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Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 9 Number Relations Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 9 Solutions Number Relations

Class 7 Maths Chapter 9 Number Relations Questions and Answers Kerala State Syllabus

Number Relations Class 7 Questions and Answers Kerala Syllabus

Page 131

Now try these problems:

Question 1.
Find the number of factors of each number below:
i) 40
ii) 54
iii) 60
iv) 100
v) 210
Answer:
i) The factors of 40, expressed as powers of prime numbers, are:,
40 = 2³ × 5 = 8 × 5
Therefore, number of factors = (3 + 1)(1 + 1) = 4 × 2 = 8

ii) The factors of 54, expressed as powers of prime numbers, are:
54 = 3³ × 2 ,
Therefore, number of factors = (3 + 1)(1 + 1) = 4 × 2 = 8

iii) The factors of 60, expressed as powers of prime numbers, are:
60 = 2² × 3 × 5
Therefore, number of factors = (2 + 1) (1 + 1) (1 + 1) = 3 × 2 × 2 = 12

iv) The factors of 100, expressed as powers of prime numbers, are:
100 = 5² × 2²
Therefore, number of factors = (2 + 1)(2 + 1) = 3 × 3 = 9

v) The factors of 210, expressed as powers of prime numbers, are:
210 = 7 × 5 × 3 × 2
Therefore, number of factors = (1 + 1) (1 + 1) (1 + 1) (1 + 1) = 2 × 2 × 2 × 2 = 16

Question 2.
From the number of factors of a number, we can deduce some peculiarities of the number. The table below lists these for number of factors up to 5. Extend it to number of factors up to 10

Answer:

6	Fifth power of a prime	p5(p, a prime)
	Product of two primes	pq2 or p2q(p, q primes)
7	Sixth power of a prime	p6(p, a prime)
8	Seventh power of a prime	p7 (p, a prime)
	Product of three primes	pqr (p, q, r primes)
	Product of two primes	p3q or pq3 (p, q primes)
9	Eight power of a prime	p8 (p, a prime)
	Product of two primes	p2q2 (p, q primes)
10	Ninth power of a prime	p9(p, a prime)
	Product of two primes	pq4 or p4q (p,q primes)

Page 134

Question 1.
For each pair of numbers given below, find the largest common factor and all other common factors:
i) 45, 75
ii) 225, 275
iii) 360, 300
iv) 210, 504
v) 336, 588
Answer:
i) Factors of 45 = 3² × 5
Factors of 75 = 3 × 5²
Common primes: 3 and 5
Take the lowest powers: 3¹ × 5¹ = 15
Largest common factor: 15
Other common factors: 1,3,5

ii) Factors of 225 = 3² × 5²
Factors of 275 = 5² × 11
Common prime: 5
Take the lowest power: 2 = 25
Largest common factor: 25
Other common factors: 1,5

iii) Factors of 360 = 2³ × 3² × 5
Factors of 300 = 2² × 3 × 5²
Common primes: 2, 3, and 5
Take the lowest powers: 2², 3¹, 5¹
Largest common factor = 2² × 3¹ × 5¹ = 4 × 3 × 5 = 60
Other common factors: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

iv) Factors of 210 = 2 × 3 × 5 × 7
Factors of 504 = 2³ × 3² × 7
Common primes: 2, 3, and 5
Take the lowest powers: 2¹, 3¹, 5¹
Largest common factor = 2¹ × 3¹ × 5¹ = 2 × 3 × 5 = 30
Other common factors: 1, 2, 3, 5, 6, 10, 15, 30

v) Factors of 336 = 2⁴ × 3 × 7
Factors of 558 = 2² × 3 × 7²
Common primes: 2, 3, and 7
Take the lowest powers: 2², 3¹, 7¹
Largest common factor = 2² × 3¹ × 7¹ = 4 × 3 × 7 = 84
Other common factors: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84

Question 2.
i) What is the largest common factor of two different prime numbers?
ii) Can the largest common factor of two composite numbers be 1?
iii) If two numbers are divided by their largest common factor, what would be the largest common factor of the quotients?
Answer:
i) The largest common factor of two different prime numbers is 1, since prime numbers have no common factors other than 1.
ii) Yes, the largest common factor of two composite numbers can be 1 if they are co prime, meaning they have no common factors other than 1.
iii) If two numbers are divided by their largest common factor, the largest common factor of the quotients will be I. This is because the largest common factor is the greatest number that divides both original numbers, and dividing by it eliminates any common factors from the quotients.

Class 7 Maths Chapter 9 Kerala Syllabus Number Relations Questions and Answers

Question 1.
Find the number of factors of each number below:
i) 36
ii) 84
iii) 144
Answer:
i) The factors of 36, expressed as powers of prime numbers, are:
36 = 2² × 3²
Therefore, number of factors = (2 + 1) (2 +1) = 3 × 3 = 9

ii) The factors of 84, expressed as powers of prime numbers, are:
84 = 2² × 3¹ × 7¹
Therefore, the number of factors = (2 + 1)(1 + 1)(1 + 1) = 3 × 2 × 2 = 12

iii) The factors of 144, expressed as powers of prime numbers, are:
144 = 2⁴ × 3²
Therefore, the number of factors = (4 + 1)(2 + 1) = 5 × 3 = 15

Question 2.
For each pair of numbers given below, find the largest common factor:
i) 48, 180
ii) 90, 150
iii) 84, 126
Answer:
i) Prime factors of 48 = 2⁴ × 3¹
Prime factors of 180 = 2² × 3² × 5¹
Common primes: 2 and 3
Take the lowest powers: 2², 3¹
Largest Common Factor = 2² × 3¹ = 4 × 3 = 12

ii) Prime factors of 90 = 2¹ × 3² × 5¹
Prime factors of 150 = 2¹ × 3¹ × 5²
Common primes: 2,3, and 5
Take the lowest powers: 2¹, 3¹ , 5¹
Largest Common Factor = 2¹ × 3¹ × 5¹ = 2 × 3 × 5 = 30

iii) Prime factors of 84 = 2² × 3¹ × 7¹
Prime factors of 126 = 2¹ × 3² × 7¹
Common primes: 2, 3, and 7
Take the lowest powers: 2¹, 3¹, 7¹
Largest Common Factor = 2¹ × 3¹ × 7¹ = 2 × 3 × 7 = 42

Class 7 Maths Chapter 9 Notes Kerala Syllabus Number Relations

In this chapter, we will explore important concepts related to numbers and their factors. You’ll learn how to determine the number of factors for prime numbers and how to identify common factors between two numbers.

In short, we can explain as,

• To find the number of factors of the product of powers of two prime numbers, we have to add one to each exponent and multiply these numbers.
• To find the common factors of two numbers, first write the factors of each number as the product of prime numbers raised to their powers. Then, identify the common prime numbers and take the lowest power of each. These will be the common factors.
• These fundamental ideas are essential for understanding the relationships between numbers and will help you solve problems with greater confidence.

Number Of Factors

Prime numbers are natural numbers that can only be divided by 1 and the number itself. Examples include 2, 3, 5, 7, and 11.
Each of these numbers has exactly two factors.,
Now, let’s examine how the number of factors changes as the power of a prime number varies.

In general,
The number of factors of a power of any prime number is one more than the exponent.
Algebraically we can say,
If p is a prime number and n is a natural number, then the number of factors of pⁿ is n + 1.

Now, let’s explore how the number of factors changes when a prime number is multiplied by another number.

For 3 × 5 = 15
Factors of 3 = 1, 3
Factors of 5 = 1, 5
Factors of 15 = 1, 3, 5, 15
Thus, the number of factors = 4
We can tabulate it like this:

1	3	(Factors of 3)
5	15	(Factors of 3 multiplied by 5)

Thus, the number of factors = 4

For 3² × 5 = 45
Factors of 3² as the power of prime numbers = 1, 3, 3² that is 1, 3, 9
Factors when each of them multiplied by 5 are:

Thus, the number of factors = 3 + 3 = 3 × 2 = 6

For 3² × 5² = 225

Thus, the number of factors = 3 × 3 = 9

For 3³ × 5³

The factors are:

Thus, the number of factors = 4 × 4 = 16

In general,
The number of factors of the product of powers of two prime numbers is got by adding one to each exponent and multiplying these numbers.
Algebraically we can say,
If p and q are two different primes and m, n are any two natural numbers, then the number of factors of p^m qⁿ is (m + 1)(n + 1).

