# NCERT Solutions for Class 7 Maths Exercise 2.3 Chapter 2 Fractions and Decimals

NCERT Solutions for Class 7 Maths Exercise 2.3 Chapter 2 Fractions and Decimals is available in downloadable PDF format. Multiplication of a fraction by a fraction is the only topic covered in this exercise of NCERT Maths Solutions for Class 7 Chapter 2. From this exercise, students can learn multiplication of proper, improper and mixed fractions. Students are advised to try solving the questions from NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals to get more marks in Maths.

## NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals â€“ Exercise 2.3

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### Access Answers to NCERT Solutions for Class 7 Maths Chapter 2 â€“ Fractions and Decimals Exercise 2.3

1. Find:

(i) Â¼ of (a) Â¼ (b) 3/5 (c) 4/3

Solution:-

(a) Â¼

We have,

= Â¼ Ã— Â¼

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= Â¼ Ã— Â¼

= (1 Ã— 1)/ (4 Ã— 4)

= (1/16)

(b) 3/5

We have,

= Â¼ Ã— (3/5)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= Â¼ Ã— (3/5)

= (1 Ã— 3)/ (4 Ã— 5)

= (3/20)

(c) (4/3)

We have,

= Â¼ Ã— (4/3)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= Â¼ Ã— (4/3)

= (1 Ã— 4)/ (4 Ã— 3)

= (4/12)

= 1/3

(ii) 1/7 of (a) 2/9 (b) 6/5 (c) 3/10

Solution:-

(a) 2/9

We have,

= (1/7) Ã— (2/9)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1/7) Ã— (2/9)

= (1 Ã— 2)/ (7 Ã— 9)

= (2/63)

(b) 6/5

We have,

= (1/7) Ã— (6/5)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1/7) Ã— (6/5)

= (1 Ã— 6)/ (7 Ã— 5)

= (6/35)

(c) 3/10

We have,

= (1/7) Ã— (3/10)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1/7) Ã— (3/10)

= (1 Ã— 3)/ (7 Ã— 10)

= (3/70)

2. Multiply and reduce to lowest form (if possible):

(i) (2/3) Ã—

Solution:-

First convert the given mixed fraction into improper fraction.

== 8/3

Now,

= (2/3) Ã— (8/3)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2 Ã— 8)/ (3 Ã— 3)

= (16/9)

=

(ii) (2/7) Ã— (7/9)

Solution:-

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2 Ã— 7)/ (7 Ã— 9)

= (2 Ã— 1)/ (1 Ã— 9)

= (2/9)

(iii) (3/8) Ã— (6/4)

Solution:-

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (3 Ã— 6)/ (8 Ã— 4)

= (3 Ã— 3)/ (4 Ã— 4)

= (9/16)

(iv) (9/5) Ã— (3/5)

Solution:-

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (9 Ã— 3)/ (5 Ã— 5)

= (27/25)

=

(v) (1/3) Ã— (15/8)

Solution:-

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1 Ã— 15)/ (3 Ã— 8)

= (1 Ã— 5)/ (1 Ã— 8)

= (5/8)

(vi) (11/2) Ã— (3/10)

Solution:-

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (11 Ã— 3)/ (2 Ã— 10)

= (33/20)

=

(vii) (4/5) Ã— (12/7)

Solution:-

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (4 Ã— 12)/ (5 Ã— 7)

= (48/35)

=

3. Multiply the following fractions:

(i) (2/5) Ã— 5 Â¼

Solution:-

First convert the given mixed fraction into improper fraction.

= 5 Â¼ = 21/4

Now,

= (2/5) Ã— (21/4)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2 Ã— 21)/ (5 Ã— 4)

= (1 Ã— 21)/ (5 Ã— 2)

= (21/10)

=

(ii) Ã— (7/9)

Solution:-

First convert the given mixed fraction into improper fraction.

== 32/5

Now,

= (32/5) Ã— (7/9)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (32 Ã— 7)/ (5 Ã— 9)

= (224/45)

=

(iii) (3/2) Ã—

Solution:-

First convert the given mixed fraction into improper fraction.

== 16/3

Now,

= (3/2) Ã— (16/3)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (3 Ã— 16)/ (2 Ã— 3)

= (1 Ã— 8)/ (1 Ã— 1)

= 8

(iv) (5/6) Ã—

Solution:-

First convert the given mixed fraction into improper fraction.

== 17/7

Now,

= (5/6) Ã— (17/7)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5 Ã— 17)/ (6 Ã— 7)

= (85/42)

=

(v) Ã— (4/7)

Solution:-

First convert the given mixed fraction into improper fraction.

== 17/5

Now,

= (17/5) Ã— (4/7)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (17 Ã— 4)/ (5 Ã— 7)

= (68/35)

=

(vi) Ã— 3

Solution:-

First convert the given mixed fraction into improper fraction.

== 13/5

Now,

= (13/5) Ã— (3/1)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (13 Ã— 3)/ (5 Ã— 1)

= (39/5)

=

(vi) Ã— (3/5)

Solution:-

First convert the given mixed fraction into improper fraction.

