NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

NCERT
SOLUTIONS
CLASS - 7TH

aglase .co

Book : Mathematics Ncert Solutions | Chapter-2 Maths

Class : 7th
Subject : Maths
Chapter : 2
Chapter Name : Fractions and Decimals

Exercise 2.1

Q1 Solve:
3
(i) 2 − 5
7
(ii) 4 + 8
3
(iii) 5 + 27
9 4
(iv) 11 − 15
7
(v) 10 + 25 + 32
(vi) 2 23 + 3 12
5
(vii) 8 12 − 3 8

Answer. (i) 2 − 35 =
2×5
5
− 35 = 10−3 5
= 75
7 4×8 7 (4×8)+7
(ii) 4 + 8 = 8 + 8 = 8
= 398
= 4 78
3×7 2×5 21+10 31
(iii) 35 + 27 = 5×7 + 7×5 = 35 = 35
9
(iv) 11 − 15 4 9×15
= 11×15 4×11
− 15×11 = 135−44
165
= 16591

7 3
(v) 10 + 25 + 2 = 10 + 2×2
7
5×2
+ 3×5
2×5
= 7+4+15
10
= 26
10
= 13
5
= 2 35
8 7 8×2 7×3 16+21 37
(vi) 2 23 + 3 12 = 3 + 2 = 3×2 + 2×3 = 6 = 6 = 6 16
5 17 29 17×4 29 68−29 39 7
(vii) 8 12 − 3 8 = 2 − 8 = 2×4 − 8 = 8 = 8 = 4 8

Page : 31 , Block Name : Exercise 2.1

Q2 Arrange the following in descending order:
8
(i) 29 , 23 , 21
3 7
(ii) 15 , 7 , 10

Answer. (i) Changing them to like fractions, we obtain
2
9
= 2×7
9×7
= 14
63
2
3
= 3×21 = 42
2×21
63
8 8×3
21
= 21×3 = 2463
Since 42 > 24 > 14 ,
8
∴ 23 > 21 > 29
(ii) Changing them to like fractions, we obtain

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

1
5
= 1×14
5×14
= 14
70
3
7
= 7×10 = 30
3×10
70
7
10
= 10×7 = 49
7×7
70
As 49 > 30 > 14
7
∴ 10 > 37 > 15

Page : 31 , Block Name : Exercise 2.1

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?

4 2 9 15
Answer. Along the �rst row, sum = 11 + 11 + 11 = 11
3 5 7 15
Along the second row, sum = 11 + 11 + 11 = 11
8 1 6 15
Along the third row, sum = 11 + 11 + 11 = 11
4 3 8 15
Along the �rst column, sum = 11 + 11 + 11 = 11
9 1 5 15
Along the second column, sum = 11 + 11 + 11 = 11
2 7 6 15
Along the third column, sum = 11 + 11 + 11 = 11
2 5 8 15
Along the �rst diagonal, sum = 11 + 11 + 11 = 11
Since the sum of the numbers in each row, in each column, and along the diagonals is the same, it is
a magic square.

Page : 31 , Block Name : Exercise 2.1

Q4 A rectangular sheet of paper is 12 12 cm long and 10 23 cm wide. Find its perimeter.

25
Answer. Length = 12 12 cm = 2
cm
32
Breadth = 10 23 cm = 3
cm
Perimeter = 2 × ( Length + Breadth )
= 2 × [ 25 ]=2×[ ] = 2 × [ 75+64 ]
(25×3)+(32×2)
2
+ 32
3 6 6
= 2 × 139
6
= 139
3
= 46 13 cm

Page : 31 , Block Name : Exercise 2.1

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

Q5 Find the perimeters of (i) △ABE (ii) the rectangle BCDE in this �gure. Whose perimeter is
greater?

