NCERT Solutions for Class 7 Maths, Chapter 2 - Fractions & Decimals are available here. Students can also download PDF of this chapter. This is chapters is one of the most important chapters of Class 7 Maths NCERT textbook.

**NCERT Solutions for Class 7 Maths: Chapter 2 - Fractions & Decimals**

NCERT Solutions for Class 7 Maths, Chapter 2 - Fractions & Decimals are available here. Students can also download PDF of this chapter. This is chapters is one of the most important chapters of Class 7 Maths NCERT textbook.

**NCERT Solutions for Class 7 Maths: Chapter 2 - Fractions & Decimals **

**EXERCISE 2.1**

**1. Solve:**

**Solution1:**

**2. Arrange the following in descending order:**

(i) (2/9), (2/3), (8/21)

(ii) (1/5), (3/7), (7/10)

**Solution****2:**

**(i) **(2/9), (2/3), (8/21)

First we convert each fraction into like fraction.

2/9 = (2 x 7)/(9 x 7) = 14/63.

2/3 = (2 x 21)/(3 x 21) = 42/63.

8/21 = (8 x 2)/(21 x 3) = 24/63.

If denominators are same, then, greater the numerator, greater will be the fraction.

Here, 42 > 24 > 14, So (2/3) > (8/21) > (2/9).

**(ii)** (1/5), (3/7), (7/10)

First we convert each fraction into like fraction.

1/5 = (1 x 14)/(5 x 14) = 14/70.

3/7 = (3 x 10)/(7 x 14) = 30/70.

7/10 = (7 x 7)/(10 x 7) = 49/70.

Now, 49 > 30 > 14

So, (7/10) > (3/7) > (1/5).

**3. 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?**

**Solution 3:**

Sum of 1^{st} column = (4/11) + (3/11) + (8/11) = (15/11)

Sum of 1^{st} row = (4/11) + (9/11) + (2/11) = (15/11)

Sum of a diagonal = (4/11) + (5/11) + (1/11) = (15/11)

Similarly sum of other rows, columns and diagonals = (15/11)

As, the sum of the numbers in each row, in each column and along the diagonals is the same, so this is magic square.

**Solution 4:**

**5. Find the perimeters of (i) Δ ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?**

**Solution 5:**

To compare the perimeters of two triangles, first, we need to convert them into like fractions.

Perimeter of triangle ABE = 177/20 = (177 x 3)/(20 x 3) = 531/60

Perimeter of rectangle, BCDE = 47/6 = (47 x 10)/(6 x 10) = 470/60

Clearly, perimeter of of triangle ABE is greater than triangle BCDE.

**Solution 6:**

**7. 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?**

**Solution 7:**

The part of the apple eaten by Ritu = 3/5

The part of the apple eaten by Somu = 1 - (3/5) = (5 - 3)/5 = 2/5.

Difference between Ritu’s and Somu’s share = (3/5) - (2/5) = 1/5.

Ritu had the larger share by 1/5.

**8. 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?**

**Solution 8:**

The time taken by Michael to finish colouring = 7/12

The time taken by Vaibahv to finish colouring = 3/4

Converting these fractions into like fractions, we have 3/4 = (3 x 3)/(4 x3) = 9/12.

Difference between their timings = (9/12) - (7/12) = 2/12 = 1/6 hour. Clearly Viabhav worked longer for 1/6 hours.

**EXERCISE 2.2**

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

**Solution 1: **

(i) In 2 x (1/5), 1/5 represents a figure divided into 5 equal parts and 1 part being shaded and 2 represents 2 such figures. So, 2 x (1/5) is represented by option(d).

(ii) 2 x (1/2) represented by option (b)

(iii) 3 x (2/3) represented by option (a)

(iv) 3 x (1/4) represented by option (c).

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

**Solution 2:**

(i) 3 x (1/5) is represented by option (c).

(ii) 2 x(1/3) is represented by option (a).

(iii) 3 x (3/4) is represented by option (b).

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

**Solution 3:**

**4. 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).

**Solution4:**

(i) There are 12 circles in the box out which we need to shade 1/2 of them i.e., 12 x (1/6) = 6.

