NCERT Solutions for Class 7 Maths - Fractions and Decimals (Exercise 2.1)

Exercise 2.1

Q1. Which of the drawings (a) to (d) show: 

Ans:
(i) 2 × (1/5) represents the addition of 2 figures, each represents 1 shaded part out of the given 5 equal parts.
∴ 2 × (1/5) is represented by fig (d).
(ii) 2 × ½ represents the addition of 2 figures, each represents 1 shaded part out of the given 2 equal parts.
∴ 2 × ½ is represented by fig (b).
(iii) 3 × (2/3) represents the addition of 3 figures, each representing 2 shaded parts out of the given 3 equal parts.
∴ 3 × (2/3) is represented by fig (a).
(iii) 3 × ¼ represents the addition of 3 figures, each represents 1 shaded part out of the given 4 equal parts.
∴ 3 × ¼ is represented by fig (c).

Q2. Some pictures (a) to (c) are given below. Tell which of them show:

Ans:

(i) 3 × (1/5) represents the addition of 3 figures, each represents 1 shaded part out of the given 5 equal parts and (3/5) represents 3 shaded parts out of 5 equal parts.
∴ 3 × (1/5) = (3/5) is represented by fig (c).
(ii) 2 × (1/3) represents the addition of 2 figures, each represents 1 shaded part out of the given 3 equal parts and (2/3) represents 2 shaded parts out of 3 equal parts.
∴ 2 × (1/3) = (2/3) is represented by fig (a).
(iii) 3 × (3/4) represents the addition of 3 figures, each represents 3 shaded part out of the given 4 equal parts and 2 ¼ represents 2 fully and 1 figure having 1 part as shaded out of 4 equal parts.
∴ 3 × (3/4) = 2 ¼ is represented by fig (b)

Q3. Multiply and reduce to lowest form and convert into a mixed fraction: 
(i)

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (7/1) × (3/5)
= (7 × 3)/ (1 × 5)
= (21/5)
=

(ii)

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (4/1) × (1/3)
= (4 × 1)/ (1 × 3)
= (4/3)
=

(iii)

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (2/1) × (6/7)
= (2 × 6)/ (1 × 7)
= (12/7)
=

(iv)

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/1) × (2/9)
= (5 × 2)/ (1 × 9)
= (10/9)
=

(v)

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (2/3) × (4/1)
= (2 × 4)/ (3 × 1)
= (8/3)
=

(vi)

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/2) × (6/1)
= (5 × 6)/ (2 × 1)
= (30/2)
= 15

(vii)

By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (11/1) × (4/7)
= (11 × 4)/ (1 × 7)
= (44/7)
=

(viii)

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (20/1) × (4/5)
= (20 × 4)/ (1 × 5)
= (80/5)
= 16

(ix)

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (13/1) × (1/3)
= (13 × 1)/ (1 × 3)
= (13/3)
=

(x)

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (15/1) × (3/5)
= (15 × 3)/ (1 × 5)
= (45/5)
= 9

Q4. Shade:

Ans: (i) From the question,
We may observe that there are 12 circles in the given box. So, we have to shade ½ of the circles in the box.
∴ 12 × ½ = 12/2
 = 6
So we have to shade any 6 circles in the box.

(ii) From the question,
We may observe that there are 9 triangles in the given box. So, we have to shade 2/3 of the triangles in the box.
∴ 9 × (2/3) = 18/3
 = 6
So we have to shade any 6 triangles in the box.

(iii) From the question,
We may observe that there are 15 squares in the given box. So, we have to shade 3/5 of the squares in the box.
∴ 15 × (3/5) = 45/5
 = 9
So we have to shade any 9 squares in the box.