Answer. (i) Perimeter of △ABE = AB + BE+ EA
= ( 52 + 2 34 + 3 35 ) = ( 52 + 11
4
+ 18
5
)
= ( 5×10
2×10
+ 11×5
4×5
+ 18×4
5×4
)
= 50+55+72
20
= 177
20
= 8 17
20
cm
(ii) Perimeter of rectangle = 2 ( Length + Breadth )
Perimeter of rectangle = 2 [ 11 + 6]7
4

2 [ 11×3
4×3
+ 7×2
6×2
] = 2 [ 33+14
12
]
2 × 47
12
= 476
= 7 56 cm
177
Perimeter of △ABE = 20 cm
Changing them to like fractions, we obtain
177
20
= 177×3
20×3
= 531
60
43 43×10 430
6
= 6×10 = 60
As 531 > 430
∴ 177
20
> 436
Perimeter ( △ABE ) > Perimeter (BCDE)

Page : 31 , Block Name : Exercise 2.1

3
Q6 Salil wants to put a picture in a frame. The picture is 7 5 cm wide. To �t in the frame the picture
3
cannot be more than 7 10 cm wide. How much should the picture be trimmed?

3 38
Answer. Width of picture = 7 5 = 5
cm
3 73
Required width = = 7 10 = 10 cm
The picture should be trimmed by = = ( 5 − 10 )
38 73

( 38×2
5×2
− 73
10
) = 76−73
10
3
= 10 cm

Page : 31 , Block Name : Exercise 2.1

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?

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

Answer. Part of apple eaten by Ritu = 3/5
Part of apple eaten by Somu = 1 - Part of apple eaten by Ritu
1 − 35 = 25
Therefore. Somu ate 2/5 part of the apple.
Since 3 > 2, Ritu had the larger share.
Difference between the 2 shares = 35 − 25 = 15
Therefore, Ritu's share is larger than the share of Somu by 1/5 .

Page : 32 , Block Name : Exercise 2.1

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

Answer. Time taken by Michael = 7/12 hr
Time taken by Vaibhav =3/4 hr
Converting these fractions into like fractions, we obtain
3 9
4
= 3×3
4×3
= 12
7
And 12
Since 9 > 7
Vaibhav worked longer.
9 7 2
Difference = 12 − 12 = 12 = 16 hour

Page : 32 , Block Name : Exercise 2.1

Exercise 2.2

Q1 Which of the drawings (a) to (d) show :
(i) 2 × 15
(ii) 2 × 12
(iii) 3 × 23
(iv) 3 × 14

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

Answer. (i) 2 × 15 represents addition of 2 �gures, each representing 1 shaded part out of 5 equal
parts. Hence, 2 × 15 is represented by (d).
(ii) 2 × 12 represents addition of 2 �gures, each representing 1 shaded part out of 2 equal parts.
Hence, 2 × 12 is represented by (b).
(iii) 3 × 23 represents addition of 3 �gures, each representing 2 shaded parts out of 3 equal parts.
Hence, 3 × 23 is represented by (a).
(iv) 3 × 14 represents addition of 3 �gures, each representing 1 shaded part out of 4 equal parts.
Hence, 3 × 14 is represented by (c).

Page : 36 , Block Name : Exercise 2.2

Q2 Some pictures (a) to (c) are given below. Tell which of them show:
(i) 3 × 15 = 35
(ii) 2 × 13 = 23
3
(iii) 3 × 4 = 2 14

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

Answer. 3 × 15 represents the addition of 3 �gures, each representing 1 shaded part out of 5 equal
parts and 3/5 represents 3 shaded parts out of 5 equal parts. Hence, 3 × 15 is represented by (c).
(ii) 2 × 13 represents the addition of 2 �gures, each representing 1 shaded part out of 3 equal parts
and 2/3 represents 2 shaded palts out of 3 equal palts. Hence, 2 × 13 is represented by (a).
3
(iii) 3 × 4 represents the addition of 3 �gures, each representing 3 shaded parts out of 4 equal parts
and 2 14 represents 2 fully shaded �gures and one �gure having 1 part as shaded out of 4 equal parts.
3
Hence, 3 × 4 is represented by (b).

