Class 7

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 3 Data Handling Ex 3.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1

1. Find the mean of the following data :

Question (i).
3, 5, 7, 9, 11, 13, 15
Answer:
Mean = \(\frac{3+5+7+9+11+13+15}{7}\)
= \(\frac {63}{7}\)
= 9

Question (ii).
40, 30, 30, 0, 26, 60
Answer:
Mean = \(\frac{40+30+30+0+26+60}{6}\)
= \(\frac {183}{6}\)
= 9

2. Find the mean of the first five whole numbers.
Answer:
The first five whole numbers are : 0, 1, 2, 3, 4
Mean = \(\frac{0+1+2+3+4}{5}\)
= \(\frac {10}{5}\)
= 2
Hence, the mean of first five whole numbers = 2

3. A batsman scored the following number of runs in six innings :
36, 35, 50, 46, 60, 55
Calculate the mean runs scored by him in an inning.
Answer:
Mean runs
= \(\frac{36+35+50+46+60+55}{6}\)
= \(\frac {282}{6}\)
Hence, the mean runs scored by batsman in an innings = 47

4. The ages in years of 10 teachers of a school are :
32, 41, 28, 54, 35, 26, 23, 33, 38, 40
(i) What is the age of the oldest teacher and that of the youngest teacher ?
(ii) What is the range of the ages of the teachers ?
(iii) What is the mean age of these teachers ?
Answer:
Arranging the ages in ascending order, we get
23, 26, 28, 32, 33, 35, 38, 40, 41, 54
(i) Age of the oldest teacher = 54 years Age of the youngest teacher 23 years
(ii) Range of the ages = 54 – 23 = 31
(iii) Mean age of the teachers

= \(\frac {350}{10}\)
= 35 years.

5. The rain fall (in mm) ip a city on 7 days of a certain week was recorded as follows :

(i) Find the range of the rainfall in the above data.
(ii) Find the mean rainfall for the week.
(iii) How many days had the rainfall less than the mean rainfall ?
Answer:
(i) Arranging the rainfall (in mm) in ascending order, we get
Highest Rainfall = 20.5 mm
Lowest Rainfall = 0.0 mm
Range of the rainfall = 20.5 mm – 0.0 mm
= 20.5 mm.

(ii) Mean Rainfall
= \(\frac{0.01+12.2+2.1+0.0+20.5+5.5+1.0}{7}\)
= \(\frac {41.31}{7}\)
= 5.9 mm.

(iii) Number of days having rainfall less than mean rainfall = 5 days.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2

1. Write the first step that you will use to separate the variable and then solve the equation.

Question (i).
x + 1 = 0
Answer:
Given equation x + 1 = 0
Subtracting 1 from both sides, we get
x + 1 – 1 = -1
or x = – 1

Question (ii).
x – 1 = 5
Answer:
Given equation is x – 1 = 5
Adding 1 to both sides we get
x – 1 + 1 = 5 + 1
or x = 6
Thus x = 6 is the solution of the given equation

Question (iii).
x + 6 = 2
Answer:
Given equation is x + 6 = 2
Subtracting 6 from both sides, we get:
x + 6 – 6 = 2 – 6
or x = – 4
Thus, x = – 4 is the solution of the given equation.

Question (iv).
y + 4 = 4
Answer:
Given equation is y + 4 = 4
Subtracting 4 from both sides we get
y + 4 – 4 = 4 – 4
or y = 0
Thus, y = 0 is the solution of the given equation.

Question (v).
y – 3 = 3
Answer:
Given equation is y – 3 = 3
Adding 3 to both sides we get
y – 3 + 3 = 3 + 3
or y = 6
Thus, y = 6 is the solution of the given equation.

2. Write the first step that you will use to separate the variable and then sotye the equation :

Question (i).
3x = 15
Answer:
Given equation is 3x = 15
Dividing both sides by 3 we get
\(\frac{3 x}{3}=\frac{15}{3}\)
or x = 5

Question (ii).
\(\frac{P}{7}\) = 4
Answer:
Given equation is \(\frac{P}{7}\) = 4
Multiplying both sides by 7, we get
7 × \(\frac{P}{7}\) = 7 × 4
or p = 28
Thus, p = 28 is the solution of the given equation.

Question (iii).
8y = 36
Answer:
Given equation is 8y = 36
Dividing both sides by 8, we get
\(\frac{8 y}{8}=\frac{36}{8}\)
or y = \(\frac {9}{2}\)

Question (iv).
20x = – 10
Answer:
Given equation is
20x = – 10
Dividing both sides by 20
\(\frac{20 x}{20}=\frac{-10}{20}\)
or x = \(\frac {-1}{2}\)

3. Give the steps you will use to separate the variable and then solve the equation.

Question (i).
5x + 7 = 17
Answer:
Given equation is 5x + 7 = 17
Subtracting 7 from both sides, we get
5x + 7 – 7 = 17 – 7
or 5x = 10
Dividing both sides by 5, we get
\(\frac{5 x}{5}=\frac{10}{5}\)
or x = 2

Question (ii).
\(\frac{20 x}{3}\) = 40
Answer:
Given equation is \(\frac{20 x}{3}\) = 40
Multiplying both sides by 3, we get
3 × \(\frac{20 x}{3}\) = 3 × 40
or 20x = 3 × 40
Dividing both sides by 20, we get
\(\frac{20 x}{20}\) = \(\frac{3 \times 40}{20}\)
or x = 6

Question (iii).
3p – 2 = 46
Answer:
Given equation is 3p – 2 = 46
Adding 2 to both sides, we get
3p – 2 + 2 = 46 + 2
or 3 p = 48
Dividing both sides by 3, we get:
\(\frac{3 p}{3}=\frac{48}{3}\)
or p = 16

4. Solve the following equations :

Question (i).
10x + 10 = 100
Answer:
Given equation is 10x + 10 = 100
Subtracting 10 from both sides, we get
10x + 10 – 10 = 100 – 10
or 10x = 90
Dividing both sides by 10, we get
\(\frac{10 x}{10}=\frac{90}{10}\)
or x = 9
Thus x = 9 is the solution of the given equation.