Q5. Find: 
(a) 1/2 of (i) 24 (ii) 46

Ans: (i) 24
We have,
= ½ × 24
= 24/2
= 12

(ii) 46
We have,
= ½ × 46
= 46/2
= 23

(b) 2/3 of (i) 18 (ii) 27

Ans: (i) 18
We have,
= 2/3 × 18
= 2 × 6
= 12

(ii) 27
We have,
= 2/3 × 27
= 2 × 9
= 18

(c) 3/4 of  (i) 16 (ii) 36

Ans: (i) 16
We have,
= ¾ × 16
= 3 × 4
= 12

(ii) 36
We have
= ¾ × 36
= 3 × 9
= 27

(d) 4/5 of (i) 20 (ii) 35

Ans: (i) 20
We have,
= 4/5 × 20
= 4 × 4
= 16

(ii) 35
We have,
= 4/5 × 35
= 4 × 7
= 28

Q6. Multiply and express as a mixed fraction: 

Ans: First convert the given mixed fraction into improper fraction.
= = 26/5
Now,= 3 × (26/5)
= 78/5
=

Ans: First convert the given mixed fraction into improper fraction.
= 6 ¾ = 27/4
Now,
= 5 × (27/4)
= 135/4
= 33 ¾ 

Ans: First convert the given mixed fraction into improper fraction.
= 2 ¼ = 9/4
Now,
= 7 × (9/4)
= 63/4
= 15 ¾ 

Ans: First convert the given mixed fraction into improper fraction.
= = 19/3
Now,
= 4 × (19/3)
= 76/3
=

Ans: First convert the given mixed fraction into improper fraction.
= 3 ¼ = 13/4
Now,
= (13/4) × 6
= (13/2) × 3
= 39/2
= 19 ½ 

Ans: First convert the given mixed fraction into improper fraction.
=  = 17/5
Now,
= (17/5) × 8
= 136/5
=

Q7. Find:

Ans: (i) First convert the given mixed fraction into an improper fraction.
= 2 ¾ = 11/4
Now,
= ½ × 11/4
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= ½ × (11/4)
= (1 × 11)/ (2 × 4)
= (11/8)
=

(ii) First convert the given mixed fraction into improper fraction.
= = 38/9
Now,
= ½ × (38/9)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= ½ × (38/9)
= (1 × 38)/ (2 × 9)
= (38/18)
= 19/9
=

Ans: (i) First convert the given mixed fraction into an improper fraction.
=  = 23/6
Now,
= (5/8) × (23/6)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/8) × (23/6)
= (5 × 23)/ (8 × 6)
= (115/48)
=

(ii) First convert the given mixed fraction into improper fraction.
=  = 29/3
Now,
= (5/8) × (29/3)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/8) × (29/3)
= (5 × 29)/ (8 × 3)
= (145/24)
=

Q8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed 2/5 of the water. Pratap consumed the remaining water. 
(i) How much water did Vidya drink? 
(ii) What fraction of the total quantity of water did Pratap drink? 

Ans: (i) From the question, it is given that,
Amount of water in the water bottle = 5 liters
Amount of water consumed by Vidya = 2/5 of 5 liters
= (2/5) × 5
= 2 liters
So, the total amount of water drank by Vidya is 2 liters

(ii) From the question, it is given that,
Amount of water in the water bottle = 5 liters
Then,
Amount of water consumed by Pratap = (1 – water consumed by Vidya)
= (1 – (2/5))
= (5-2)/5
= 3/5
∴ Total amount of water consumed by Pratap = 3/5 of 5 liters
 = (3/5) × 5
 = 3 liters
So, the total amount of water drank by Pratap is 3 liters

Deleted Exercise

Q1. Solve:
(i)

Ans: For subtraction of two unlike fractions, first change them to the like fractions.
LCM of 1, 5 = 5
Now, let us change each of the given fraction into an equivalent fraction having 5 as the denominator.
= [(2/1) × (5/5)] = (10/5)
= [(3/5) × (1/1)] = (3/5)
Now,
= (10/5) - (3/5)
= [(10 – 3)/5]
= (7/5)

(ii)

Ans: For the addition of two unlike fractions, first change them to the like fractions.
LCM of 1, 8 = 8
Now, let us change each of the given fraction into an equivalent fraction having 8 as the denominator.
= [(4/1) × (8/8)] = (32/8)
= [(7/8) × (1/1)] = (7/8)
Now,
= (32/8) + (7/8)
= [(32 + 7)/8]
= (39/8)

(iii)  

Ans: For the addition of two unlike fractions, first change them to the like fractions.
LCM of 5, 7 = 35
Now, let us change each of the given fraction into an equivalent fraction having 35 as the denominator
= [(3/5) × (7/7)] = (21/35)
= [(2/7) × (5/5)] = (10/35)
Now,
= (21/35) + (10/35)
= [(21 + 10)/35]
= (31/35)