Page : 36 , Block Name : Exercise 2.2

Q3 Multiply and reduce to lowest form and convert into a mixed fraction:

3
Answer. (i) 7 × 5 = 21
5
= 4 15
(ii) 4 × 13 = 43 = 1 13
6 5
(iii) 2 × 7 = 12
7
= 17
10
(iv) 5 × 29 = 9
= 1 19
8
(v) 23 × 4 = 3 = 2 23
5
(vi) 2 × 6 = 15
(vii) 11 × 47 = 44 7
= 6 27
(viii) 20 × 45 = 16
13
(ix) 13 × 13 = 3 = 4 13
3
(x) 15 × 5 = 9

Page : 36 , Block Name : Exercise 2.2

Q4 Shade: (i) 1/2 of the circles in box (a) (ii) 2 3 of the triangles in box (b) (iii) 3 5 of the squares in
box (c).

Answer. (i) It can be observed that there are 12 circles in the given box. We have to shade 1/2 of the
circles in it. As 12 × 12 = 6 , therefore, we will shade any 6 circles of it.

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

(ii) It can be observed that there are 9 triangles in the given box. We have to shade 2/3 of the
triangle in it. As 9 × 23 = 6 , therefore, we will shade any 6 triangles of it.

(iii) It can be observed that there are 15 squares in the given box. We have to shade 3/5
of the squares in it. As 35 × 15 = 9 , therefore, we will shade any 9 squares of it.

Page : 37 , Block Name : Exercise 2.2

Q5 Find:
(a) 1/2 of (i) 24 (ii) 36
(b) 2/3 of (i) 18 (ii) 27
(c) 3/4 of (i) 16 (ii) 36
(d) 4 5 of (i) 20 (ii) 35

Answer. (a) (i) 12 × 24 = 12
(ii) 12 × 46 = 23
(b) (i) 23 × 18 = 12
(ii) 23 × 27 = 18
(c) (i) 34 × 16 = 12
3
(ii) 4 × 36 = 27
(d) (i) 45 × 20 = 16
(ii) 45 × 35 = 28

Page : 37 , Block Name : Exercise 2.2

Q6 Multiply and express as a mixed fraction :
(a) 3 × 5 15
3
(b) 5 × 6 4
(c) 7 × 2 14

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(d) 4 × 6 13
(e) 3 14 × 6
(f) 3 25 × 8

78
26
Answer. (a) 3 × 5 15 = 3 × 5 = 5
= 15 35
3 27 135 3
(b) 5 × 6 4 = 5 × 4 = 4 = 33 4
9 63 3
(c) 7 × 2 14 = 7 × 4 = 4 = 15 4
19 76
(d) 4 × 6 13 = 4 × 3 = 3 = 25 13
13 78 39
(e) 3 14 × 6 = 4 × 6 = 4 = 2 = 19 12
17 136
(f) 3 25 × 8 = 5 × 8 = 5 = 27 15

Page : 37 , Block Name : Exercise 2.2

Q7 Find:
(a) 1/2 of (i) 2 34 (ii) 4 29
5
(b)5/8 of (i) 3 6 (ii) 9 23

3 3
Answer. (a) (i) 12 × 2 4 = 12 × 11
4
= 11
8
= 18
38
19
(ii) 12 × 4 29 = 12 × 9 = 9
= 2 19
5 5 5 23 115 19
(b) (i) 8 × 3 6 = 8 × 6 = 48 = 2 48
5 5 29 145
(ii) 8 × 9 23 = 8 × 3 = 24 = 6 24 1

Page : 37 , Block Name : Exercise 2.2

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?

Answer. (i) Water consumed by Vidya = 2/5 of 5 litres
= 25 × 5 = 2
3
(ii) Water consumed by Pratap = 1 − 25 = 5 of the total water.