Now, let’s consider the situation with three different prime numbers.
Examine the expression 3³ × 5² × 11
First, tabulate the factors of 3³ × 5²

Here, the number of factors for 3³ × 5² = 4 × 3 = 12
Then, tabulate the factors of 3³ × 5² × 11

Thus, all together 3³ × 5² × 11 = 12 × 2 = 24

In general,
To compute the number of factors of a number, we write it a product of powers of different prime numbers, and find the product of the numbers got by adding one to each exponent.

Common Factors
Let’s understand what a common factor of two numbers means:
For that consider the two numbers 180 and 270.
Factors of 180 = 2² × 3² × 5
Factors of 270 = 2 × 3³ × 5
Here, the prime common for both numbers are 2, 3, 5
The smaller power of these primes in the factorizations is 2, 32, 5
Here, the common factors are the factors of 2 × 3² × 5
We can tabulate it as follows:

In this case, the largest common factor of 180 and 270 is 90.
In short,
To find the common factors of two numbers, first write each number as the product of prime numbers raised to their powers. Then, identify the common prime numbers and take the lowest power of each. These will be the common factors.

• The number of factors of a power of any prime number is one more than the exponent.
• If p is a prime number and n is a natural number, then the number of factors of pn is n + 1.
• The number of factors of the product of powers of two prime numbers is got by adding one to each exponent and multiplying these numbers.
• If p and q are two different primes and m, n are any two natural numbers, then the number of factors of pmqn is (m + 1)(n + 1).
• To find the common factors of two numbers, first write each number as the product of prime numbers raised to their powers. Then, identify the common prime numbers and take the lowest power of each. These will be the common factors.

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 8 Repeated Multiplication Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 8 Solutions Repeated Multiplication

Class 7 Maths Chapter 8 Repeated Multiplication Questions and Answers Kerala State Syllabus

Repeated Multiplication Class 7 Questions and Answers Kerala Syllabus

Page 112

Question 1.
Write each number below either as a power of a single prime or as a product of powers of different primes:
i) 125
ii) 72
iii) 100
iv) 250
v) 3600
vi) 10800
Answer:
i) 125
125 = 5 × 5 × 5 = 5³

ii) 72
72 = 2 × 2 × 2 × 3 × 3
= 2³ × 3²

iii) 100
100 = 2 × 2 × 5 × 5
= 2² × 5²

iv) 250
250 = 2 × 5 × 5 × 5
= 2 × 5³

v) 3600
3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
= 2⁴ × 3² × 5²

(vi) 10800
10800 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5
= 2⁴ × 3³ × 5²

Page 115

Question 1.
Calculate the powers below as fractions:
(i) \(\left(\frac{2}{3}\right)^2\)
Answer:
\(\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)=\frac{4}{9}\)

(ii) \(\left(1 \frac{1}{2}\right)^2\)
Answer:
\(\left(1 \frac{1}{2}\right)=\left(\frac{3}{2}\right)\)
= \(\left(\frac{3}{2}\right)\left(\frac{3}{2}\right)=\left(\frac{9}{4}\right\)

(iii) \(\left(\frac{2}{5}\right)^3\)
Answer:
\(\left(\frac{2}{5}\right)\left(\frac{2}{5}\right)\left(\frac{2}{5}\right)\)
= \(\frac{8}{125}\)

(iv) \(\left(2 \frac{1}{2}\right)^3\)
Answer:
\(\left(2 \frac{1}{2}\right)=\left(\frac{5}{2}\right)\)
\(\left(\frac{5}{2}\right)\left(\frac{5}{2}\right)\left(\frac{5}{2}\right)=\frac{125}{8}\)

Question 2.
Calculate the powers below in decimal form:
(i) (0.5)²
(ii) (1.5)²
(iii) (0.1)³
(iv) (0.01)³
Answer:
i) (0.5)²
= 0.5 × 0.5
= 0.25

ii) (1.5)²
= (1.5)(1.5)
= 2.25

iii) (0.1)³
= (0.1)(0.1)(0.1)
= 0.001

iv) (0.01)³
= (0.01)(0.01)(0.01)
= 0.000001

Question 3.
Using 15³ = 3375 calculate the powers below:
i) (1.5)³
ii) (0.15)³
iii) (0.015)³
Answer:
i) (1.5)³ = 1.5 × 1.5 × 1.5 = 3.375
ii) (0.15)³ = 0.15 × 0.15 × 0.15 = 0.003375
iii) (0.015)³ = 0.015 × 0.015 × 0.015 = 0.000003375

Page 118

Question 1.
Write each product below as the product of powers of different primes:
i) 72 × 162
ii) 225 × 135
iii) 105 × 175
iv) 25 × 45 × 75
Answer:
i) 72 × 162
72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3²
162 = 3 × 3 × 3 × 3 × 2 = 3⁴ × 2
72 × 162 = (2³ × 3²)(3⁴ × 2)
= 2⁴ × 3⁶

ii) 225 × 135
225 = 3 × 3 × 5 × 5 = 3² × 5²
135 = 3 × 3 × 3 × 5 = 3³ × 5¹
225 × 135 = (3² × 5²)( 3³ × 5¹)
= 3⁵ × 5³

iii) 105 × 175
105 = 3 × 5 × 7 = 3¹ × 5¹ × 7¹
175 = 5 × 5 × 7 = 5² × 7¹
105 × 175 = (3¹ × 5¹ × 7¹)(5² × 7¹)
= 3 × 5³ × 7²

iv) 25 × 45 × 75
25 = 5¹ × 5¹
45 = 3² × 5¹
75 =3¹ × 5²
25 × 45 × 75 = (5¹ × 5¹)(3² × 5¹)(3¹ × 5²)
= 5⁵ × 3³

Question 2.
Write the product of the numbers from 1 to 15 as the product of powers of different primes.
Answer:
1 (no primes)
2 = 2¹
3 = 3¹
4 = 2²
5 = 5¹
6 = 2¹ × 3¹
7 = 7¹
8 = 2³
9 = 3²
10 = 2¹ × 5¹
11 = 11¹
12 = 2² × 3¹
13 = 1³¹
14 = 2¹ × 7¹
15 = 3¹ × 5¹
∴ The product of the numbers from 1 to 15 as the product of powers of different primes are
1 × 2¹¹ × 3⁶ × 5³ × 7² × 11¹ × 13¹

Question 3.
Consider the numbers from 1 to 25
i) Which of them are divisible by 2, but not by 4?
ii) Which of them are divisible by 4, but not by 8?
iii) Which of them are divisible by 8, but not by 16?
iv) Which of them are divisible by 16?
v) What is the highest power of 2 that divides the product of the numbers from 1 to 25 without remainder?
Answer:
i) The numbers from 1 to 25 that are divisible by 2 but not by 4 are:
2, 6, 10, 14, 18, 22

ii) The numbers from 1 to 25 that are divisible by 4 but not by 8 are:
4, 12, 20

iii) The numbers from 1 to 25 that are divisible by 8 but not by 16 are:
8, 24

iv) The numbers from 1 to 25 that are divisible by 16 are:
16

v) Numbers that are divisible by 2 from 1 to 25 are:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24

Thus, the total numbers that can be divisible by 2 are 12.
Number that are divisible by 4 from 1 to 25 are:
4, 8, 12, 16, 20, 24

Thus, the total number numbers that can be divisible by 4 are 6.
Number that are divisible by 8 from 1 to 25 are:
8, 16, 24

Thus, the total number numbers that can be divisible by 8 are 3.
Number that are divisible by 16 from 1 to 25 are:
16

Thus, the total number numbers that can be divisible by 16 are 1.
From this, the highest power of 2 = 12 + 6 + 3 + 1 = 22
So, the highest power of 2 that divides the product of number from 1 to 25 is 2²².

Question 4.
Consider the product of the numbers from 1 to 25
i) What is the highest power of 5 which divides this product without remainder?
ii) And the highest power of 10 dividing this product without remainder?
iii) How many zeros does this product end with?
Answer:
i) The numbers divisible by 5 are 5, 10, 15, 20, 25 that is 5 number.
The number divisible by 25 is 25 that is 1 number.
Thus, the total contribution from multiple of 5 = 5 + 1 = 6
Thus, the highest power of 5 that divides the product between 1 to 25 is 56.

ii) Since 10 = 2 × 5. We need to count how many times both 2 and 5 appear as factors in the product.
Thus, the smallest count between the 2 and 5 will give us the highest powers of 10.
The highest power of 2, which divides the product of numbers from 1 to 25 is 22 and
the highest power of 5, which divides the product of numbers from 1 to 25 is 6.
Thus, the highest power of 10 = smallest count of the highest power of 2 and 5
Therefore, the highest power of 10 dividing the product from 1 to 25 is 106.

iii) Since we found that 106 divides the product of numbers from 1 to 25.
This means that the product ends with 6 zeros.

