14\text{Position from left end} = 41 -
14Position from left end=41−14
Position from left end=27\text{Position from left end} = 27Position from left end=27
Therefore, Manik's
position from the left end of the row is 27\boxed{27}27.
In a row of boys
facing the north, A is 16th from the left end and C is 16th from the right end.
B,
who is 4th to the
right of A, is 5th to the left of C, in a row. How many boys are there in a
row?
4. In a row of
students, Anil is 7th from left, while Sunil is 18th from right. Both of them
interchanged their
positions such that Anil becomes 21st from left. What will be the total number
of
students in the class?
To determine the total number of boys in the row based on the
given information:
Given:
- A is
16th from the left end.
- C is
16th from the right end.
- B is
4th to the right of A.
- B is
5th to the left of C.
Let's solve step by step:
1.
Positions of A and C:
o A is 16th
from the left end.
o C is 16th
from the right end.
Total number of boys TTT in the row:
T=Position of A from left+Position of C from right−1T
= \text{Position of A from left} + \text{Position of C from right} -
1T=Position of A from left+Position of C from right−1
T=16+16−1T = 16 + 16 - 1T=16+16−1 T=31T = 31T=31
So, there are 31 boys in the row.
2.
Position of B relative to A and C:
o B is 4th to
the right of A.
o B is 5th to
the left of C.
Let's calculate B's position:
o Position of
B from left end: Position of B=Position of A+4\text{Position
of B} = \text{Position of A} +
4Position of B=Position of A+4
Position of B=16+4\text{Position of B} = 16 +
4Position of B=16+4 Position of B=20\text{Position of B} =
20Position of B=20
o Position of
B from right end: Position of B=Position of C−5\text{Position
of B} = \text{Position of C} -
5Position of B=Position of C−5
Position of B=16−5\text{Position of B} = 16 -
5Position of B=16−5 Position of B=11\text{Position of B} =
11Position of B=11
Both calculations confirm that B is indeed 20th from the left
and 11th from the right.
Therefore, the total number of boys in the row is
31\boxed{31}31.
Mohan is taller than
Shyam but shorter than Ramesh. Ramesh is taller than Rajat but shorter than
Gautam. If Shyam is
taller than Rajat, then who is the shortest among all?
Based on the given information:
- Mohan
is taller than Shyam but shorter than Ramesh.
- Ramesh
is taller than Rajat but shorter than Gautam.
- Shyam
is taller than Rajat.
Let's analyze the relationships step by step:
1.
Height Order Deduction:
o From the
information:
§ Ramesh >
Mohan > Shyam
§ Gautam >
Ramesh > Rajat
2.
Comparing Heights:
o Ramesh is
taller than Mohan and Shyam.
o Gautam is
taller than Ramesh and Rajat.
3.
Identifying the Shortest:
o Shyam is
taller than Rajat.
From the given comparisons:
- Gautam
> Ramesh > Mohan > Shyam > Rajat
Therefore, Rajat is the shortest among all based on
the given conditions.
Unit 13: Direction Sense Test
13.1 General Directions
13.2 Casting of Shadows
13.3 Distance Related Questions
13.4 Left Right Movement
13.5 Direction Puzzles
13.6 Problems on Direction Sense Test
13.1 General Directions
Definition:
- General
Directions involve the cardinal directions: North, South, East, and West.
- Intermediate
directions include Northeast (NE), Southeast (SE), Southwest (SW), and
Northwest (NW).
- Questions
typically involve understanding movements or positions relative to these
directions.
Skills Tested:
- Ability
to identify and apply directions correctly.
- Logical
reasoning to determine movements or positions based on given directions.
13.2 Casting of Shadows
Definition:
- Casting
of Shadows refers to understanding the direction in which shadows fall
based on the position of the sun or light source.
- Questions
involve determining the direction of shadows at specific times of the day.
Skills Tested:
- Understanding
the concept of shadow formation.
