These Class 7 Maths Extra Questions Chapter 2 Fractions and Decimals will help students prepare well for the exams.

## Class 7 Maths Chapter 2 Extra Questions Fractions and Decimals

### Class 7 Maths Fractions and Decimals Extra Questions – 2 Marks

Question 1.

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

1. i) \(\frac { 7 }{ 3 }\) × 4

Solution:

\(\frac { 7 }{ 3 }\) × 4 = \(\frac { 28 }{ 3 }\) = 9\(\frac { 1 }{ 3 }\)

ii) 25 × \(\frac { 4 }{ 5 }\)

Solution:

25 × \(\frac { 4 }{ 5 }\) = 5 × 4 = 20

2. i) 16 × \(\frac { 9 }{ 4 }\)

Solution:

16 × \(\frac { 9 }{ 4 }\) = 4 × 9 = 36

ii) 17 × \(\frac { 2 }{ 5 }\)

Solution:

17 × \(\frac { 2 }{ 5 }\) = \(\frac { 34 }{ 5 }\) = 6\(\frac { 4 }{ 5 }\)

3. i) 8 × \(\frac { 2 }{ 5 }\)

Solution:

8 × \(\frac { 2 }{ 5 }\) = \(\frac { 16 }{ 5 }\) = 3\(\frac { 1 }{ 5 }\)

ii) 32 × \(\frac { 9 }{ 8 }\)

Solution:

32 × \(\frac { 9 }{ 8 }\) = 4 × 9 = 36

4. i) 12 × \(\frac { 1 }{ 7 }\)

Solution:

12 × \(\frac { 1 }{ 7 }\) = \(\frac { 12 }{ 7 }\) = 1\(\frac { 5 }{ 7 }\)

ii) 8 × \(\frac { 4 }{ 3 }\)

Solution:

8 × \(\frac { 4 }{ 3 }\) = \(\frac { 32 }{ 3 }\) = 10\(\frac { 2 }{ 3 }\)

Question 2.

How much less is 18 km, then 24.6 km ?

Solution:

6.6 km less is 18 km , then 24.6 km.

Question 3.

Write the place value of 3 in the following decimal numbers

i) 3.65

ii) 32.97

Solution:

i) The place value of 3 in 3.65 is ones

ii) The place value of 3 in 32.97 is Tens.

Question 4.

Express in kg .

i) 350 g

Solution:

350 g = \(\frac { 350 }{ 1000 }\)kg = \(\frac { 35 }{ 100 }\) kg = 0.035 kg

ii) 4120 g

Solution:

4120g = \(\frac { 4120 }{ 1000 }\)kg = \(\frac { 412 }{ 100 }\) kg = 4.12kg

Question 5.

Write the following decimal number is the expanded form.

i) 30.13

Solution:

30.13 = 3 × 10 + 0 × 1 + 1 × \(\frac { 1 }{ 10 }\) + \(\frac { 3 }{ 100 }\)

ii) 321.5

Solution:

321.5 = 3 × 100 + 2 × 10 + 1 × 1 + \(\frac { 5 }{ 10 }\)

Question 6.

Find

i) 0.4 × 100

Solution:

0.4 × 100 = \(\frac { 4 }{ 10 }\) × 100 = 40

ii) 1.3 × 1000

Solution:

1.3 × 1000 = \(\frac { 13 }{ 10 }\) × 1000 = 1300

Question 7.

7. Find

i) 35.5 ÷ 10

Solution:

35.5 ÷ 10 = 35.5 × \(\frac { 1 }{ 10 }\)

= \(\frac { 355 }{ 10 }\) × \(\frac { 1 }{ 10 }\) = \(\frac { 355 }{ 100 }\) = 3.55

ii) 235 ÷ 0.1

Solution:

235 ÷ 0.1 = 235 × \(\frac { 1 }{ 0.1 }\)

= 235 × \(\frac { 1 }{ 10 }\) = 235 × \(\frac { 10 }{ 1 }\) = 2350

Question 8.