Question (ii).
\(\frac{-p}{3}\) = 5
Answer:
Given equation is \(\frac{-p}{3}\) = 5
Multiplying both sides by – 3, we get
– 3 × \(\frac{-p}{3}\) = -3 × 5
or p = -15
Thus p = – 15 is the solution of the given equation.

Question (iii).
3x + 12 = 0
Answer:
Given equation is 3x + 12 = 0
Subtracting 12 from both sides, we get
3x + 12 – 12 = – 12
or 3x = – 12
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{-12}{3}\)
or x = -4
Thus x = – 4 is the solution of the given equation.

Question (iv).
2q – 6 = 0
Answer:
The given equation is 2q – 6 = 0
Adding 6 to both sides, we get
2q – 6 + 6 = 0 + 6
or 2q = 6
Dividing both sides by 2, we get
\(\frac{2 q}{2}=\frac{6}{2}\)
or q = 3
Thus, q = 3 is the solution of the given equation.

Question (v).
3p = 0
Answer:
The given equation is 3p = 0
Dividing both sides by 3, we get
\(\frac{3 p}{3}=\frac{0}{3}\)
or p = 0
Thus, p = 0 is the solution of the given equation.

Question (vi).
3s = -9
Answer:
The given equation is
3s = -9
Dividing both sides by 3, we get
\(\frac{3 s}{3}=-\frac{9}{3}\)
or s = – 3
Thus, s = – 3 is the solution of the given equation.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 2 Fractions and Decimals MCQ Questions

Multiple Choice Questions :

Question 1.
Shaded area of given circle represent the fraction.
(a) \(\frac {1}{4}\)
(b) \(\frac {2}{3}\)
(c) \(\frac {3}{4}\)
(d) \(\frac {1}{2}\)
Answer:
(c) \(\frac {3}{4}\)

Question 2.
2 – \(\frac {3}{5}\) = …………….
(a) 7
(b) -7
(c) \(\frac {7}{5}\)
(d) –\(\frac {7}{5}\)
Answer:
(c) \(\frac {7}{5}\)

Question 3.
Place value of 5 in 17.56 is :
(a) 5
(b) \(\frac {5}{10}\)
(c) \(\frac {5}{100}\)
(d) 50
Answer:
(b) \(\frac {5}{10}\)

Question 4.
1.31 × 10 = ?
(a) 0.131
(b) 131
(c) 13.1
(d) 1.31.
Answer:
(c) 13.1

Question 5.
2.7 ÷ 10 is :
(a) 27
(b) 0.27
(c) 0.027
(d) None of these.
Answer:
(b) 0.27

Fill in the blanks :

Question 1.
Equivalent fraction of \(\frac {2}{5}\) is ………….
Answer:
\(\frac {4}{10}\)

Question 2.
\(\frac {2}{3}\) of 18 is ………….
Answer:
12

Question 3.
Expanded form is 40.38 is :
Answer:
40 + 3 × \(\frac {1}{10}\) + 8 × \(\frac {1}{100}\)

Question 4.
The product of decimal number and zero is always.
Answer:
0

Question 5.
The average of decimal numbers 1.1, 2.1 and 3.1 is ………….
Answer:
2.1

Write True or False :

Question 1.
The place value of 2 in 2.56 is 20 (True/False)
Answer:
False

Question 2.
The value of 15.37 × 100 is 1537. (True/False)
Answer:
True

Question 3.
When a decimal number is multiplied by 100, the decimal point in the product is shifted to the right by two places. (True/False)
Answer:
True

Question 4.
The value of 1.5 × 8 is 12 (True/False)
Answer:
True

Question 5.
On dividing a decimal number by 1000, the decimal point is shifted to the left by three places. (True/False)
Answer:
True

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals Ex 2.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7

1. Solve, dividing decimal number by 10, 100 or 1000 in the following :

Question (i).
2.7 ÷ 10
Answer:
2.7 ÷ 10 = \(\frac{27}{10} \times \frac{1}{10}\)
= \(\frac{27}{100}\)
= 0.27

Question (ii).
3.35 ÷ 10
Answer:
3.35 ÷ 10 = \(\frac{335}{100} \times \frac{1}{10}\)
= \(\frac {335}{1000}\)
= 0.335

Question (iii).
0.15 ÷ 10
Answer:
0.15 ÷ 10 = \(\frac{15}{100} \times \frac{1}{10}\)
= \(\frac {15}{1000}\)
= 0.015

Question (iv).
32.7 ÷ 10
Answer:
32.7 ÷ 10 = \(\frac{327}{10} \times \frac{1}{10}\)
= \(\frac {327}{100}\)