Page 122

Question 1.
Calculate the following quotients:
i) 512 ÷ 64
ii) 3125 ÷ 125
iii) 243 ÷ 27
iv) 1125 ÷ 45
Answer:
i) 512 ÷ 64
512 = 29 = 26 × 23
64 = 26 = 23 × 23
512 ÷ 64 = ( 2⁶ × 2³) ÷ (2³ × 2³)
= 2⁶ ÷ 2³
= 2^6-3
= 2³
= 8

ii) 3125 ÷ 125
3125 = 5⁵ = 5² × 5³
125 = 53 = 5² × 5¹
3125 ÷ 125 = (5² × 5³) ÷ (5² × 5¹)
= 5³ ÷ 5¹
= 5^3-1
= 5²
=25

iii) 243 ÷ 27
243 = 3⁵ = 3² × 3³
27 = 3³ = 3² × 3¹
243 ÷ 27 = (3² × 3³) – (3² × 3¹)
= 3³ – 3¹
= 3^3-1
= 3²
= 9

iv) 1125 ÷ 45
1125 = 5³ × 3²
45 = 5¹ × 3²
1125 ÷ 45 = (53 × 32) ÷ (51 × 32)
= 5^3-1
= 5²
= 25

Question 2.
i) Write half of 2¹⁰ as a power of 2.
ii) Write one-third of 3¹² as a power of 3.
Answer:
i) Half of 2¹⁰ = 2¹⁰ ÷ 2¹
= 2^10-1
= 2⁹

ii) One-third of 3¹² = 3¹² ÷ 3¹
= 3¹¹

There is another way for doing this type of problems, and it can be stated as a general principle, using algebra:
\(\frac{x^m}{x^n}=\frac{1}{x^{n-m}}\), for all natural numbers x ≠ 0 and for all natural numbers m < n
For example, let’s factorize the numerator and denominator, and calculate 64 ÷ 512.

Page 123

Question 1.
Can’t you simplify the fractions below like this?
i) \(\frac{27}{243}\)
Answer:

ii) \(\frac{125}{3125}\)
Answer:

iii) \(\frac{48}{64}\)
Answer:

iv) \(\frac{54}{81}\)
Answer:

Page 126 

Question 1.
Calculate the products below in head:
i) 5² × 4²
ii) 5³ × 6³
iii) 25³ × 4³
iv) 125² × 8²
Answer:
i) 5² × 4² = (5 × 4)²
= 20²
= 400

ii) 5³ × 6³ = (5 × 6)³
= 30³
= 27000

iii) 25³ × 4³ = (25 × 4)³
= (100)³
= 1000000

iv) 125² × 8² = (125 × 8)²
= (1000)²
= 1000000

Question 2.
Write each number below as a product of powers of different primes;
i) 15²
ii) 30³
iii) 12² × 21²
iv) 12² × 21³
Answer:
i) 15² = (3 × 5)³
= 3² × 5²

ii) 30³ = (2 × 3 × 5)³
= 2³ × 3³ × 5³

iii) 12² × 21²
12² = (2 × 2 × 3)²
= 2² × 2² × 3²

21² = (7 × 3)²
= 7² × 3²

12² × 21² = 2² × 2² × 3² × 7² × 3²
= 2⁴ × 3⁴ × 7²

iv) 12² × 21³
12² = (2 × 2 × 3)² = 2² × 2² × 3²
21³ = (7 × 3)³ = 7³ × 3³
12² × 21³ = 2² × 2² × 3² × 7³ × 3³
= 2⁴ × 3⁵ × 7³

Class 7 Maths Chapter 8 Kerala Syllabus Repeated Multiplication Questions and Answers

Question 1.
Write each number below either as a power of a single prime or as a product of powers of different primes.
i) 3125
ii) 200
iii) 1600
Answer:
i) 3125 = 5 × 5 × 5 × 5 × 5 = 5s
ii) 200 = 2 × 2 × 2 × 5 × 5 = 2³3 × 5²
iii) 1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 = 2⁶ × 5²

Question 2.
Calculate the powers below as fractions:
i) \(\left(\frac{3}{2}\right)^3\)
Answer:
\(\left(\frac{3}{2}\right)^3=\left(\frac{3}{2}\right) \times\left(\frac{3}{2}\right) \times\left(\frac{3}{2}\right)\)
= \(\left(\frac{27}{8}\right)\)

ii) \(\left(\frac{3}{5}\right)^2\)
Answer:
\(\left(\frac{3}{5}\right)^2=\left(\frac{3}{5}\right) \times\left(\frac{3}{5}\right)\)
= \(\left(\frac{9}{25}\right)\)

iii) \(\left(2 \frac{3}{2}\right)^2\)
Answer:
\(\left(2 \frac{3}{2}\right)^2=\left(\frac{7}{2}\right)^2\)
= \(\left(\frac{49}{4}\right)\)

Question 3.
Write each product below as the product of powers of different primes:
i) 75 × 45
ii) 96 × 144
iii) 72 × 175
Answer:
i) 75 × 45
75 = 3 × 5²
45 = 3² × 5
5 × 45 = (3 × 5²) × (3² × 5)
= 3³ × 5³

ii) 96 × 144
96 = 2⁵ × 3¹
144 = 122 = (2² × 3)² = 2⁴ × 3²
96 × 144 = (2⁵ × 3¹) × (2⁴ × 3²)
= 2⁹ × 3³

iii) 72 × 175
7² = 8 × 9 = 2³ × 3²
17⁵ = 25 × 7 = 5² × 7¹
7² × 17⁵ = (2³ × 3²) × (5² × 7¹)

Question 4.
Calculate the following quotients:
(i) \(\frac{1440}{120}\)
(ii) \(\frac{729}{27}\)
Answer:
i) 1440 = 12² × 10
= (2² × 3)² × (2¹ × 5¹)
= 2⁵ × 3²2 × 5¹
120 = 12 × 10 = (22 × 3) × (2 × 5)
120 = 23 × 31 × 51
= \(\frac{1440}{120}=\frac{2^5 \times 3^2 \times 5^1}{2^3 \times 3^1 \times 5^1}\)
= 2^5-3 x 3^2-1 x 5^1-1
= 2² × 3¹ × 5⁰
=4 × 3 × 1
= 12

(ii) \(\frac{729}{27}\)
729 = 3⁶
27 = 3³
\(\frac{729}{27}=\frac{3^6}{3^3}\) = 3^6-3 = 3³ = 27

Question 5.
Write each number below as a product of powers of different primes:
i) 28
ii) 45²
iii) 18² × 30²
iv) 20³ × 27¹
Answer:
i) 28 = 2² × 7¹
ii) 45² = (3² × 5¹)² = 3⁴ × 5²
iii) 18² × 30² = (2¹ × 3²)² × (2¹ × 3¹ × 5¹)² = 2²⁺² × 3⁴⁺² × 5² = 2⁴ × 3⁶ × 5²
iv) 20³ × 27¹ = (2² × 5¹)³ × (3³)
= 2⁶ × 5³ × 3³

Class 7 Maths Chapter 8 Notes Kerala Syllabus Repeated Multiplication

In this chapter, we will explore the concept of repeated multiplication, which is a fundamental idea in mathematics. Repeated multiplication occurs when we multiply a number by itself multiple times. In this chapter we discuss about Factors, Power of Fractions, Product of Powers, Quotient of Powers, Multiples And Powers. So,

• In the first topic ‘Factors’, we will discuss about exponents, powers and its related problems.
• And in the second topic ‘Power of fractions’, we discuss about how powers of fractions increases and decreases.
• In the third topic, we discuss about product of powers, here we discuss products and its powers.
• In the fourth topic, we discuss about Quotient of Powers, here we studied how we deal with powers in division.
• And in the last topic, Multiples and Powers, here we discuss about how we use powers in multiplication.

Understanding repeated multiplication helps us simplify complex calculations and is essential for learning about powers and roots.

Factors
Numbers can be split into products of prime numbers.
For example,128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
We can write this product in a shortened form as 27 (“read as, two to the seventh power”)
27 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Therefore, Repeated multiplication of the same number is in this form Similarly, we write repeated addition as multiplication.
For example,
5 + 5 = 2 × 5
5 × 5 = 5²
5 + 5 + 5 = 3 × 5
5 × 5 × 5 = 5³
5 + 5 + 5 + 5 = 4 × 5
5 × 5 × 5 × 5 = 5⁴
The operation of multiplying a number by itself repeatedly is called exponentiation.
The number showing how many are multiplied together is called exponent.
We write the exponent in a smaller size, to the right and slightly above the number multiplied.
The number got by repeatedly multiplying a number by itself are called powers.