Page : 37 , Block Name : Exercise 2.2

Exercise 2.3

Q1 Find:

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

Answer. (i) (a) 14 × 14 = 16
1

3 3
(b) 14 × 5 = 20
(c) 14 × 43 = 13
(ii) (a) 17 × 29 = 63
2

6 6
(b) 17 × 5 = 35
3 3
(c) 17 × 10 = 70

Page : 41 , Block Name : Exercise 2.3

Q2 Multiply and reduce to lowest form (if possible) :

Answer. (i) 23 × 2 23 = 23 × 3 =
8 16
9
= 1 79
7
(ii) 27 × 9 = 29
6 9
(iii) 38 × 4 = 16
9 3 27 2
(iv) 5 × 5 = 25 = 1 25
15 5
(v) 13 × 8 = 8
3 33 13
(vi) 11
2
× 10 = 20 = 1 20
48 13
(vii) 45 × 12
7
= 35 = 1 35

Page : 42 , Block Name : Exercise 2.3

Q3 Multiply the following fractions:

Answer. (i) 25 × 5 14 = 25 × 21
4
= 21
10
1
This is an improper fraction and it can be written as a mixed fraction as 2 10
7
(ii) 6 25 × 9 =
32
5
× 79 = 224
45

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

This is an improper fraction and it can be written as a mixed fraction as 4 44
45
3 3 16
(iii) 2 × 5 13 = 2 × 3 = 8
This is a whole number.
5 3 5 17 85
(iv) 6 × 2 7 = 6 × 7 = 42
1
This is an improper fraction and it can be written as a mixed fraction as 2 42
17
(v) 3 25 × 47 = 5
× 47 = 68
35
33
This is an improper fraction and it can be written as a mixed fraction as 1 35
13
3
(vi) 2 5 × 3 = 5
× 3 = 39
5
This is an improper fraction and it can be written as a mixed fraction as 7 45
25
(vii) 3 47 × 35 = 7
× 35 = 15
7
1
This is an improper fraction and it can be written as a mixed fraction as 22 7

Page : 42 , Block Name : Exercise 2.3

Q4 Which is greater:
(i) 2/7 of 3/4 or 3/5 of 5/8
(ii) 1/2 of 6/7 or 2/3 of 3/7

Answer. (i) 27 × 34 = 14
3

3
5
× 58 = 38
Converting these fractions into like fractions,
3 3×4
14
= 14×4 = 12 56
3 3×7 21
8
= 8×7 = 56
Since 21 56
> 12
56
3 3
∴ 8 > 14
5
Therefore, 35 of 8 is greater.
6
(ii) 12 × 7 = 37
2
3
× 37 = 27
Since 3 > 2,
∴ 37 > 27
6
Therefore, 12 of 7 is greater.

Page : 42 , Block Name : Exercise 2.3

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 �rst and the last sapling.

Answer.

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

From the �gure, it can be observed that gaps between 1st and last sapling = 3
Length of 1 gap = 3/4 m
9
Therefore, distance between I and IV sapling = 3 × 34 = 4 = 2 14 m

Page : 42 , Block Name : Exercise 2.3

3
Q6 Lipika reads a book for 1 4 hours everyday. She reads the entire book in 6 days. How many hours
in all were required by her to read the book?

7
Answer. Number of hours Lipika reads the book per day = 1 34 = 4
Number of days = 6

7
Total number of hours required by her to read the book = 4 × 6
= 21
2
= 10 12

Page : 42 , Block Name : Exercise 2.3

3
Q7 A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2 4 litres of petrol.

Answer. Number of kms a car can run per litre petrol = 16 km
Quantity of petrol = 2 34 L = 11
4
L
Number of kms a car can run for 11/4 litre petrol = 11
4
× 16
3
It will cover 44 km distance by using 2 4 litres of petrol.

Page : 42 , Block Name : Exercise 2.3

Q8

Answer.