Question (v).
5.72 ÷ 100
Answer:
5.72 ÷ 100 = \(\frac{572}{100} \times \frac{1}{100}\)
= \(\frac {572}{10000}\)
= 0.0572

Question (vi).
23.75 ÷ 100
Answer:
23.75 ÷ 100 = \(\frac{2375}{100} \times \frac{1}{100}\)
= \(\frac {2375}{10000}\)
= 0.2375

Question (vii).
532.73 ÷ 100
Answer:
532.73 ÷ 100 = \(\frac{53273}{100} \times \frac{1}{100}\)
= \(\frac {53273}{10000}\)
= 5.3273

Question (viii).
1.321 ÷ 100
Answer:
1.321 ÷ 100 = \(\frac{1321}{1000} \times \frac{1}{100}\)
= \(\frac {1321}{10000}\)
= 0.01321

Question (ix).
2.5 ÷ 1000
Answer:
2.5 ÷ 1000 = \(\frac{25}{10} \times \frac{1}{1000}\)
= \(\frac {25}{10000}\)
= 0.0025

Question (x).
53.83 ÷ 1000
Answer:
53.83 ÷ 1000 = \(\frac{5383}{100} \times \frac{1}{1000}\)
= \(\frac {5383}{100000}\)
= 0.05383

Question (xi).
217.35 ÷ 1000
Answer:
217.35 ÷ 1000 = \(\frac{21735}{100} \times \frac{1}{1000}\)
= \(\frac {21735}{100000}\)
= 0.21735

Question (xii).
0.2 ÷ 1000
Answer:
0.2 ÷ 1000 = \(\frac{2}{10} \times \frac{1}{1000}\)
= \(\frac {2}{10000}\)
= 0.0002

2. Solve, dividing decimal number by whole number.

Question (i).
7.5 ÷ 5
Answer:
7.5 ÷ 5 = \(\frac{75}{10} \times \frac{1}{5}\)
= \(\frac {15}{10}\)
= 1.5

Question (ii).
16.9 ÷ 13
Answer:
16.9 ÷ 13 = \(\frac{169}{10} \times \frac{1}{13}\)
= \(\frac {13}{10}\)
= 1.3

Question (iii).
65.4 ÷ 6
Answer:
65.4 ÷ 6 = \(\frac{654}{10} \times \frac{1}{6}\)
= \(\frac {109}{10}\)
= 10.9

Question (iv).
0.121 ÷ 11
Answer:
0.121 ÷ 11 = \(\frac{121}{1000} \times \frac{1}{11}\)
= \(\frac {11}{1000}\)
= 0.011

Question (v).
11.84 ÷ 4
Answer:
11.84 ÷ 4 = \(\frac{1184}{100} \times \frac{1}{4}\)
= \(\frac {296}{100}\)
= 2.96

Question (vi).
47.6 ÷ 7
Answer:
47.6 ÷ 7 = \(\frac{476}{10} \times \frac{1}{7}\)
= \(\frac {68}{10}\)
= 6.8

3. Solve, dividing the decimal number by decimal number

Question (i).
3.25 ÷ 0.5
Answer:
3.25 ÷ 0.5 = \(\frac{325}{100} \div \frac{5}{10}\)
= \(\frac{325}{100} \times \frac{10}{5}\)
= \(\frac {65}{10}\)
= 6.5

Question (ii).
5.4 ÷ 1.2
Answer:
5.4 ÷ 1.2 = \(\frac{54}{10} \div \frac{12}{10}\)
= \(\frac{54}{10} \times \frac{10}{12}\)
= \(\frac {9}{2}\)
= 4.5

Question (iii).
26.32 ÷ 3.5
Answer:
26.32 ÷ 3.5 = \(\frac{2632}{100} \div \frac{35}{10}\)
= \(\frac{2632}{100} \times \frac{10}{35}\)
= \(\frac {752}{100}\)
= 7.52

Question (iv).
2.73 ÷ 13
Answer:
2.73 ÷ 13 = \(\frac{273}{100} \times \frac{10}{13}\)
= \(\frac {21}{10}\)
= 2.1

Question (v).
12.321 ÷ 11.1
Answer:
12.321 ÷ 11.1 = \(\frac{12321}{1000} \div \frac{111}{10}\)
= \(\frac{12321}{1000} \times \frac{10}{111}\)
= \(\frac {111}{100}\)
= 1.11

Question (vi).
0.0018 ÷ 0.15
Answer:
0.0018 ÷ 0.15 = \(\frac{18}{10000} \div \frac{15}{100}\)
= \(\frac{18}{10000} \times \frac{100}{15}\)
= \(\frac {12}{1000}\)
= 0.012

4. 25 steel chairs were purchased by a school for ₹ 11,883.75. Find the cost of one steel chair.
Answer:
Cost Price of 25 steel chairs = ₹ 11,883.75
Cost Price of 1 steel chair = ₹ 11,883.75 ÷ 15
= ₹ \(\frac{11,88375}{100} \times \frac{1}{15}\)
= ₹ \(\frac {47535}{100}\)
= ₹ 475.35

5. A car covers a distance of 276.75 km in 4.5 hours. What is the average speed of the car ?
Answer:
Total Distance covered = 276.75 km
Time taken = 4.5 hours
Average speed of car = \(\frac{Distance}{Time}\)
= \(\frac{276.75}{4.5}\)
= \(\frac{27675}{100} \times \frac{10}{45}\)
= \(\frac {615}{10}\)
= 61.5 km/hr.

