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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

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 1 Parallel Lines Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 1 Solutions Parallel Lines

Class 7 Maths Chapter 1 Parallel Lines Questions and Answers Kerala State Syllabus

Parallel Lines Class 7 Questions and Answers Kerala Syllabus

Page 17

Question 1.
Draw the parallelogram below with the given measures.

Calculate the other three angles.
Answer:
Draw a horizontal line segment AB that is 5 cm long. This will be one side of the parallelogram.

At point A, use a protractor to measure and mark an angle of 60 degrees from line AB.

From point A, draw a line segment AC that is 3 cm long, making sure it forms the 60° angle with AB.

From point B, draw a line segment BD that is 3 cm long, parallel to AC. You can use a ruler and a set square to ensure parallelism.
From point C, draw a line segment CD that is 5 cm long, parallel to AB. The points C and D should connect to complete the parallelogram.

Now finding the remaining angles,
We know that, ∠A + ∠C = 180°
∠C = 180° – 60° = 120°
Similarly, ∠A + ∠B = 180°
∠B = 180° – 60° = 120°
And, ∠B + ∠D = 180°
∠D = 180° – 120° = 60°

Question 2.
The top and bottom blue lines in the figure are parallel. Find the angle between the green lines.

Answer:
Draw a horizontal line parallel to the other two parallel lines

From the figure, we get,
∠EAB = ∠ABD = 40°
∠DBC – ∠BGF = 50°
:. ∠ABC = ∠ABD + ∠DBC
= 40° + 50° = 90°

Question 3.
In the figure, the pair of lines slanted to the left are parallel; and also the pair of lines slanted to the right.
Draw this figure:

Answer:
Draw a line segment AB measuring 2 cm.

Draw a perpendicular line from point A.

Measure 20° on the protector, draw two lines on both sides of the perpendicular line, and mark C and D.

Using a set square, draw a line parallel to line AC starting from point B

Using a set square, draw a line parallel to line AD starting from point B and mark the intersection point as E.

Page 21

Question 1.
Draw the triangle with the given measures.

Answer:
First off all we have to find the third angle of the given triangle
⇒ Third angle = 180° – (60° + 40°) = 180° – 100° = 80°
Now we can start drawing the triangle, for that
Draw a horizontal line segment that is 5 cm long.

At left end of the line, use a protractor to measure and mark an angle of 40° from the line

At right end of the line, use a protractor to measure and mark an angle of 40°from the line

Construct line segments forming angles of 40° and 80°, extending each until the lines intersect

Question 2.
The figure shows a triangle drawn in a rectangle. Calculate the angles of the triangle.

Answer:
Consider the figure

Since, ABCD is rectangle ∠A, ∠B, ∠C, ∠D equals to 90°
Now, it is given that one part of ∠A is 40°, so the remaining part of A is 50°
Similarly, one part of ∠D is 25°, so the remaining part of ∠D is 65°
It implies that two angles of the triangle are 50° and 65°.
Therefore, the third angle of the triangle is = 180° – (50° + 65°)
= 180° – (105°)
= 75°

So, the angles of triangles are 50°, 65°, 75o

Question 3.
The top and bottom lines in the figure are parallel. Calculate the third angle of the bottom triangle and all angles of the top triangle.

Answer:

Consider the above figure,
Given that AB and CD are parallel lines
The two given angles of the bottom triangle are 35° and 45°
Therefore, third angle of bottom triangle = 180° – (35° +45°)
= 180° – (80°)
= 100°
In bottom triangle, ∠B and in top triangle, ∠C are Alternate angles
Therefore, ∠A and ∠D are equal. ∠D of top triangle is = 35°
Similarly, in bottom triangle, ∠B and in top triangle, ∠C are Alternate angles
Therefore, B and C are equal, ∠C of top triangle is = 45°
So, the third angle of top triangle = 180° – (35° + 45o)
= 180° – (80°)
= 100°
All angles of bottom triangle = 35°, 45°,
All angles of top triangle = 35°, 45°, 100°.

Question 4.
The left and right sides of the large triangle are parallel to the left and right sides of the small triangle. Calculate the other two angles of the large triangle and all angles of the small triangle.

Answer:

Consider the above figure,
Given that AB and CD are parallel, and AC and DE are also parallel,
It’s given that the middle angle is 70°.
Considering the parallel lines AB and CD, AC is a slanting line intersecting the parallel lines.
Therefore, the middle angle and ∠A of the larger triangle are alternate angles,
Therefore, these two angles are equal
i.e., ∠A = 70°
Now, we can calculate the third angle of the larger triangle
i.e., ∠B = 180° – (70° +60°)
= 180° – (130°) = 50°

All three angles of larger triangle are 50°, 60°, 70°
Similarly, considering the parallel lines AC and DE, CD is a slanting line intersecting the parallel lines.
Considering the parallel lines AC and DE, CD is slanting line through the parallel lines.
Here, middle angle and the ∠D of smaller triangle are alternate angles,
Therefore, these two angles are equal
i.e., ∠D = 70°
Now, ZC of larger triangle, middle angle.and ∠C of the smaller triangle form a straight line.
Therefore, ∠C of smaller triangle is = 180° − (60° + 70°)
= 180° – 130°
= 50°
Now we can find the third angle of smaller triangle (∠E) = 180° – (50° + 70°)
= 180° – (120°)
= 60°
All three angle of smaller triangle are 50°, 60°, 70°

Question 5.
A triangle is drawn inside a parallelogram.
Calculate the angles of the triangle.
Answer:

Consider the above figure,
∠D and ∠B are opposite angles of parallelogram.
Therefore, D = ∠B = 110°
But, one part of ∠D is 60o, so remaining part of ∠D = 110° – 60° = 50°
Now, considering the angles A and <D, their sum is equals to 180°
i.e., ∠A + ∠D = 180°
∠A = 180° – ∠D
= 180° – 110° = 70°
But, one part of ∠A is 30°, so remaining part of ∠A = 70° – 30° = 40°
It means, we get two angles of the triangle which are 50° and 40°
So, third angle of the triangle = 180° – (50° + 40°)
= 180° – (90°)
= 90°

Hence, all three angles of the triangles are 50°, 40°, 90°

Intext Questions and Answers

Question 1.
Find the unknown angle in the following figures?

Answer:
i. By extending the parallel lines and slanting line we get the figure below

From the figure, it is clear that two small angles are formed when the slanting line intersects the top and bottom parallel lines. Since these angles are equal in measure.
Therefore, the unknown angle is 45°

ii. By extending the parallel lines and slanting line we get the figure below

From the figure, it is clear that two small angles are formed when the slanting line intersects the top and bottom parallel lines. Since these angles are equal in measure.
Therefore, the unknown angle is 40°.

iii. By drawing a vertical line, as shown in the figure below, we can observe that this vertical line is parallel to the other parallel lines and bisects the middle angle. Therefore, the angles on the left side of the vertical line are 30°

Similarly, since the angle on the right side is also 30°, then the unknown angle will also be 30o.

When two parallel lines are cut by a third line (called a slanting line), different pairs of angles are formed.

Question 2.
One angle of a triangle is 72°. The other two angles are of equal measure. What are their measures?
Answer:
Sum of two other angles = 180° – 72° = 108°
Since the other two angles are equal, each angle is = \(\frac{108^0}{2}\) = 54

Question 3.
When a line crosses another line, how many angles are formed between them ?

Answer:
When a line crosses another line, then 4 angles are formed between them.

When two lines cross perpendicular, then all angles are 90°.

Consider the figure below,

Here, 4 angles are 21, 22, 23, 24, where 21 and 22 are small angles, and 23 and 24 are large angles Then we can say that,
The two small angles are of the same measure.
i.e., ∠1 = ∠2

The two large angles are of the same measure.
i.e., ∠3 = ∠4

The sum of a small angle and a large angle is 180°.
i.e., ∠1 + ∠4 = 180° and ∠3 + ∠4 = 180°.

Consider the figure below, which shows two parallel lines intersected by another line, forming eight angles.

Then we can say that,
∠1 = ∠5
∠2 = ∠6
∠3 = ∠7
∠4 = ∠8
For example,
In the given figure, one of the angles is given as 50°

Then, the remaining angles are shown below

A line intersects two parallel lines at angles of the same measure

Question 4.
Can you calculate the other seven angles which the parallel lines make with the slanting line?

Answer:

Class 7 Maths Chapter 1 Kerala Syllabus Parallel Lines Questions and Answers

Question 1.
Draw the parallelogram below with the given measures.

Answer:
Draw a horizontal line segment AB that is 6 cm long. This will be one side of the parallelogram.

At point A, use a protractor to measure and mark an angle of 50° from line AB.