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Page : 42 , Block Name : Exercise 2.3

Exercise 2.4

Q1 Find:

3
Answer. (i) 12 ÷ 4 = 12 × 43 = 16
5 6 84
(ii) 14 ÷ 6 = 14 × 5 = 5
7 3
(iii) 8 ÷ 3 = 8 × 7 = 24
7
8
(iv) 4 ÷ 3 = 4 × 38 = 32
7 3 9
(v) 3 ÷ 2 13 = 3 ÷ 3 = 3 × 7 = 7
25 7 7
(vi) 5 ÷ 3 47 = 5 ÷ 7 = 5 × 25 = 5

Page : 46 , Block Name : Exercise 2.4

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

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(i) 3/7
(ii) 5/8
(iii) 9/7
(iv) 6/5
(v) 12/7
(vi) 1/8
(vii) 1/11

Answer. A proper fraction is the fraction which has its denominator greater than its numerator
while improper fraction is the fraction which has its numerator greater than its
denominator. Whole numbers are a collection of all positive integers including 0.
(i) 3/7
Reciprocal = 7/3
Therefore, it is an improper fraction.
(ii) 5/8
Reciprocal = 8/5
Therefore, it is an improper fraction.
(iii) 9/7
Reciprocal = 7/9
Therefore, it is a proper fraction.
(iv) 6/5
Reciprocal = 5/6
Therefore, it is a proper fraction.
(v) 12/7
Reciprocal = 7/12
Therefore, it is a proper fraction.
(vi) 1/8
Reciprocal = 8/1
Therefore, it is a whole number.
(vii)1/11
Reciprocal = 11/1
Therefore, it is a whole number.

Page : 46 , Block Name : Exercise 2.4

Q3 Find:
7
(i) 3 ÷ 2
(ii) 49 ÷ 5
6
(iii) 13 ÷ 7
(iv) 4 13 ÷ 3
(v) 3 12 ÷ 4
3
(vi) 4 7 ÷ 7

7 7 7
Answer. (i) 3 ÷ 2 = 3 × 12 = 6
(ii) 49 ÷ 5 = 49 × 15 = 45
4

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6 6 6
(iii) 13 ÷ 7 = 13 × 17 = 91
13
(iv) 4 13 ÷ 3 = 3
÷ 3 = 13
3
× 13 = 13
9
7 7 7
(v) 3 12 ÷ 4 = 2 ÷ 4 = 2 × 14 = 8
3 31 31
(vi) 4 7 ÷ 7 = 7 × 17 = 49

Page : 46 , Block Name : Exercise 2.4

Q4 Find:
(i) 25 ÷ 12
(ii) 49 ÷ 23
3 8
(iii) 7 ÷ 7
3
(iv) 2 13 ÷ 5
8
(v) 3 12 ÷ 3
(vi) 25 ÷ 1 12
(vii) 3 15 ÷ 1 23
(viii) 2 15 ÷ 1 15

Answer. (i) 25 ÷ 12 = 25 × 2 = 45
(ii) 49 ÷ 23 = 49 × 32 = 23
3 8 3 7 3
(iii) 7 ÷ 7 = 7 × 8 = 8
7 7 535
(iv) 2 13 ÷ 35 = 3 ÷ 35 = 3 × 3 = 9
8 7 8 7 3
(v) 3 12 ÷ 3 = 2 ÷ 3 = 2 × 8 = 21 16
2 1 2 3 2 2 4
(vi) 5 ÷ 1 2 = 5 ÷ 2 = 5 × 3 = 15
16 5 16 3 48
(vii) 3 15 ÷ 1 23 = 5 ÷ 3 = 5 × 5 = 25
(viii) 2 15 ÷ 1 15 = 115
÷ 65 = 11
5
× 56 = 11
6

Page : 46 , Block Name : Exercise 2.4

Exercise 2.5

Q1 Which is greater?
(i) 0.5 or 0.05 (ii) 0.7 or 0.5 (iii) 7 or 0.7 (iv) 1.37 or 1.49 (v) 2.03 or 2.30 (vi) 0.8 or 0.88

Answer. (i) Converting these decimal numbers into equivalent fractions,
5 5×10 50 5
0.5 = 10 = 10×10 = 100 and 0.05 = 100
It can be observed that both fractions have the same denominator.
As 50 > 5,
Therefore, 0.5 > 0.05
(ii) Converting these decimal numbers into equivalent fractions,