6. Multiple Choice Questions :

Question (i).
27.5 ÷ 10 = ?
(a) 275
(b) 0.275
(c) 2.75
(d) None of these.
Answer:
(c) 2.75

Question (ii).
The value of 1.5 ÷ 3 is :
(a) 5
(b) 0.05
(c) 0.5
(d) 4.5.
Answer:
(c) 0.5

Question (iii).
The average of decimal number 1.1, 2.1 and 3.1 is :
(a) 2.5
(b) 1.1
(c) 2.1
(d) 6.3.
Answer:
(c) 2.1

7. On dividing a decimal number by 100, the decimal point is shifted to the left by one place. (True/False)
Answer:
False.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals Ex 2.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

1. Find the product of each of the following:

Question (i).
1.31 × 10
Answer:
1.31 × 10
= \(\frac {131}{100}\) × 10
= \(\frac {131}{10}\)
= 13.1

Question (ii).
1.31 × 10
Answer:
25.7 × 10
= \(\frac {257}{10}\) × 10
= 257

Question (iii).
1.01 × 100
Answer:
1.01 × 100
= \(\frac {101}{100}\) × 100
= 101

Question (iv).
0.45 × 100
Answer:
0.45 × 100
= \(\frac {45}{100}\) × 100
= 45

Question (v).
9.7 × 100
Answer:
9.7 × 100
= \(\frac {97}{10}\) × 100
= 970

Question (vi).
3.87 × 10
Answer:
3.87 × 10
= \(\frac {387}{100}\) × 100
= \(\frac {387}{10}\)
= 38.7

Question (vii).
0.07 × 10
Answer:
0.07 × 10
= \(\frac {7}{100}\) × 10
= \(\frac {7}{100}\)
= 0.70

Question (viii).
0.3 × 100
Answer:
0.3 × 100
= \(\frac {3}{10}\) × 100
= 30

Question (ix).
5.37 × 1000
Answer:
5.37 × 1000
= \(\frac {537}{10}\) × 100
= 53700

Question (x).
0.02 × 1000
Answer:
0.02 × 1000
= \(\frac {2}{100}\) × 100
= 20

2. Find the product of each of the following :

Question (i).
1.5 × 3
Answer:
1.5 × 3 = \(\frac {15}{10}\) × 3
= \(\frac {45}{10}\)
= 4.5

Question (ii).
2.71 × 12
Answer:
2.71 × 12 = \(\frac {271}{100}\) × 12
= \(\frac {3252}{100}\)
= 32.52

Question (iii).
7.05 × 4
Answer:
7.05 × 4 = \(\frac {705}{100}\) × 4
= \(\frac {2820}{100}\)
= 28.2

Question (iv).
0.05 × 12
Answer:
0.05 × 12 = \(\frac {5}{100}\) × 12
= \(\frac {60}{100}\)
= 0.6

Question (v).
112.03 × 8
Answer:
112.03 × 8 = \(\frac {89624}{100}\) × 8
= 896.24

Question (vi).
3 × 7.53
Answer:
3 × 7.53 = 3 × \(\frac {753}{100}\)
= \(\frac {2259}{100}\)
= 22.59

3. Evaluate the following :

Question (i).
3.7 × 0.4
Answer:
3.7 × 0.4 = \(\frac{37}{10} \times \frac{4}{10}\)
= \(\frac {148}{100}\)
= 1.48

Question (ii).
2.75 × 1.1
Answer:
2.75 × 1.1 = \(\frac{275}{100} \times \frac{11}{10}\)
= \(\frac {3025}{1000}\)

Question (iii).
0.07 × 1.9
Answer:
0.07 × 1.9 = \(\frac{7}{100} \times \frac{19}{10}\)
= \(\frac {133}{1000}\)
= 0.133

Question (iv).
0.5 × 31.83
Answer:
0.5 × 31.83 = \(\frac{5}{10} \times \frac{3183}{100}\)
= \(\frac {15915}{1000}\)
= 15.915

Question (v).
7.5 × 5.7
Answer:
7.5 × 5.7 = \(\frac{75}{10} \times \frac{57}{10}\)
= \(\frac {4275}{100}\)
= 42.75

Question (vi).
10.02 × 1.02
Answer:
10.02 × 1.02 = \(\frac{1002}{100} \times \frac{102}{100}\)
= \(\frac {102240}{10000}\)
= 10.2204

Question (vii).
0.08 × 0.53
Answer:
0.08 × 0.53 = \(\frac{8}{10} \times \frac{53}{100}\)
= \(\frac {424}{10000}\)
= 0.0424

Question (viii).
21.12 × 1.21
Answer:
21.12 × 1.21 = \(\frac{2112}{100} \times \frac{121}{100}\)
= \(\frac {255552}{10000}\)
= 25.5552

Question (ix).
1.06 × 0.04
Answer:
1.06 × 0.04 = \(\frac{106}{100} \times \frac{4}{100}\)
= \(\frac {424}{1000}\)
= 0.0424

4. A piece of wire is divided into 15 equal parts. If length of one part is 2.03 m, then find the total length of the wire.
Answer:
Length of one part = 2.03 m
Length of 15 parts = 15 × 2.03 m
= 30.45 m