From point A, draw a line segment AC that is 4 cm long, making sure it forms the 50° angle with AB.

From point B, draw a line segment BD that is 4 cm long, parallel to AC. You can use a ruler and a set square to ensure parallelism.

From point C, draw a line segment CD that is 6 cm long, parallel to AB. The points C and D should connect to complete the parallelogram.

Question 2.
The top and bottom lines in the figure are parallel. Find the unknown angle shown in the figure.

Answer:
Draw a horizontal line parallel to the other two parallel lines

From the figure, we get,
∠EAB = ∠ABD = 30°
∠DBC = ∠BCF = 60°
∴ ∠ABC = ∠ABD + ∠DBC
= 30° + 60° = 90°

Question 3.
Two vertical lines in the figure are parallel. Find the unknown angle shown in the figure.

Answer:
Consider the figure below

Given that AB and CD are parallel line
Draw a line PQ parallel to both AB and CD
∠A and ∠AQP are alternate angles
Therefore, ∠A = ∠AQP = 30°

Similarly,
∠C and ∠CQP are alternate angles
Therefore, ∠C = ∠CQP = 40°
But, ∠Q = ∠AQP + ∠CQP
= 30° + 40° = 70°

Question 4.
The figure shows a triangle drawn in a rectangle. Calculate the angles of the triangle.

Answer:
Consider the figure below,

Since, ABCD is rectangle ∠P, ∠Q, ∠R, ∠S equals to 90°
Now, it is given that one part of ∠Q is 50°, so the remaining part of∠A is 40°
Similarly, one part of ∠R is 30°, so the remaining part of ∠R is 60°
It implies that two angles of the triangle are 40° and 60o.
Therefore, the third angle of the triangle is = 180° – (40° + 60°)
= 180° — (100°) = 80°
So, the angles of triangles are 40°, 60°, 80°

Question 5.
A triangle is drawn inside a parallelogram. Calculate the angles of the triangle.

Answer:
Consider the figure below

∠D and ∠B are opposite angles of parallelogram.
Therefore, ∠D = ∠B = 105°
But, one part of ∠D is 50°, so remaining part of ∠D = 105° – 50° = 55°
Now, considering the angles ∠A and ∠D, their sum is equals to 180°
i.e., ∠A + ∠D = 180°
∠A = 180° – ∠D
= 180° – 105° = 75o
But, one part of ∠A is 20°, so remaining part of ∠A = 75° – 20o = 55°
It means, we get two angles of the triangle which are 55° and 55°
So, third angle of the triangle = 180° – (55° + 55°)
= 180° – (110°) = 70°
Hence, all three angles of the triangles are 55°, 55°, 70°.

Practice Questions

Question 1.
Draw the parallelogram below with the given measures.

Calculate the other three angles.
Answer:
The other three angles of Parallelogram 60°, 60°, 120°

Question 2.
Given that PQ and QR are parallel lines, find the angle shown in the figure?

Answer:
90°

Question 3.
In the figure, the pair of lines slanted to the left are parallel; and also the pair of lines slanted to the right.

Draw this figure:
Answer:
third angle of the bottom triangle = 90°

Question 4.
The top and bottom lines in the figure are parallel.

Calculate the third angle of the bottom triangle and all angles of the top triangle.
Answer:
all angles of the top triangle = 90°, 40o, 50°,

Question 5.
The left and right lines in the figure are parallel.

Calculate the third angle of the larger triangle and all angles of the smaller triangle.
Answer:
Third angle of the larger triangle = 80°

Question 6.
A triangle is drawn inside a parallelogram.

Calculate the angles of the triangle.
Answer:
All angles of smaller triangle = 80°, 55°, 45°

Question 7.
The figure shows a triangle drawn in a rectangle.

Calculate the angles of the triangle.
Answer:
All angles of triangle = 70°, 70°, 40°

Question 8.
Draw the triangle with the given measures.

Answer:
90°, 55°, 35° 70°.

Class 7 Maths Chapter 1 Notes Kerala Syllabus Parallel Lines

Understanding geometry begins with recognising the relationships between lines and angles. In this chapter, we delve into the fascinating world of parallel lines and the angles they create when intersected by other lines. This exploration will lay the foundation for more advanced geometric concepts and develop your ability to visualise and solve geometric problems.

Lines and Angles
We start by examining lines and the angles formed when they intersect. A special focus is given to pairs of angles that are formed on the same side of the slanting line, both above and below the parallel lines. These pairs of angles are equal in measure, helping us understand the special properties of angles created when parallel lines are intersected by a slanting line.

Matching Angles
Next, we explore the idea of matching angles. This section introduces alternate angles and co-interior angles more explicitly. Alternate angles are pairs of angles that lie on opposite sides of the transversal but inside the two lines. These angles are crucial in identifying and proving the parallel nature of lines. Co-interior angles are also revisited here, reinforcing their properties and significance.

Triangle Sum
Finally, we turn our attention to triangles, a fundamental shape in geometry. One of the most important properties of triangles is the sum of their interior angles. You will learn and prove that the total sum of all three angles in any triangle is always 180°. This section will include various exercises to solidify your understanding and application of this essential geometric principle.

By the end of this chapter, you will have a solid understanding of how parallel lines interact with transversals, the relationships between the resulting angles, and the fundamental properties of triangles. These concepts are not only crucial for your current studies but also form the bedrock of many advanced topics in geometry.

Matching Angles
Of the angles made when two parallel lines are cut by a slanting line,

• • the small angles are of the same measure.
  • the large angles are of the same measure.
  • a small angle and a large angle add up to 180°
• If the intersecting line is perpendicular to one of the parallel lines, it would be perpendicular to the other line too, and all angles would be right angles.

Interior Angles:
These are the angles on the inside of the parallel lines.
When you look at the pairs of interior angles on the same side of the slanting line, they are called co-interior angles.
The sum of each pair of co-interior angles is always 180°.

Exterior Angles:
These are the angles on the outside of the parallel lines.
Similarly, when you look at the pairs of exterior angles on the same side of the slanting line, they are called co-exterior angles.
The sum of each pair of co-exterior angles is also 180°.

Alternate Interior Angles:
Alternate interior angles are pairs of angles that lie between the two parallel lines and on opposite sides of the slanting line.
These angles are same measure

Alternate Exterior Angles:
Alternate exterior angles are pairs of angles that lie outside the two parallel lines and on opposite sides of the slanting line.
Like alternate interior angles, alternate exterior angles are also equal.

Corresponding Angles:
These angles are in the same relative position at each intersection where a slanting line crosses the parallel lines.
Each pair of corresponding angles has the same measure.

Find out the other angles in the given parallelogram?

Answer:
First, consider the given angle (55° angle). To determine other angles, observe the intersections made by the left side with the top and bottom parallel lines.
The 55° angle and the angle above it form a pair of angles that add up to 180°. Therefore, the top angle is:
180° – 55° = 125°
Next, look at the angle to the right of the marked 55° angle.
To find this angle, examine the angles formed by the left and right parallel sides with the bottom line. The 55° angle and the angle to its right also form a pair of angles that add up to 180°.
Thus, this angle is also 125°, as calculated earlier.

Triangle Sum
The sum of all angles of a triangle is 180°

• A line intersects two parallel lines at angles of the same measure
• When two parallel lines are cut by a slanting line
  • the small angles are of the same measure.
  • the large angles are of the same measure.
  • a small angle and a large angle add up to 180°.
• If the intersecting line is perpendicular to one of the parallel lines, would be perpendicular to the other line too, and all angles would be right angles.

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 4 Reciprocals Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 4 Solutions Reciprocals

Class 7 Maths Chapter 4 Reciprocals Questions and Answers Kerala State Syllabus

Reciprocals Class 7 Questions and Answers Kerala Syllabus

Page 64

Question 1.
Suma has 16 rupees with her. Safeer has 4 rupees.
(i) What part of Suma’s money does Safeer have?
(ii) How many times Safeer’s money does Suma have?
Answer:
Suma’s amount = 16
Safeer’s amount = 4
(i) Part = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{4}{16}\)
= \(\frac{1}{4}\)
So, Safeer has of Suma’s money.

(ii) Times = \(\frac{\text { large number }}{\text { small number }}\)
= \(\frac{16}{4}\)
= 4
So, Suma has 4 times that of Safeer’s money.

Question 2.
A large bag contains 9 kilograms of sugar. A small bag contains 6 kilograms.
(i) The weight of sugar in the heavier bag is how much times that in the lighter bag?
(ii) The weight of sugar in the lighter bag is what part of that in the heavier bag?
Answer:
(i) Times = \(\frac{\text { large number }}{\text { small number }}\)
= \(\frac{9}{6}\)
= \(\frac{3}{2}\)
So, the weight of sugar in the heavier bag is \(\frac{3}{2}\) times that in the lighter bag.