(iv)

Ans: For subtraction of two, unlike fractions, first, change them to the like fractions.
LCM of 11, 15 = 165
Now, let us change each of the given fraction into an equivalent fraction having 165 as the denominator.
= [(9/11) × (15/15)] = (135/165)
= [(4/15) × (11/11)] = (44/165)
Now,
= (135/165) - (44/165)
= [(135 – 44)/165]
= (91/165)

(v)

Ans: For the addition of two unlike fractions, first change them to the like fractions.
LCM of 10, 5, 2 = 10
Now, let us change each of the given fraction into an equivalent fraction having 35 as the denominator.
= [(7/10) × (1/1)] = (7/10)
= [(2/5) × (2/2)] = (4/10)
= [(3/2) × (5/5)] = (15/10)
Now,
= (7/10) + (4/10) + (15/10)
= [(7 + 4 + 15)/10]
= (26/10)
= (13/5)

(vi)

Ans: First convert mixed fraction into improper fraction,
= = 8/3
= 3 ½
= 7/2
For the addition of two unlike fractions, first, change them to the like fractions.
LCM of 3, 2 = 6
Now, let us change each of the given fraction into an equivalent fraction having 6 as the denominator.
= [(8/3) × (2/2)] = (16/6)
= [(7/2) × (3/3)] = (21/6)
Now,
= (16/6) + (21/6)
= [(16 + 21)/6]
= (37/6)

(vii)

Ans: First convert mixed fraction into improper fraction,
= 8 ½ = 17/2
=  = 29/8
For the Subtraction of two, unlike fractions, first, change them to the like fractions.
LCM of 2, 8 = 8
Now, let us change each of the given fraction into an equivalent fraction having 35 as the denominator.
= [(17/2) × (4/4)] = (68/8)
= [(29/8) × (1/1)] = (29/8)
Now,
= (68/8) - (29/8)
= [(68 - 29)/8]
= (39/8)

Q2. Arrange the following in descending order:

(ii)

Ans: (i)
[Converting into like fractions]
⇒
[Arranging in descending order]
⇒
Therefore,

(ii)  
[Converting into like fractions]
⇒  
[Arranging in descending order]
⇒    
Therefore,

Q3. In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square?

Ans: Sum of first row 
Sum of second row
Sum of third row
Sum of first column
Sum of second column
Sum of third column
Sum of first diagonal [left to right)
Sum of second diagonal [left to right]
Since the sum of fractions in each row, in each column and along the diagonals are same, therefore it is a magic square.

Q4. A rectangular sheet of paper is  cm wide. Find its perimeter. 

Ans: Given: The sheet of paper is in rectangular form.
Length of sheet =  cm and Breadth of sheet cm
Perimeter of rectangle = 2 (length + breadth)

Thus, the perimeter of the rectangular sheet is

Q5. Find the perimeter of 
(i) ΔABE 
(ii) the rectangle BCDE in this figure. 
Whose perimeter is greater?

Ans: (i) In ΔABE
The perimeter of ΔABE = AB + BE + AE

=
Thus, the perimeter of ΔABE iscm

(ii) In rectangle BCDE,
Perimeter of rectangle = 2 [length + breadth]

Thus, the perimeter of rectangle BCDE is
Comparing the perimeter of triangle and that of rectangle,

Therefore, the perimeter of triangle ABE is greater than that of rectangle BCDE.

Q6. Salil wants to put a picture in a frame. The picture is  cm wide. To fit in the frame the picture cannot be more than cm wide. How much should the picture be trimmed?

Ans: Given: The width of the picture
and the width of picture frame
Therefore, the picture should be trimmed

Thus, the picture should be trimmed by

Q7. Ritu ate 3/5 part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? Who had the larger share? By how much? 

Ans: The part of an apple eaten by Ritu = 3/5
The part of an apple eaten by Somu
Comparing the parts of apple eaten by both Ritu and Somu
The larger share will be more b
Thus, Ritu's part is 1/5 more than Somu’s part.

Q8. Michael finished colouring a picture in 7/12 hour. Vaibhav finished colouring the same picture in 3/4 hour. Who worked longer? By what fraction was it longer? 