5. The cost of 1 metre cloth is ₹ 75.80. Find the cost of 4.75 metre cloth.
Answer:
Cost of 1 metre cloth = ₹ 75.80
Cost of 4.75 metre cloth = ₹ 75.80 × 4.75
= ₹ 360.05

6. Multiple choice questions :

Question (i).
1.25 × 10 = ?
(a) 0.125
(b) 125
(c) 12.5
(d) 1.25
Answer:
(c) 12.5

Question (ii).
If x × 100 = 135.72 then value of x is equal to
(a) 13.572
(b) 1.3572
(c) 135.72
(d) 13572.
Answer:
(b) 1.3572

Question (iii).
The value of 1.5 × 8 is :
(a) 1.2
(b) 120
(c) 12
(d) 0.12.
Answer:
(c) 12

7.
Question (i).
The product of a decimal number and zero is always zero. (True/False)
Answer:
True

Question (ii).
On multiplying a decimal number by 10, the decimal point is shifted to the left by one place. (True/False)
Answer:
False.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals Ex 2.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.3

1. Solve the following :

Question (i).
\(\frac {1}{3}\) of
(a) \(\frac {1}{5}\)
(b) \(\frac {2}{7}\)
(c) \(\frac {3}{2}\)
Answer:

Question (ii).
\(\frac {3}{4}\) of
(a) \(\frac {2}{9}\)
(b) \(\frac {4}{7}\)
(c) \(\frac {8}{3}\)
Answer:

2. Multiply the fractions and reduce it to the lowest form (If possible) :

Question (i).
\(\frac{2}{7} \times \frac{7}{9}\)
Answer:
\(\frac{2}{7} \times \frac{7}{9}\)
= \(\frac {2}{9}\)

Question (ii).
\(\frac{1}{3} \times \frac{15}{8}\)
Answer:
\(\frac{1}{3} \times \frac{15}{8}\)
= \(\frac {5}{8}\)

Question (iii).
\(\frac{12}{27} \times \frac{3}{9}\)
Answer:
\(\frac{12}{27} \times \frac{3}{9}\)
= \(\frac {4}{27}\)

Question (iv).
\(\frac{2}{5} \times \frac{6}{4}\)
Answer:
\(\frac{2}{5} \times \frac{6}{4}\)
= \(\frac {3}{5}\)

Question (v).
\(\frac{81}{100} \times \frac{6}{7}\)
Answer:
\(\frac{81}{100} \times \frac{6}{7}\)
= \(\frac {243}{350}\)

Question (vi).
\(\frac{3}{5} \times \frac{5}{27}\)
Answer:
\(\frac{3}{5} \times \frac{5}{27}\)
= \(\frac {1}{9}\)

3. Multiply the following fractions :

Question (i).
\(\frac{3}{2} \times 5 \frac{1}{3}\)
Answer:
\(\frac{3}{2} \times 5 \frac{1}{3}\)
= \(\frac{3}{2} \times \frac{16}{3}\)
= 8

Question (ii).
\(\frac{1}{7} \times 5 \frac{2}{3}\)
Answer:
\(\frac{1}{7} \times 5 \frac{2}{3}\)
= \(\frac{1}{7} \times \frac{17}{3}\)
= \(\frac {17}{21}\)

Question (iii).
\(2 \frac{5}{6} \times 4\)
Answer:
\(2 \frac{5}{6} \times 4\)
= \(\frac {17}{6}\) × 4
= \(\frac {34}{3}\)
= 11\(\frac {1}{3}\)

Question (iv).
\(4 \frac{1}{3} \times 9 \frac{1}{4}\)
Answer:
\(4 \frac{1}{3} \times 9 \frac{1}{4}\)
= \(\frac{13}{3} \times \frac{37}{4}\)
= \(\frac {481}{12}\)

Question (v).
\(2 \frac{2}{3} \times 3 \frac{5}{8}\)
Answer:
\(2 \frac{2}{3} \times 3 \frac{5}{8}\)
= \(\frac{8}{3} \times \frac{29}{8}\)
= 9\(\frac {2}{3}\)

Question (vi).
\(3 \frac{1}{5} \times 2 \frac{1}{4}\)
Answer:
\(3 \frac{1}{5} \times 2 \frac{1}{4}\)
= \(\frac{16}{5} \times \frac{9}{4}\)
= \(\frac {36}{5}\)
= 7\(\frac {1}{5}\)

4. Which fraction is greater in the following fractions ?

Question (i).
\(\frac {3}{2}\) of \(\frac {2}{7}\) or \(\frac {5}{2}\) of \(\frac {3}{8}\)
Answer:
\(\frac {3}{2}\) of \(\frac {2}{7}\)
= \(\frac{3}{2} \times \frac{2}{7}\)
= \(\frac {3}{7}\)
or
\(\frac {5}{2}\) of \(\frac {3}{8}\)
= \(\frac{5}{2} \times \frac{3}{8}\)
= \(\frac {15}{16}\)
Since, \(\frac {15}{16}\) > \(\frac {3}{7}\)
So, \(\frac {5}{2}\) of \(\frac {3}{8}\) is greater.