For example,
3 × 3 × 3 = 3³ Third power of three
5 × 5 × 5 × 5 = 5⁴ Fourth power of five
7 × 7= 7² Second power of seven

We can consider, any number as the first power of itself.
How do we split 576 as a product of primes?
576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
= 2⁶ × 3³
Any natural number greater than 1 can be written either as the power of a prime or as the product of powers of different primes.

Powers of Fractions
We know that area of a square of side 4 meters is 4 × 4 as 4²
Then what is the area of a square of side \(\frac{1}{4}\) meter?
i.e, \(\frac{1}{4} \times \frac{1}{4}=\left(\frac{1}{4}\right)^2\)
\(\frac{1}{4} \times \frac{1}{4}=\frac{1}{4 \times 4}\)
\(\left(\frac{1}{4}\right)^2=\frac{1}{4^2}\)

Similarly, what is the area of a square of sides 0.33 metres?
(0.33)² = 0.33 × 0.33
= 0.1089 square metres

In general, the area of a square with sides of length x is x².
Find the volume of a cube with lengths of edges 0.5 metre
(0.5)³ = 0.5 × 0.5 × 0.5
= 0.125 cubic metre

Using algebra, the volume of a cube with the length of the edges x is x³.
Then, Find the volume of a cube of edges 2\(\frac{1}{4}\) metres?
(2\(\frac{1}{4}\)) = \(\left(\frac{9}{4}\right)^3\)
= \(\frac{9}{4} \times \frac{9}{4} \times \frac{9}{4}\)
= \(\frac{9 \times 9 \times 9}{4 \times 4 \times 4}\)
= \(\frac{729}{64}\)

We can split this into quotient and reminder
\(\frac{729}{64}\) = 11\(\frac{25}{64}\)

Or use a calculator to compute
\(\frac{729}{64}\) = 11.390625

1 metre = 100 centimetres
= 10² centimetres

1 cubic metres = 100 × 100 × 100 cubic centimetres
= 10⁶ cubic centimetres

11.390625 cubic metres = 11. 390625 × 10⁶ cubic centimetres
= 11390625 cubic centimetres

Powers of 2
2² = 2 × 2 = 4
2³ = 4 × 2 = 8
2⁴ = 8 × 2 = 16
2⁵ = 16 × 2 = 32
By knowing the powers of 2, we can easily compute the powers of \(\frac{1}{2}\).

Each multiplication by 2 doubles, while each multiplication by \(\frac{1}{2}\) halves.
In general, Powers of numbers greater than 1 steadily increase. Powers of numbers greater than 0 and less than 1 steadily decrease. All powers of 1 remain 1 and all powers of 0 remain 0.

Multiples And Powers
Let’s see how we can add 2 times 4 and 2 times 6.

We know,
2 times 4 = 4 + 4
2 times 6 = 6 + 6
Adding this we get,
(2 times 4) + (2 times 6) = (4 + 6) + (4 + 6)
= 10 + 10 = 2 times 10

In short, we can write,
(2 × 4) + (2 × 6) = 2 × (4 + 6)

Using power, we can do the same as:
2^nd power of 4 = 4 × 4
2^nd power of 6 = 6 × 6

Multiplying,
(2^nd power of 4) × (2^nd power of 6)
= (4 × 6) × (4 × 6)
= 2nd power of (4 + 6)

That is,
4² × 6² = (4 × 6)²
In general,
The product of the same powers of two numbers is equal to the same power of the product of these numbers

And by using algebra,
xⁿyⁿ = (xy)ⁿ for all numbers x, y and for all natural numbers

Now let’s check what is 5 times 2 times 7?
We can calculate like this:
5 times 2 is 5 × 2 = 10
10 times 7 = 10 × 7 = 70

Also, we can calculate like this:
2 times 7 = 2 × 7 = 14
5 times 14 is 5 × 14 = 70

Let’s see how the second computation works:
2 times 7 = 7 + 7 = 14
5 times 14 is (7 + 7) (7+ 7) (7+ 7) (7+ 7) (7+ 7) = 10 times 7

Thus, we can write the computation as:
5 × (2 × 7) = (5 × 2) × 7

Using power, we can do the same as:
That is 5th power of 7 = 7 × 7
5th power of (7 × 7) = (7 × 7)(7 × 7)(7 × 7)(7 × 7)(7 × 7)
= 10th power of 7

In short using exponent we can write this as,
(7 × 7)⁵ = (7²)⁵ = 7¹⁰
Thus, we get the relation, that is,
In computing a power of a power of a nmnber the exponents should be multiplied
This can be written using algebra as an equation.
(x^m)ⁿ = x^mn for all numbers x and for all natural numbers m and n

Products Of Powers
Let’s check how we can find 3² × 3⁴
We know that
3² = 3 × 3
3⁴ = 3 × 3 × 3

Multiplying

Thus 3² × 3⁴ = 3²⁺⁴ = 3⁶
In general, we can say:
In multiplying two powers of a number, the exponents should be added

Using algebra, we can write it as follows:
x^m × xⁿ = x^m+n for all numbers x and all natural numbers m and n

We should note two things here:
(i) The product of two powers of the same number is a power of that number
(ii) The exponent of the product is the sum of the exponents of the numbers multiplied
For example, compute 5² × 5³ × 54⁴
5² × 5³ × 5⁴ = (5² × 5³) × 54⁴
= 5⁵ × 5⁴
= 5⁹

This method is also used for prime factorization of a product.
For example consider, 96 × 144
96 = 2 × 2 × 2 × 2 × 2 × 3 = 2⁵ × 3
144 = 2 × 2 × 2 × 2 × 3 × 3 = 2⁴ × 3²

and then compute the product as
96 × 144 = (2⁵ × 3) × (2⁴ × 3²)
= (2⁵ × 2⁴) × (3 × 3²)
= 2⁹ × 3³

Quotient Of Powers
Now let’s see how we can find the quotient using the powers.
For that consider an example, 288 ÷ 36
First factorize the numbers 288 and 36
288 = 2⁵ × 3²
36 = 2² × 3²

We can remove common factors. So that we get,
(2⁵ × 3²) ÷ (2² × 3²) = 2⁵ ÷ 2²

Now we can do the division easily
2⁵ ÷ 2² = 2^5-2 = 2³

Write all these steps together we get:
288 ÷ 36 = (2⁵ × 3²) ÷ (2² × 3²)
= 2⁵ ÷ 2²
= 2³
= 8
As a general principle we can state this as:

In dividing the larger power of a non-zero number by a smaller power of the same number, the exponents should be subtracted
And we can state this using algebra like this:
\(\frac{x^m}{x^n}\) = x^m-n, for all numbers x ≠ 0 and for all natural numbers m > n

• The operation of multiplying a number by itself repeatedly is called exponentiation.
• The number showing how many are multiplied together is called exponent.
• The number got by repeatedly multiplying a number by itself are called powers.
• Any natural number greater than 1 can be written either as the power of a prime or as the product of powers of different primes.

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 6 Ratio Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 6 Solutions Ratio

Class 7 Maths Chapter 6 Ratio Questions and Answers Kerala State Syllabus

Ratio Class 7 Questions and Answers Kerala Syllabus

Page 87

Question 1.
Write down the ratio of the height to width of each of the following rectangles using the smallest possible natural numbers.
(i) Height 8 centimetres, width 10 centimetres
(ii) Height 8 metres; width 12 metres
(iii) Height 20 centimetres, width 1 metre
(iv) Height 40 centimetres; width 1 metre
(v) Height 1;5 centimetres; width 2 centimetres
Answer:
(i) Height 8 centimetres, width 10 centimetres.
The ratio of the height to width = 8 : 10 = 4 : 5

(ii) Height 8 metres; width 12 metres.
The ratio of the height to width = 8 : 12 = 2 : 3

(iii) Height 20 centimetres, width 1 metre.
i.e, Height 20 centimetres, width 100 centimetres.
The ratio of the height to width = 20 : 100 = 1:5

(iv) Height 40 centimetres; width 1 metre.
i.e, Height 40 centimetres, width 100 centimetres.
The ratio of the height to width = 40 : 100 = 2:5

(v) Height 1.5 centimetres; width 2 centimetres.
The ratio of the height to width – 1.5 : 2 = 3 : 4

Question 2.
In the table below, the height, width and their ratio of some rectangles are given, but only two of each. Can you calculate the third and complete the table.

