**NCERT Answers for Class 7 Maths**

Chapter 2 Fractions and Decimals Exercise 2.3

Chapter 2 Fractions and Decimals Exercise 2.3

Chapter 2 Exercise 2.1 |

**Page 41-42**

**Ex 2.3 Class 7 Maths Question 1. Find**

**Answer**

**(a) 1/4**

= ¼ × ¼

By the rule Multiplication of fraction,

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

Then,

= ¼ × ¼

= (1 × 1)/ (4 × 4)

= (1/16)

**(b) 3/5**

= ¼ × (3/5)

By the rule Multiplication of fraction,

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

Then,

= ¼ × (3/5)

= (1 × 3)/ (4 × 5)

= (3/20)

**(c) 4/3**

= ¼ × (4/3)

By the rule Multiplication of fraction,

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

Then,

= ¼ × (4/3)

= (1 × 4)/ (4 × 3)

= (4/12)

= 1/3

**(a) 2/9**

= (1/7) × (2/9)

By the rule Multiplication of fraction,

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

Then,

= (1/7) × (2/9)

= (1 × 2)/ (7 × 9)

= (2/63)

**(b) 6/5**

= (1/7) × (6/5)

By the rule Multiplication of fraction,

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

Then,

= (1/7) × (6/5)

= (1 × 6)/ (7 × 5)

= (6/35)

**(c) 3/10**

= (1/7) × (3/10)

By the rule Multiplication of fraction,

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

Then,

= (1/7) × (3/10)

= (1 × 3)/ (7 × 10)

= (3/70)

**Ex 2.3 Class 7 Maths Question 2. Multiply and reduce to lowest form (if possible) :**

**Answer**

= = 8/3

Now,

= (2/3) × (8/3)

By the rule Multiplication of fraction,

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

Then,

= (2 × 8)/ (3 × 3)

= (16/9)

=

**(ii) (2/7) × (7/9)**

**Answer**

By the rule Multiplication of fraction,

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

Then,

= (2 × 7)/ (7 × 9)

= (2 × 1)/ (1 × 9)

= (2/9)

**(iii) (3/8) × (6/4)**

**Answer**

By the rule Multiplication of fraction,

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

Then,

= (3 × 6)/ (8 × 4)

= (3 × 3)/ (4 × 4)

= (9/16)

**(iv) (9/5) × (3/5)**

**Answer**

By the rule Multiplication of fraction,

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

Then,

= (9 × 3)/ (5 × 5)

= (27/25)

=

**(v) (1/3) × (15/8)**

**Answer**

By the rule Multiplication of fraction,

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

Then,

= (1 × 15)/ (3 × 8)

= (1 × 5)/ (1 × 8)

= (5/8)

**(vi) (11/2) × (3/10)**

**Answer**

By the rule Multiplication of fraction,

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

Then,

= (11 × 3)/ (2 × 10)

= (33/20)

=

**(vii) (4/5) × (12/7)**

**Answer**

By the rule Multiplication of fraction,

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

Then,

= (4 × 12)/ (5 × 7)

= (48/35)

=

**Ex 2.3 Class 7 Maths Question 3. Multiply the following fractions:**

**Answer**

= = 8/3

Now,

= (2/3) × (8/3)

By the rule Multiplication of fraction,

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

Then,

= (2 × 8)/ (3 × 3)

= (16/9)

=

**(ii) (2/7) × (7/9)**

**Answer**

By the rule Multiplication of fraction,

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

Then,

= (2 × 7)/ (7 × 9)

= (2 × 1)/ (1 × 9)

= (2/9)

**(iii) (3/8) × (6/4)**

**Answer**

By the rule Multiplication of fraction,

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

Then,

= (3 × 6)/ (8 × 4)

= (3 × 3)/ (4 × 4)

= (9/16)

**(iv) (9/5) × (3/5)**

**Answer**

By the rule Multiplication of fraction,

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

Then,

= (9 × 3)/ (5 × 5)

= (27/25)