(ii) Part = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{6}{9}\)
= \(\frac{2}{3}\)
So, the weight of sugar in the lighter bag is \(\frac{2}{3}\) part of that in the heavier bag.

Question 3.
The weight of an iron block is 6 kilograms. The weight of another block is 26 kilograms.
(i) The weight of the lighter block is what fraction of that of the heavier block?
(ii) The weight of the heavier block is how much times that of the lighter block?
Answer:
(i) Part = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{6}{26}\)
= \(\frac{3}{13}\)
So, the weight of the lighter block is \(\frac{3}{13}\) fraction of that of the heavier block.

(ii) Times = \(\frac{\text { large number }}{\text { small number }}\)
= \(\frac{26}{6}\)
= \(\frac{13}{3}\)
So, the weight of the heavier block is \(\frac{13}{3}\) times that of the lighter block.

Question 4.
The length of a ribbon is 2 times the length of a smaller ribbon. What part of the length of the large ribbon is the length of the small ribbon?
Answer:
Times = \(\frac{\text { length of the large ribbon }}{\text { length of the smaller ribbon }}\)
2\(\frac{2}{3}=\frac{\text { length of the large ribbon }}{\text { length of the smaller ribbon }}\)
\(\frac{8}{3}=\frac{\text { length of the large ribbon }}{\text { length of the smaller ribbon }}\)

8 × length of the smaller ribbon = 3 × length of the large ribbon
\(\frac{\text { length of the smaller ribbon }}{\text { length of the large ribbon }}=\frac{3}{8}\)
Part = \(\frac{3}{8}\)
So, length of the smaller ribbon is \(\frac{3}{8}\) part of the length of the large ribbon.

Page 67

Question 1.
27 students of a class got A plus in Maths. They form of the entire class. How many students are there in the class?
Answer:
\(\frac{3}{4}\) × entire class = 27
entire class = 27 ÷ \(\frac{3}{4}\)
= 27 × \(\frac{4}{3}\)
= 9 × 4
= 36
Thus, there are 36 students in the class.

Question 2.
\(\frac{2}{3}\) of a bottle was filled with \(\frac{1}{2}\) litre of water. How many litres of water will the bottle hold?
Answer:
\(\frac{2}{3}\) x entire bottle = \(\frac{1}{2}\) litres
entire bottle = \(\frac{1}{2} \div \frac{2}{3}\)
= \(\frac{1}{2} \times \frac{3}{2}\)
= \(\frac{3}{4}\)
Thus, the bottle will hold – litres of water.

Question 3.
\(\frac{3}{4}\) of a vessel holds 1\(\frac{1}{2}\) litres of water. What is the capacity of the vessel in litres if it is completely filled with water?
Answer:
\(\frac{3}{4}\) × entire vessel = 1\(\frac{1}{2}\) = \(\frac{3}{2}\) litres
entire vessel = \(\frac{3}{2} \div \frac{3}{4}\)
= \(\frac{3}{2} \times \frac{4}{3}\)
= \(\frac{4}{2}\)
= 2
Thus, the capacity of the vessel if it is completely filled with water is 2 litres.

Question 4.
Two of the three ribbons of the same length and half the third ribbon were placed end to end. It came to 1 metre. What is the length of a ribbon in centimetres?
Answer:
1 metre = 100 cm
2.5 ribbons = 100 cm
\(\frac{5}{2}\) × length of a ribbon = 100 cm
length of a ribbon = 100 ÷ \(\frac{5}{2}\)
= 100 × \(\frac{2}{5}\)
= 20 × 2
= 40 cm

Page 68

Question 1.
A 16 metres long rod is cut into pieces of length \(\frac{2}{3}\) metre. How many such pieces will be there?
Answer:
Total length of the rod = 16 m
Length of a piece = \(\frac{2}{3}\) m
Number of pieces = Total length of the rod ÷ Length of a piece
= 16 ÷ \(\frac{2}{3}\)
= 16 × \(\frac{3}{2}\)
= 8 × 3
= 24.

Question 2.
How many \(\frac{3}{4}\) litre bottles are needed to fill 5\(\frac{1}{4}\) litres of water?
Answer:
Total water = 5\(\frac{1}{4}=\frac{21}{4}\) litres
Amount of water in a bottle = \(\) litres
Number of bottles = Total water ÷ Amount of water in a bottle
= \(\frac{21}{4} \times \frac{4}{3}\)
= \(\frac{21}{4} \times \frac{4}{3}\)
= \(\frac{21}{3}\)
= 7

Question 3.
13\(\frac{1}{2}\) kilograms of sugar is to be packed into bags with 2\(\frac{1}{4}\) kilograms sugar each. How many bags are needed?
Answer:
Total sugar = 13\(\frac{1}{2}\) = \(\frac{27}{2}\) kg
Amount of sugar in one bag = 2\(\frac{1}{4}=\frac{9}{4}\)kg
Number of bags = Total sugar Amount of sugar in one bag
= \(\frac{27}{2} \div \frac{9}{4}\)
= \(\frac{27}{2} \times \frac{4}{9}\)
= 3 × 2
= 6.

Question 4.
The area of a rectangle is 22\(\frac{1}{2}\) square centimetres, and one side is 3\(\frac{3}{4}\) centimetres long. What is the length of the other side?
Answer:
Area of a rectangle = 22\(\frac{1}{2}\) = \(\frac{45}{2}\)sq.cm
length of one side = 3\(\frac{3}{4}=\frac{15}{4}\) cm
length of the other side = area of a rectangle ÷ length of one side
= \(\frac{45}{2} \div \frac{15}{4}\)
= \(\frac{45}{2} \times \frac{4}{15}\)
= 3 × 2
= 6 cm

Question 5.
How many pieces, each of length 2\(\frac{1}{2}\) metres, can be cut off from a rope of length 11\(\frac{1}{2}\) metres? How many metres of rope will be left?
Answer:
Total length of the rope = 11\(\frac{1}{2}\) = \(\frac{23}{2}\) m
Length of a piece = 2\(\frac{1}{2}=\frac{5}{2}\)m
Number of pieces = Total length of the rope ÷ Length of a piece
= \(\frac{23}{2} \div \frac{5}{2}\)
= \(\frac{23}{2} \times \frac{2}{5}\)
= \(\frac{23}{5}\)
= 4\(\frac{3}{5}\)

So, we can cut 4 pieces of length 2\(\frac{1}{2}\) m.
4 × 2\(\frac{1}{2}\)
= 4 × \(\frac{5}{2}\)
= 2 × 5
= 10 m
Length of the remaining rope =11\(\frac{1}{2}\) – 10 = \(\frac{23}{2}\) – 10 = \(\frac{3}{2}\)m

Page 68

Question 1.
A rod is 36 metres long. How many pieces each of length 2\(\frac{1}{2}\) metres can be cut off from it? What is the length of the rod left?
Appu did the problem this way.
36 ÷ 2\(\frac{1}{2}\) = 36 ÷ \(\frac{5}{2}\)
= 36 ÷ \(\frac{2}{5}\)
= \(\frac{72}{5}\)
When we divide 72 by 5, the quotient is 14 and remainder is 2. So we get 14 pieces. The remaining rod is of length 2 metres.
Ammu used another idea.
2 pieces, each of length 2\(\frac{1}{2}\) metres makes 5 metres.
7 × 5 = 35
So 7 × 2 = 14 pieces can be cut off. The remainder is 36 – 35 = 1 metre. Whose answer is right?
Solution:
In Appu’s case, he find that 14 pieces of length 2\(\frac{1}{2}\) metres can be cut off from it.
This 14 pieces together forms;
14 × 2\(\frac{1}{2}\) = 14 × \(\frac{5}{2}\)
= 7 × 5
= 35 m
So, the remaining length = 36 – 35 = 1 m.
Thus, Ammu’s answer is the right one.

Class 7 Maths Chapter 4 Kerala Syllabus Reciprocals Questions and Answers

Question 1.
Find the reciprocal of \(\frac{3}{7}\)
Answer:
reciprocal of \(\frac{3}{7}=\frac{7}{3}\)

Question 2.
Simplify \(\frac{3}{4} \div \frac{5}{6}\)
Answer:
\(\frac{3}{4} \div \frac{5}{6}=\frac{3}{4} \times \frac{6}{5}=\frac{18}{20}=\frac{9}{10}\)

Question 3.
A recipe requires cup of sugar to make 12 cookies. How much sugar is needed to make 36 cookies?
12 കുക്കീസ് ഉണ്ടാക്കുവാനായി \(\frac{3}{4}\) കപ്പ് പഞ്ചസാര വേണം. അങ്ങനെയങ്കിൽ 36 കുക്കീസ് ഉണ്ടാക്കുവാൻ എത്ര കപ്പ് പഞ്ചസാര വേണം?
Answer:
\(\frac{3}{4}\) cup of sugar gives 12 cookies.
12 × 3 gives 36.
So, for 36 cookies we need \(\frac{3}{4}\) × 3 = \(\frac{9}{4}\) cups of sugar.

