NCERT Solutions for Class 7 Maths Chapter 2 – Fractions and Decimals(New Syllabus)

Here are the NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals. From here students can practice online or download these files to practice different types of questions related to this chapter and score maximum marks in their examinations. Students have learned about addition and subtraction of fractions and addition and subtraction of decimals in previous classes.

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1 NCERT Solutions for Class 7 Maths Chapter 2 Exercise 2.1

In this class, students will learn about multiplication and division of fractions and multiplication and division of decimals. Subject experts have prepared these NCERT Solutions for Class 7 Maths for Fractions and Decimals based on the new syllabus to help the students prepare for their exams.

NCERT Solutions for Class 7 Maths Chapter 2  Exercise 2.1

Question 1. Which of the drawings (a) to (d) show:

(i) 2 × (1/5) (ii) 2 × ½ (iii) 3 × (2/3) (iv) 3 × ¼

NCERT Solutions for Class 7 Maths Chapter 2

Solution:-

(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 represents 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).

Question 2. Some pictures (a) to (c) are given below. Tell which of them show:
(i) 3 × (1/5) = (3/5) (ii) 2 × (1/3) = (2/3) (iii) 3 × (3/4) = 2 ¼

NCERT Solutions for Class 7 Maths Chapter 2

Solution:-

(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 parts 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).

Question 3. Multiply and reduce to lowest form and convert into a mixed fraction:

(i) 7 × (3/5)

Solution:-

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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 24

(ii) 4 × (1/3)

Solution:-

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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 25

(iii) 2 × (6/7)

Solution:-

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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 26

(iv) 5 × (2/9)

Solution:-

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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 27

(v) (2/3) × 4

Solution:-

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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 28

(vi) (5/2) × 6

Solution:-

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) 11 × (4/7)

Solution:-

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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 29

(viii) 20 × (4/5)

Solution:-

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) 13 × (1/3)

Solution:-

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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 30

(x) 15 × (3/5)

Solution:-

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

4. Shade:
(i) ½ of the circles in box (a) (b) 2/3 of the triangles in box (b)
(iii) 3/5 of the squares in the box (c)

 NCERT Solutions for Class 7 Maths Chapter 2

Solution:-

(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.

NCERT Solutions for Class 7 Maths Chapter 2

(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.

NCERT Solutions for Class 7 Maths Chapter 2

(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.

NCERT Solutions for Class 7 Maths Chapter 2

5. Find:
(a) ½ of (i) 24 (ii) 46
Solution:-

(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

Solution:-

(i) 18

We have,

= 2/3 × 18

= 2 × 6

= 12

(ii) 27

We have,

= 2/3 × 27

= 2 × 9

= 18

(c) ¾ of (i) 16 (ii) 36

Solution:-

(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

Solution:-

(i) 20

We have,

= 4/5 × 20

= 4 × 4

= 16

(ii) 35

We have,

= 4/5 × 35

= 4 × 7

= 28

6. Multiply and express as a mixed fraction:

(a) 3 × NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 35

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 36

= 26/5

Now,

= 3 × (26/5)

= 78/5

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 37

(b) 5 × 6 ¾

Solution:-

First convert the given mixed fraction into improper fraction.

= 6 ¾ = 27/4

Now,

= 5 × (27/4)

= 135/4

= 33 ¾

(c) 7 × 2 ¼

Solution:-

First convert the given mixed fraction into improper fraction.

= 2 ¼ = 9/4

Now,

= 7 × (9/4)

= 63/4

= 15 ¾

(d) 4 × NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 38

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 39

= 19/3

Now,

= 4 × (19/3)

= 76/3

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 40

(e) 3 ¼ × 6

Solution:-

First convert the given mixed fraction into improper fraction.

= 3 ¼ = 13/4

Now,

= (13/4) × 6

= (13/2) × 3

= 39/2

= 19 ½

(f) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 41

 × 8

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 42

= 17/5

Now,

= (17/5) × 8

= 136/5

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 43

7. Find:

(a) ½ of (i) 2 ¾ (ii) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 44

Solution:-

(i) 2 ¾

First convert the given mixed fraction into 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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 45

(ii)
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 46

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 47

= 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

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 48

(b) 5/8 of (i) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 49

 (ii) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 50

Solution:-

(i)
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 51

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 52

= 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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 53

(ii)
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 54

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 55

= 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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 56

8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 liters 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

Solution:-

(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.