Ans: Time is taken by Michael to colour the picture = 7/2 hour
Time is taken by Vaibhav to colour the picture = 3/4 hour
Converting both fractions in like fractions,
Here,  
Thus, Vaibhav worked longer time.
Vaibhav worked longer time by
Thus, Vaibhav took 1/6 hour more than Michael.

Exercise 2.2

Fractions and Decimals 

Q1. Find: 
(i) 1/4 of 
(a) 1/4 
(b) 3/5 
(c) 4/3
(ii) 1/7 of 
(a) 2/9 
(b) 6/5 
(c) 3/10

Ans: 
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
(i)
 (a)
(b)
(c)

(ii)
 (a)  
(b)
(c)

Q2: Multiply and reduce to lowest form (if possible): 
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)

Ans:
(i) First, convert the given mixed fraction into an improper fraction.
== 8/3,

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

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

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

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

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


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

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

Multiplying FractionQ3. Multiply the following fractions: 
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)

Ans: 
(i) 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,

(ii) First, convert the given mixed fraction into an improper fraction.
== 32/5Now,= (32/5) × (7/9)

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

(iii) First, convert the given mixed fraction into an improper fraction.
== 16/3
Now,
= (3/2) × (16/3)
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
Then,

(iv) First convert the given mixed fraction into an improper fraction.
== 17/7
Now,
= (5/6) × (17/7)
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
 

(v) First convert the given mixed fraction into improper fraction.
== 17/5
Now,
= (17/5) × (4/7)

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

(vi) First convert the given mixed fraction into improper fraction.
== 13/5
Now,
= (13/5) × (3/1)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
 

(vii) First convert the given mixed fraction into improper fraction.
== 25/7
Now,
= (25/7) × (3/5)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,  

Q4. Which is greater: 
(i)
(ii)

Ans: 
(i)
⇒
⇒
⇒

(ii)
⇒
⇒
⇒

Q5. 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.

Ans: The distance between two adjacent saplings = 3/4 m
Saili planted 4 saplings in a row, then number of gap in saplings = 3

Therefore,
The distance between the first and the last saplings
Thus the distance between the first and the last saplings is

Q6. Lipika reads a book forhours every day. She reads the entire book in 6 days. How many hours in ail were required by her to read the book?

Ans: Time is taken by Lipika to read a book = hours
She reads the entire book in 6 days.
Now, the total hours are taken by her to read the entire book
Thus, 10 hours were required by her to read the book.

Q7. A car runs 16 km using 1 litre of petrol. How much distance will it cover using  of petrol?

Ans: In 1 litre of petrol, car covers the distance = 16 km
In litres of petrol, car covers the distance =  

Thus, the car will cover 44 km distance.

Q8. 
(a) 
(i) Provide the number in the box 

(ii) The simplest form of the number obtained in

(b)
 (i) Provide the number in the box
 
(ii) The simplest form of the number obtained in

Ans:
 (a)

(ii) The simplest form
(b)

(ii) The simplest form

Exercise 2.3 

Q1.
Find: 
(i)
(ii)
(iii)
(iv)
(v)
(vi)

Ans: 
(i) We have,
= 12 × reciprocal of ¾

(ii) We have,
= 14 × reciprocal of (5/6)

(iii) We have,
= 8× reciprocal of (7/3)
  

(iv) We have,
= 4 × reciprocal of (8/3)

(v) While dividing a whole number by a mixed fraction, first convert the mixed fraction into improper fraction We have,
== 7/3
Then,
= 3 ÷ (7/3)
= 3 × reciprocal of (7/3)

(vi) While dividing a whole number by a mixed fraction, first convert the mixed fraction into improper fraction We have,
== 25/7
Then,
= 5 ÷ (25/7)
= 5 × reciprocal of (25/7)   

Q2. Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.