(ii) We need to shade (2/3) x 9 = 6 triangle.

(iii) We need to shade (3/5) x 15 = 9 squares.

The respective figures are given below

**5. Find:**

(a) 1/2 of (i) 24 (ii) 46

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

(c) 3/4 of (i) 16 (ii) 36

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

**Solution 5:**

**6. Multiply and express as a mixed fraction:**

**Solution 6:**

**7. Find:**

**Solution7:**

**8. 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?

**Solution8:**

(i) Water drunk by Divya = (2/5) of 5 litres = (2/5) x 5 = 2 litres

(ii) Water drunk by Pratap = 1 - (2/5) = (3/5) litres of the total water

**EXERCISE 2.3**

**1. 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

Solution:

*(i)*

**( a)** (1/4) x (1/4) = 1/16

**(**(1/4) x (3/5) = 3/20

*b*)**(**(1/4) x (4/3) =1/3

*c*)*(ii) *

**( a)** (1/7) x (2/9) = 2/63

**(**(1/7) x (6/5) = 6/35

*b*)**(**(1/7) x (3/10) = 3/70

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

**Solution2:**

**3. Multiply the following fractions:**

**Solution 3:**

**4. Which is greater:**

**Solution4:**

**5. 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.**

**Solution:**

There are three uniform gaps of length 3/4 metre in 4 saplings.

Distance between 1^{st} and 4^{th} sampling = (3/4) x 3 = 9/4 metre

Or 2 ¼ metre

**Solution 6:**

In one day, she read 7/4 hours.

In 6 days, she read (7/4) x 6 = 21/2 hours

**Solution 7:**

In 1 litre, the car can travel a distance of 16 km

In 11/4 litre, the car can travel a distance of (11/4) x 16 = 44 km.

**Solution 8:**

(a) (i) 5/10

(ii) 1/2

(b) (i) 8/15

(ii) 8/15

**EXERCISE 2.4**

**1. Find:**

**Solution 1:**

(i) 12 x (4/3) = 14

(ii) 14 x (6/5) = 84/5

(iii) 8 x (3/7) = 24/7

(iv) 4 x (3/8) = 3/2

(v) 3 ÷ (7/3) = 3 x (3/7) = 9/7

(vi) 5 ÷ (25/7) = 5 x (7/25) = 7/5

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

(i) 3/7

(ii) 5/8

(iii) 9/7

(iv) 6/5

(v) 12/7

(vi) 1/8

(vii) 1/11

**Solution:**

(i) 7/3, improper fraction

(ii) 8/5, improper fraction

(iii) 7/9, proper fraction

(iv) 5/6, proper fraction

(v) 7/12, proper fraction

(vi) 8/1, whole number

(vii) 11/1, whole number

**3. Find:**

**Solution:**

**4. Find:**

**Solution:**

**EXERCISE 2.5**

**1. 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.

**Solution1:**

First we will convert these decimal numbers into equivalent fractions

(i) 0.5 = 5/10 = (5 x10)/(10 x 10) = 50/100 = 5/100 & 0.05 = 5/100

When two fractions have same denominator then fraction with greater numerator is greater. Here, 50 is greater than 5. So, 0.5 > 0.05.

(ii) 0.7 or 0.5

0.7 = 7/10 & 0.5 = 5/10.

Clearly, 0.7 > 0.5.

(iii) 7 or 0.7

7 = (7 x 10)/(1 x 10) = 70/10

Also, 0.7 = 7/10.

So, 7 > 0.7

(iv) 1.37 = 137/100 & 1.49 = 149/100

So, 1.49 > 1.37.

(v) 2.03 or 2.30

2.03 = 203/100 & 2.30 = 230/100

Clearly, 2.03 < 2.30

(vi)

0.8 = 8/10 = [(8x10)/(10 x 10)] =80/100

Also, 0.88 = 88/100.

Clearly, 0.88 > 0.8.

**2. Express as rupees using decimals:**

(i) 7 paise

(ii) 7 rupees 7 paise

(iii) 77 rupees 77 paise

(iv) 50 paise

(v) 235 paise

**Solution2:**

In 1 rupee there are 100 paise.