NCERT Solutions for Class 7 Maths Chapter 2 

 EXERCISE 2.2

Question 1. Find:
(i) ¼ of (a) ¼ (b) 3/5 (c) 4/3
Solution:-

(a) ¼

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

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)

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 (b) 6/5 (c) 3/10
Solution:-

(a) 2/9

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

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

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):
(i) (2/3) × NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 57
Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 58

= 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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 59

(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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 60

(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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 61

(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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 62

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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 63

(ii) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 64

 × (7/9)

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 65

= 32/5

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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 66

(iii) (3/2) × NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 67

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 68

= 16/3

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

(iv) (5/6) × NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 69

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 70

= 17/7

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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 71

(v) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 72

 × (4/7)

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 73

= 17/5

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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 74

(vi)NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 75

 × 3

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 76

= 13/5

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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 77

(vi) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 78

 × (3/5)

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 79

= 25/7

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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 80

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,

(3/14) < (3/8)

(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],

Clearly,

(3/7) > (2/7)

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 (5/20)

(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

NCERT Solutions for Class 7 Maths Chapter 2

EXERCISE 2.3

Question 1. Find:
(i) 12 ÷ ¾
Solution:-

We have,

= 12 × reciprocal of ¾

= 12 × (4/3)

= 4 × 4

= 16

(ii) 14 ÷ (5/6)

Solution:-

We have,

= 14 × reciprocal of (5/6)

= 14 × (6/5)

= 84/5

(iii) 8 ÷ (7/3)

Solution:-

We have,

= 8 × reciprocal of (7/3)

= 8 × (3/7)

= (24/7)

(iv) 4 ÷ (8/3)

Solution:-

We have,

= 4 × reciprocal of (8/3)

= 4 × (3/8)

= 1 × (3/2)

= 3/2

(v) 3 ÷ NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 81

Solution:-

While dividing a whole number by a mixed fraction, first convert the mixed fraction into improper fraction

We have,

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 82

= 7/3

Then,

= 3 ÷ (7/3)

= 3 × reciprocal of (7/3)

= 3 × (3/7)

= 9/7

(vi) 5 ÷ NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 83

Solution:-

While dividing a whole number by a mixed fraction, first convert the mixed fraction into improper fraction

We have,

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 84

= 25/7

Then,

= 5 ÷ (25/7)

= 5 × reciprocal of (25/7)

= 5 × (7/25)

= 1 × (7/5)

= 7/5

Question 2. Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.
(i) 3/7
Solution:-

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

Solution:-

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

Solution:-

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

Solution:-

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

Solution:-

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

Solution:-

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

Solution:-

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.

3. Find:
(i) (7/3) ÷ 2
Solution:-

We have,

= (7/3) × reciprocal of 2

= (7/3) × (1/2)

= (7 × 1) / (3 × 2)

= 7/6

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 85

(ii) (4/9) ÷ 5

Solution:-

We have,

= (4/9) × reciprocal of 5

= (4/9) × (1/5)

= (4 × 1) / (9 × 5)

= 4/45

(iii) (6/13) ÷ 7

Solution:-

We have,

= (6/13) × reciprocal of 7

= (6/13) × (1/7)

= (6 × 1) / (13 × 7)

= 6/91

(iv) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 86

 ÷ 3

Solution:-

First convert the mixed fraction into improper fraction.

We have,

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 87

= 13/3

Then,

= (13/3) × reciprocal of 3

= (13/3) × (1/3)

= (13 × 1) / (3 × 3)

= 13/9

(iv) 3 ½ ÷ 4

Solution:-

First convert the mixed fraction into improper fraction.

We have,

= 3 ½ = 7/2

Then,

= (7/2) × reciprocal of 4

= (7/2) × (1/4)

= (7 × 1) / (2 × 4)

= 7/8

(iv) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 88

 ÷ 7

Solution:-

First convert the mixed fraction into improper fraction.