=

**(v) (1/3) × (15/8)**

**Answer**

By the rule Multiplication of fraction,

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

Then,

= (1 × 15)/ (3 × 8)

= (1 × 5)/ (1 × 8)

= (5/8)

**(vi) (11/2) × (3/10)**

**Answer**

By the rule Multiplication of fraction,

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

Then,

= (11 × 3)/ (2 × 10)

= (33/20)

=

**(vii) (4/5) × (12/7)**

**Answer**

By the rule Multiplication of fraction,

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

Then,

= (4 × 12)/ (5 × 7)

= (48/35)

=

**Ex 2.3 Class 7 Maths Question 4. Which is greater:**

**Answer**

= (2/7) × (3/4) and (3/5) × (5/8)

By the rule Multiplication of fraction,

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

Then,

= (2/7) × (3/4)

= (2 × 3)/ (7 × 4)

= (1 × 3)/ (7 × 2)

= (3/14) … [i]

And,

= (3/5) × (5/8)

= (3 × 5)/ (5 × 8)

= (3 × 1)/ (1 × 8)

= (3/8) … [ii]

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

LCM of 14 and 8 is 56

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

[(3/14) × (4/4)] = (12/56)

[(3/8) × (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)**

**Answer**

= (1/2) × (6/7) and (2/3) × (3/7)

By the rule Multiplication of fraction,

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

Then,

= (1/2) × (6/7)

= (1 × 6)/ (2 × 7)

= (1 × 3)/ (1 × 7)

= (3/7) … [i]

And,

= (2/3) × (3/7)

= (2 × 3)/ (3 × 7)

= (2 × 1)/ (1 × 7)

= (2/7) … [ii]

By comparing [i] and [ii],

Clearly,

(3/7) > (2/7)

**Ex 2.3 Class 7 Maths Question 5. Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is 3/4 m. Find the distance between the first and the last sapling.**

**Answer**

The distance between two adjacent saplings = ¾ m

Number of saplings planted by Saili in a row = 4

Then, number of gap in saplings = ¾ × 4

= 3

∴ The distance between the first and the last saplings = 3 × ¾

= (9/4) m

= 2 ¼ m

Hence, the distance between the first and the last saplings is 2 ¼ m.

**Ex 2.3 Class 7 Maths Question 6. Lipika reads a book for 1 ¾ hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book?**

**Answer**

Lipika reads the book for = 1 ¾ hours every day = 7/4 hours

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

∴ Total number of hours required by her to complete the book = (7/4) × 6

= (7/2) × 3

= 21/2

= 10 ½ hours

Hence, the total number of hours required by her to complete the book is 10 ½ hours.

**Ex 2.3 Class 7 Maths Question 7. ****A car runs 16 km using 1 litre of petrol. How much distance will it cover using 3 2/4 litres of petrol.**

**Answer**

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

Then,

Total quantity of petrol = 2 ¾ liter = 11/4 liters

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

= 11 × 4

= 44 km

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

**Ex 2.3 Class 7 Maths Question 8. **

**(a) ****(i) Provide the number in the box ☐ , such that 2/3 x ☐ = 10/30**

**(ii) The simplest form of the number obtained in ☐ is _____.**

**(b) (i) Provide the number in the box ☐ , such that 3/5 x ☐ = 24/75.**

**(ii) The simplest form of the number obtained in ☐ is _____.**

**Answer**

Let the required number be x,

Then,

= (2/3) × (x) = (10/30)

By cross multiplication,

= x = (10/30) × (3/2)

= x = (10 × 3) / (30 × 2)

= x = (5 × 1) / (10 × 1)

= x = 5/10

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

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

**Answer**

The number in the box is 5/10

Then,

The simplest form of 5/10 is ½

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

**Answer**

Let the required number be x,

Then,

= (3/5) × (x) = (24/75)

By cross multiplication,

= x = (24/75) × (5/3)

= x = (24 × 5) / (75 × 3)

= x = (8 × 1) / (15 × 1)

= x = 8/15

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

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

**Answer**

The number in the box is 8/15

Then, the simplest form of 8/15 is 8/15