NCERT Solutions Class 7th Math Chapter – 2 Fractions and Decimals Exercise – 2.2

1. Find:

(i) ¼ of

(a) ¼

Solution:

We have,
= 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

Solution:

We have,
= 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

Solution:

We have,
= 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

(ii) 1/7 of

(a) 2/9

Solution:

We have,
= (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

Solution:

We have,
= (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

Solution:

We have,
= (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

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

Solution:

First convert the given mixed fraction into improper fraction.

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution: 

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

Solution:

By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (4 × 12)/ (5 × 7)
= (48/35)

3. Multiply the following fractions:

(i) (2/5) × 5 ¼

Solution:

First convert the given mixed fraction into improper fraction.
= 5 ¼ = 21/4
Now, = (2/5) × (21/4)

By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)

Then,
= (2 × 21)/ (5 × 4)
= (1 × 21)/ (5 × 2)
= (21/10)

Solution:

First convert the given mixed fraction into improper fraction.

Now
= (32/5) × (7/9)

By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)

Then,
= (32 × 7)/ (5 × 9)
= (224/45)

Solution:

First convert the given mixed fraction into improper fraction.

Now,
= (3/2) × (16/3)

By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)

Then,
= (3 × 16)/ (2 × 3)
= (1 × 8)/ (1 × 1)
= 8

Solution:

First convert the given mixed fraction into improper fraction.

Now,
= (5/6) × (17/7)

By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)

Then,
= (5 × 17)/ (6 × 7)
= (85/42)

Solution:

First convert the given mixed fraction into improper fraction.

Now,
= (17/5) × (4/7)

By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)

Then,
= (17 × 4)/ (5 × 7)
= (68/35)

Solution:

First convert the given mixed fraction into improper fraction.

Now,
= (13/5) × (3/1)

By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)

Then,
= (13 × 3)/ (5 × 1)
= (39/5)

Solution:

First convert the given mixed fraction into improper fraction.

Now,
= (25/7) × (3/5)

By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)

Then,
= (25 × 3)/ (7 × 5)
= (5 × 3)/ (7 × 1)
= (15/7)

4. Which is greater:

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

Solution:

We have,
= (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,
(3/5) of (5/8)

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

Solution:

We have,
= (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],
(3/7) > (2/7)
(1/2) of (6/7)

5. Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is ¾ 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 = 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.

6. Lipika reads a book for 1 ¾ 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 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.

7. A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2 ¾ 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 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.

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

Solution:

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 (5/10)

(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 ½

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

Solution:

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

Solution:

The number in the box is 8/15
Then,
The simplest form of 8/15 is 8/15