Find the average of 9.35, 8.12 and 6.15 .

Solution:

Average = \(\frac{9.35+8.12+6.15}{3}\)

= \(\frac{23.62}{3}\) = 7.87 (approximately)

Question 9.

A car can cover a distance of 80.5 km in 2 hours. How much it travel in 1 hour with same speed ?

Solution:

Distance travelled in 2 hours = 805 km.

Distance travelled in 1 hour = \(\frac{80.5}{2}\) = 40.25 km.

Question 10.

Find

i) 4 × \(\frac{9}{8}\)

Solution:

4 × \(\frac{9}{8}\) = \(\frac{9}{2}\) = 4\(\frac{1}{2}\)

ii) 36 × \(\frac{7}{4}\)

Solution:

36 × \(\frac{7}{4}\) = 9 × 7 = 63

Question 11.

Find

i) 0.5 × \(\frac{1}{2}\)

Solution:

0.5 × \(\frac{1}{2}\) = \(\frac{5}{10}\) × \(\frac{1}{2}\) = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)

ii) \(\frac{1}{2}\) × 10

Solution:

\(\frac{1}{2}\) × 10 = \(\frac{80}{7}\) = 11\(\frac{3}{7}\)

Question 12.

Udayan has brought 2\(\frac{1}{2}\)kg tomatoes, 3\(\frac{3}{4}\) potatoes and 2\(\frac{1}{4}\)kg of Brinjal from vegetable market. Find the total quantity of vegetables.

Solution:

Question 13.

Multiply the reciprocals of \(\frac{3}{4}\) and 7\(\frac{1}{3}\). What is the value of product is mixed fraction?

Solution:

Reciprocal of \(\frac{3}{4}\) = \(\frac{4}{3}\)

Reciprocal of 7\(\frac{1}{3}\) or \(\frac{22}{3}\) = \(\frac{3}{22}\)

Product = \(\frac{4}{3}\) × \(\frac{3}{22}\) = \(\frac{4}{22}\) = \(\frac{2}{11}\)

Question 14.

Which is greater \(\frac{7}{2}\) of \(\frac{4}{3}\) (or) \(\frac{5}{2}\) of \(\frac{8}{5}\)

Solution:

\(\frac{7}{2}\) of \(\frac{4}{3}\) = \(\frac{7}{2}\) × \(\frac{4}{3}\) = \(\frac{14}{3}\) = 4\(\frac{2}{3}\)

\(\frac{5}{2}\) of \(\frac{8}{5}\) = \(\frac{5}{2}\) × \(\frac{8}{5}\) = 4

4\(\frac{2}{3}\) > 4

∴ \(\frac{7}{2}\) of \(\frac{4}{3}\) is greater than \(\frac{5}{2}\) of \(\frac{8}{5}\)

Question 15.

Which is greater \(\frac{2}{3}\) of \(\frac{6}{7}\) (or) \(\frac{1}{4}\) of \(\frac{16}{7}\)

Solution:

\(\frac{2}{3}\) of \(\frac{6}{7}\) = \(\frac{2}{3}\) × \(\frac{6}{7}\) = \(\frac{2 \times 3}{7}\) = \(\frac{6}{7}\)

\(\frac{1}{4}\) of \(\frac{16}{7}\) = \(\frac{2}{3}\) × \(\frac{6}{7}\) = \(\frac{2 \times 3}{7}\) = \(\frac{6}{7}\)

\(\frac{1}{4}\) of \(\frac{16}{7}\) = \(\frac{1}{4}\) × \(\frac{16}{7}\) = \(\frac{4}{7}\)

\(\frac{6}{7}\) > \(\frac{4}{7}\)

∴ \(\frac{2}{3}\) > \(\frac{6}{7}\) is greater than \(\frac{1}{4}\) of \(\frac{16}{7}\)

### Fractions and Decimals Extra Questions Class 7 – 3 Marks

Question 1.