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7 5
0.7 = 10 and 0.5 = 10
It can be observed that fractions have the same denominator.
As 70 > 0.7
Therefore, 7 > 0.7
(iii) Converting these decimal numbers into equivalent fractions,
7 = 71 = 7×10
1×10
= 70
10
7
and 0.7 = 10
It can be observed that both fractions have the same denominator.
As 70 > 7,
Therefore, 7 > 0.7
(iv) Converting these decimal numbers into equivalent fractions,
1.37 = 137
100
and 1.49 = 149
100
It can be observed that both fractions have the same denominator.
AS 137 < 149,
Therefore, 1.37 < 1.49
(v) Converting these decimal numbers into equivalent fractions,
2.03 = 203
100
and 2.30 = 230
100
It can be observed that both fractions have the same denominator.
As 203 < 230,
Therefore, 2.03 < 2.30
(vi) Converting these decimal numbers into equivalent fractions,
8 8×10 80 88
0.8 = 10 = 10×10 = 100 and 0.88 = 100
It can be observed that both fractions have the same denominator.
As 80 < 88,
Therefore, 0.8 < 0.88

Page : 47 , Block Name : Exercise 2.5

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.

Answer. There are 100 paise in 1 rupee. Therefore, if we want to convert paise into rupees, we have
to divide paise by 100.
7
(i) 7 paise = Rs 100 = Rs 0.07
7
(ii) 7 Rs 7 paise = Rs7 + Rs 100 = Rs 7.07
77
(iii) 77 Rs 77 paise = = Rs 77 + Rs 100 = Rs 77.77
50
(iv) 50 paise = Rs 100 = Rs 0.50
235
(v) 235 paise = 100 rupees = Rs2.35

Page : 47 , Block Name : Exercise 2.5

Q3 (i) Express 5 cm in metre and kilometre (ii) Express 35 mm in cm, m and km

Answer. (i) 5 cm
5
5cm = 100 m = 0.05m

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5
5cm = 100000 km = 0.00005km
(ii) 35 mm
35mm = 35
10
cm = 3.5cm
35
35mm = 1000 m = 0.035m
35
35mm = 1000000 km = 0.000035km

Page : 47 , Block Name : Exercise 2.5

Q4 Express in kg: (i) 200 g (ii) 3470 g (iii) 4 kg 8 g

200
Answer. (i) 200 g = 1000 kg = 0.2kg
3470
(ii) 3470 g = 1000 kg = 3.470kg
8
(iii) 4 kg 8 g = 4kg + 1000 kg = 4.008kg

Page : 48 , Block Name : Exercise 2.5

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

1 1
Answer. (i) 20.03 = 2 × 10 + 0 × 1 + 0 × 10 + 3 × 100
1 1
(ii) 2.03 = 2 × 1 + 0 × 10 + 3 × 100
1 1
(iii) 200.03 = 2 × 100 + 0 × 10 + 0 × 1 + 0 × 10 + 3 × 100
1 1 1
(iv) 2.034 = 2 × 1 + 0 × 10 + 3 × 100 + 4 × 1000

Page : 48 , Block Name : Exercise 2.5

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.

Answer. (i) 2.56
Ones
(ii) 21.37
Tens
(iii) 10.25
Tenths
(iv) 9.42
Hundredths
(v) 63.352
Thousandths

Page : 48 , Block Name : Exercise 2.5

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

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11.8 km from D. Who travelled more and by how much?

Answer. Distance travelled by Dinesh = AB + BC = (7.5 + 12.7) km
7.5
+12.7
20.2
Therefore, Dinesh travelled 20.2 km.
Distance travelled by Ayub = AD + DC = (9.3 + 11.8) km
9.3
+11.8
21.1
Therefore, Ayub travelled 21.1 km.
Hence, Ayub travelled more distance.
Difference = (21.1 — 20.2) km
21.1
−20.2
0.9
Therefore. Ayub travelled 0.9 km more than Dinesh

Page : 48 , Block Name : Exercise 2.5

Q8 Shyama 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?