Question (ii).
\(\frac {1}{2}\) of \(\frac {6}{5}\) or \(\frac {1}{3}\) of \(\frac {4}{5}\)
Answer:
\(\frac {1}{2}\) of \(\frac {6}{5}\)
= \(\frac {1}{2}\) × \(\frac {6}{5}\)
= \(\frac {3}{5}\)
or
\(\frac {1}{3}\) of \(\frac {4}{5}\)
= \(\frac {1}{3}\) × \(\frac {4}{5}\)
= \(\frac {4}{15}\)
Since, \(\frac {3}{5}\) > \(\frac {4}{15}\)
So, \(\frac {1}{2}\) of \(\frac {6}{5}\) is greater.

5. If the speed of a car is 105 \(\frac {1}{3}\) km/hr, find the distance covered by it m 3\(\frac {2}{3}\) hours.
Answer:
Distance travelled in 1 hour= 105\(\frac {1}{3}\) km
= \(\frac {316}{3}\) km
Distance travelled in 3\(\frac {2}{3}\) hours i.e. in \(\frac {11}{3}\) hours
= \(\frac {11}{3}\) × \(\frac {316}{3}\) km
= \(\frac {3476}{9}\) km
= 386\(\frac {2}{9}\)km

6. The length of a rctangular plot of land is 29\(\frac {3}{7}\) m. If its breadth is 12\(\frac {8}{11}\)m, find its area.
Answer:
Length of rectangular plot (l)
= 29\(\frac {3}{7}\)m = \(\frac {206}{7}\)m
Breadth of rectangular plot (b)
= 12\(\frac {8}{11}\)m = \(\frac {140}{11}\)m
Area of rectangular plot
= l × b = \(\frac{206}{7} \mathrm{~m} \times \frac{140}{11} \mathrm{~m}\)
= \(\frac{4120}{11}\) sq.m
= 374\(\frac{6}{11}\) sq.m

7. If a cloth costs ₹ 120 \(\frac {1}{4}\) per metre, find the cost of 4\(\frac {1}{3}\) metre of this cloth.
Answer:
Per m cost of cloth = ₹ 120\(\frac {1}{4}\)
= ₹ \(\frac {481}{4}\)
Cost of 4\(\frac {1}{3}\) metre of this cloth
= \(\frac {1}{3}\) × ₹ \(\frac {481}{4}\)
= ₹ \(\frac {6253}{12}\)
= ₹ 521\(\frac {1}{12}\)

8. Multiple choice questions :

Question (i).
\(\frac {1}{4}\) of \(\frac {8}{3}\) is :
(a) \(\frac {9}{7}\)
(b) \(\frac {8}{4}\)
(c) \(\frac {2}{3}\)
(d) 1
Answer:
(c) \(\frac {2}{3}\)

Question (ii).
\(\frac {3}{2}\) × \(\frac {2}{3}\) = ?
(a) 1
(b) \(\frac {5}{6}\)
(c) 3
(d) \(\frac {6}{5}\)
Answer:
(a) 1

Question (iii).
The value of the product of two proper fractions is :
(a) greater than both of the proper fractions.
(b) lesser than both of the proper fractions.
(c) lie between the given two fractions.
(d) none of these.
Answer:
(b) lesser than both of the proper fractions.

9. Check if the following equations are ‘True’ or ‘False’ :

Question (i).
\(1 \frac{2}{3} \times 4 \frac{5}{7}=4 \frac{10}{21} ?\) (True/False)
Answer:
False

Question (ii).
\(\frac{3}{4} \times \frac{2}{3}=\frac{2}{3} \times \frac{3}{4} ?\) (True/False)
Answer:
True

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals Ex 2.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.2

1. Match the following :

Answer:
(a) (ii),
(b) (iii),
(c) (iv),
(d) (i).

2. Multiply whole number to a fraction and reduce it to the lowest form. Also convert into a mixed fraction :

Question (i).
4 × \(\frac {1}{3}\)
Answer:
4 × \(\frac {1}{3}\) = \(\frac{4 \times 1}{3}\)
= \(\frac {4}{3}\)
= 1\(\frac {1}{3}\)

Question (ii).
11 × \(\frac {4}{7}\)
Answer:
11 × \(\frac {4}{7}\) = \(\frac{11 \times 4}{7}\)
= \(\frac {44}{7}\)
= 6\(\frac {2}{7}\)

Question (iii).
\(\frac {3}{4}\) × 6
Answer:
\(\frac {3}{4}\) × 6 = \(\frac{3 \times 6}{4}\)
= \(\frac {9}{2}\)
= 4\(\frac {1}{2}\)

Question (iv).
\(\frac {9}{7}\) × 5
Answer:
\(\frac {9}{7}\) × 5 = \(\frac {45}{7}\)
= 6\(\frac {3}{7}\)

Question (v).
2\(\frac {5}{6}\) × 4
Answer:
2\(\frac {5}{6}\) × 4 = \(\frac{17}{6} \times 4\)
= \(\frac{17 \times 4}{6}\)
= \(\frac{17 \times 2}{3}\)
= \(\frac {34}{3}\)
= 11\(\frac {1}{3}\)

Question (vi).
10\(\frac {5}{6}\) × 5
Answer:
10\(\frac {5}{6}\) × 5
= \(\frac {65}{6}\) × 5
= \(\frac{65 \times 5}{6}\)
= \(\frac {325}{6}\)
= 54\(\frac {1}{6}\)

Question (vii).
5 × 6\(\frac {3}{4}\)
Answer:
5 × 6\(\frac {3}{4}\) = 5 × \(\frac {27}{4}\)
= \(\frac{5 \times 27}{4}\)
= \(\frac {135}{4}\)
= 33\(\frac {3}{4}\)