(i) 3/7
Ans: Reciprocal of (3/7) is (7/3) [∵ ((3/7) × (7/3)) = 1]
So, it is an improper fraction.
Improper fraction is that fraction in which numerator is greater than its denominator.
(ii) 5/8
Ans: Reciprocal of (5/8) is (8/5) [∵ ((5/8) × (8/5)) = 1]
So, it is an improper fraction.
Improper fraction is that fraction in which numerator is greater than its denominator.
(iii) 9/7
Ans: Reciprocal of (9/7) is (7/9) [∵ ((9/7) × (7/9)) = 1]
So, it is a proper fraction.
A proper fraction is that fraction in which denominator is greater than the numerator of the fraction.
(iv) 6/5
Ans: Reciprocal of (6/5) is (5/6) [∵ ((6/5) × (5/6)) = 1]
So, it is a proper fraction.
A proper fraction is that fraction in which denominator is greater than the numerator of the fraction.
(v) 12/7
Ans: Reciprocal of (12/7) is (7/12) [∵ ((12/7) × (7/12)) = 1]
So, it is a proper fraction.
A proper fraction is that fraction in which denominator is greater than the numerator of the fraction.
(vi) 1/8
Ans: Reciprocal of (1/8) is (8/1) or 8 [∵ ((1/8) × (8/1)) = 1]
So, it is a whole number.
Whole numbers are collection of all positive integers including 0.
(vii) 1/11
Ans: Reciprocal of (1/11) is (11/1) or 11 [∵ ((1/11) × (11/1)) = 1]
So, it is a whole number.
Whole numbers are collection of all positive integers including 0.

Q3. Find: 
(i)
(ii)
(iii)
(iv)
(v)
(vi)

Ans: 
(i) We have,
= (7/3) × reciprocal of 2
= (7/3) × (1/2)
 
(ii) We have,
= (4/9) × reciprocal of 5
= (4/9) × (1/5)
 

(iii) We have,
= (6/13) × reciprocal of 7
= (6/13) × (1/7)
 

(iv) First convert the mixed fraction into improper fraction. We have,
== 13/3 Then,
= (13/3) × reciprocal of 3

(v) First convert the mixed fraction into improper fraction.
We have,
= 3 ½ = 7/2
Then,
= (7/2) × reciprocal of 4

(vi) First convert the mixed fraction into improper fraction.
We have,

== 31/7
Then,
= (31/7) × reciprocal of 7

Q4.  Find 
(i)

(ii)  

(iii)  

(iv)  

(v)  

(vi)  

(vii)  

(viii)  

Ans: 
(i) We have,
= (2/5) × reciprocal of ½
= (2/5) × (2/1)
 
(ii) We have,
= (4/9) × reciprocal of (2/3)
= (4/9) × (3/2)
 
(iii) We have,
= (3/7) × reciprocal of (8/7)
= (3/7) × (7/8)
 
(iv) First convert the mixed fraction into improper fraction.
We have,
== 7/3
Then,
= (7/3) × reciprocal of (3/5)
 
(v) First convert the mixed fraction into improper fraction.
We have,
= 3 ½ = 7/2
Then,
= (7/2) × reciprocal of (8/3)
= (7/2) × (3/8)

 
(vi) First convert the mixed fraction into improper fraction.
We have,
= 1 ½ = 3/2
Then,
= (2/5) × reciprocal of (3/2)

= (2/5) × (2/3)

(vii) First convert the mixed fraction into improper fraction.
We have,
== 16/5

== 5/3Then,= (16/5) × reciprocal of (5/3)= (16/5) × (3/5)

 
(viii) First convert the mixed fraction into improper fraction.
We have,
== 11/5

== 6/5
Then,
= (11/5) × reciprocal of (6/5)
= (11/5) × (5/6)
 

Exercise 2.4 

Q1. 
Find: 
(i) 0.2 x 6 
(ii) 8 x 4.6 
(iii) 2.71 x 5 
(iv) 20.1 x 4 
(v) 0.05 x 7 
(vi) 211.02 x 4 
(vii) 2 x 0.86 

Ans: 
(i) 0.2 x 6 = 1.2
(ii) 8 x 4.6 = 36.8
(iii) 2.71 x 5 = 13.55
(iv) 20.1 x 4 = 80.4
(v) 0.05 x 7 = 0.35
(vi) 211.02 x 4 = 844.08
(vii) 2 x 0.86 = 1.72

Q2. Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm. 

Ans: Given: Length of rectangle = 5.7 cm and
Breadth of rectangle = 3 cm
Area of rectangle = Length x Breadth = 5.7 x 3 = 17.1 cm²
Thus, the area of rectangle is 17.1 cm².