Question 4.
If of the cake is eaten, what fraction of the whole cake remains?
ഒരു കേക്കിന്റെ \(\frac{2}{5}\) ഭാഗം കഴിച്ചെങ്കിൽ ബാക്കി എത്ര ഭാഗം ഉണ്ട്?
Answer:
The whole cake is \(\frac{5}{5}\) = 1
Remaining cake = \(\frac{5}{5}-\frac{2}{5}=\frac{3}{5}\)
Thus, \(\frac{3}{5}\) part of the whole cake remains.

Question 5.
A ribbon is cut into 12 equal parts. What fraction of the ribbon is 3 parts?
ഒരു റിബൺ 12 തുല്യ ഭാഗങ്ങളായി മുറിച്ചു. അതിലെ 3 ഭാഗം ആകെയുള്ള റിബണിന്റെ എത്ര ഭാഗമാണ്?
Answer:
Part (or fraction) = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{3}{12}\)
= \(\frac{1}{4}\)

Practice Questions

Question 1.
Solve \(\frac{4}{5} \div \frac{5}{4}\)
Answer:
\(\frac{16}{25}\)

Question 2.
Calculate \(\frac{3}{7} \div \frac{7}{3}\)
Answer:
1

Question 3.
A company produces \(\frac{2}{5}\) of its daily output in the morning and \(\frac{3}{8}\) in the afternoon. What fraction of the total daily output is produced in the afternoon?
Answer:
\(\frac{15}{31}\)

Question 4.
Calculate the reciprocal of \(\frac{2}{5}\). Which is larger?
Answer:
\(\frac{5}{2}\), reciprocal is the largest

Question 5.
When a tank is \(\frac{1}{4}\) full, it contains 80 litres of water. What will it contain when it is \(\frac{3}{8}\) full?
Answer:
120 litres

Class 7 Maths Chapter 4 Notes Kerala Syllabus Reciprocals

We are already familiar with a number of mathematical operations, such as addition, subtraction, multiplication and division. This chapter introduces a new type of mathematical operation named “reciprocal”. Following are the topics discussed in this chapter.

Times and parts
If we are given a large number and a small number, we can say the large number is how many times the small number. Always,
Times = \(\frac{\text { large number }}{\text { small number }}\)

If we are given a large number and a small number, we can say the small number is what part of the large number. Always,
Part = \(\frac{\text { small number }}{\text { large number }}\)

Topsy – Turvy
A fraction obtained by interchanging the numerator and denominator of the given fraction is called the reciprocal of the given fraction.
If we multiply a natural number by its reciprocal, we will get one as the result.
Eg:
The reciprocal of \(\frac{2}{3}\) is \(\frac{3}{2}\)

If we multiply a fraction by its reciprocal, we will get one as the result.
Eg:
Consider the fraction \(\frac{3}{4}\). Its reciprocal is \(\frac{4}{3}\). Now, \(\frac{3}{4} \times \frac{4}{3}\) = 1

Zero does not have a reciprocal because any number multiplied by zero is zero, not 1.

Fraction division
Division of natural numbers is the same as multiplication by the reciprocal.
Dividing a fraction (or a number) by another fraction is exactly same as multiplying the first fraction (or number) by the reciprocal of the second fraction.
Eg:
\(\frac{1}{2} \div \frac{3}{4}=\frac{1}{2} \times \frac{4}{3}=\frac{4}{6}=\frac{2}{3}\)
8 ÷ \(\frac{2}{5}\) = 8 × \(\frac{5}{2}\) = 4 × 5 = 20

If we are given a large number and a small number, we can say;
the large number is how many times the small number
the small number is what part of the large number large number
Times = \(\frac{\text { large number }}{\text { small number }}\)
Part = \(\frac{\text { small number }}{\text { large number }}\)

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 2 Fractions Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 2 Solutions Fractions

Class 7 Maths Chapter 2 Fractions Questions and Answers Kerala State Syllabus

Fractions Class 7 Questions and Answers Kerala Syllabus

Page 24

Do the following problems mentally. Write each as how many times and also as a product.

Question 1.
Each piece of a pumpkin weighs a quarter kilogram. What is the weight of two pieces together? What is the weight of four such pieces? Six pieces?
Answer:
Two times quarter kilogram is half kilogram.
Four times quarter kilogram is one kilogram.
Six times quarter kilogram is one and a half kilogram.
Mathematically,
Each piece of a pumpkin weighs \(\frac{1}{4}\) kilogram.
The weight of two pieces is two times \(\frac{1}{4}\), which is \(\frac{1}{2}\) kilogram.
The weight of four pieces is 4 times \(\frac{1}{4}\) which is 1 kilogram.
The weight of six pieces is 6 times \(\frac{1}{4}\) which is 1\(\frac{1}{2}\) kilogram.

Using products,
The weight of two pieces = 2 × \(\frac{1}{4}=\frac{2}{4}\) = \(\frac{1}{2}\) kilogram
The weight of four pieces= 4 × \(\frac{1}{4}=\frac{4}{4}\) = 1 kilogram
The weight of six pieces = 6 × \(\frac{1}{4}\) = 1\(\frac{1}{2}\) kilogram.

Question 2.
We can fill a cup with one-third of a litre of milk. How much milk is needed to fill two cups? Four cups?
Answer:
Two times one-third of a litre is two-third of a litre.
Four times one-third of a litre is four-third of a litre. Mathematically,
Each cup contains one-third of a litre of milk.
So, milk needed to fill two cups is 2 times \(\frac{1}{3}\) which is \(\frac{2}{3}\) litre.
Milk needed to fill four cups is 4 times \(\frac{1}{3}\), which is 1\(\frac{1}{3}\) litre.
Using products,
Milk needed to fill two cups = 2 × \(\frac{1}{3}=\frac{2}{3}\) litre
Milk needed to fill four cups = 4 × \(\frac{1}{3}\) = 1\(\frac{1}{3}\) litre.

Question 3.
What is the total length of four pieces of ribbons, each of length three fourths of a metre? What about five pieces?
Answer:
Four times three fourth of a metre is three metre.
Five times three fourth of a metre is three and three-fourth of a metre.
Mathematically,
Length of each piece is three-fourth of a metre.
So, total length of four pieces of ribbon is 4 times \(\frac{3}{4}\), which is 3 metre.
Total length of five pieces of ribbon is 5 times \(\frac{3}{4}\), which is 3\(\frac{3}{4}\) metre.
Using products,
Total length of four pieces of ribbon = 4 × \(\frac{3}{4}\) = 3 metre
Total length of five pieces of ribbon = 5 × \(\frac{1}{2}\) = 3\(\frac{3}{4}\) metre

Question 4.
It takes hour to walk around a play ground once.
(i) How much time does it take to walk 4 times around at this speed?
(ii) What about 7 times?
Answer:
(i) 4 times \(\frac{1}{4}\) is 1 hour.
Using products,
Time taken to walk 4 time around = 4 × \(\frac{1}{4}\) = 1 hour

(ii) 7 times \(\frac{1}{4}\) is 1\(\frac{3}{4}\) hour.
Using products,
Time taken to walk 7 times around = 7 × \(\frac{1}{4}\) = 1\(\frac{3}{4}\)hour.

Page – 25,26

Question 1.
The weight of an iron block is kilogram.
(i) What is the total weight of such 15 blocks?
(ii) 16 blocks?
Answer:
(i) Weight of one block = \(\frac{1}{4}\) kilogram.
Weight of 15 blocks = 15 × \(\frac{1}{4}=\frac{15 \times 1}{4}\)
= \(\frac{15}{4}=\frac{12+3}{4}\)
= \(\frac{12}{4}+\frac{3}{4}\)
= 3\(\frac{3}{4}\)

(ii) Weight of 16 blocks = 16 × \(\frac{1}{4}=\frac{16}{4}\) = 4 kilogram.