Answer:

Page 89

Question 1.
Amina has 105 rupees with her and Mercy has 175 rupees. What is the ratio of the smaller amount to the larger?
Answer:
Amount with Amina =105 rupees
Amount with Mercy = 175 rupees
The ratio of the smaller amount to the larger = 105 : 175
= 3:5

Question 2.
96 women and 144 men attended a meeting. Find the ratio of the number of women to the number of men.
Answer:
Number of women = 96
Number of men =144
The ratio of the number of women to the number of men = 96 : 144
= 2:3

Question 3.
Of two pencils, the shorter is 4.5 centimetres long and the longer 7.5 centimetres. What is the ratio of the length of the longer to the shorter?
Answer:
Length of the longer pencil = 7.5 centimetres
Length of the shorter pencil = 4.5 centimetres
The ratio pf the length of the longer to the shorter = 7.5 : 4.5
= 5:3

Question 4.
When a rope was used to measure the sides of a rectangle, the width was \(\frac{1}{4}\) of the rope and the height was \(\frac{1}{3}\) of the rope. What is the ratio of the height to the width?
Answer:
Width = \(\frac{1}{4}\) of the rope
Height = \(\frac{1}{3}\) of the rope

The ratio of the height to the width = \(\frac{1}{3}: \frac{1}{4}\)
\(\frac{1}{4}=\frac{1 \times 3}{4 \times 3}=\frac{3}{12}\)
\(\frac{1}{3}=\frac{1 \times 4}{3 \times 4}=\frac{4}{12}\)

∴ Ratio = \(\frac{4}{12}: \frac{3}{12}\) = 4 : 3

Question 3.
3\(\frac{1}{2}\) glasses of water would fill a large bottle and 2\(\frac{1}{2}\) glasses of water could fill a smaller bottle. What is the ratio of the capacities of the larger bottle to the smaller bottle?
Answer:
Capacity of the large bottle = 3\(\frac{1}{2}\) glasses of water
= \(\frac{7}{2}\) glasses of water.

The capacity of the small bottle = 2\(\frac{1}{2}\) glasses of water
= \(\frac{5}{2}\) glasses of water

The ratio of the capacities of the larger bottle to the smaller bottle = \(\frac{7}{2}: \frac{5}{2}\)
= 7:5

Page 90,91

Question 1.
To make dosa, we need to take 2 cups of black gram for every 6 cups of rice. For 9 cups of rice, how many cups of black gram shall be taken?
Answer:
2 cups of black gram for every 6 cups of rice.
Ratio of black gram to rice = 2:6 = 1: 3
For 9 cups of rice,
Black gram : 9 = 1: 3
\(\frac{\text { Black gram }}{9}=\frac{1}{3}\)
Black gram = \(\frac{1}{3}\) × 9 = 3 cups
So, 3 cups of black gram should be taken for 9 cups of rice.

Question 2.
To plaster the walls of a house, cement and sand are mixed in the ratio 1: 5.45 sacks of cement were bought. How many sacks of sand are needed?
Answer:
The ratio of cement to sand = 1: 5 ,
45 sacks of cement were bought,
45: sand = 1 :5
\(\frac{45}{\text { sand }}=\frac{1}{5}\)
sand = 45 × \(\frac{1}{5}\) = 225 sacks
So, 225 sacks of sand are needed for 45 sacks of cement.

Question 3.
12 litres of paint was mixed with 8 litres of turpentine while painting the house. How many litres of turpentine should be mixed with 15 litres of paint?
Answer:
12 litres of paint was mixed with 8 litres of turpentine.
∴ The ratio of”turpentine to paint = 8: 12 = 2: 3
For 15 litres of paint,
Turpentine: 15 = 2:3
\(\frac{\text { Turpentine }}{15}=\frac{2}{3}\)
Turpentine = \(\frac{2}{3}\) × 15 = 10 litres.
So, 10 litres of turpentine should be mixed with 15 litres of paint.

Question 4.
In a ward of a panchayat, the women and men are in the ratio 11 : 10. There are 1793 women in the ward. How many men are there in the ward? What is the total number of women and men?
Answer:
The ratio of women to men = 11: 10.
There are 1793 women in the ward.
\(\frac{1793}{\mathrm{men}}=\frac{11}{10}\)
∴ Men = 1793 × \(\frac{10}{11}\) = 1630
Total number of women and men = 1793 + 1630 = 3423

Page 92

Question 1.
Suhara and Sita started a business. Suhara invested 40000 rupees and Sita 50000 rupees. They made a profit of 9000 rupees. It was divided in the ratio of their
Answer:
Amount invested by Suhara = 40000 rupees
Amount invested by Sita = 50000 rupees
Total profit = 9000 rupees
Ratio of investment of Suhara to Sita = 40000: 50000 = 4: 5
So, Suhara got \(\frac{4}{9}\) of the total profit and Sita got \(\frac{5}{9}\) of the total profit.
∴ Profit earned by Suhara = 9000 × \(\frac{4}{9}\) = 4000 rupees
Profit earned by Sita = 9000 × \(\frac{5}{9}\) = 5000 rupees.

Question 2.
Ramesan and John took up a contract for a work. Ramesan worked the first 6 days and John 7 days. They got 6500 rupees. They divided it in the ratio of the number of days each worked. How much did each get?
Answer:
Number of days Ramesan worked = 6 days
Number of days John worked = 7 days
Total amount they got = 6500 rupees
The ratio of the number of days worked by Ramesan to John = 6:7
So, amount Ramesan get is \(\frac{6}{13}\) of the toatal amount and John get \(\frac{7}{13}\) of the total amount.
Amount Ramesan get = 6500 × \(\frac{6}{13}\) = 3000 rupees
Amount John get = 6500 × \(\frac{7}{13}\) = 3500 rupees

Question 3.
When Ramu and Raju divided a sum of money in the ratio 3 : 2, Ramu got 480 rupees.
(i) How much did Raju get?
(ii) What was the sum that was divided?
Answer:
(i) The ratio of amount got by Ramu to Raju = 3:2
480 : Raju = 3: 2
\(\frac{480}{\text { Raju }}=\frac{3}{2}\)
∴ Raju = 480 × \(\frac{2}{3}\) = 320 rupees.
So, the amount Raju get = 320 rupees.

(ii) Total amount = 480 + 320 = 800 rupees.

Question 4.
Draw a line AB, 9 centimetres long. Mark a point P on it. The lengths of AP and PB should J>e in the ratio 1:2. How far away from A should we mark P ? Calculate and mark it.
Answer:
Length of AB = 9 cm .
AP : PB = 1:2
So, the length of AP is \(\frac{1}{3}\) of AB and that of PB is \(\frac{2}{3}\) of AB.,
∴ AP = 9 × \(\frac{1}{3}\) = 3 cm
PB = 9 × \(\frac{2}{3}\) = 6 cm
So, we have to mark P at a distance of 3 cm from A

Question 5.
Draw a line 15 centimetres long. Mark a point on it that divides the line in the ratio 2 : 3. Calculate the length and mark the point.
Answer:
Let AB = 15 cm and P be the point thet divide AB in the ratio 2:3.
i.e., AP: PB = 2: 3
So, the length of AP is j of AB and that of PB is \(\frac{3}{5}\) of AB.
∴ AP = 15 × \(\frac{2}{5}\) = 6 cm
PB = 15 × \(\frac{3}{5}\) = 9 cm

Question 6.
Draw a rectangle of perimeter 30 centimetres and sides of length in the ratio 1:2.
(i) With the same perimeter draw two more rectangles with sides in the ratio 2 : 3 and 3 : 7.
(ii) Calculate the areas of the three rectangles. Which rectangle has the greatest area?
Answer:
Perimeter of the rectangle = 30 cm
2 (length + breadth) =30
Length+breadth =15
The ratio of sides =1:2
∴ Length = \(\frac{2}{3}\) × 15 = 10 cm
Breadth = \(\frac{1}{3}\) × 15 = 5 cm

(i) The ratio of sides = 2:3
∴ Length = \(\frac{3}{5}\) × 15 = 9 cm
Breadth = \(\frac{2}{5}\) × 15 = 6 cm

The ratio of sides = 3:7
∴ Length = \(\frac{7}{10}\) × 15 = 10.5 cm
Breadth = \(\frac{3}{10}\) × 15 = 4.5 cm

(ii) Area of first rectangle = 10 × 5 = 50 cm²
Area of second rectangle = 9 × 6 = 54 cm²
Area of third rectangle = 10.5 × 4.5 = 47.25cm²
So, second rectangle has the greatest area.