We have,

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 89

= 31/7

Then,

= (31/7) × reciprocal of 7

= (31/7) × (1/7)

= (31 × 1) / (7 × 7)

= 31/49

4. Find:

(i) (2/5) ÷ (½)

Solution:-

We have,

= (2/5) × reciprocal of ½

= (2/5) × (2/1)

= (2 × 2) / (5 × 1)

= 4/5

(ii) (4/9) ÷ (2/3)

Solution:-

We have,

= (4/9) × reciprocal of (2/3)

= (4/9) × (3/2)

= (4 × 3) / (9 × 2)

= (2 × 1) / (3 × 1)

= 2/3

(iii) (3/7) ÷ (8/7)

Solution:-

We have,

= (3/7) × reciprocal of (8/7)

= (3/7) × (7/8)

= (3 × 7) / (7 × 8)

= (3 × 1) / (1 × 8)

= 3/8

(iv) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 90

 ÷ (3/5)

Solution:-

First convert the mixed fraction into improper fraction.

We have,

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 91

= 7/3

Then,

= (7/3) × reciprocal of (3/5)

= (7/3) × (5/3)

= (7 × 5) / (3 × 3)

= 35/9

(v) 3 ½ ÷ (8/3)

Solution:-

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)

= (7 × 3) / (2 × 8)

= 21/16

(vi) (2/5) ÷ 1 ½

Solution:-

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)

= (2 × 2) / (5 × 3)

= 4/15

(vii) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 92

 ÷ NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 93

Solution:-

First convert the mixed fraction into improper fraction.

We have,

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 94

= 16/5

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 95

= 5/3

Then,

= (16/5) × reciprocal of (5/3)

= (16/5) × (3/5)

= (16 × 3) / (5 × 5)

= 48/25

(viii) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 96

 ÷ NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 97

Solution:-

First convert the mixed fraction into improper fraction.

We have,

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 98

= 11/5

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 99

= 6/5

Then,

= (11/5) × reciprocal of (6/5)

= (11/5) × (5/6)

= (11 × 5) / (5 × 6)

= (11 × 1) / (1 × 6)

= 11/6

NCERT Solutions for Class 7 Maths Chapter 2 

EXERCISE 2.4

Question 1 Find:

(i) 0.2 × 6

Solution:-

We have,

= (2/10) × 6

= (12/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 1.2

(ii) 8 × 4.6

Solution:-

We have,

= (8) × (46/10)

= (368/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 36.8

(iii) 2.71 × 5

Solution:-

We have,

= (271/100) × 5

= (1355/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 13.55

(iv) 20.1 × 4

Solution:-

We have,

= (201/10) × 4

= (804/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 80.4

(v) 0.05 × 7

Solution:-

We have,

= (5/100) × 7

= (35/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.35

(vi) 211.02 × 4

Solution:-

We have,

= (21102/100) × 4

= (84408/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 844.08

(vii) 2 × 0.86

Solution:-

We have,

= (2) × (86/100)

= (172/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 1.72

2. Find the area of rectangle whose length is 5.7cm and breadth is 3 cm.
Solution:- 

length = 5.7 cm

breadth = 3 cm

NCERT Solutions for Class 7 Maths Chapter 2 – Fractions and Decimals

Question 3. Find:

(i) 1.3 × 10

Solution:-

On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 1.3 × 10 = 13

(ii) 36.8 × 10

Solution:-

On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 36.8 × 10 = 368

(iii) 153.7 × 10

Solution:-

On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 153.7 × 10 = 1537

(iv) 168.07 × 10

Solution:-

On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 168.07 × 10 = 1680.7

(v) 31.1 × 100

Solution:-

On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 31.1 × 100 = 3110

(vi) 156.1 × 100

Solution:-

On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 156.1 × 100 = 15610

(vii) 3.62 × 100

Solution:-

On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 3.62 × 100 = 362

(viii) 43.07 × 100

Solution:-

On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 43.07 × 100 = 4307

(ix) 0.5 × 10

Solution:-

On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 0.5 × 10 = 5

(x) 0.08 × 10

Solution:-

On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 0.08 × 10 = 0.8

(xi) 0.9 × 100

Solution:-

On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 0.9 × 100 = 90

(xii) 0.03 × 1000

Solution:-

On multiplying a decimal by 1000, the decimal point is shifted to the right by three places.

We have,

= 0.03 × 1000 = 30

Question 4. 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?

Solution:-

From the question, it is given that,

Distance covered by two-wheeler in 1 litre of petrol = 55.3 km

Then,

Distance covered by two wheeler in 10L of petrol = (10 × 55.3)

= 553 km

∴ The two-wheeler covers a distance of 553 km in 10L of petrol.