- Applying
knowledge of sun's position and its impact on shadow direction.
13.3 Distance Related Questions
Definition:
- Distance
Related Questions involve determining distances between points or
locations in specific directions.
- Questions
may require calculating shortest or alternative routes based on
directional changes.
Skills Tested:
- Calculation
of distances in different directions.
- Application
of direction sense to determine optimal paths or routes.
13.4 Left Right Movement
Definition:
- Left
Right Movement questions assess the ability to determine movements or
positions based on left or right turns from a starting point.
- Questions
may involve sequential movements or changes in direction.
Skills Tested:
- Ability
to visualize and execute directional changes.
- Logical
deduction to follow sequential instructions involving left and right
turns.
13.5 Direction Puzzles
Definition:
- Direction
Puzzles are complex problems that require applying direction sense in
solving puzzles or riddles.
- These
may involve multiple steps or conditions affecting directional movements.
Skills Tested:
- Critical
thinking to decode and solve directional puzzles.
- Integration
of multiple directional clues to arrive at the correct solution.
13.6 Problems on Direction Sense Test
Definition:
- Problems
on Direction Sense Test combine various aspects of direction sense,
including general directions, distance calculations, shadow casting,
left-right movements, and puzzles.
- These problems
simulate real-world scenarios requiring logical reasoning and spatial
orientation skills.
Skills Tested:
- Comprehensive
application of direction sense concepts.
- Problem-solving
under time constraints or complex conditions.
Key Learning Points
- Understanding
General and Intermediate Directions:
- Recognition
and application of cardinal and intermediate directions in various
contexts.
- Shadow
Casting and Distance Calculation:
- Understanding
how shadows are cast and calculating distances based on directional
movements.
- Left-Right
Movements and Direction Puzzles:
- Ability
to interpret and execute movements involving left and right turns, and
solving complex directional puzzles.
Applications
- Educational
Context:
- Essential
for competitive exams, aptitude tests, and assessments requiring spatial
reasoning and directional sense.
- Real-Life
Applications:
- Valuable
in navigation, logistics, and spatial planning.
- Enhances
critical thinking and problem-solving abilities in everyday scenarios
involving directions.
Conclusion
Unit 13 on Direction Sense Test equips learners with
essential skills in understanding and applying directional concepts. Mastery of
these skills through practice and understanding of the underlying principles
will enhance proficiency and confidence in tackling related challenges
effectively. Regular practice and exposure to various types of
direction-related problems will reinforce spatial reasoning and logical
deduction abilities, crucial for both academic and practical applications.
Summary: Unit 13 - Direction Sense Test
Key Concepts Learned
1.
Understanding Distance Related Questions
o Definition: Distance
related questions involve calculating distances between points or locations in
specific directions.
o Skills
Learned:
§ Ability to
calculate distances based on given directional cues.
§ Application
of direction sense to determine optimal routes or shortest paths.
2.
Left Right Movement
o Definition: Left Right
Movement questions assess the ability to follow directions involving left and
right turns from a starting point.
o Skills
Learned:
§ Visualizing
and executing directional changes based on sequential instructions.
§ Logical
deduction to navigate through given movements effectively.
3.
Shadow Type Questions of Directions Test
o Definition: Shadow
type questions require understanding the direction of shadows cast based on the
position of the sun or a light source.
o Skills
Learned:
§ Understanding
how shadows are formed and determining their direction at different times of
the day.
§ Applying
knowledge of the sun's position relative to objects or individuals.
4.
Puzzle Type of Direction Sense Test
o Definition: Puzzle
type questions in direction sense involve solving complex problems or riddles
that require applying directional clues.
o Skills
Learned:
§ Critical
thinking to decode and solve puzzles involving directional clues.
§ Integration
of multiple directional cues to arrive at a logical solution.
Applications of Learning
- Educational
Context:
- Essential
for competitive exams, aptitude tests, and assessments that evaluate
spatial reasoning and directional sense.