(i) 7 paise = Rs. (7/100) = Rs. 0.07

(ii) 7 rupees 7 paise = Rs. 7 + Rs. 7/100 = Rs. 7.07

(iii) 77 rupees 77 paise = Rs. 77 + Rs. 77/100 = Rs. 77.77

(iv) 50 paise = Rs. 50/100 = Rs. 0.50

(v) 235 paise = Rs. 235/100 = Rs. 2.35

**3. **(i) Express 5 cm in metre and kilometer

(ii) Express 35 mm in cm, m and km

**Solution 3:**

(i) 5 cm = (5/100) m = 0.05 m

(ii) 5 cm = (5/100000) km = 0.00005 km

**4. Express in kg:**

(i) 200 g

(ii) 3470 g

(iii) 4 kg 8 g

**Solution 4:**

(i) 200 g = (200/1000) kg = 0.2 kg

(ii) 3470 g =( 3470/1000) kg = 3.470 kg

(iii) 4 kg 8 g = 4 kg + (8/1000) kg = 4.008 kg

**5. Write the following decimal numbers in the expanded form:**

(i) 20.03

(ii) 2.03

(iii) 200.03

(iv) 2.034

**Solution:**

(i) 20.03 = (2 × 10) + (0 x 1) + [0 x (1/10)] + [3 x (1/100)]

(ii) 2.03 = (2 x 1) + [0 x (1/10)] + [3 x (1/100)]

(iii) 200.03 = (2 x 100) + (0 x 10) + (0 x 1) + [0 x (1/10)] + [3 x (1/100)]

(iv) 2.034 = (2 x 1) + [0 x (1/10)] + [3 x (1/100)] + [4 x (1/1000)]

**6. 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

**Solution 6.**

(i) Ones

(ii) Tens

(iii) Tenths

(iv) Hundredths

(v) Thousandths

**7. 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?**

**Solution 7:**

Distance traversed by Dinesh = length of path AB + length of path BC = (7.5 + 12.7) km = 20.2 km

Distance traversed by Ayub = length of path AD + length of path DC = (9.3 + 11.8) km = 21.1 km

Difference between distances = 21.1 - 20.2 = 0.9 km

Clearly Ayub travelled more distance by 0.9 km.

**8. 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?**

**Solution 8. **

Net weight of fruits bought by Shyama = 5 kg 300 g + 3 kg 250 g = 8 kg 550 gm or 8.550 kg.

Net weight of fruits bought by Sarla = 4 kg 800 g + 4 kg 150 g = 8 kg 950 gm or 8.950 kg.

Clearly, Sarla bought more fruits.

**9. How much less is 28 km than 42.6 km?**

**Solution 9:**

We can find it by subtracting 28 km &42.6 km.

42.6 - 28.0 = 14.6

So, 28 km is 14.6 km less than 42.6 km.

**EXERCISE 2.6**

**1. 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

**Solution:**

(i) 0.2 × 6 = (2/10) x 6 = 12/10 1.2

(ii) 8 × 4.6 = 8 x (46/10) = 368/10 = 36.8

(iii) 2.71 × 5 = (271/100) x 5 = 1355/100 = 13.55

(iv) 20.1 × 4 = (201/10) x 4 = 804/10 = 80.4

(v) 0.05 × 7 = (5/100) x 7 = 35/100 = 0.35

(vi) 211.02 × 4 = (21102/100) x 4 = (84408/100) = 844.08

(vii) 2 × 0.86 = 2 x (86/100) = 172/100 = 1.72

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

**Solution:**

Given,

Length = 5.7

Breadth = 3 cm

Area of the rectangle = 5.7 x 3 = 17.1 cm^{2}

**3. 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

**Solution3:**

(i) 1.3 × 10 = 13

(ii) 36.8 × 10 = 368

(iii) 153.7 × 10 = 1537

(iv) 168.07 × 10 = 16807

(v) 31.1 × 100 = 311

(vi) 156.1 × 100 = 15610

(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

**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 4:**

Distance covered by the two-wheeler in 1 litre= 55.3

Distance covered by the two-wheeler in 10 litres = 55.3 x 10 = 553 km.