== 25/7

Now,

= (25/7) Ã— (3/5)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (25 Ã— 3)/ (7 Ã— 5)

= (5 Ã— 3)/ (7 Ã— 1)

= (15/7)

=

4. Which is greater?

(i) (2/7) of (3/4) or (3/5) of (5/8)

Solution:-

We have,

= (2/7) Ã— (3/4) and (3/5) Ã— (5/8)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2/7) Ã— (3/4)

= (2 Ã— 3)/ (7 Ã— 4)

= (1 Ã— 3)/ (7 Ã— 2)

= (3/14) â€¦ [i]

And,

= (3/5) Ã— (5/8)

= (3 Ã— 5)/ (5 Ã— 8)

= (3 Ã— 1)/ (1 Ã— 8)

= (3/8) â€¦ [ii]

Now, convert [i] and [ii] into like fractions,

LCM of 14 and 8 is 56

Now, let us change each of the given fractions into an equivalent fraction having 56 as the denominator.

[(3/14) Ã— (4/4)] = (12/56) [(3/8) Ã— (7/7)] = (21/56)

Clearly,

(12/56) < (21/56)

Hence,

(3/14) < (3/8)

(ii) (1/2) of (6/7) or (2/3) of (3/7)

Solution:-

We have,

= (1/2) Ã— (6/7) and (2/3) Ã— (3/7)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1/2) Ã— (6/7)

= (1 Ã— 6)/ (2 Ã— 7)

= (1 Ã— 3)/ (1 Ã— 7)

= (3/7) â€¦ [i]

And,

= (2/3) Ã— (3/7)

= (2 Ã— 3)/ (3 Ã— 7)

= (2 Ã— 1)/ (1 Ã— 7)

= (2/7) â€¦ [ii]

By comparing [i] and [ii],

Clearly,

(3/7) > (2/7)

5. Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is Â¾ m. Find the distance between the first and the last sapling.

Solution:-

From the question, it is given that,

The distance between two adjacent saplings = Â¾ m

Number of saplings planted by Saili in a row = 4

Then, number of gap in saplings = Â¾ Ã— 4

= 3

âˆ´ The distance between the first and the last saplings = 3 Ã— Â¾

= (9/4) m

= 2 Â¼ m

Hence, the distance between the first and the last saplings is 2 Â¼ m.

6. Lipika reads a book for 1 Â¾ hours every day. She reads the entire book in 6 days. How many hours in all were required by her to read the book?

Solution:-

From the question, it is given that,

Lipika reads the book for = 1 Â¾ hours every day = 7/4 hours

Number of days she took to read the entire book = 6 days

âˆ´ Total number of hours required by her to complete the book = (7/4) Ã— 6

= (7/2) Ã— 3

= 21/2

= 10 Â½ hours

Hence, the total number of hours required by her to complete the book is 10 Â½ hours.

7. A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2 Â¾ litres of petrol.

Solution:-

From the question, it is given that,

The total number of distance travelled by a car in 1 liter of petrol = 16 km

Then,

Total quantity of petrol = 2 Â¾ liter = 11/4 liters

Total number of distance travelled by car in 11/4 liters of petrol = (11/4) Ã— 16

= 11 Ã— 4

= 44 km

âˆ´ Total number of distance travelled by car in 11/4 liters of petrol is 44 km.

8. (a) (i) provide the number in the box [ ], such that (2/3) Ã— [ ] = (10/30)

Solution:-

Let the required number be x,

Then,

= (2/3) Ã— (x) = (10/30)

By cross multiplication,

= x = (10/30) Ã— (3/2)

= x = (10 Ã— 3) / (30 Ã— 2)

= x = (5 Ã— 1) / (10 Ã— 1)

= x = 5/10

âˆ´ The required number in the box is (5/20)

(ii) The simplest form of the number obtained in [ ] is

Solution:-

The number in the box is 5/10

Then,

The simplest form of 5/10 is Â½

(b) (i) provide the number in the box [ ], such that (3/5) Ã— [ ] = (24/75)

Solution:-

Let the required number be x,

Then,

= (3/5) Ã— (x) = (24/75)

By cross multiplication,

= x = (24/75) Ã— (5/3)

= x = (24 Ã— 5) / (75 Ã— 3)

= x = (8 Ã— 1) / (15 Ã— 1)

= x = 8/15

âˆ´ The required number in the box is (8/15)

(ii) The simplest form of the number obtained in [ ] is

Solution:-

The number in the box is 8/15

Then,

The simplest form of 8/15 is 8/15

### Access Other Exercises of NCERT Solutions for Class 7 Maths Chapter 2 â€“ Fractions and Decimals

Exercise 2.1 Solutions

Exercise 2.2 Solutions

Exercise 2.4 Solutions

Exercise 2.5 Solutions

Exercise 2.6 Solutions

Exercise 2.7 Solutions

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NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7Â

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