- Provides
foundational skills for logical thinking and problem-solving abilities.
- Real-Life
Applications:
- Valuable
in navigation, logistics, and spatial planning.
- Enhances
everyday decision-making by improving directional awareness and
problem-solving skills.
Conclusion
Unit 13 - Direction Sense Test covers essential concepts in
understanding and applying directional sense. Mastery of these skills through
practice and application in various scenarios enhances proficiency in spatial
reasoning and logical deduction. Regular practice with distance calculations,
left-right movements, shadow type questions, and puzzle-solving reinforces
these skills, preparing individuals for both academic and real-world challenges
requiring directional awareness and problem-solving abilities.
Keywords
1.
Distance
o Definition: Distance
refers to the measurement of how far apart two points are.
o Usage: In
direction sense tests, distance often involves calculating the length or space
between locations or objects based on given directional cues.
2.
Directions
o Definition: Directions
indicate the path or course along which someone or something moves, faces, or
points.
o Usage: Directions
include cardinal directions (North, South, East, West) and intermediate
directions (North-East, South-East, South-West, North-West). They are crucial
in navigation and understanding spatial relationships.
3.
East
o Definition: East is
the direction towards the point on the horizon where the sun rises.
o Usage: It is one
of the primary cardinal directions and is often used to indicate a specific
orientation or position relative to other directions.
4.
West
o Definition: West is
the direction towards the point on the horizon where the sun sets.
o Usage: Like East,
West is a cardinal direction and is used to denote a specific orientation or
location relative to other directions.
5.
South
o Definition: South is
the direction towards the Earth's South Pole.
o Usage: It is a
cardinal direction used to describe a particular orientation or position
relative to other directions.
6.
North
o Definition: North is
the direction towards the Earth's North Pole.
o Usage: North is a
cardinal direction that indicates a specific orientation or position relative
to other directions.
7.
South-East
o Definition: South-East
is the direction halfway between South and East.
o Usage: It is an
intermediate direction that denotes a position or movement towards the
southeastern quadrant.
8.
North-East
o Definition: North-East
is the direction halfway between North and East.
o Usage: It is an
intermediate direction indicating a position or movement towards the
northeastern quadrant.
9.
South-West
o Definition: South-West
is the direction halfway between South and West.
o Usage: It is an
intermediate direction describing a position or movement towards the
southwestern quadrant.
10. North-West
o Definition: North-West
is the direction halfway between North and West.
o Usage: It is an
intermediate direction denoting a position or movement towards the northwestern
quadrant.
Application
- Educational
Context:
- Crucial
for competitive exams, aptitude tests, and assessments requiring spatial
reasoning and directional awareness.
- Fundamental
in understanding navigation and geographical positioning.
- Real-Life
Applications:
- Essential
in everyday navigation, travel, and logistics.
- Improves
decision-making by enhancing directional awareness and spatial
orientation skills.
Conclusion
Understanding these keywords in the context of direction
sense tests is essential for mastering spatial reasoning and logical deduction.
Regular practice with these concepts strengthens one's ability to interpret and
apply directions effectively in various scenarios, from academic assessments to
real-world navigation challenges.
Satish starts from his
house and takes two right turns and then one left turn. Now he is moving
towards south. In
which direction Satish started from his house?
analyze the movements of Satish based on the given
information:
1.
Satish starts from his house.
2.
He takes two right turns.
3.
After taking these right turns, he takes one left
turn.
4.
Now, he is moving towards south.
To determine the initial direction Satish started from his
house, we need to trace back his movements:
- First
Right Turn: If Satish starts facing North and takes a right turn,
he would face East.
- Second
Right Turn: Another right turn from East would make him face
South.
- Left
Turn: A left turn from facing South would then make him face
East again.
Therefore, based on his final direction after taking the
turns (South), and retracing the path backward:
- Satish
started from his house facing East.