Question 5. Find:

(i) 2.5 × 0.3

Solution:-

We have,

= (25/10) × (3/10)

= (75/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.75

(ii) 0.1 × 51.7

Solution:-

We have,

= (1/10) × (517/10)

= (517/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 5.17

(iii) 0.2 × 316.8

Solution:-

We have,

= (2/10) × (3168/10)

= (6336/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 63.36

(iv) 1.3 × 3.1

Solution:-

We have,

= (13/10) × (31/10)

= (403/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 4.03

(v) 0.5 × 0.05

Solution:-

We have,

= (5/10) × (5/100)

= (25/1000)

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

Then,

= 0.025

(vi) 11.2 × 0.15

Solution:-

We have,

= (112/10) × (15/100)

= (1680/1000)

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

Then,

= 1.680

(vii) 1.07 × 0.02

Solution:-

We have,

= (107/100) × (2/100)

= (214/10000)

On dividing a decimal by 10000, the decimal point is shifted to the left by four places.

Then,

= 0.0214

(viii) 10.05 × 1.05

Solution:-

We have,

= (1005/100) × (105/100)

= (105525/10000)

On dividing a decimal by 10000, the decimal point is shifted to the left by four places.

Then,

= 10.5525

(ix) 101.01 × 0.01

Solution:-

We have,

= (10101/100) × (1/100)

= (10101/10000)

On dividing a decimal by 10000, the decimal point is shifted to the left by four places.

Then,

= 1.0101

(x) 100.01 × 1.1

Solution:-

We have,

= (10001/100) × (11/10)

= (110011/1000)

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

Then,

= 110.011

NCERT Solutions for Class 7 Maths Chapter 2 – Fractions and Decimals : EXERCISE 2.5

Question 1. Find:

(i) 0.4 ÷ 2

Solution:-

We have,

= (4/10) ÷ 2

Then,

= (4/10) × (1/2)

= (2/10) × (1/1)

= (2/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 0.2

(ii) 0.35 ÷ 5

Solution:-

We have,

= (35/100) ÷ 5

Then,

= (35/100) × (1/5)

= (7/100) × (1/1)

= (7/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.07

(iii) 2.48 ÷ 4

Solution:-

We have,

= (248/100) ÷ 4

Then,

= (248/100) × (1/4)

= (62/100) × (1/1)

= (62/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.62

(iv) 65.4 ÷ 6

Solution:-

We have,

= (654/10) ÷ 6

Then,

= (654/10) × (1/6)

= (109/10) × (1/1)

= (109/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 10.9

(v) 651.2 ÷ 4

Solution:-

We have,

= (6512/10) ÷ 4

Then,

= (6512/10) × (1/4)

= (1628/10) × (1/1)

= (1628/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 162.8

(vi) 14.49 ÷ 7

Solution:-

We have,

= (1449/100) ÷ 7

Then,

= (1449/100) × (1/7)

= (207/100) × (1/1)

= (207/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 2.07

(vii) 3.96 ÷ 4

Solution:-

We have,

= (396/100) ÷ 4

Then,

= (396/100) × (1/4)

= (99/100) × (1/1)

= (99/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.99

(viii) 0.80 ÷ 5

Solution:-

We have,

= (80/100) ÷ 5

Then,

= (80/100) × (1/5)

= (16/100) × (1/1)

= (16/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.16

Question 2. Find:

(i) 4.8 ÷ 10

Solution:-

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 4.8 ÷ 10

= (4.8/10)

= 0.48

(ii) 52.5 ÷ 10

Solution:-

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 52.5 ÷ 10

= (52.5/10)

= 5.25

(iii) 0.7 ÷ 10

Solution:-

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 0.7 ÷ 10

= (0.7/10)

= 0.07

(iv) 33.1 ÷ 10

Solution:-

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 33.1 ÷ 10

= (33.1/10)

= 3.31

(v) 272.23 ÷ 10

Solution:-

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 272.23 ÷ 10

= (272.23/10)

= 27.223

 

(vi) 0.56 ÷ 10

Solution:-

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 0.56 ÷ 10

= (0.56/10)

= 0.056

 