Question 2.
Some 2 metre long rods are cut into 5 pieces of equal length.
(i) What is the length of each piece?
(ii) What is the total length of 4 pieces?
(iii) of 10 pieces?
Answer:
(i) Total length of the rod = 2 metre.
Length of each piece = \(\frac{2}{5}\) metre

(ii) Total length of 4 pieces = 4 × \(\frac{2}{5}=\frac{4 \times 2}{5}=\frac{8}{5}\)
= \(\frac{5+3}{5}=\frac{5}{5}+\frac{3}{5}\)
= 1\(\frac{1}{5}\) metre

Question 3.
5 litres of milk is filled in 6 bottles of the same size.
(i) How many litres of milk does each bottle hold?
(ii) How many litres in 3 bottles together?
(iii) In 4 bottles?
Answer:
Total milk = 5 litre
Number of bottles = 6
(i) Milk in each bottle = \(\frac{5}{6}\) litre

(ii) Milk in 3 bottles together = 3 × \(\frac{5}{6}=\frac{3 \times 5}{6}=\frac{15}{6}\)
= \(\frac{12+3}{6}=\frac{12}{6}+\frac{3}{6}\)
= 2\(\frac{1}{2}\) litre.

(iii) Milk in 4 bottles = 4 × \(\frac{5}{6}=\frac{4 \times 5}{6}=\frac{20}{6}\)
= \(\frac{18+2}{6}=\frac{18}{6}+\frac{2}{6}\)
= 3\(\frac{1}{3}\) litre

Page – 28

Do these problems in head. Then write each as a part and also as a product of numbers.

Question 1.
Nine litres of milk is divided equally among three children. How many litres will each get? What if there are four children?
Answer:
Each will get a third of nine litres, that is three litres.
As a part,
Each will get \(\frac{1}{3}\) of 9, which is 3 litres.

As a product,
Each will get \(\frac{1}{3}\) × 9 = \(\frac{9}{3}\) = 3 litres.
If there are four children,
Each will get a fourth of nine litres. A fourth of 8 litre is 2 litre and then a fourth of the remaining one litre. So, two and one-fourth of a litre.

As a part,
Each will get \(\frac{1}{4}\) of 9, which is 2\(\frac{1}{4}\) litres.

As a product,
Each will get \(\frac{1}{4}\) × 9 = \(\frac{9}{4}\) = 2\(\frac{1}{4}\) litres.

Question 2.
Six kilograms of rice was packed in five bags of the same size. How many kilograms of rice in each bag? What if it is packed in four bags?
Answer:
Each bag has one fifth of six kilograms. One fifth of 5 kilogram is 1 kilogram and then one fifth of remaining one kilogram. So, one and one-fifth of a kilogram.

As a part,
Each bag has \(\frac{1}{5}\) of 6, which is 1\(\frac{1}{5}\) kilograms.

As a product,
Each bag has \(\frac{1}{4}\) × 6 = \(\frac{6}{5}\) = 1\(\frac{1}{5}\) kilograms.
If it is packed in four bags,
Each bag has one-fourth of six kilograms. One-fourth of four kilograms is 1 kilogram and then one- fourth of remaining two kilograms. So, one and a half kilograms.

As a part,
Each bag has \(\frac{1}{4}\) of 6, which is 1\(\frac{1}{2}\) kilograms.
As a product,
Each bag has \(\frac{1}{4}\) × 6 = \(\frac{6}{4}\) = 1\(\frac{1}{2}\) kilograms.

Question 3.
A seven metre long string is divided into six equal pieces. What is the length of each piece? What if it is divided into three equal pieces?
Answer:
Each piece is one-sixth of seven metres. One-sixth of six metres is 1 metre and then one-sixth of the remaining one metre. So, one and one-sixth of a metre.

As a part,
Length of each piece is \(\frac{1}{6}\) of 7, which is 1\(\frac{1}{6}\) metres.

As a product,
Length of each piece = \(\frac{1}{6}\) × 7 = \(\frac{7}{6}\) = 1 \(\frac{1}{6}\) metres.
If it is divided into three equal pieces,
Each piece is one-third of seven metres. One-third of six metres is 2 metres and one-third of remaining one metre. So, two and one-third of a metre.

As a part,
Length of each piece is \(\frac{1}{3}\) of 7, which is 2\(\frac{1}{3}\) metres.
As a product,
Length of each piece \(\frac{1}{3}\) × 7 = \(\frac{7}{3}\) = 2\(\frac{1}{3}\) metres.
The calculations of the types of problems above can be done as follows:

Question 4.
We have to cut off \(\frac{3}{5}\) of a 7 metre long string. How long is this piece?
Answer:
Here we have to calculate \(\frac{3}{5}\) of 7.
\(\frac{3}{5}\) x 7 = \(\frac{3 \times 7}{5}=\frac{21}{5}=\frac{20+1}{5}\)
\(\frac{20}{5}+\frac{1}{5}\) = 4 + \(\frac{1}{5}\)
= 4\(\frac{1}{5}\)
So, we need to cut off 4\(\frac{1}{5}\) metres.

Page – 29

Question 1.
There are 35 children in a class. of them are girls. How many girls are there in the class?
Answer:
Number of girls = \(\frac{3}{5}\) × 35
= 3 × \(\frac{35}{5}\)
= 3 × 7
= 21
Or,
Number of girls = \(\frac{3}{5}\) × 35
= \(\frac{3 \times 35}{5}\)
= \(\frac{105}{5}\)
= 21

Question 2.
10 kilograms of rice is filled equally in 8 bags. If the rice in 3 such bags are taken together, how many kilograms would that be?
Answer:
Rice in 3 bags are taken together.
That is, we have to find- of 10 kilograms.
So, amount of rice = \(\frac{3}{8}\) × 10
= \(\frac{30}{8}=\frac{24+6}{8}\)
= \(\frac{24}{8}+\frac{6}{8}\)
= 3\(\frac{3}{4}\) kilograms.

Question 3.
The area of the rectangle in the figure is 27 square centimetres. It is divided into 9 equal parts.

What is the area of the darker part in square centimetres?
Answer:
5 of the 9 equal parts are darker.
So, we have to find \(\frac{5}{9}\) of 27 square centimetres.

Area of darker part = \(\frac{5}{9}\) × 27 = 5 × \(\frac{27}{9}\)
= 5 × 3 = 15 square centimeres.

Area of darker part = \(\frac{5}{9}\) × 27 = \(\frac{5 \times 27}{9}=\frac{135}{9}\)
= 15 square centimetres.

Page – 32

Question 1.
Draw rectangles and find these products.
(i) \(\frac{1}{2} \times \frac{1}{4}\)
(ii) \(\frac{1}{3} \times \frac{1}{6}\)
(iii) \(\frac{1}{5} \times \frac{1}{8}\)
Answer:
(i) \(\frac{1}{2} \times \frac{1}{4}\)
Draw a rectangle a divide it into 4 equal parts. Then divide one part into half.

Now extend the horizontal line.

So, \(\frac{1}{2} \times \frac{1}{4}\) = \(\frac{1}{8}\)

(ii) \(\frac{1}{3} \times \frac{1}{6}\)
Draw a rectangle and divide it into 6 equal parts. Then divide each part into 3 equal parts.

Now extend the horizontal line.

So, \(\frac{1}{3} \times \frac{1}{6}=\frac{1}{18}\)

(iii) \(\frac{1}{5} \times \frac{1}{8}\)
Draw a rectangle a divide it into 8 equal parts. Then divide one part into 5 equal parts.

Now extend the horizontal line.

So, \(\frac{1}{5} \times \frac{1}{8}=\frac{1}{40}\)

Question 2.
A one metre long string is divided into five equal parts. How long is half of each part in metres? In centimetres?
Answer:
When one metre long string divided into five equal parts each part is one- fifth of one metre.
Now half of each part is \(\frac{1}{2}\) of \(\frac{1}{5}\)of one metre.
∴ Length of each part = \(\frac{1}{2} \times \frac{1}{5}\) = \(\frac{1}{10}\) metre
= \(\frac{1}{10}\) × 100
= 10 centimetre.

Question 3.
One litre of milk is filled in two bottles of equal size. A quarter of the milk in one bottle was used to make tea. How many litres of milk were used for tea? In millilitres?
Answer:
When one litre of milk is filled in two bottles of equal size, each bottle contain half of a litre.
A quarter of the milk in one bottle was used to make tea, which is of of one litre.
Milk used for tea \(\frac{1}{4}\) of \(\frac{1}{2}\) one litre
= \(\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}\) litre
= \(\frac{1}{8}\) × 1000 = 125 millilitre.