Question (viii).
3 \(\frac {2}{5}\) × 8
Answer:
3 \(\frac {2}{5}\) × 8 = \(\frac {17}{5}\) × 8
= \(\frac{17 \times 8}{5}\)
= \(\frac {136}{5}\)
= 27\(\frac {1}{5}\)

3. Solve :

Question (i).
\(\frac {1}{2}\) of 46
Answer:
\(\frac {1}{2}\) of 46
= \(\frac {1}{2}\) × 46
= 23

Question (ii).
\(\frac {2}{3}\) of 27
Answer:
\(\frac {2}{3}\) of 27
= \(\frac {2}{3}\) × 27
= 18

Question (iii).
\(\frac {1}{3}\) of 36
Answer:
\(\frac {1}{3}\) of 36
= \(\frac {1}{3}\) × 36
= 12

Question (iv).
\(\frac {3}{4}\) of 16
Answer:
\(\frac {3}{4}\) of 16
= \(\frac {3}{4}\) × 16
= 12

Question (v).
\(\frac {5}{7}\) of 35
Answer:
\(\frac {5}{7}\) of 35
= \(\frac {5}{7}\) × 35
= 25

4. Shade :

(i) \(\frac {1}{3}\) of the rectangles in box (a)
(ii) \(\frac {2}{3}\) of the circles in box (b)
(iii) \(\frac {1}{2}\) of the triangles in box (c)

Answer:

5. Rahul earns ₹ 44,000 per month. He spends \(\frac {3}{4}\) th of his income every month and saves the rest of his earning. Find his monthly savings ?
Answer:
Rahul earns per month = ₹ 44000
He spends = \(\frac {3}{4}\)th of ₹ 44000
= ₹ \(\frac {3}{4}\) × 44000
= ₹ 33000
His savings = ₹ 44,000 – ₹ 33000
= ₹ 11000

6. The cost of a book is ₹ 117\(\frac {1}{2}\). Find the cost of 8 books ?
Answer:
Cost of one book = ₹ 117 \(\frac {1}{2}\)
= ₹ \(\frac {235}{2}\)
Cost of 8 books = 8 × ₹ \(\frac {235}{2}\)
= ₹ 4 × 235
= ₹ 940

7. Multiple choice questions :

Question (i).
\(\frac {1}{2}\) × 8 is equal to
(a) 8
(6) 2
(c) 4
(d) 1
Answer:
(c) 4

Question (ii).
\(\frac {3}{2}\) of 16 is :
(a) 48
(b) 8
(c) 3
(d) 24
Answer:
(d) 24

Question (iii).
What fraction of an hour is 40 minutes ?
(a) \(\frac {2}{3}\)
(b) 40
(c) \(\frac {1}{4}\)
(d) \(\frac {1}{2}\)
Answer:
(a) \(\frac {2}{3}\)

Question (iv).
What fraction does the shaded part of figure represent ?

(a) \(\frac {1}{3}\)
(b) \(\frac {2}{3}\)
(c) \(\frac {3}{4}\)
(d) \(\frac {1}{2}\)
Answer:
(c) \(\frac {3}{4}\)

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals Ex 2.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

1. Which is greater decimal number ?

Question (i).
0.9 or 0.4
Answer:
0.9 or 0.4
Here in 0.9 tenth place is greater than tenth place of 0.4.
9 > 4
∴ 0.9 > 0.4

Question (ii).
1.35 or 1.37
Answer:
1.35 or 1.37
Whole number parts of both number are equal
So, we have to compare decimal part
Also digits at tenths place are also equal.
Hundredths part of 1.37 is greater than hundredth the part of 1.35
∴ 1.37 > 1.35

Question (iii).
10.10 or 10.01
Answer:
10.10 or 10.01
Whole number parts of both numbers are equal
So, we have to compare decimal part
Tenths part of 10.10 is greater than tenths part of 10.01
∴ 10.10 > 10.01

Question (iv).
1735.101 or 1734.101
Answer:
1735.101 or 1734.101
Whole number part of 1735.101 is greater than whole number part of 1734.101
∴ 735.101 > 1734.101

Question (v).
0.8 or 0.88.
Answer:
0.8 or 0.88
Here tenths place in both the number is same and hundredths place in 0.88 is greater than the hundredths place in 0.8.
∴ 0.88 > 0.8

2. Write following decimal number in the expanded form :

Question (i).
40.38
Answer:
40.38 = 40 + 0 + .3+ .08
= 4 × 10 + 0 + 3 × \(\frac {1}{10}\) + 8 × \(\frac {1}{100}\)

Question (ii).
4.038
Answer:
4.038 = 4 + 0.0 + 0.03 + 0.008
4 + 0 × \(\frac {1}{10}\) + 3 × \(\frac {1}{100}\) + 8 × \(\frac {1}{1000}\)

Question (iii).
0.1038
Answer:
0.4038 = 0 + 0.4 + 0.00 + 0.003 + 0.0008
= 0 + 4 × \(\frac {1}{10}\) + 0 × \(\frac {1}{100}\) + 3 × \(\frac {1}{1000}\) + 8 × \(\frac {1}{10000}\)

Question (iv).
4.38.
Answer:
4.38 = 4 + 0.3 + 0.08
= 4 + 3 × \(\frac {1}{10}\) + 8 × \(\frac {1}{100}\)