(vii) 3.97 ÷10

Solution:-

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 3.97 ÷ 10

= (3.97/10)

= 0.397

Question 3. Find:

(i) 2.7 ÷ 100

Solution:-

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 2.7 ÷ 100

= (2.7/100)

= 0.027

(ii) 0.3 ÷ 100

Solution:-

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 0.3 ÷ 100

= (0.3/100)

= 0.003

(iii) 0.78 ÷ 100

Solution:-

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 0.78 ÷ 100

= (0.78/100)

= 0.0078

(iv) 432.6 ÷ 100

Solution:-

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 432.6 ÷ 100

= (432.6/100)

= 4.326

 

(v) 23.6 ÷100

Solution:-

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 23.6 ÷ 100

= (23.6/100)

= 0.236

(vi) 98.53 ÷ 100

Solution:-

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 98.53 ÷ 100

= (98.53/100)

= 0.9853

4. Find:

(i) 7.9 ÷ 1000

Solution:-

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 7.9 ÷ 1000

= (7.9/1000)

= 0.0079

 

(ii) 26.3 ÷ 1000

Solution:-

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 26.3 ÷ 1000

= (26.3/1000)

= 0.0263

 

(iii) 38.53 ÷ 1000

Solution:-

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 38.53 ÷ 1000

= (38.53/1000)

= 0.03853

 

(iv) 128.9 ÷ 1000

Solution:-

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 128.9 ÷ 1000

= (128.9/1000)

= 0.1289

 

(v) 0.5 ÷ 1000

Solution:-

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 0.5 ÷ 1000

= (0.5/1000)

= 0.0005

Question 5. Find:

(i) 7 ÷ 3.5

Solution:-

We have,

= 7 ÷ (35/10)

= 7 × (10/35)

= 1 × (10/5)

= 2

(ii) 36 ÷ 0.2

Solution:-

We have,

= 36 ÷ (2/10)

= 36 × (10/2)

= 18 × 10

= 180

(iii) 3.25 ÷ 0.5

Solution:-

We have,

= (325/100) ÷ (5/10)

= (325/100) × (10/5)

= (325 × 10)/ (100 × 5)

= (65 × 1)/ (10 × 1)

= 65/10

= 6.5

 

(iv) 30.94 ÷ 0.7

Solution:-

We have,

= (3094/100) ÷ (7/10)

= (3094/100) × (10/7)

= (3094 × 10)/ (100 × 7)

= (442 × 1)/ (10 × 1)

= 442/10

= 44.2

(v) 0.5 ÷ 0.25

Solution:-

We have,

= (5/10) ÷ (25/100)

= (5/10) × (100/25)

= (5 × 100)/ (10 × 25)

= (1 × 10)/ (1 × 5)

= 10/5

= 2

 

(vi) 7.75 ÷ 0.25

Solution:-

We have,

= (775/100) ÷ (25/100)

= (775/100) × (100/25)

= (775 × 100)/ (100 × 25)

= (155 × 1)/ (1 × 5)

= (31 × 1)/ (1 × 1)

= 31

 

(vii) 76.5 ÷ 0.15

Solution:-

We have,

= (765/10) ÷ (15/100)

= (765/10) × (100/15)

= (765 × 100)/ (10 × 15)

= (51 × 10)/ (1 × 1)

= 510

(viii) 37.8 ÷ 1.4

Solution:-

We have,

= (378/10) ÷ (14/10)

= (378/10) × (10/14)

= (378 × 10)/ (10 × 14)

= (27 × 1)/ (1 × 1)

= 27

(ix) 2.73 ÷ 1.3

Solution:-

We have,

= (273/100) ÷ (13/10)

= (273/100) × (10/13)

= (273 × 10)/ (100 × 13)

= (21 × 1)/ (10 × 1)

= 21/10

= 2.1

6. 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?

Solution:-

From the question, it is given that,

Total distance covered by vehicle in 2.4 litres of petrol = 43.2 km

Then,

Distance covered in 1 litre of petrol = 43.2 ÷ 2.4

= (432/10) ÷ (24/10)

= (432/10) × (10/24)

= (432 × 10)/ (10 × 24)

= (36 × 1)/ (1 × 2)

= (18 × 1)/ (1 × 1)

= 18 km

∴ Total distance covered in 1 liter of petrol is 18 km.

 

 

 

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