A man is facing west.
He turns 45° in the clockwise direction and then another 180° in the same
direction and then
270° in the anti-clockwise direction. Which direction is he facing now?
1.
Initial Direction: The man is facing West.
2.
First Turn (Clockwise 45°):
o Turning 45°
clockwise from West:
§ West →
North-West
3.
Second Turn (Clockwise 180°):
o Turning 180°
clockwise from North-West:
§ North-West →
South-East
4.
Third Turn (Anti-clockwise 270°):
o Turning 270°
anti-clockwise from South-East:
§ South-East →
East (after 270° anti-clockwise is equivalent to 90° clockwise)
Therefore, after performing all the turns, the man is facing East.
Sumi ran a distance of
40 m towards South. She then turned to the right and ran for about 15 m,
turned right again and
ran 50 m. Turning to right then ran for 15 m. Finally, she turned to the left
an angle of 45° and
ran. In which direction was she running finally?
Sumi's movements step by step:
1.
Sumi starts by running 40 m towards South.
o Initial
direction: South
2.
She turns to the right and runs 15 m.
o Right turn
from South: West
o Distance
covered: 15 m
3.
Next, she turns right again and runs 50 m.
o Right turn
from West: North
o Distance
covered: 50 m
4.
Then, she turns right again and runs 15 m.
o Right turn
from North: East
o Distance
covered: 15 m
5.
Finally, she turns left at an angle of 45° and runs.
o Left turn
from East: North-East
Therefore, after all these movements, Sumi is running in the North-East
direction
.
child is looking for
his father. He went 90 metres in the east before turning to his right. He
went 20 metres before
turning to his right again to look for his father at his uncle’s
place 30 metres
from this point. His
father was not there. From there, he went 100 metres to his north before
meeting his father in
a street. How far did the son meet his father from the starting point?
1.
Initial Movement:
o Child goes
90 meters towards the East.
o Starting
point to this point: East (90 meters)
2.
First Turn:
o Child turns
right (South).
o Distance
covered: 20 meters
o New
position: South (20 meters from the starting point)
3.
Second Turn:
o Child turns
right again (West) towards his uncle's place.
o Distance
covered: 30 meters
o New
position: West (30 meters from the second turn point)
4.
Movement towards Father:
o Child goes
100 meters to the North.
o New
position: North (100 meters from the second turn point)
Now, let's calculate the final position relative to the
starting point:
- From
the starting point, the child moved 90 meters East, then 20 meters South,
and finally 30 meters West before heading North 100 meters.
- The net
effect of these movements is that the child ends up 90 meters East and 20
meters South of the starting point.
To find the direct distance from the starting point to where
the child meets his father:
Distance=(90 m)2+(20 m)2\text{Distance} = \sqrt{(90
\text{ m})^2 + (20 \text{ m})^2}Distance=(90 m)2+(20 m)2
Distance=8100+400\text{Distance} = \sqrt{8100 + 400}Distance=8100+400
Distance=8500\text{Distance} = \sqrt{8500}Distance=8500
Distance≈92.2 meters\text{Distance} \approx 92.2 \text{
meters}Distance≈92.2 meters
Therefore, the child meets his father approximately 92.2
meters from the starting point.
Deepa moved a distance
of 75 metres towards the north. She then turned to the left and walking
for about 25 metres,
turned left again and walked 80 metres. Finally, she turned to the right at an
angle of 45°. In which
direction was she moving finally?
1.
Deepa moves 75 meters towards the North.
o Initial
direction: North
2.
She turns to the left and walks 25 meters.
o Left turn
from North: West
o Distance
covered: 25 meters
3.
Next, she turns left again and walks 80 meters.
o Left turn
from West: South
o Distance
covered: 80 meters
4.
Finally, she turns to the right at an angle of 45°.
o Right turn
from South: South-West
Therefore, after all these movements, Deepa is moving in the South-West
direction.