Answer. Total fruits bought by Shyama = 5 kg 300 g + 3 kg 250 g
= 8 kg 550 g
(8 + 1000
550
) kg
= 8.550 kg
Total fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g
= 8 kg 950 g
= (8 + 1000
950
) kg
= 8.90 kg
Sarala bought more fruits.

Page : 48 , Block Name : Exercise 2.5

Q9 How much less is 28 km than 42.6 km?

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42.6
Answer. −28.0
14.6
Therefore, 28 km is 14.6 km less than 42.6 km.

Page : 48 , Block Name : Exercise 2.5

Exercise 2.6

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

2
Answer. (i) 0.2 × 6 = 10 × 6 = 12
10
= 1.2
46 368
(ii) 8 × 4.6 = 8 × 10 = 10
= 36.8
271 1355
(iii) 2.71 × 5 = 100 × 5 = 100 = 13.55
201 804
(iv) 20.1 × 4 = 10 × 4 = 10 = 80.4
5 35
(v) 0.05 × 7 = 100 × 7 = 100 = 0.35
21102 84408
(vi) 211.02 × 4 = 100 × 4 = 100 = 844.08
86 172
(vii) 2 × 0.86 = 2 × 100 = 100 = 1.72

Page : 52 , Block Name : Exercise 2.6

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

Answer. Length = 5.7 cm
Breadth = 3 cm
Area = Length x Breadth
= 5.7 x 3 = 17.1 cm2

Page : 52 , Block Name : Exercise 2.6

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

Answer. We know that when a decimal number is multiplied by 10, 100, 1000, the decimal point
in the product is shifted to the right by as many places as there are zeroes. Therefore,
these products can be calculated as
(i) 1.3 × 10 = 13
(ii) 36.8 × 10 = 368
(iii) 153.7 × 10 = 1537
(iv) 168.07 × 10 = 1680.7
(v) 31.1 × 100 = 3110
(vi) 156.1 × 100 = 15610

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

(vii) 3.62 × 100 = 362
(viii) 43.07 × 100 = 4307
(ix) 0.5 × 10 = 5
(x) 0.08 × 10 = 0.8
(xi) 0.9 × 100 = 90
(xii) 0.03 × 1000 = 30

Page : 52 , Block Name : Exercise 2.6

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?

Answer. Distance covered in 1 litre of petrol = 55.3 km
Distance covered in 10 Iitre of petrol = 10 x 55.3 = 553 km
Therefore, it will cover 553 km distance in 10 litre petrol.

Page : 52 , Block Name : Exercise 2.6

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

253 75
Answer. (i) 2.5 × 0.3 = 10 × 10 = 100 = 0.75
1 517 517
(ii) 0.1 × 51.7 = 10 × 10 = 100 = 5.17
3168
2
(iii) 0.2 × 316.8 = 10 × 10
= 6336
100
= 63.36
13 31 403
(iv) 1.3 × 3.1 = 10 × 10 = 100 = 4.03
5 5 25
(v) 0.5 × 0.05 = 10 × 100 = 1000 = 0.025
15
(vi) 11.2 × 0.15 = 112
10
× 100 = 1680
1000
= 1.680 = 1.68
107 2 214
(vii) 1.07 × 0.02 = 100 × 100 = 10000 = 0.0214
1005 105 105525
(viii) 10.05 × 1.05 = 100 × 100 = 10000 = 10.5525
10101
(ix) 101.01 × 0.01 = 100 × 100 1
= 10101
10000
= 1.0101
100101 11 1101111
(x) 1001.01 × 1.1 = 100 × 10 = 1000 = 1.101.111