Q3. Find: 
(i) 1.3 x 10 
(ii) 36.8 x 10 
(iii) 153.7 x 10 
(iv) 168.07 x 10 
(v) 31.1 x 100 
(vi) 156.1 x 100
(vii) 3.62 x 100 
(viii) 43.07 x 100 
(ix) 0.5 x 10
(x) 0.08 x 10 
(xi) 0.9 x 100 
(xii) 0.03 x 1000 

Ans:
(i) 1.3 x 10 = 13.0
(ii) 36.8 x 10 = 368.0
(iii) 153.7 x 10 = 1537.0
(iv) 168.07 x 10 = 1680.7
(v) 31.1 x 100 = 3110.0
(vi) 156.1 x 100 = 15610.0
(vii) 3.62 x 100 = 362.0
(viii) 43.07 x 100 = 4307.0
(ix) 0.5 x 10 = 5.0
(x) 0.08 x 10 = 0.80
(xi) 0.9 x 100 = 90.0
(xii) 0.03 x 1000 = 30.0

Q4. A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol? 

Ans: ∵ In one litre, a two-wheeler covers a distance = 55.3 km
∴ In 10 litres, a two- wheeler covers a distance = 55.3x10 = 553.0 km
Thus, 553 km distance will be covered by it in 10 litres of petrol.

Q5. Find: 
(i) 2.5 x 0.3 
(ii) 0.1 x 51.7 
(iii) 0.2 x 316.8 
(iv) 1.3 x 3.1 
(v) 0.5 x 0.05 
(vi) 11.2 x 0.15 
(vii) 1.07 x 0.02 
(viii) 10.05 x 1.05 
(ix) 101.01 x 0.01 
(x) 100.01 x 1.1 

Ans: 
(i) 2.5 x 0.3 = 0.75
(ii)  0.1 x 51.7 = 5.17
(iii) 0.2 x 316.8 = 63.36
(iv) 1.3 x 3.1=4.03
(v) 0.5 x 0.05 =0.025
(vi) 11.2 x 0.15 = 1.680
(vii) 1.07 x 0.02 = 0.0214
(viii) 10.05 x 1.05 = 10.5525
(ix) 101.01 x 0.01 = 1.0101
(x) 100.01 x1.1=110.11

Exercise 2.5 

Q1. Find: 
(i) 0.4 ÷ 2
(ii) 0.35 ÷ 5
(iii) 2.48 ÷ 4
(iv) 65.4 ÷ 6
(v)  651.2 ÷ 4
(v)  14.49 ÷ 7
(vii) 3.96 ÷ 4
(viii) 0.80 ÷ 5

Ans: 
(i) 
(ii)
(iii)
(iv)
(v)
(vi)  
(vii) 
(viii)

Q2. Find: 
(i) 4.8 ÷ 10
(ii) 52.5 ÷ 10
(iii) 0.7 ÷ 10
(iv) 33.1 ÷ 10
(v) 272.23 ÷ 10
(vi) 0.56 ÷ 10
(vii) 3.97 ÷ 10

Ans: 
(i)
(ii) 
(iii)
(iv) 
(v)
(vi)
(vii)

Q3. Find: 
(i) 2.7 ÷ 100
(ii)  0.3 ÷ 100
(iii) 0.78 ÷ 100
(iv) 432.6 ÷ 100
(v) 23.6 ÷ 100
(vi) 98.53 ÷ 100

Ans: 
(i)
(ii) 
(iii)
(iv) 
(v)  
(vi)

Q4. Find: 
(i) 7.9 ÷ 1000 
(ii)  26.3 ÷ 1000 
(iii) 38.53 ÷ 1000
(iv) 128.9 ÷ 1000 
(v)  0.5 ÷ 1000

Ans: 
(i)
(ii) 
(iii) 
(iv)
(v)

Q5. Find: 
(i) 7÷ 3.5 
(ii)  36 ÷ 0.2 
(iii)  3.25  ÷  0.5
(iv)  30.94 ÷ 0.7 
(v)  0.5 ÷ 0.25 
(vi)  7.75  ÷  0.25
(vii) 76.5 ÷ 0.15 
(viii)  37.8 ÷1.4 
(ix)  2.73  ÷  1.3

Ans: 
(i)
(ii)
(iii)
(iv)  
(v)  
(vi)
(vii)  
(viii)  
(ix) 

Q6. A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it cover in one litre petrol?