Consider another type of problem:
Find of \(\frac{4}{5}\) of \(\frac{2}{3}\).
Answer:

Page – 34

Question 1.
A rope 2 metres long is cut into 5 equal pieces. What is the length of three quarters of one of the pieces in metres? In centimetres?
Answer:
When 2 metre long rope cut into 5 equal pieces, length of each piece is \(\frac{1}{5}\) of 2 metre, which is \(\frac{2}{5}\) metre.
Length of three quarters of one of the pieces is \(\frac{3}{4}\) of \(\frac{2}{5}\)
Required length of the piece = \(\frac{3}{4} \times \frac{2}{5}\)
= 3 × \(\frac{1}{4} \times \frac{1}{5}\) × 2
= 3 × \(\frac{1}{20}\) × 2
= \(\frac{6}{20}=\frac{3}{10}\)

Question 2.
4 bottles of the same size were filled with 3 litres of water. One of these was used to fill 5 cups of the same size. How much water is there in one such cup, in litres? And in millilitres?
Answer:
When 4 bottles of the same size were filled with 3 litres of water, each bottle has of 3 litre, which is \(\frac{3}{4}\) litre.
One of these was used to fill 5 cups of the same size. Then, amount of water in one cup is \(\frac{1}{5}\) of \(\frac{3}{4}\).
Amount of water in one cup = \(\frac{1}{5} \times \frac{3}{4}\)
= \(\frac{1}{5} \times \frac{1}{4}\) × 3
= \(\frac{1}{20}\) × 3
= \(\frac{3}{20}\)

Question 3.
A watermelon weighing four kilograms was cut into five equal pieces. One piece was again halved. What is the weight of each of these two pieces in kilograms? And in grams?
Answer:
When watermelon weighing 4 kg cut into five equal pieces, weight of each piece is \(\frac{1}{5}\) of 4 kg, which is \(\frac{4}{5}\)kg.
Each piece is again halved.
Then weight of each of these two pieces is \(\frac{1}{2}\) of \(\frac{4}{5}\)
Required weight = \(=\frac{1}{2} \times \frac{4}{5}\)
= \(=\frac{1}{2} \times \frac{1}{5}\) × 4
= \(\frac{1}{10}\) × 4
= \(\frac{4}{10}=\frac{2}{5}\) kilograms

Question 4.
A vessel full of milk is used to fill three bottles of the same size. Then the milk in each bottle was used to fill four cups of the same size. What fraction of the milk in the first vessel does each cup contain?
Answer:
When a vessel full of milk used to fill three bottles of same size, each bottle has one-third of milk.
When milk in each bottle was used to fill four cups of same size, each cup has \(\frac{1}{4}\) of \(\frac{1}{3}\) of the milk.
∴ fraction of milk in each cup = \(\frac{1}{4} \times \frac{1}{3}=\frac{1}{12}\)

Question 5.
Draw a line AB of length 12 centimetres. Mark AC as \(\frac{2}{3}\) of AB. Mark AD as \(\frac{1}{4}\) of AC. What part of AB is AD?
Answer:

AD = \(\frac{1}{4}\) of AC
= \(\frac{1}{4}\) of \(\frac{2}{3}\) of AB
= \(\frac{1}{4} \times \frac{2}{3}\) of AB
= \(\frac{1}{4} \times \frac{1}{3}\) × 2 of AB
= \(\frac{1}{12}\) × 2 of AB
= \(\frac{2}{12}=\frac{1}{6}\) of AB

Question 6.
Calculate the following using multiplication:
(i) \(\frac{3}{7}\) of \(\frac{2}{5}\)
Answer:
\(\frac{3}{7}\) of \(\frac{2}{5}\) = \(\frac{3}{7} \times \frac{2}{5}\)
= 3 × \(\frac{1}{7} \times \frac{1}{5}\) × 2
= 3 × \(\frac{1}{35}\) × 2
= \(\frac{6}{35}\)

(ii) \(\frac{2}{3}\) of \(\frac{3}{4}\)
Answer:
\(\frac{2}{3}\) of \(\frac{3}{4}\) = \(\frac{2}{3} \times \frac{3}{4}\)
= 2 × \(\frac{1}{3} \times \frac{1}{4}\) × 3
= 2 × \(\frac{1}{12}\) × 3
= 6 × \(\frac{1}{12}=\frac{6}{12}=\frac{1}{2}\)

(iii) \(\frac{3}{5}\) of \(\frac{2}{7}\)
Answer:
\(\frac{3}{5}\) of \(\frac{2}{7}\) = \(\frac{3}{5} \times \frac{2}{7}\)
= 3 × \(\frac{1}{5} \times \frac{1}{7}\) × 2
= 3 × \(\frac{1}{35}\) × 2
= \(\frac{6}{35}\)

(iv) \(\frac{5}{6}\) of \(\frac{3}{10}\)
Answer:
\(\frac{5}{6}\) of \(\frac{3}{10}\) = \(\frac{5}{6} \times \frac{3}{10}\)
= 5 × \(\frac{1}{6} \times \frac{1}{10}\) × 3
= 5 × \(\frac{1}{60}\) × 3
= 15 × \(\frac{1}{60}=\frac{15}{60}=\frac{1}{4}\)

Page – 37

Question 1.
One and a half metres of cloth is needed for a shirt. How much cloth is required for five such shirts?
Answer:
Cloth needed for a shirt = 1 metres
Cloth needed for 5 shirts = 5 × 1\(\frac{1}{2}\)
= 5 × (1 + \(\frac{1}{2}\))
= (5 × 1) + (5 × \(\frac{1}{2}\))
= 5 + 2\(\frac{1}{2}\)
= 7\(\frac{1}{2}\) metres.
Or,
5 × 1\(\frac{1}{2}\) = 5 × \(\frac{3}{2}\)
= \(\frac{15}{2}\) = 7\(\frac{1}{2}\)

Question 2.
The price of one kilogram of okra is thirty rupees. What is the price of two and a half kilograms?
Answer:
Price of one kilogram of okra = 30 rupees
Price of two and a half kilograms of okra = 30 × 2\(\frac{1}{2}\)
= 30 × (2 + \(\frac{1}{2}\))
= (30 × 2) + (30 × \(\frac{1}{2}\))
= 60 + 15
= 75 rupees
Or,
30 × 2\(\frac{1}{2}\) = 30 × \(\frac{5}{2}\)
= \(\frac{30}{2}\) × 5
= 15 × 575 rupees

Question 3.
A person walks two and a half kilometres in an hour. At the same speed, how far will he walk in one and a half hours?
Answer:
Distance he walks in one hour = 2 \(\frac{1}{2}\) km
Distance he walks in one and a half hours = 1 \(\frac{1}{2}\) x 2\(\frac{1}{2}\)
= \(\frac{3}{2} \times \frac{5}{2}\)
= \(\frac{3 \times 5}{2 \times 2}=\frac{15}{4}\)
= 3 \(\frac{3}{4}\) km

Question 4.
Roni has 36 stamps with her. Sahira says she has 2 times this. How many stamps does Sahira have?
Answer:
Number of stamps with Roni = 36
Number of stamps with Sahira = 2\(\frac{1}{2}\) × 36
= (2 + \(\frac{1}{2}\)) × 36
= (2 × 36) +(\(\frac{1}{2}\) × 36)
= 72 + 18
= 90
Or
2\(\frac{1}{2}\) × 36 = \(\frac{5}{2}\) × 36
= 5 × \(\frac{36}{2}\)
= 5 × 18
= 90

Question 5.
Joji works 4\(\frac{1}{2}\) hours each day. How many hours does he work in 6 days?
Answer:
Number of hours Joji works each day = 4 hours
Number of hours Joji work in 6 days = 6 × 4\(\frac{1}{2}\)
= 6 × (4 + \(\frac{1}{2}\))
= (6 × 4) + (6 × \(\frac{1}{2}\))
= 24 + 3
= 27 hours

Question 6.
Calculate the following:
(i) 4 times 5\(\frac{1}{3}\)
Answer:
4 times 5\(\frac{1}{3}\) = 4 × 5\(\frac{1}{3}\)
= 4 × (5 + \(\frac{1}{3}\))
= (4 × 5) + (4 × \(\frac{1}{3}\))
= 20 + 1\(\frac{1}{3}\)
= 21 \(\frac{1}{3}\)

(ii) 4\(\frac{1}{3}\) times 5
Answer:
4\(\frac{1}{3}\) times 5 = 4\(\frac{1}{3}\) × 5
= (4 + \(\frac{1}{3}\)) × 5
= (4 × 5) + (\(\frac{1}{3}\) × 5)
= 20 + 1\(\frac{2}{3}\)
= 21\(\frac{2}{3}\)

(iii) 1\(\frac{1}{2}\) times \(\frac{2}{3}\)
Answer:
1\(\frac{1}{2}\) times \(\frac{2}{3}\) = 1\(\frac{1}{2}\) × \(\frac{2}{3}\)
= \(\frac{3}{2} \times \frac{2}{3}\)
= \(\frac{3 \times 2}{2 \times 3}\)
= 1