Raunak will read 2\(\frac{1}{2}\) pages of book consists of 120 pages. Then with a week how many pages he will complete on an average and how many pages will be left over.

Solution:

Number of pages to be read in 1 day = 2\(\frac{1}{2}\)

Total pages of book = 120

Number of days in a week = 7

Number of pages in the week = 7 × 2\(\frac{1}{2}\)

= 7 × \(\frac{5}{2}\) = \(\frac{35}{2}\) = 17.5

Number of pages left = 120 – 17.5 = 102.5 pages

Question 2.

Give an example of

i) Proper fraction

ii) Improper fraction

iii) Mixed fraction

Solution:

i) Proper fraction : \(\frac{4}{7}\)

ii) Improper fraction : \(\frac{11}{9}\)

iii) Mixed fraction : 3\(\frac{1}{7}\)

Question 3.

Find

i) 5\(\frac{1}{3}\) ÷ 3\(\frac{1}{2}\)

Solution:

5\(\frac{1}{3}\) ÷ 3\(\frac{1}{2}\) = \(\frac{16}{3}\) ÷ \(\frac{7}{2}\) = \(\frac{16}{3}\) × \(\frac{2}{7}\) = \(\frac{32}{21}\) = 1\(\frac{11}{21}\)

ii) \(\frac{1}{2}\) ÷ 1\(\frac{1}{2}\)

Solution:

\(\frac{1}{2}\) ÷ 1\(\frac{1}{2}\) = \(\frac{1}{2}\) ÷ \(\frac{3}{2}\) = \(\frac{1}{2}\) × \(\frac{2}{3}\) = \(\frac{1}{3}\)

Question 4.

Write the expanded form

i) 5215.312

Solution:

5215.312 = 5 × 1000 + 2 × 100 + 1 × 10 + 5 × 1 + \(\frac{3}{10}\) + \(\frac{1}{100}\) + \(\frac{2}{1000}\)

ii) 850.21

Solution:

850.21 = 8 × 100 + 5 × 10 + 0 × 1 + \(\frac{2}{10}\) + \(\frac{1}{100}\)

Question 5.

Find 0.75 – 1.5 + 3.18 + 9.52 – 8.5

Solution:

0.75 – 1.5 + 3.18 + 9.52 – 8.5

= 0.75 + 3.18 + 9.52 – 1.5 – 8.5

= 13.45 – 10 = 3.45

Question 6.

Each side of a regular polygon is 3.5 m in length. The perimeter of the polygon is 12.25 cm. How many sides does the polygon have?

Solution:

Side of polygon = 3.5 cm

Perimeter = 12.25 cm

Number of sides = \(\frac{12.25}{3.5}\)

= \(\frac{1225}{100 \times \frac{35}{10}}\) = \(\frac{1225}{350}\) = \(\frac{350}{10}\) = 3.5 cm

Question 7.

The average of 2.7, 8.4, 9.1, 3.2 and x is 10.1. Find x.

Solution:

Average = 10.1

\(\frac{2.7+8.4+9.1+3.2+x}{5}\) = 10.1

23.4 + x = 50.5

x = 50.5 – 23.4

x = 27.1

Question 8.

The average of 0.5, 0.75, 0.35, 0.12, 0.72 and P is 1 , then find P.

Solution:

Average = 1

\(\frac{0.5+0.75+0.35+0.12+0.72+P}{6}\) = 1

2.44 + P = 6

P = 6 – 2.44 = 3.56

Question 9.

Find

i) 42.25 ÷ 0.5

Solution:

42.25 ÷ 0.5 = \(\frac{4225}{100}\) ÷ \(\frac{5}{10}\)

= \(\frac{4225}{100}\) × \(\frac{10}{5}\) = \(\frac{845}{10}\) = 84.5

ii) 83.75 ÷ 5

Solution:

83.75 ÷ 5 = \(\frac{8375}{100}\) × \(\frac{1}{5}\) = \(\frac{1675}{100}\) = 16.75

Question 25.