Ans:
∵ In 2.4 litres of petrol, distance covered by the vehicle = 43.2 km
∴ In 1 litre of petrol, distance covered by the vehicle = 43.2 ÷2.4

Thus, it covered 18 km distance in one litre of petrol.

Deleted Exercise 

Q1: Which is greater: 
(i) 0.5 or 0.05 
(ii) 0.7or 0.5 
(iii) 7 or 0.7
(iv) 1.37 or 1.49 
(v)  3.03 or 2.30 
(vi) 0.8 or 0.88
Ans: 
(i)  0.5 > 0.05    
(ii)  0.7 > 0.5
(iii) 7 >0.7
(iv) 1.37 <1.49
(v)  2.03 < 2.30
(vi) 0.8 <0.88

Q2: Express as rupees using decimals: 
(i) 7 paise 
(ii) 7 rupees 7 paise 
(iii) 77 rupees 77 paise 
(iv) 50 paise 
(v) 235 paise 
Ans: We know that,
∵
∴
(i)
(ii)
(iii)
(iv)
(v)

Q3:
 (i) Express 5 cm in meter and kilometer. 
(ii) Express 35 mm in cm, m and km. 
Ans: 
(i) Express 5 cm in meter and kilometer.
We know that, 100 cm = 1 meter
∴ 1 cm = 1/100 meter
⇒ 5 cm = 5/100 = 0.05 meter
Now,
∴ 1000 meters = 1 kilometers
∴ 1 meters = 1/1000 kilometer
⇒ 0.05 meter = 0.05/1000 = 0.00005 kilometer
(ii) Express 35 mm in cm, m and km.
∵ 10 mm = 1 cm
∴ 1 mm  = 1/10 cm
⇒ 35 mm = 35/100 = 3.5 cm
Now, ∵ 100 cm = 1 meter
∴ 1 cm = 1/100 meter
⇒ 3.5 cm  = 3.5/100 = 0.035 meter
Again,
∵ 1000 meters = 1 kilometers
∴ 1 meter = 1/1000 kilometers
⇒ 0.035 meter = 0.035/1000 = 0.000035 kilometer

Q4:  Express in kg. 
(i) 200 g 
(ii) 3470 g 
(iii) 4 kg 8 g 
Ans: 
Let us consider, 1000 g = 1 kg

⇒
(i) 
(ii)
(iii) 4kg 8 g = = 5 kg + 0.008 kg = 4.005 kg

Q5: Write the following decimal numbers in the expanded form: 
(i) 20.03 
(ii) 2.03 
(iii) 200.03 
(iv) 2.034 
Ans: 
(i)
(ii)
(iii)
(iv)

Q6: Write the place value of 2 in the following decimal numbers: 
(i) 2.56 
(ii) 21.37 
(iii) 10.25 
(iv) 9.42
(v) 63.352
Ans: 
(i) Place value of 2 in 2.56 = 2 x 1 = 2 ones
(ii)  Place value of 2 in 21.37 = 2 x 10 = 2 tens
(iii) Place value of 2 in 10.25
(iv) Place value of 2 in 9.42
(v) Place value of 2 in 63.352 

Q7: Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much? 

Ans: 

Distance travelled by Dinesh when he went from place A to place B = 7.5 km and from place B to C = 12.7 km.
Total distance covered by Dinesh = AB + BC
= 7.5 + 12.7
= 20.2 km
Total distance covered by Ayub = AD + DC
= 9.3 + 11.8
= 21.1 km
On comparing the total distance of Ayub and Dinesh,
21.1 km > 20.2 km
Therefore, Ayub covered more distance by 21.1 – 20.2 = 0.9 km = 900 m
∴Ayub travelled 0.9 km more than Dinesh.

Q8: Shyam bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits? 
Ans: From the question, it is given that, Fruits bought by Shyama = 5 kg 300 g

= 5 kg + (300/1000) kg
= 5 kg + 0.3 kg
= 5.3 kg
Fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g
= (4 + (800/1000)) + (4 + (150/1000))
= (4 + 0.8) kg + (4 + .150) kg
= 4.8 kg + 4.150kg
= 8.950 kg
So, Sarala bought more fruits.

Q9: How much less is 28 km than 42.6 km? 
Ans: 
We have to find the difference between 42.6 km and 28 km.
Difference = 42.6 – 28.0 = 14.6 km
Therefore,14.6 km less is 28 km than 42.6 km.