Intext Questions And Answers

Question 1.
The walls of Aji’s house needs painting. First, 25 litres of green paint and 20 litres of white paint were mixed. To get the same shade of green, how many litres of green should be mixed with 16 litres of white?
Ans:
The ratio of green to white = 25: 20 = 5: 4
That is, 5 litres of green for every 4 litres of white.
4 times 4 litres is 16 litres.
So, 4 times 5 litres, 20 litres of green should be mixed with 16 litres of white.
In terms of ratio,
Green: White = 5: 4
For 16 litres of white,
Green: 16 = 5 : 4
\(\frac{\text { Green }}{16}=\frac{5}{4}\)
∴ Green = \(\frac{5}{4}\) × 16 = 20 litres

Question 2.
The school needs a vegetable garden. A rectangular plot is to be roped off for this. The length of the rope is 32 metres. They decided to have width and length in the ratio 3:5. What should be the width and length?
Answer:
Length of the rope = 32 metres.
So, perimeter of rectangular garden = 32 metres
2 (length + width) = 32
Length + width = 16
The ratio of width to length = 3: 5
∴ Length = 16 × \(\frac{3}{8}\) = 6 metres
Breadth = 16 × \(\frac{5}{8}\) =10 metres

Class 7 Maths Chapter 6 Kerala Syllabus Ratio Questions and Answers

Question 1.
Angles of a linear pair are in the ratio 4: 5. What is the measure of each angle?
Answer:
Ratio of angles in the linear pair = 4: 5
Sum of angles = 180
In 180, \(\frac{4}{9}\) is one angle and – is the other angle.
So, one angle = 180 × \(\frac{4}{9}\) = 80°
Other angle = 180 × \(\frac{5}{9}\) = 100°

Question 2.
Sita and Soby divided some money in the ratio 1: 2 and sita got 400 rupees. What is the total amount they divided?
Answer:
Ratio in which money divided =1:2
Amount Sita got = 400 rupees
400: Sita =1: 2
\(\frac{400}{\text { Sita }}=\frac{1}{2}\)
∴ Sita = 400 × \(\frac{2}{1}\) = 800 rupees
So, total amount = 400 + 800 = 1200 rupees.

Question 3.
Ramesh’s father divided his saving as follows:

• \(\frac{2}{7}\) of his savings to Ramesh
• \(\frac{5}{7}\) of his savings to his mother.

Find the ratio of this division.
Answer:
Ratio = \(\frac{2}{7}: \frac{5}{7}\) = 2: 5

Question 4.
What does it mean to say that the width to length ratio of a rectangle is 1:1? What sort of rectangle is it?
Answer:
The width to length ratio of a rectangle is 1:1,
Which means, length = width
When the sides of a rectangle are equal, it will be a square.

Question 5.
Santha decided to give her salary to her children Ravi and Shinu in the ratio 3:2. If Ravi gets Rs.4500, then.
a) Find the amount Shinu gets?
b) How much salary does Santha get?
Answer:
(a) Ratio of amount got by Ravi to Shinu = 3:2
Amount Ravi got = 4500 rupees
4500: Shinu = 3:2
\(\frac{4500}{\text { Shinu }}=\frac{3}{2}\)
∴ Shinu = 4500 × \(\frac{2}{3}\) = 3000 rupees.

(b) Santha’s salary = 4500 + 3000 = 7500 rupees

Question 6.
Raju has an amount of 120 rupees and Mary has 180 rupees.
A) What is the ratio of the amounts Mary and Raju have?
(a) 3:2
(b) 2:3
(c) 6:5
(d) 5:9

B) Mother gave 60 rupees more to Mary. How much more money Raju needed to make the same ratio?

C) If they divide 800 rupees in the same ratio, how much amount did each get?
Answer:
A) Ratio of the amount Mary and Raju have = 180 : 120 = 3:2
B) Mother gave 60 rupees more to Mary.
Now the amount with Mary = 180 + 60 = 240 rupees.
240 : Raju = 3: 2
\(\frac{240}{\text { Raju }}=\frac{3}{2}\)
∴ Raju = 240 × \(\frac{2}{3}\) = 160
So, more money Raju needed = 160 – 120 = 40 rupees

C) Amount Raju got = 800 × \(\frac{2}{5}\) = 320 rupees
Amount Mary got = 800 × \(\frac{3}{5}\)
= 480 rupees.

Question 7.
Last year the ratio of the female teachers and male teachers of Ramapuram UP School was 6:1.
a) If the number of male teachers is 6, what is the number of female teachers? Instead of the female teachers who got transfer from the school, male teachers joined there. Now the ratio of the female teachers and male teachers is 11:10.
b) How many female teachers got transfer this year?
c) This year some female teachers and male teachers will be 1:1. If so, how many female teachers will retire this year?
Answer:
a) Ratio of female teachers to male teachers = 6:1
Number of male teachers = 6
Number of female teachers = 6 × 6 = 36

b) Total number of teachers = 36 + 6 = 42
Ratio = 11: 10
Number of female teachers = 42 × \(\frac{11}{21}\) = 22
So, number of female teachers transferred = 36 – 22 = 14

c) Number of male teachers = 42 × \(\frac{10}{21}\) = 20
So, inorder to become ratio 1:1, number of male teachers = number of female teachers
∴ The number of female teachers retire this year = 22 – 20 = 2

Practice Questions

Question 1.
Express the width and length in ratios.
(i) Width = 3 cm Length= 9 cm
(ii) Width = 6 cm Length= 14 cm
Answer:
(i) 1:3
(ii) 3:7

Question 2.
The perimeter of a rectangular garden is 28 metres. The ratio of length and width is given by 3:4. Find its length and width.
Answer:
6 cm, 8 cm

Question 3.
The capacity of small tank is 500 litres and big tank is 1500 litres.
(a) Find the ratio of the capacities of the small tank and the big tank.
(b) Small tank is fully filles with water and big tank is half filled. Then, find the ratio of the capacities of the small tank and the big tank.
(c) 1500 litres of water is distributed to the houses of Appu and Muthu in the ratio 3:2. How much water will each house get?
Answer:
(a) 1:3
(b) 2:3
(c) 900 litres, 600 litres

Question 4.
24 litres of curd was mixed with 96 litres of water to make buttermilk.
(a) What is the ratio of water and curd used?
(b) How many litres of water is needed to mix with 96 litres of curd to make buttermilk in the same ratio?
(c) How many litres of curd is needed to make 600 litres of buttermilk?
Answer:
(a) 4:1
(b) 384 litres
(c) 120 litres

Question 5.
Anu and Manu divided ah amount in the ratio 3:2. If Anu got 100 rupees more, what is the total amount they divide?
Answer:
500 rupees

Question 6.

(a) What part of the circle is shaded in the picture?
(b) What part of the circle is unshaded?
(c) What is the ratio of shaded to unshaded?
Answer:
(a) \(\frac{1}{4}\)
(b) \(\frac{3}{4}\)
(c) 1:3

Question 7.
For making idlis, rice and Urad are to be taken in the ratio 2:1. In 9 cups of such a mxiture of rice and urad, how may cup of rice and urad are taken?
Answer:
6 cups,3 cups

Question 8.
In a co-operative society, there are 600 members are male and 400 are female. An executive committee of 30 members is to be formed with the same male to female ratio as in the society. How many male and female members are to be there in the committee?
Answer:
8, 12

Question 9.
A rectangular piece of land is to be marked on the school ground for a vegetable garden.
Hari and Mary started making rectangle with a 24 metre long rope. Vimala teacher said it would be nice, if the sides are in the ratio 3:5. What should be length and width of the rectangle? .
Answer:
7.5 m , 4.5 m

Question 10.
When measuring the length of two buses with one rope the length of the first bus is \(\frac{2}{3}\) of the length of the rope and the length of the second bus was \(\frac{3}{5}\) of the length of the rope. What is the ratio between the length of the buses.
Answer:
10:9

Class 7 Maths Chapter 6 Notes Kerala Syllabus Ratio

When quantities like length are measured using a definite unit, we may not always get counting numbers; and it is this fact which led to the idea of fractions. In comparing the sizes of two quantities, one question is whether both can be given as counting numbers, using a suitably small unit of measurement. It is this question that leads to the idea of ratio. We can use ratio to compare parts of a whole also

• Generally, when we state ratios, we avoid fractions and decimals. In other words, the smallest possible counting number is used to express ratios. By multiplying or dividing the ratio by the same number, we can make it in terms of the smallest possible counting number.
• We can use ratios to express any two measures, not just lengths, as multiples and parts.
• In any mixture, components are in fixed ratio. So, by knowing the ratio one can find the quantity of one component if other is given.
• If we are given the ratio in which a quantity is divided, we can find how much each part is using this ratio.

