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 cm2 = 22.36 cm2
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 cm2. If one of its side is 2.5 cm, then find the other side. Write your inference.
Solution:
Area of rectangle = 6.25 cm2
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 cm2 and its breadth is 15 cm find its other side also perimeter.
Solution:
Given area of rectangle = 33.75 cm2
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.