Unit 14: Blood Relation
14.1 Types of Questions on Blood Relation
14.2 Family or Blood Relations Test
14.3 Family/Blood Relations Described General
14.4 Relationships Involving the Term ‘-in-law’ General
14.5 Father-in-law, Mother-in-law, Son-in-law and
Daughter-in-law
14.6 Brother-in-law and Sister-in-law
14.7 Half Sibling and Step Relations
1.
Types of Questions on Blood Relation
o Definition: Questions
on blood relations typically involve identifying relationships between
individuals based on familial connections.
o Examples: These
questions may include determining parent-child relationships, sibling
relationships, and extended family relationships.
2.
Family or Blood Relations Test
o Definition: A test or
set of questions designed to assess one's ability to understand and analyze
familial connections.
o Purpose: Tests
understanding of terminology, hierarchy, and roles within a family structure.
3.
Family/Blood Relations Described General
o Definition: General
descriptions of family relations focus on clarifying terminology and
understanding roles.
o Examples: Includes
explanations of terms like parents, siblings, cousins, etc., and their
respective positions in the family tree.
4.
Relationships Involving the Term ‘-in-law’ General
o Definition: Explains
relationships formed through marriage using the suffix "-in-law."
o Examples:
Father-in-law, mother-in-law, son-in-law, daughter-in-law, etc., denote
relationships formed by marriage rather than by blood.
5.
Father-in-law, Mother-in-law, Son-in-law and
Daughter-in-law
o Definition: Specific
relationships formed through marriage:
§ Father-in-law: The father
of one's spouse.
§ Mother-in-law: The mother
of one's spouse.
§ Son-in-law: The
husband of one's daughter.
§ Daughter-in-law: The wife
of one's son.
6.
Brother-in-law and Sister-in-law
o Definition:
Relationships formed through marriage between siblings:
§ Brother-in-law: The
husband of one's sister or the brother of one's spouse.
§ Sister-in-law: The wife
of one's brother or the sister of one's spouse.
7.
**Half S
ibling and Step Relations**
- Definition:
Differentiates between blood-related siblings and those who share only one
biological parent (half siblings), as well as individuals related through
the marriage of a parent (step relations).
- Examples:
- Half
Sibling: Someone who shares one biological parent with another
sibling.
- Step
Sibling: Someone who becomes a sibling through the marriage of
a parent, where no biological relationship exists.
Understanding these concepts in blood relations helps in
accurately interpreting and solving questions that involve family
relationships. Each category within Unit 14 provides clarity on terminology and
the structural dynamics of familial connections, crucial for reasoning and
analytical tests involving human relationships.
Summary: Unit 14 - Blood Relation
1.
Family or Blood Relationship
o Key Concepts
Learned:
§ Understanding
and identifying familial connections based on blood relations.
§ Recognizing
roles such as parents, siblings, cousins, etc., in a family structure.
§ Ability to
deduce relationships based on direct biological connections.
2.
Relationships Involving the Term ‘-in-law’ General
o Key Concepts
Learned:
§ Defining
relationships formed through marriage using the suffix "-in-law".
§ Examples
include father-in-law, mother-in-law, son-in-law, daughter-in-law, etc.
§ Understanding
these relationships as connected through marriage rather than blood ties.
3.
Relationships Involving Father-in-law, Mother-in-law,
Son-in-law, and Daughter-in-law
o Key Concepts
Learned:
§ Specific
definitions and identification of roles in marriage-related relationships:
§ Father-in-law: The father
of one's spouse.
§ Mother-in-law: The mother
of one's spouse.
§ Son-in-law: The
husband of one's daughter.
§ Daughter-in-law: The wife
of one's son.
§ Understanding
the dynamics and positions these individuals hold within a family.
4.
Relationships Involving Brother-in-law and
Sister-in-law
o Key Concepts
Learned:
§ Recognizing
relationships formed through marriage between siblings:
§ Brother-in-law: The
husband of one's sister or the brother of one's spouse.