**5. 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

**Solution:**

(i) 2.5 × 0.3 = (25/10) x (3/10) = 0.75

(ii) 0.1 × 51.7 = (1/10) x (517/10) = (517/100) = 5.17

(iii) 0.2 × 316.8 = (2/10) x (3168/10) = 6336/100 = 63.36

(iv) 1.3 × 3.1 = (13/10) x (31/10) = 403/100 = 4.03

(v) 0.5 × 0.05 = (5/10) x (5/100) = 25/1000 = 0.025

(vi) 11.2 × 0.15 = (112/10) x (15/100) = (1680/10000) = 1.680

(vii) 1.07 × 0.02 = (107/100) x (2/100) = 214/100000 = 0.0214

(viii) 10.05 × 1.05 = (1005/100) x (105/100) = 105525/10000 = 10.5525

(ix) 101.01 × 0.01 = (10101/100) x (1/100) = 10101/10000 = 1.0101

(x) 100.01 × 1.1 = (10001/100) x (11.10) = 110011/1000 = 110.011.

**EXERCISE 2.7**

**1. 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

**Solution1:**

(i) 0.4 ÷ 2 = (4/10) x (1/2) = 0.2

(ii) 0.35 ÷ 5 = (35/100) x (1/5) = 7/100 = 0.07

(iii) 2.48 ÷ 4 = (248/100) x (1/4) = 62/100 = 0.62

(iv) 65.4 ÷ 6 = (654/10) x (1/6) = 109/10 = 10.9

(v) 651.2 ÷ 4 = (6512/10) x (1/4) = (1628/10) = 162.8

(vi) 14.49 ÷ 7 = (1449/100) x (1/7) = (207/100) = 2.07

(vii) 3.96 ÷ 4 = (396/100) x (1/4) =(99/100) = 0.99

(viii) 0.80 ÷ 5 =(80/100) ÷ 5 = (80/100) x (1/5) = (16/100) = 0.16

**2. 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

**Solution 2.**

(i) 4.8 ÷ 10 = 0.48

(ii) 52.5 ÷ 10 = 5.25

(iii) 0.7 ÷ 10 = 0.07

(iv) 33.1 ÷ 10 = 3.31

(v) 272.23 ÷ 10 = 27.223

(vi) 0.56 ÷ 10 = 0.056

(vii) 3.97 ÷10 = 0.0397

**3. 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

**Solution 3.**

(i) 2.7 ÷ 100 = 0.027

(ii) 0.3 ÷ 100 = 0.003

(iii) 0.78 ÷ 100 = 0.0078

(iv) 432.6 ÷ 100 =4.326

(v) 23.6 ÷100 = 0.236

(vi) 98.53 ÷ 100 = 0.9853

**4. Find:**

(i) 7.9 ÷ 1000

(ii) 26.3 ÷ 1000

(iii) 38.53 ÷ 1000

(iv) 128.9 ÷ 1000

(v) 0.5 ÷ 1000

**Solution:**

(i) 7.9 ÷ 1000 = 0.0079.

(ii) 26.3 ÷ 1000 = 0.0263.

(iii) 38.53 ÷ 1000 0= 0.03853.

(iv) 128.9 ÷ 1000 = 0.1289.

(v) 0.5 ÷ 1000 = 0.0005.

**5. 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 ÷ 0.1536

(viii) 37.8 ÷ 1.4

(ix) 2.73 ÷ 1.3

**Solution:**

(i) 7 ÷ 3.5 = 7 ÷ (35/10) = 7 x (10/35) = 2.

(ii) 36 ÷ 0.2 = 36 ÷ (2/10) = 36 x (10/2) = 180.

(iii) 3.25 ÷ 0.5 = (325/100) ÷ (05/10) = (325/100) x (10/5) = 65/10 = 6.5

(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 = (378/10) ÷ (14/10) = (378/10) x (10/14) = 27.

(ix) 2.73 ÷ 1.3 = (273/100) ÷ (13/10) = (273/100) x (10/13) = 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 6:**

Distance covered by the vehicle in 2.4 litres = 43.3

Distance covered by the vehicle in in 1 litre = 43.3/2.4 = 18.

So, the vehicle will cover 18 kilometer is one liter.