(iv) \(\frac{2}{5}\) times 2\(\frac{1}{2}\)
Answer:
\(\frac{2}{5}\) times 2\(\frac{1}{2}\) = \(\frac{2}{5}\) × 2\(\frac{1}{2}\)
= \(\frac{3}{2} \times \frac{2}{3}\)
= \(\frac{3 \times 2}{2 \times 3}\)
= 1

(v) 2\(\frac{1}{2}\) times 5\(\frac{1}{2}\)
Answer:
2\(\frac{1}{2}\) times 5\(\frac{1}{2}\) = 2\(\frac{1}{2}\) × 5\(\frac{1}{2}\)
= \(\frac{5}{2} \times \frac{11}{2}\)
= \(\frac{5 \times 11}{2 \times 2}\)
= \(\frac{55}{4}\)
= 13 \(\frac{3}{4}\)

Page – 42

Question 1.
The length and breadth of some rectangles are given below. Find the area of each:
(i) 3\(\frac{1}{4}\) centimetres, 4\(\frac{1}{2}\) centimetres
Answer:
3\(\frac{1}{4}\) centimetres, 4\(\frac{1}{2}\) centimetres
Area of the rectangle = 3\(\frac{1}{4}\) × 4\(\frac{1}{2}\)
= \(\frac{13}{4} \times \frac{9}{2}\)
= \(\frac{13 \times 9}{4 \times 2}\)
= \(\frac{117}{8}\)
= 14\(\frac{5}{8}\) square centimetres

(ii) 5\(\frac{1}{3}\) metres, 6\(\frac{3}{4}\) metres
Answer:
5\(\frac{1}{3}\) metres, 6\(\frac{3}{4}\) metres
Area of the rectangle = 5\(\frac{1}{3}\) × 6\(\frac{3}{4}\)
= \(\frac{16}{3} \times \frac{27}{4}\)
= \(\frac{16}{4} \times \frac{27}{3}\)
= 4 × 3 = 12

Question 2.
What is the area of a square of side 1 metres?
Answer:
Area of the square = 1\(\frac{1}{2}\) × 1\(\frac{1}{2}\)
= \(\frac{3}{2} \times \frac{3}{2}\)
= \(\frac{3 \times 3}{2 \times 2}\)
= \(\frac{9}{4}\)
= 2\(\frac{1}{4}\) square metre

Question 3.
The perimeter of a square is 14 metres. What is its area?
Answer:
Perimeter of the square = 14 metres
4 × side = 14
∴ Side = \(\frac{14}{4}\)= 3\(\frac{1}{2}\) metre
Area = 3\(\frac{1}{2}\) × 3\(\frac{1}{2}\)
= \(\frac{7}{2} \times \frac{7}{2}\)
= \(\frac{7 \times 7}{2 \times 2}\)
= \(\frac{49}{4}\)
= 12\(\frac{1}{4}\) square metres

Intext Questions and Answers

Question 1.
A bottle holds a quarter litre of water. How much water is needed to fill three such bottles?
Answer:
Each bottle holds a quarter litre of water. So, water needed to fill three such bottles is three times a quarter litre, which is three-quarter litres.
This can be calculated as,
3 times \(\frac{1}{4}\) is \(\frac{3}{4}\) is \(\frac{3}{4}\)
As a product,
3 × \(\frac{1}{4}=\frac{3}{4}\)

Question 2.
The calculations in the type of problems above can be done easily as follows:
\(\frac{3}{4}\) litres of milk in a bottle; how many litres in 7 such bottles?
Answer:
Amount of milk in one bottle = \(\frac{3}{4}\) litres
Amount of milk in 7 such bottles = 7 times \(\frac{3}{4}\)
7 × \(\frac{3}{4}=\frac{7 \times 3}{4}=\frac{21}{4}\)

Split 21 as a multiple of 4.
\(\frac{21}{4}=\frac{20+1}{4}=\frac{20}{4}+\frac{1}{4}\)
= 5 + \(\frac{1}{4}\)
= 5\(\frac{1}{4}\)

Question 3.
A five metre long string is cut into three equal pieces. What is the length of each piece?
Answer:
3 metres cut into three equal pieces, each piece is 1 metre. The remaining 2 metre cut into three equal pieces, each piece must be two-third of a metre. So, the length of each piece is 1 metres.
In terms of number alone,
\(\frac{1}{3}\) of 5 is 1\(\frac{2}{3}\)
Writing it as a product,
\(\frac{1}{3}\) × 5 = \(\frac{5}{3}\) = 1\(\frac{2}{3}\)

Question 4.
Draw rectangle and find \(\frac{1}{3} \times \frac{1}{2}\)
Answer:
First, draw a rectangle and divide it into two equal part. Then each part is \(\frac{1}{2}\)

Then divide one part into three equal parts. Here, each part is \(\frac{1}{3} \times \frac{1}{2}\)

Now extend the horizontal line.

The rectangle is divided into 6 equal parts and so each part is
Hence, \(\frac{1}{3} \times \frac{1}{2}\) = \(\frac{1}{6}\)

Question 5.
Find 3 × 2\(\frac{1}{4}\)
Answer:
3 × 2\(\frac{1}{4}\) = 3 × (2 + \(\frac{1}{4}\))
= (3 × 2) + (3 × \(\frac{1}{4}\))
= 6 + \(\frac{3}{4}\)
= 6\(\frac{3}{4}\)

Another way,
3 × 2\(\frac{1}{4}\) = 3 × \(\frac{9}{4}\)
= \(\frac{27}{4}\)
= 6\(\frac{3}{4}\)
Consider another problem.

Question 6.
Find 3\(\frac{1}{2}\) × 2\(\frac{1}{4}\)
Answer:
3\(\frac{1}{2}\) × 2\(\frac{1}{4}\)
= \(\frac{7}{2} \times \frac{9}{4}\)
= \(\frac{7 \times 9}{2 \times 4}\)
= \(\frac{63}{8}=\frac{56+7}{8}=\frac{56}{8}+\frac{7}{8}\)
= 7\(\frac{7}{8}\)

Question 7.
Find the area of the rectangle having length 5\(\frac{1}{2}\) cm and breadth 3\(\frac{1}{3}\) cm.
Answer:

Now, divide the length 5\(\frac{1}{2}\) cm into 11 equal parts of length \(\frac{1}{2}\) cm.
Divide the breadth 3\(\frac{1}{3}\) cm into 10 equal parts of length \(\frac{1}{3}\) cm.

So, 11 × 10 = 110 rectangles in all, each of area \(\frac{1}{6}\) square centimetres.
∴ Area of rectangle = 110 × \(\frac{1}{6}\) = 18\(\frac{1}{3}\)
Now, 5\(\frac{1}{2}\) × 3\(\frac{1}{3}\)
= \(\frac{11}{2} \times \frac{10}{3}\)
= 11 × \(\frac{1}{2}\) × 10 × \(\frac{1}{3}\)
= 110 × \(\frac{1}{6}\)
= 18\(\frac{1}{3}\)
So, even if the lengths are in fractions, the area of a rectangle is still the product of the lengths of sides.

Class 7 Maths Chapter 2 Kerala Syllabus Fractions Questions and Answers

Question 1.
Calculate the following:
(i) \(\frac{2}{3}\) of 16
Answer:
\(\frac{2}{3}\) of 16
= \(\frac{2}{3}\) × 16
= \(\frac{2 \times 16}{3}=\frac{32}{3}\)
= 10\(\frac{2}{3}\)

(ii) \(\frac{4}{7}\) of 25
Answer:
\(\frac{4}{7}\) of 25
= \(\frac{4}{7}\) × 25
= \(\frac{4 \times 25}{7}=\frac{100}{7}\)
= 14\(\frac{2}{3}\)

(iii) \(\frac{2}{7}\) of \(\frac{1}{4}\)
Answer:
\(\frac{2}{7}\) of \(\frac{1}{4}\)
= \(\frac{2}{7}\) × \(\frac{1}{4}\)
= \(\frac{2}{7} \times \frac{1}{4}\)
= \(\frac{2 \times 1}{7 \times 4}=\frac{2}{28}\)
= \(\frac{1}{14}\)

(iv) 1\(\frac{1}{2}\) × 6\(\frac{2}{3}\)
Answer:
1\(\frac{1}{2}\) × 6\(\frac{2}{3}\)
= \(\frac{3}{2} \times \frac{20}{3}=\frac{3 \times 20}{2 \times 3}\)
= \(\frac{60}{6}\)
= 10