§ Sister-in-law: The wife
of one's brother or the sister of one's spouse.
§ Differentiating
between these roles based on marital connections rather than biological ties.
5.
Relationships Half Sibling and Step Relations
o Key Concepts
Learned:
§ Understanding
the distinction between:
§ Half Sibling:
Individuals who share one biological parent.
§ Step Sibling:
Individuals who become siblings through the marriage of a parent, without
biological relation.
§ Identifying
these relationships within blended families and their implications in familial
dynamics.
By mastering these concepts, one gains proficiency in solving
complex reasoning problems involving family relationships. These skills are
essential for various competitive exams and reasoning tests that assess logical
thinking and relationship analysis.
Keywords: Blood Relationship, Relationships based Puzzles,
Coded Relations
1.
Blood Relationship
o Definition: Refers to
the familial connections based on biological ties.
o Key Points:
§ Understanding
Family Structures: Recognizing relationships like parents, siblings,
cousins, etc., based on direct biological descent.
§ Problem
Solving: Solving problems that involve identifying relationships
using clues about direct biological connections.
§ Examples: Questions
typically involve determining who is related to whom in a family tree, based on
given relationships.
2.
Relationships based Puzzles
o Definition: Puzzles or
logical problems that involve deciphering family relationships.
o Key Points:
§ Logical
Deduction: Requires logical deduction and reasoning skills to solve.
§ Types of
Puzzles: Could involve scenarios where relationships are given
indirectly or through clues, requiring deduction to establish connections.
§ Examples: Solving
puzzles like "A is the father of B, who is married to C," and
deducing who C's in-laws are.
3.
Coded Relations
o Definition: Relations that
are not straightforward and may involve hidden or indirect information.
o Key Points:
§ Deciphering
Codes: Involves decoding relationships from coded information or
indirect references.
§ Logical
Analysis: Requires logical analysis and inference to understand the
relationships.
§ Examples: Questions
where relationships are described in coded language or symbolic
representations, and one needs to decode them to understand the family
connections.
By mastering these keywords, one gains proficiency in
analyzing and solving complex problems related to family relationships, often
tested in competitive exams and reasoning assessments. Understanding these
concepts helps in developing logical thinking and problem-solving abilities
related to familial structures and relationships.
Pointing to Ajay,
Radha said, “His father is the only son of my
grandfather”. How is Radha Related to Ajay?
Based on the statement "His father is the only son of my
grandfather" made by Radha, we can deduce the relationship between Radha
and Ajay as follows:
- Radha's
grandfather has only one son.
- Ajay's
father is that only son of Radha's grandfather.
Therefore, Ajay's father is Radha's uncle. Consequently, Ajay
is Radha's cousin. Hence, Radha is related to Ajay as his cousin.
Lalita said to Tina, “You are the daughter-in-law of the grandmother of my father’s only son.”
statement to determine the relationship between Lalita and
Tina:
Lalita said to Tina, "You are the daughter-in-law of the
grandmother of my father’s only son."
Here’s the analysis:
- Lalita's
Father's Only Son: This refers to Lalita's brother.
- Grandmother
of Lalita's Father's Only Son: This is Lalita and her
brother's grandmother.
- Tina:
According to the statement, Tina is the daughter-in-law of Lalita and her
brother's grandmother.
So, Tina is married to the son of Lalita and her brother's
grandmother. Therefore, Tina is the daughter-in-law of Lalita's brother, making
Lalita and Tina related through marriage (sisters-in-law).
Pointing to a
photograph, Amar said, “I have no brother or sister but
that man’s father is my father’s
son.” Whose photograph, was it? How is Lalita related
to Tina?
Let's analyze Amar's statement to determine whose photograph
he is referring to:
Amar said, "I have no brother or sister but that man’s
father is my father’s son."