Page : 52 , Block Name : Exercise 2.6

Exercise 2.7

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

4 4
Answer. (i) 0.4 ÷ 2 = 10 ÷ 2 = 10 × 12 = 10
2
= 0.2
35 35 7
(ii) 0.35 ÷ 5 = 100 ÷ 5 = 100 × 15 = 100 = 0.07
248 248 62
(iii) 2.48 ÷ 4 = 100 ÷ 4 = 100 × 14 = 100 = 0.62

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

654
(iv) 65.4 ÷ 6 = 10
÷ 6 = 654
10
× 16 = 109
10
= 10.9
6512 6512 1 1628
(v) 651.2 ÷ 4 = 10 ÷ 4 = 10 × 4 = 10 = 162.8
1449 1449 207
(vi) 14.49 ÷ 7 = 100 ÷ 7 = 100 × 17 = 100 = 2.07
396 396 99
(vii) 3.96 ÷ 4 = 100 ÷ 4 = 100 × 14 = 100 = 0.99
80 80 16
(viii) 0.80 ÷ 5 = 100 ÷ 5 = 100 × 15 = 100 = 0.16

Page : 55 , Block Name : Exercise 2.7

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

Answer. We know that when a decimal number is divided by a multiple of 10 only (i.e., 10, 100,
1000, etc.), the decimal point will be shifted to the left by as many places as there are
zeroes. Since here we are dividing by 10, the decimal will shift to the left by 1 place.
(i) 0.48
(ii) 5.25
(iii) 0.07
(iv) 3.31
(v) 27.223
(vi) 0.056
(vii) 0.397

Page : 55 , Block Name : Exercise 2.7

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

Answer. We know that when a decimal number is divided by a multiple of 10 only (i.e., 10, 100,
1000, etc.), the decimal point will be shifted to the left by as many places as there are
zeroes. Since here we are dividing by 1000, the decimal will shift to the left by 2 places.
(i) 0.027
(ii) 0.003
(iii) 0.0078
(iv) 4.326
(v) 0.236
(vi) 0.9853

Page : 55 , Block Name : Exercise 2.7

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

Answer. We know that when a decimal number is divided by a multiple of 10 only (i.e., 10, 100,
1000, etc.), the decimal point will be shifted to the left by as many places as there are
zeroes. Since here we are dividing by 1000, the decimal will shift to the left by 3 places.
(i) 0.0079
(ii) 0.0263

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

(iii) 0.03853
(iv) 0.1289
(v) 0.0005

Page : 55 , Block Name : Exercise 2.7

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

35 10
Answer. (i) 7 ÷ 3.5 = 7 ÷ 10 = 7 × 35 = 2
2 10
(ii) 36 ÷ 0.2 = 36 ÷ 10 = 36 × 2 = 180
325 5 325 10 65
(iii) 3.25 ÷ 0.5 = 100 ÷ 10 = 100 × 5 = 10 = 6.5
3094
(iv) 30.94 ÷ 0.7 = 100
÷ 107
= 3094
100
× 10 7
= 442
10
= 44.2
5 25 5 100
(v) 0.5 ÷ 0.25 = 10 ÷ 100 = 10 × 25 = 2
775 25 775 100
(vi) 7.75 ÷ 0.25 = 100 ÷ 100 = 100 × 25 = 31
765 15 765 100
(vii) 76.5 ÷ 0.15 = 10 ÷ 100 = 10 × 15 = 510
378
(viii) 37.8 ÷ 1.4 = 10 ÷ 14 10
= 378
10
× 1014
= 27
273 13 273 10
(ix) 2.73 ÷ 1.3 = 100 ÷ 10 = 100 × 13 = 21 10
= 2.1

Page : 55 , Block Name : Exercise 2.7

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 of petrol?

Answer. Distance covered in 2.4 litres of petrol = 43.2 km
∴ Distance covered in 1 litre of petrol = = 43.2 ÷ 2.4 = 432
10
÷ 24
10
= 432
10
× 10
24
= 18
Therefore, the vehicle will cover 18 km in 1 litre petrol.

Page : 55 , Block Name : Exercise 2.7

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