Find

i) 8352 ÷ 4

Solution:

8352 ÷ 4 = 8352 × \(\frac{1}{4}\) = 208

ii) 855 ÷ \(\frac{1}{5}\)

Solution:

855 ÷ \(\frac{1}{5}\) = 855 × \(\frac{1}{5}\) = 4275

Question 10.

Find –\(\frac{5}{9}\)–\(\frac{7}{12}\)+\(\frac{1}{18}\)

Solution:

–\(\frac{5}{9}\)–\(\frac{7}{12}\)+\(\frac{1}{18}\)

–\(\frac{5}{9}\) × \(\frac{4}{4}\) + \(\frac{7}{12}\) × \(\frac{3}{3}\) + \(\frac{1}{18}\) × \(\frac{2}{2}\)

\(\frac{-20}{36}\)–\(\frac{21}{36}\)+\(\frac{2}{36}\)

\(\frac{-20-21+2}{36}\) = \(\frac{-41+2}{36}\) = \(\frac{-39}{36}\) = \(\frac{-13}{12}\)

Question 11.

The product of two rational number is \(8 \frac{3}{5}\). If one of the number is \(\frac{3}{2}\) find other.

Solution:

Other number = \(8 \frac{3}{2}\) ÷ \(\frac{3}{2}\) = \(\frac{43}{5}\) ÷ \(\frac{3}{2}\)

= \(8 \frac{43}{5}\) × \(\frac{2}{3}\) = \(\frac{86}{15}\) = \(5 \frac{11}{15}\)

Question 12.

Find the reciprocal of

i) 0.5

Solution:

0.5 = \(\frac{5}{10}\) = \(\frac{1}{2}\)

Reciprocal of \(\frac{1}{2}\) = \(\frac{2}{1}\) = 2

∴ Reciprocal of 0.5 = 2

ii) \(3\frac{1}{2}\)

Solution:

\(3\frac{1}{2}\) = \(\frac{7}{2}\)

Reciprocal of \(\frac{7}{2}\) = \(\frac{2}{7}\)

Question 13.

Express \(\frac{-247}{228}\) in simplest form

Solution:

\(\frac{-247}{228}\) = \(\frac{(-247) \div 19}{228 \div 19}\) = \(\frac{-13}{12}\)

Question 14.

What should be added to \(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{5}\) to get 4 ?

Solution:

\(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{5}\) = \(\frac{1}{2}\) × \(\frac{15}{15}\) + \(\frac{1}{3}\) × \(\frac{10}{10}\) + \(\frac{1}{5}\) × \(\frac{6}{6}\)

= \(\frac{15}{30}\) + \(\frac{1}{30}\) + \(\frac{6}{30}\)

= \(\frac{15+1+6}{30}\) = \(\frac{22}{30}\) = \(\frac{11}{15}\)

According to the problem

\(\frac{11}{15}\) + = 4

= 4 – \(\frac{11}{15}\)

= \(\frac{4 \times 15}{15}\) – \(\frac{11}{15}\)

= \(\frac{60}{15}\) – \(\frac{11}{15}\) = \(\frac{60-11}{15}\) = \(\frac{49}{15}\)

### Extra Questions of Fractions and Decimals Class 7 – 5 Marks

Question 1.

Multiply by 10

Question 2.

Fill in the correct number by division.

Question 3.

A rope of 18.44 length by is divided into 4 equal pieces. Find the length of each piece and also length of 6 pieces.

Solution:

Length of rope = 18.44 m

Number of equal pieces = 4.

Length of each equal piece = \(\frac{18.44}{4}\) = 4.61 m.

Length of 6 such pieces = 6 × 4.61 = 27.66 m

Question 4.

Nidhi Agarwal has baught 14.5 L of oil for ₹ 797.50. Find the price of 1 L of oil.

Solution:

Total cost quantitiy of oil = ₹797.50

Number of litres of oil = 14.5L

Cost of each litre oil = \(\frac{797.5}{14.5}\)

= \(\frac{7975}{10 \times \frac{145}{10}}\)

= \(\frac{7975}{145}\) = ₹55

∴ Cost of 1 L oil = ₹55

Question 5.