Here’s the breakdown:
- Amar
has no brother or sister: This means Amar is talking
about himself and any reference to siblings is about absence.
- That
man’s father is my father’s son: This statement implies that the
man's father is Amar himself, because in familial terms, Amar's father’s
son would be Amar.
Therefore, the photograph Amar is pointing to is of Amar's
own son. This conclusion arises because Amar states that the man's father
(in the photograph) is Amar's father’s son, which directly refers to Amar
himself.
Regarding Lalita and Tina's relationship, as previously
discussed:
- Lalita
said to Tina, "You are the daughter-in-law of the grandmother of my
father’s only son."
- This
means Tina is married to Lalita's brother (Lalita's father’s only son).
- Therefore,
Lalita and Tina are sisters-in-law.
Looking at the
portrait of a man, Ashok said, „His mother is the wife
of my father’s son. Brothers and sisters, I have none‟. At whose portrait was Ashok looking?
Ashok said, "His mother is the wife of my father’s son.
Brothers and sisters, I have none."
Here’s the analysis:
- Ashok's
father’s son: This refers to Ashok himself, as Ashok is
stating that he has no brothers or sisters.
- His
mother is the wife of my father’s son: This statement means
that the man's mother is Ashok's wife.
Therefore, the portrait Ashok is looking at is of his own
son. Ashok is saying that the man's mother (in the portrait) is Ashok's
wife, which directly points to the fact that the man in the portrait is Ashok's
son.
So, Ashok was looking at the portrait of his own son.
Ahmad said to Saira, „Your only brother’s son is my wife’s brother‟. How is Saira related to
the Ahmad’s wife?
Ahmad said to Saira, "Your only brother’s son is my
wife’s brother."
Here’s the analysis:
- Your
only brother’s son: This refers to Saira's nephew (her brother's
son).
- My
wife’s brother: This means that Ahmad's wife's brother is
Saira's nephew.
Therefore, Saira's nephew is Ahmad's wife's brother. This relationship
can be understood as follows:
- Saira's
nephew is Ahmad's wife's brother.
- Saira
and Ahmad's wife are sisters-in-law.
So, Saira is Ahmad's sister-in-law.
Pointing to a gentleman,
Abdul said, “His only brother is the father of my
daughter’s father”. How is the
gentleman related to Abdul?”
statement to determine how the gentleman is related to Abdul:
Abdul said, “His only brother is the father of my daughter’s
father.”
Here’s the breakdown:
- Abdul’s
daughter’s father: This refers to Abdul himself.
- His
only brother: Refers to the gentleman's brother.
- So, the
gentleman's brother is Abdul.
Therefore, the gentleman is Abdul's father.
So, Abdul is pointing to his own father and saying that the
gentleman's (his father's) only brother is Abdul himself.
Pointing to a man in a
photograph, Malati tells, “His brother’s
father is the only son of my grandfather.” How is
Malati related to the man in the photograph?
Malati said, "His brother’s father is the only son of my
grandfather."
Here’s the analysis:
- His
brother’s father: This refers to the father of the man in the
photograph.
- Only
son of my grandfather: This means Malati's father, because if Malati's
grandfather has only one son, that son is Malati's father.
Therefore, the man in the photograph is Malati's father.
So, Malati is the daughter of the man in the photograph.
Pointing to
Dharmendra, Hema said, “He is the son of my father’s only son.” How is Dharmendra’s mother related to Hema?
Dharmendra's mother is related to her:
Hema said, "He is the son of my father’s only son."
Here’s the breakdown:
- My
father’s only son: This refers to Hema's brother.
- He is
the son of my father’s only son: This means that Dharmendra
is Hema's nephew (her brother's son).
Now, let's determine how Dharmendra's mother is related to
Hema:
- Dharmendra's
mother is Hema's sister-in-law.
Therefore, Dharmendra's mother is Hema's brother's wife. They
are related by the fact that Dharmendra is Hema's nephew through her brother.
