NCERT Solution class 7 Mathematics EXERCISE- 2.3

NCERT Solution class 7 Mathematics EXERCISE- 2.3

 

                   CHAPTER – 2

                                                   EXERCISE – 2.3

Q1. Find:

(i) ¼ of

(a) ¼

SOLUTION:-

= 1/16

(B) 3/5

SOLUTION:-

= 3/20

(C) 4/3

SOLUTION:-

= 4/12

(ii) 1/7 of

(a) 2/9

SOLUTION:-

2/63

(B) 6/5

SOLUTION:-

= 6/35

(C) 3/10

SOLUTION:-

= 3/70

Q2. Multiply and reduce to lowest form

(i) 2/3 x 8/3

SOLUTION:-

=16/9

(ii) 2/7 x 7/9

SOLUTION:-

= 2/9

(iii) 3/8 x 6/4

SOLUTION:-

= 18/32

= 9/16

(iv) 9/5 x 3/5

SOLUTION:-

= 27/25

(v) 1/3 x 15/8

SOLUTION:-

= 15/24

= 5/8

(vi) 11/2 x 3/10

SOLUTION:-

= 33/20

(vii) 4/5 x 12/7

SOLUTION:-

= 49/35

Q3. Multiply the following fractions:

(i) 2/5 x 21/4

SOLUTION:-

= 42/20

= 6/5

(ii)  32/5 x 7/9

SOLUTION:-

= 224/45

(iii) 3/2 x 16/3

SOLUTION:-

= 48/6

= 8

(iv) 5/6 x 17/7

SOLUTION:-

= 85/42

(v) 17/5 x 4/7

SOLUTION:-

= 68/35

(vi) 13/5 x 3

= 39/5

(vii) 25/7 x 3/5

SOLUTION:-

= 75/35

= 15/7

Q4. Which

is greater :

(i) 2/7 of ¾ or 3/5 of 5/8

SOLUTION:-

We have,

= (2/7) x (3/4) and (3/5) x (5/8)

By the rule multiplication of fraction,

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

Then,

= (2/7) x (3/4)

= (2 x3) / (7×4)

= (1x 3) /(7×2)

= (3/14) [i]

And,

= (3/5) x (5/8)

= (3/5) / (5×8)

= (3 x 1) / (1×8)

= (3/8) [ii]

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

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) x (4/4)] = (12/56)

[(3/8) x (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) x (6/7) and (2/3) x (3/7)

By the rule of multiplication of fraction

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

Then,

= (1/2) x (6/7)

= (1 x 6) / (2 x 7)

= (1 x 3 )/(1 x 7)

= (3/7)   [i]

And,

= (2/3) x (3/7)

= (2 x 3 ) / (3 x 7)

= ( 2 x 1) / (1 x 7)

= (2/7)    [ii]

By comparing [i] and [ii]

Clearly,

(3/7) > (2/7)

Q5. Saili plants 4 sampling , in a row , in her garden. The distance between two adjacent sampling is ¾ m. Find the distance between the first and last sapling .

SOLUTION:-

From the question , it is the given that,

The distance between two adjacent sapling = ¾ m.

Numbers of saplings planted by saili in a row = 4

Then, numbers of gap in sapling = ¾ x 4

= 3

The distance between the first and last saplings = 3 x ¾

= (9/4) m

=2 ¼ m

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

Q6. 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 given question ,it is given that,

Lipika reads the book for = 1¾ hours every day = 7/4 hours

Numbers of day she took to read the entire book = 6 days

Total numbers of hours required by her to complete the book = (7/4) x 6

= (7/2) x 3

= 21/2

= 10 ½ hours.

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

Q7. A car runs 16km using 1 liter of petrol . How much distance will it cover using 2¾ liters of petrol.

SOLUTION:-

From the question , It is given that,

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

Then,

Total quantity of petrol = 2¾  = 11/4 liters

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

= 11 x4

= 44km

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

Q8. (A) (i) Provide the numbers in the box [] such that 2/3 x [] = 10/30.

SOLUTION:-

Let the required numbers be x,

Then,

= (2/3) x (x) = (10/30)

By cross multiplication ,

= x= (10/30) x (3/2)

= x= (10 x3 )/ (30 x 2)

= x= ( 5 x 1) / (10 x 1)

=x = 5/10

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

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

SOLUTION:-

The numbers 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) x [] = (24/75)

SOLUTION:-

Let the required number be x,

Then,

= (3/5) x (x) = (24/75)

By cross multiplication,

= x= (24/75) x (5/3)

=x= (24×5)/(75x 3)

= x= (8 x 1 ) / (15 x 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

Leave a Comment