1.150 L of medicine is packed in 23 bottles. Find the quantity of medicine in each bottle.

Solution:

Total quantity of medicine = 1.150L

Number of bottles to pack = 23.

Quantity of medicine in each bottle = 1.150 ÷ 23 = 0.05 L = 50ml.

Question 6.

Do the following :

Question 7.

Multiply the following:

Question 8.

Find

i) \(\frac { 1 }{ 2 }\) of 118

Solution:

\(\frac { 1 }{ 2 }\) of 118 = \(\frac { 1 }{ 2 }\) × 118 = 59

ii) \(\frac { 1 }{ 3 }\) of 99

Solution:

\(\frac { 1 }{ 3 }\) of 99 = \(\frac { 1 }{ 3 }\) × 99 = 33

iii) \(\frac { 2 }{ 3 }\) of 180

Solution:

\(\frac { 2 }{ 3 }\) of 180 = \(\frac { 2 }{ 3 }\) × 180 = 2 × 60 = 120

iv) \(\frac { 3 }{ 4 }\) of 20

Solution:

\(\frac { 3 }{ 4 }\) of 20 = \(\frac { 3 }{ 4 }\) × 20 = 3 × 5 = 15

v) 7 × 3\(\frac { 1 }{ 2 }\)

Solution:

7 × 3\(\frac { 1 }{ 2 }\) = 7 × \(\frac { 7 }{ 2 }\) = \(\frac { 49 }{ 2 }\)

Question 9.

Find

i) 351 ÷ 100

Solution:

351 ÷ 100 = \(\frac { 351 }{ 100 }\) = 3.51

ii) 8.51 ÷ 10

Solution:

8.51 ÷ 10 = \(\frac{8.51}{10}\) = 0.851

iii) 3.81222 ÷ 1000

Solution:

3.81222 ÷ 1000 = 0.00381222

iv) 752101 ÷ 1000

Solution:

752101 ÷ 1000 = 752.101

v) 0.85 × 100

Solution:

0.85 × 100 = 85

Question 10.

Fill in the boxes.

Question 11.

Jithin reads \(\frac{2}{3}\) part of a book in 1 hour. How many times of the will be read in \(\frac{1}{2}\) day ?

Solution:

1 day = 24 hours

\(\frac{1}{2}\) day = \(\frac{24}{2}\) = 12 hours.

Part of book read in 1hr = \(\frac{2}{3}\) × 12 = 8

He will read 8 times in half a day

Question 12.

Find

i) 2\(\frac{1}{5}\) ÷ 2\(\frac{1}{5}\)

Solution:

2\(\frac{1}{5}\) ÷ 2\(\frac{1}{5}\) = \(\frac{11}{5}\) ÷ \(\frac{11}{5}\) = \(\frac{11}{5}\) × \(\frac{5}{11}\) = 1

ii) 2\(\frac{1}{3}\) ÷ \(\frac{1}{2}\)

Solution:

2\(\frac{1}{3}\) ÷ \(\frac{1}{2}\) = \(\frac{7}{3}\) ÷ \(\frac{1}{2}\) = \(\frac{7}{3}\) × \(\frac{2}{1}\) = \(\frac{14}{3}\)

iii) 3\(\frac{1}{5}\) ÷ 1\(\frac{1}{5}\)

Solution:

3\(\frac{1}{5}\) ÷ 1\(\frac{1}{5}\) = \(\frac{16}{5}\) ÷ \(\frac{6}{5}\) = \(\frac{16}{5}\) × \(\frac{5}{6}\) = \(\frac{8}{3}\)

iv) 4\(\frac{3}{7}\) ÷ 8

Solution:

4\(\frac{3}{7}\) ÷ 8 = \(\frac{31}{7}\) ÷ 8 = \(\frac{31}{7}\) × \(\frac{1}{8}\) = \(\frac{31}{56}\)

v) 0 ÷ 12\(\frac{1}{2}\)

Solution:

0 ÷ 12\(\frac{1}{2}\) = 0

Question 13.