(v) 2\(\frac{3}{4}\) × \(\frac{5}{8}\)
Answer:
2\(\frac{3}{4}\) × \(\frac{5}{8}\)
= \(\frac{11}{4} \times \frac{5}{8}\)
= \(\frac{11 \times 5}{4 \times 8}=\frac{55}{32}\)
= 1\(\frac{23}{32}\)

(vi) \(\frac{4}{7}\) of \(\frac{3}{5}\)
Answer:
\(\frac{4}{7}\) of \(\frac{3}{5}\)
= \(\frac{4}{7}\) × \(\frac{3}{5}\)
= \(\frac{4 \times 3}{7 \times 5}\)
= \(\frac{12}{35}\)

Question 2.
12 metre long rope cut into 4 equal pieces.
(i) What is the length of each piece?
(ii) What if it is cut into 5 equal pieces?
Answer:
(i) Length of each piece = \(\frac{12}{4}\) = 3 metres

(ii) If it is cut into 5 equal pieces,
Length of each piece = \(\frac{12}{5}\) = 2\(\frac{2}{5}\) metres

Question 3.
Each piece of rope is \(\frac{7}{4}\) metres long. What is the total length of 8 such pieces?
Answer:
Lenth of each piece = \(\frac{7}{4}\) metres
Length of 8 pieces = \(\frac{7}{4}\) × 8 = 7 × \(\frac{8}{4}\)
= 7 × 2
= 14 metres

Question 4.
If 4 strings of length \(\frac{1}{3}\) metre were laid end to end, what would be the total length?
Answer:
Total length = \(\frac{1}{3}\) × 4 = \(\frac{4}{3}\)
= 1\(\frac{1}{3}\) metres

Question 5.
Suhara has 1 metre long silk ribbon. She gave half of it to Soumya. She in turn gave half of this to Reena. What is the length of the piece Reena got?
Answer:
Length of the piece Reena got = \(\frac{1}{2}\) of \(\frac{1}{2}\) of 1 metre
= \(\frac{1}{2}\) × \(\frac{1}{2}\)
= \(\frac{1}{4}\) metre

Question 6.
Half the children in a class are girls. A third of them are in the Math Club. What fraction of the total children are they?
Answer:
Number of girls in Math club = \(\frac{1}{3}\) of \(\frac{1}{2}\) of total children
= \(\frac{1}{3} \times \frac{1}{2}\) of total children
= \(\frac{1}{6}\) of total children

Question 7.
The length and breadth of some rectangles are given below. Calculate their areas.
(i) 4\(\frac{1}{2}\) cm, 3\(\frac{1}{4}\) cm
Answer:
Area = 4\(\frac{1}{2}\) × 3\(\frac{1}{4}\)
= \(\frac{9}{2} \times \frac{13}{4}=\frac{9 \times 13}{2 \times 4}\)
= \(\frac{117}{8}\)
= 14\(\frac{5}{8}\) cm²

(ii) 6\(\frac{3}{4}\) cm, 5\(\frac{1}{3}\) cm
Answer:
Area = 6\(\frac{3}{4}\) × 5\(\frac{1}{3}\)
= \(\frac{27}{4} \times \frac{16}{3}\)
= \(\frac{27}{3} \times \frac{16}{4}\)
= 9 × 4
= 36 cm²

Question 8.
What is the area of a square of side 1 metre?
Answer:
Area 1\(\frac{1}{2}\) x 1\(\frac{1}{2}\)
= \(\frac{3}{2} \times \frac{3}{2}\)
= \(\frac{9}{4}\)
= 2\(\frac{1}{4}\) m²

Practice Questions

Question 1.
Each bottle has \(\frac{3}{4}\) litres of water. What is the quantity of water in 14 such bottles?
Answer:
\(\frac{21}{2}\) = 10\(\frac{1}{2}\) litres

Question 2.
A farmer plants the sapling of a plant at a uniform distance of cm. If he plants 27 such saplings in a row, find the total distance between the first and the last sapling.
Answer:
45 cm

Question 3.
In a class of 60 students, two-thirds are boys. How many girls are there in the class?
Answer:
20

Question 4.
The weight of an iron block is \(\frac{3}{4}\) kilogram. What is the total weight of such 17 blocks?
Answer:
\(\frac{51}{4}\) = 12\(\frac{3}{4}\) kilogram

Question 5.
There are some cans, each containing 3 litres of milk. The milk in each vessel is used to fill 5 identical bottles.
(i) How much milk is there in each bottle?
Answer:
\(\frac{3}{5}\) litres

(ii) How much milk in 3 such bottles?
Answer:
\(\frac{9}{5}\) = 1 \(\frac{4}{5}\) litres

(iii) In 10 bottles?
Answer:
6 litres

Question 6.
Calculate the following:
(i) \(\frac{1}{8} \times \frac{1}{5}\)
Answer:
\(\frac{1}{40}\)

(ii) \(\frac{1}{6} \times \frac{1}{7}\)
Answer:
\(\frac{1}{42}\)

(iii) \(\frac{2}{5} \times \frac{7}{9}\)
Answer:
\(\frac{14}{45}\)

(iv) \(\frac{4}{5}\) of \(\frac{2}{3}\)
Answer:
\(\frac{8}{15}\)

(v) \(\frac{2}{9}\) of \(\frac{3}{2}\)
Answer:
\(\frac{1}{3}\)

(vi) 1\(\frac{3}{4}\) of 4
Answer:
7

(vii) \(\frac{3}{8}\) of 2\(\frac{1}{2}\)
Answer:
\(\frac{15}{16}\)

(viii) 3\(\frac{1}{4}\) × 5\(\frac{2}{9}\)
Answer:
\(\frac{611}{36}\)

(ix) 4\(\frac{1}{7}\) × 3\(\frac{1}{8}\)
Answer:
\(\frac{725}{56}\)

Question 7.
If three litres of milk is equally divided among four persons, how much would each get?
Answer:
\(\frac{3}{4}\) litres

Question 8.
Six kilogram of rice is packed into four identical bags.
(i) How much rice is in each bag?
Answer:
\(\frac{3}{2}\) Kilograms

(ii) What if it is packed into two bags?
Answer:
3 Kilograms

Class 7 Maths Chapter 2 Notes Kerala Syllabus Fractions

Fractions are a way to represent parts of a whole. A fraction consists of a numerator and a denominator. The numerator is the number written above the fraction line, which tells how many equal parts you consider, and the denominator is the number written below the fraction line, which tells the total number of equal parts the whole thing has been cut into.

Multifold multiplication
If we have to find 4 times three-quarter, we calculate it in following way:
\(\frac{3}{4}+\frac{3}{4}+\frac{3}{4}+\frac{3}{4}\) = 3
This can be easily calculated as:
4 × \(\frac{3}{4}=\frac{12}{4}\) = 3

Part and multiplication
If three litres of milk divided among four people, each will get \(\frac{1}{4}\) of 3 litres.
i.e, \(\frac{1}{4}\) of 3 = \(\frac{1}{4}\) × 3 = \(\frac{3}{4}\) litres.

Part of part
If we have to calculate \(\frac{1}{5}\) of \(\frac{1}{4}\), it can be done as as follows:
\(\frac{1}{5} \times \frac{1}{4}=\frac{1}{5 \times 4}=\frac{1}{20}\)
Now, if have to calculate \(\frac{3}{5}\) of \(\frac{2}{7}\),
\(\frac{3}{5} \times \frac{2}{7}=3 \times \frac{1}{5} \times \frac{1}{7} \times 2=\frac{3 \times 2}{5 \times 7}=\frac{6}{35}\)

To calculate 2\(\frac{1}{4}\) × 5,
2\(\frac{1}{4}\) × 5 = (2 + \(\frac{1}{4}\)) × 5
= (2 × 5) + (\(\frac{1}{4}\) × 5)
= 10 + 1\(\frac{1}{4}\)
= 11\(\frac{1}{4}\)

To calculate 1\(\frac{1}{2}\) × 3\(\frac{3}{4}\)
1\(\frac{1}{2}\) × 3\(\frac{3}{4}\) = \(\frac{3}{2} \times \frac{15}{4}\)
= \(\frac{3 \times 15}{2 \times 4}\)
= \(\frac{45}{8}=\frac{40+5}{8}=\frac{40}{8}+\frac{5}{8}\)
= 5 \(\frac{5}{8}\)
Even if the lengths are in fractions, the area of a rectangle is still the product of the lengths of sides.

Share And Fraction
If 4 litres of milk is divided equally among 3 persons, how much milk will each one get? First give 1 litre to each one. If the remaining 1 litre is divided among 3 persons, each will get \(\frac{1}{3}\) litre. So, in total each get 1\(\frac{1}{3}\) litres.
Numerically,
4 ÷ 3 = \(\frac{4}{3}\) = 1\(\frac{1}{3}\)