Rectangle Problems
Consider the following rectangle.

In both the rectangles, the width is 3 times the height. Or we can say that the height is ^ times the width.
So, the ratio of width to height is 3 : 1 and the ratio of height to width is 1 : 3 (The ratio of width to height = 6 : 2 = 3 : 1 or 4.5 : 1.5 = 3 : 1)
If we extend the height and width by the same multiple or shorten them by the same fraction, the ratio remains the same.

Other Measures
We can use ratios to express any two measures (not just lengths) as multiples and parts. For example, .
Suppose we have a 15 litre bucket and a 25 litre bucket.
The small bucket holds \(\frac{15}{25}=\frac{3}{5}\) of what large bucket holds.
Or
we can say that,
The ratio of capacity of small bucket to large bucket = 15 : 25 = 3:5

• If we extend the height and width by the same multiple or shorten them by the same fraction, the ratio remains the same.
• We can use ratios to express any two measures as multiples and parts.
• In a mixture, the components are mixed in fixed ratio.
• We can use ratio to compare parts of a whole also.

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 7 Shorthand Math Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 7 Solutions Shorthand Math

Class 7 Maths Chapter 7 Shorthand Math Questions and Answers Kerala State Syllabus

Shorthand Math Class 7 Questions and Answers Kerala Syllabus

Page 96

Question 1.
Write the following statements using the language of algebra.
1) Zero added to any number gives the same number.
2) Zero subtracted from any number gives the same number.
3) Any number subtracted from the same number gives zero.
4) Any number multiplied by zero gives zero.
5) Any number divided by the same number gives 1.
6) Twice a number added to the number makes three times the number.
7) Twice a number subtracted from thrice the number gives the number.
8) A number added to a number, and then the added number subtracted gives the original number.
Answer:
Let n be the number.
1) n + 0 = n
2) n – 0 = n
3) n – n = 0
4) n × 0 = 0
6) n + 2n = 3n
7) 3n – 2n = n
8) Let m be the other number, n + m – m = n

Page 98

Question 1.
Do the following problems mentally:
(i) 49 + 125 + 75
(ii) 3\(\frac{1}{2}\) + 8\(\frac{3}{4}\) + \(\frac{1}{4}\)
(iii) 15.5 + 0.25 + 0.75
(iv) 38 + 27
(v) 136 + 64
Answer:
(i) 49 + 125 + 75 = 49 + 100 + 25 + 75
= 49 + 100+ 100
= 49 + 200
= 249

(ii) 3\(\frac{1}{2}\) + 8\(\frac{3}{4}\) + \(\frac{1}{4}\)
= \(\frac{7}{2}+\frac{36}{4}\)
= \(\frac{7}{2}+\frac{18}{2}\)
= \(\frac{25}{2}\)
= 12\(\frac{1}{2}\)

(iii) 15.5 + 0.25 + 0.75 = 15.5 + 1.0
= 16.5

(iv) 38 + 27 = 38 + 2 + 25
= 40 + 25
= 65

(v) 136 + 64 = 136 + 4 + 60
= 140 + 60
= 200

Question 2.
Do the following proble mentally:
(i) (135 – 73) – 27
(ii) (37 – 1\(\frac{1}{2}\)) – \(\frac{1}{2}\)
(iii) (298 – 4.5) – 3.5
(iv) 78 – 29
(v) 140 – 51
Answer:
(i) (135 – 73) – 27 = 135 – (73 + 27)
= 135 – 100
= 35

(ii) (37 – 1\(\frac{1}{2}\)) – \(\frac{1}{2}\) = (37 – 1.5) – 0.5
= 37 – (1.5 + 0.5)
= 37 – 2
= 35

(iii) (298 – 4.5) – 3.5
= 298 – (4.5 + 3.5)
= 298 – 8
= 290

(iv) 78 – 29 = 78 – (30 – 1)
= 78 – 30 + 1
= 48 + 1
= 49

(v) 140 – 51 = 140 – (50 + 1)
= 140 – 50 – 1
= 90 – 1
= 89

Page 99

Do the following problems mentally:
(i) (136 + 29) – 19
(ii) (3\(\frac{1}{2}\) + 5\(\frac{3}{4}\)) – 2\(\frac{1}{4}\)
(iii) (298 + 14.5) – 12.5
(iv) 23 + (35 – 18)
(v) 65 + 98
Answer:
(i) (136 + 29) – 19 = 136 + (29 – 19)
= 136 + 10
= 146

(ii) (3\(\frac{1}{2}\) + 5\(\frac{3}{4}\)) – 2\(\frac{1}{2}\) = (3.5 + 5.75) – 2.25
= 3.5+ (5.75 – 2.25)
= 3.5 + 3.5
= 7

(iii) (298 + 14.5) – 12.5 = 298 + (14.5 – 12.5)
= 298 + 2
= 300

(iv) 23 + (35 – 18) = (23 + 35) – 18
= 58 – 18
= 40

(v) 65 + 98 = 65 + (100 – 2)
= (65 + 100) – 2
= 165 – 2
= 163

Page 101

Do the following problems mentally:
(i) (135 – 73) + 23
(ii) (38 – 8\(\frac{1}{2}\)) + \(\frac{1}{2}\)
(iii) (19 – 6.5) + 2.5
(iv) 135 – (35 – 18)
(v) 240 – (40 – 13)
Answer:
(i) (135 -73) + 23 = 135 -(73 – 23)
= 135 – 50
= 85

(ii) (38 – 8\(\frac{1}{2}\)) + \(\frac{1}{2}\) = (38 – 8.5) + 0.5
= 38 – (8.5 – 0.5)
= 38 – 8
= 30

(iii) (19 – 6.5) + 2.5 = 19 – (6.5 – 2.5)
= 19 – 4
= 15

(iv) 135 – (35 – 18) = (135 – 35) + 18
= 100 + 18
= 118

(v) 240 – (40 – 13) = (240 – 40) + 13
= 200 + 13
= 213

Page 102

Do the following problems mentally:
(i) 103 × 15
(ii) 98 × 25
(iii) (63 × 12) + (37 × 12)
(iv) (65 × 11) – (55 × 11)
(v) (15 × \(\frac{3}{4}\)) + (5 × \(\frac{3}{4}\))
(vi) (5\(\frac{1}{2}\) × 23) – (4\(\frac{1}{2}\) × 23)
Answer:
(i) 103 × 15 = (100+ 3) × 15
= (100 × 15) +(3 × 15)
= 1500 + 45
=1545

(ii) 98 × 25 = (100 – 2) × 25
= (100 × 25)-(2 × 25)
= 2500 – 50
= 2450

(iii) (63 × 12) + (37 × 12)
= (63 + 37) × 12
= 100 × 12
= 1200

(iv) (65 × 11) – (55 × 11)
= (65 – 55) × 11
= 10 × 11
= 110

(v) (15 × \(\frac{3}{4}\)) + (5 × \(\frac{3}{4}\))
= \(\frac{3}{4}\) × (15 + 5)
= \(\frac{3}{2}\) × 15
= \(\frac{45}{2}\)

(vi) (5\(\frac{1}{2}\) × 23) – (4\(\frac{1}{2}\) × 23)
= 23 × (5\(\frac{1}{2}\) + 4\(\frac{1}{2}\))
= 23 × (5.5 + 4.5)
= 23 × 10
= 230

Class 7 Maths Chapter 7 Kerala Syllabus Shorthand Math Questions and Answers

Question 1.
Write the following statements using the language of algebra.
i. When 15 is added to a number, then it is equal to three times that number.
ii. If 25 is added to three times a number, the result is 70.
iii. One third of a number when added to 1 gives 15
Answer:
i. n + 15 = 3n
ii. 3n + 25 = 70
iii. 1 + \(\frac{n}{3}\) = 15

Question 2.
Find 24 + 16 + 34
Answer:
24 + 16 + 34 = 24 + (16 + 34)
= 24 + 50
= 74

Question 3.
Find 79 – 52 – 18
Answer:
79 – 52 – 18 = 79 – (52 + 18)
= 79 – 70
= 9

Question 4.
Find 3 × 13 + 3 × 7
Answer:
3 × 13 + 3 × 7 = 3(13 + 7)
= 3 × 20
= 60

Question 5.
Find 7 × 48
Answer:
7 × 48 = 7(50 – 2)
= 350 – 14
= 336

Class 7 Maths Chapter 7 Notes Kerala Syllabus Shorthand Math

Algebra is one of the main branches of mathematics. In algebra, we make use of letters to solve mathematics. This chapter introduces some basic concepts of algebra. Following are the main topics discussed in this chapter.