Sibal went from place P to Q and from there to place Ras shown in the figure. Peter walks directly from P to R. Who travelled short distance by how much ? (Not exact to the scale, the distances are measured).

Solution:

Sibal :

Distance from P to Q = 3\(\frac{3}{4}\) km

Distance from Q to R = 8\(\frac{1}{2}\) km

Total distance = 3\(\frac{3}{4}\) + 8\(\frac{1}{2}\) = \(\frac{15}{4}\) + \(\frac{17}{2}\)

= \(\frac{15}{4}\) + \(\frac{17}{2}\) × \(\frac{2}{2}\) = \(\frac{15}{4}\) + \(\frac{34}{4}\)

= \(\frac{49}{4}\)km = 12\(\frac{1}{4}\) km.

Peter:

Distance from P to R = 11\(\frac{3}{4}\) km

12\(\frac{1}{4}\) km > 11\(\frac{3}{4}\) km

∴ Sibal travelled more than peter difference = 12\(\frac{1}{4}\) – 11\(\frac{3}{4}\)

= \(\frac{49}{4}\) – \(\frac{47}{4}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\) km.

Question 14.

Find

Question 15.

The side of an equilateral triangle is 3\(\frac{3}{4}\)cm. Find its perimeter?

Solution:

Given side of equilateral triangle = 3\(\frac{3}{4}\)cm

Perimeter = 3 × 3\(\frac{3}{4}\) cm

= 3 × \(\frac{15}{4}\) = \(\frac{45}{4}\) cm = 11\(\frac{1}{4}\)cm

Question 16.

The perimeter of an equilateral triangle is 12.42 cm what is its side ?

Solution:

Given perimeter of an equilateral triangle = 12.42 cm

Each side = \(\frac{12.42 \mathrm{~cm}}{3}\) = 4.14 cm

Question 17.

The sides of a rectangle are 5.2 cm and 4.3 cm. Find its area and perimeter.

Solution:

In a rectangle l = 5.2 cm; b = 4.3 cm.

Area = l × b = 5.2 × 4.3 cm^{2} = 22.36 cm^{2}

Perimeter = 2(l +b)

= 2(5.2 + 4.3) cm

= 2(9.5) cm = 19 cm

Question 18.

The area of a rectangle is 6.25 cm^{2}. If one of its side is 2.5 cm, then find the other side. Write your inference.

Solution:

Area of rectangle = 6.25 cm^{2}

One side = 2.5 cm

Other side = 6.25 ÷ 2.5

= \(\frac { 625 }{ 100 }\) ÷ \(\frac { 25 }{ 10 }\) = \(\frac { 625 }{ 100 }\) × \(\frac { 10 }{ 25 }\) = 2.5 cm

Both sides are equal

∴ It is a square.

Question 19.

The area of a rectangle is 33.75 cm^{2} and its breadth is 15 cm find its other side also perimeter.

Solution:

Given area of rectangle = 33.75 cm^{2}

One side = 15 cm

Other side = \(\frac{33.75}{15}\) = 2.25 cm

Perimeter = 2(l + b)

= 2(15 + 2.25)

= 2(17.25) = 34.5 cm

Question 20.

Find the product of the following and arrange them in ascending order.

i) 2.5 × 4

Solution:

2.5 × 4 = \(\frac{25}{10}\) × 4 = \(\frac{100}{10}\) = 10

ii) 1.7 × 1.3

Solution:

1.7 × 1.3 = \(\frac{17}{10}\) × \(\frac{13}{10}\) = \(\frac{221}{100}\) = 2.21

iii) 2.9 × 1.9

Solution:

2.9 × 1.9 = \(\frac{29}{10}\) × \(\frac{19}{10}\) = \(\frac{491}{100}\) = 4.91

Ascending order 2.21 < 4.91 < 10.