Numbers and letters

• The method of stating relations between measures or numbers using letters is called algebra.
• In algebra, we write products without multiplication signs.
• In products with numbers and letters, we write the number first in algebra
• We can use any letter to denote the number or the measure in algebra
• We write division as a fraction in algebra
• We need to use more than one letter when we talk about several numbers in algebra

One by one and altogether

• While adding three numbers in a row, we can use the following result.
  (x + y) + z = x + (y + z), for any three numbers x, y, z.
• While subtracting three numbers in a row, we can use the following result.
  (x – y) – z — x – (y + z) for any three numbers x, y, z.

Addition and Subtraction
Adding a larger number to the given number and then subtracting a smaller number gives the same result as adding the difference between, the larger number and the smaller number to the given number. That is,
(x + y) – z = x + (y – z), for any three numbers x, y, z with y > z

Subtracting a larger number from the given number and then adding a smaller number gives the same result as subtracting the difference between the larger number and the smaller number from the given number. That is,
(x – y) + z = x – (y – z), for any x, y,-z with y > z
We must use brackets to show the order of operations clearly.

Addition and subtraction and then multiplication

• Multiplying a sum by the given number gives the same result as multiplying each number in the
  sum separately by the given number and then adding them. That is,
  (x + y)z = xz + yz, for any three numbers x, y and z
• Multiplying a difference by a number gives the same result as multiplying each number in the difference and then subtracting them. That is,
  (x – y)z = xz – yz, for any three numbers x, y, z

Numbers And Letters
The method of stating relations between measures or numbers using letters is called algebra. Eg:
Consider the following fact: “A number added to itself is twice the number.”
In the language of math, we write it as follows; a number + the same number = 2 × number Let’s denote the number by n.
Then, n + n = 2n for any number n, which is the algebraic representation of the given fact.

Some features of algebra:

• Write products without multiplication sign.
  Eg:
  3 × n can be written as 3n.
• In products with numbers and letters, write the number first.
  Eg;
  5 × m is written as 5m, not m5.
• We can use any letter to denote the number or the measure.
  Eg:

Consider the fact; “A number added to itself is twice the number.”

• If we denote the number as n, the algebraic form is n + n = 2n.
• If we denote the number as x, the algebraic form is x + x = 2x.

We write division as a fraction.
Eg:
Consider the fact: “Five times a number divided by five gives that number.”
If we denote the number as n, the algebraic form is \(\frac{5n}{5}\) = n.

We need to use more than one letter when we talk about several numbers.
Eg:
Consider the fact: If we add a number to another number and then subtract the original number, we get the added number.
Let n be the original number and m be the added number. Then the fact becomes; n + m – n = m

The letters that we use to represent numbers or measures in algebra are generally know as variables

One By One And Altogether
Adding two numbers one after another to a number, or adding their sum, gives the same result. Algebraically, (x + y) + z = x + (y + z), for any three numbers x, y, z.
Eg:
Consider the numbers 1, 2 and 3.
(1 + 2) + 3 = 3 + 3 = 6
1 + (2 + 3) = 1 + 5 = 6

There are situations where adding the sum of these numbers is easier than adding one after another. .
Eg:
15 + 28 + 2 =15 + 30
= 45

There are situations where adding one after another is easier than adding the sum.
Eg:
25 + 18 = 25 + 5 + 13
= 30 + 13
= 43

Subtracting two numbers one after the other from a number or subtracting the sum of these two numbers from the first number both give the same result. Algebraically,
(x – y) – z = x – (y + z) for any three numbers x, y, z.
Eg:
Consider the numbers 10, 7 and 2.
(10 – 7) – 2 = 3 – 2 = 1
10 – (7 + 2) = 10 – 9 = 1

There are situations where subtracting the sum is easier than subtracting the numbers one after the other.
Eg:
35 – 17 – 3 = 35 – (17 + 3)
= 35 – 20
= 15

There are situations where subtracting the numbers one after the other is easier than subtracting the sum.
Eg:
500 – 201 = 500 – 200 – 1
= 300 – 1
= 299

Addition And Subtraction
(x + y) – z = x + (y – z), for any three numbers x, y, z with y > z
Eg:
Consider the numbers 5, 4 and 3.
(5 + 4) – 3 = 9 – 3 = 6
5 + (4 – 3) = 5 + 1 = 6
Therefore, (5 + 4) – 3 = 5 + (4 – 3)

Sometimes, it is more convenient to apply it in the reverse.
Eg:
25 + 99 = 25 + (100 – 1)
= (25 + 100) – 1
= 125 – 1
= 124

(x – y) + z = x – (y – z), for any x, y, z with y > z Eg:
Consider the numbers 10, 7 and 4.
(10 – 7) + 4 = 3 + 4 = 7
10 – (7 – 4) = 10 – 3 = 7
Therefore, (10 – 7) + 4 = 10 – (7 – 4)

We must use brackets to show the order of operations clearly.

Explanation:
Adding 7 and 4 and then adding 2 gives 13.
Adding 4 and 2 first and then adding this to 7 also gives 13.
That is, (7 + 4) + 2 = 7 + (4 + 2)

So, we may write this sum as 7 + 4 + 2 without brackets.
But if we subtract 4 from 7 and then subtract 2, we get 1.

That is, (7 – 4) – 2 = 1. Whereas if we
subtract 2 from 4 and then subtract it from 7, we get 5.

That is 7 – (4 – 2) = 5.
So, if we just write 7 – 4 – 2, the answer will be different depending on which operation we do first.
So, we must use brackets to show the order of operations clearly.

Addition And Subtraction And Then Multiplication

(x + y)z = xz + yz, for any three numbers x, y and z Eg:
Consider the numbers 1, 2, 3.
(1 + 2) × 3 = 3 × 3 = 9
(1 × 3) + (2 × 3) = 3 + 6 = 9
Therefore, (1 + 2) × 3 = (1 × 3) + (2 × 3)

Reading these in reverse is also useful in some problems. That is, xz + yz = (x + y)z
Eg:
32 + 56 =(4 × 8)+ (7 × 8)
= 8 × (4 + 7)
= 8 × 11
= 88

(x – y)z = xz – yz, for any three numbers x, y, z
Eg:
Consider the numbers 7, 5 and 3. .
(7 – 5) × 3 = 2 × 3 = 6
(7 × 3) – (5 × 3) = 21 – 15 = 6
Therefore, (7 – 5) × 3 = (7 × 3) – (5 × 3)

Reading these in reverse is also useful in some problems, That is, xz – yz = (x – y)z
Eg:
(\(\frac{1}{2}\) × 35) – (\(\frac{1}{2}\) × 15) = \(\frac{1}{2}\) (35 – 15)
= \(\frac{1}{2}\) × 20
= 10

The method of stating relations between measures or numbers using letters is called algebra.

Features of algebra:

• Write products without multiplication sign.
• In products with numbers and letters, write the number first.
• We can use any letter to denote the number or the measure.
• We write division as a fraction.
• We need to use more than one letter when we talk about several numbers.

Adding two numbers one after another to a number, or adding their sum, gives the same result.
Algebraically, (x + y) + z = x + (y + z), for any three numbers x, y, z.

• There are situations where adding the sum of these numbers is easier than adding one after another.
• There are situations where adding one after another is easier than adding the sum.

Subtracting two numbers one after the other from a number or subtracting the sum of these two numbers from the first number both give the same result. Algebraically,
(x – y) – z = x – (y + z) for any three numbers x, y, z. ,

There are situations where subtracting the numbers one after the other is easier than subtracting the sum.
(x + y) – zfc x + (y – z), for any three numbers x, y, z with y > z

Sometimes, it is more convenient to apply it in the reverse. That is, x + (y – z) = (x + y) – z

• We must use brackets to show the order of operations clearly.
• (x – y) + z = x – (y – z), for any x, y, z with y > z
• (x + y)z = xz + yz, for any three numbers x, y, z

Reading these in reverse is also useful in some problems. That is,
xz + yz = (x + y)z ,

• (x – y)z = xz – yz, for any three numbers x, y, z
• Reading these in reverse is also useful in some problems. That is, xz – yz = (x – y)z