3. Write the place value of 5 in the following decimal numbers :

Question (i).
17.56
Answer:
Place value of 5 in 17.56 = 0.5
= \(\frac {5}{10}\)

Question (ii).
1.253
Answer:
Place value of 5 in 1.253 = 0.05
= \(\frac {5}{100}\)

Question (iii).
10.25
Answer:
Place value of 5 in 10.25 = 0.05
= \(\frac {5}{100}\)

Question (iv).
5.62.
Answer:
Place value of 5 in 5.62 = 5

4. Express in rupees using decimals :

Question (i).
55 paise
Answer:
55 paise = ₹ \(\frac {55}{100}\)
= ₹ 0.55

Question (ii).
55 rupees 5 paise
Answer:
55 rupees 5 paise = 55 rupees + 5 paise
= ₹ 55 + ₹ \(\frac {5}{100}\)
= ₹ 55 + ₹ 0.5
= ₹ 55.05

Question (iii).
347 paise
Answer:
347 paise = ₹ \(\frac {347}{100}\)
= ₹ 3.47

Question (iv).
2 paise.
Answer:
2 paise = ₹ \(\frac {2}{100}\)
= ₹ 0.02.

5. Express in km :

Question (i).
350 m
Answer:
350 m = \(\frac {350}{1000}\) km
= 0.350 km
[Since 1000 m = 1 km,
∴ 1 m =\(\frac {1}{1000}\) km]

Question (ii).
4035 m
Answer:
4035 m = \(\frac {4035}{1000}\) km
= 4.035 km

Question (iii).
2 km 5 m
Answer:
2 km 5 m = 2 km + 5 m
= 2 km + \(\frac {5}{1000}\) km
= 2.05 km

6. Multiple Choice Questions :

Question (i).
Place value of 2 in 3.02 is :
(a) 2
(b) 20
(c) \(\frac {2}{10}\)
(d) \(\frac {2}{100}\)
Answer:
(d) \(\frac {2}{100}\)

Question (ii).
The correct ascending order of 0.7, 0.07, 7 is :
(a) 7 < 0.07 < 0.7
(b) 0.07 < 0.7 < 7
(c) 0.7 < 0.07 < 7
(d) 0.07 < 7 < 0.7.
Answer:
(b) 0.07 < 0.7 < 7

Question (iii).
Decimal expression of 5 kg 20 gram is :
(a) 5.2 kg
(b) 5.20 kg
(c) 5.02 kg
(d) None of these.
Answer:
(c) 5.02 kg

Question (iv).
Expanded form of 2.38 is
(a) 2 + \(\frac {38}{10}\)
(b) 2 + 3 + \(\frac {8}{10}\)
(c) \(\frac {238}{100}\)
(d) 2 + \(\frac {3}{10}\) + \(\frac {8}{100}\)
Answer:
(d) 2 + \(\frac {3}{10}\) + \(\frac {8}{100}\)

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals Ex 2.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4

1. Find the reciprocal of each of the following fraction.
(i) \(\frac {2}{7}\)
(ii) \(\frac {3}{2}\)
(iii) \(\frac {5}{7}\)
(iv) \(\frac {1}{9}\)
(v) \(\frac {2}{3}\)
(vi) \(\frac {7}{8}\)
Answer:
(i) \(\frac {7}{2}\)
(ii) \(\frac {2}{3}\)
(iii) \(\frac {7}{5}\)
(iv) 9
(v) \(\frac {3}{2}\)
(vi) \(\frac {8}{7}\)

2. Solve the following (Division of a fraction by a non zero whole number)

Question (i).
\(\frac {19}{6}\) ÷ 10
Answer:
\(\frac {19}{6}\) ÷ 10
= \(\frac{19}{6} \times \frac{1}{10}\)
= \(\frac {19}{60}\)

Question (ii).
\(\frac {4}{9}\) ÷ 5
Answer:
\(\frac {4}{9}\) ÷ 5
= \(\frac{4}{9} \times \frac{1}{5}\)
= \(\frac {4}{45}\)

Question (iii).
\(\frac {8}{9}\) ÷ 8
Answer:
\(\frac {8}{9}\) ÷ 8
= \(\frac{8}{9} \times \frac{1}{8}\)
= \(\frac {1}{9}\)

Question (iv).
3\(\frac {1}{2}\) ÷ 4
Answer:
3\(\frac {1}{2}\) ÷ 4
= \(\frac{7}{2} \times \frac{1}{4}\)
= \(\frac {7}{8}\)

Question (v).
16\(\frac {1}{2}\) ÷ 5
Answer:
16\(\frac {1}{2}\) ÷ 5
= \(\frac{33}{2} \times \frac{1}{5}\)
= \(\frac {33}{10}\)
= 3\(\frac {3}{10}\)

Question (vi).
4\(\frac {1}{3}\) ÷ 3
Answer:
4\(\frac {1}{3}\) ÷ 3
= \(\frac{13}{3} \times \frac{1}{3}\)
= \(\frac {33}{9}\)
= 1\(\frac {4}{9}\)

3. Solve the following (Division of a whole number by a fraction)

Question (i).
8 ÷ \(\frac {7}{3}\)
Answer:
8 ÷ \(\frac {7}{3}\)
= 8 × \(\frac {3}{7}\)
= \(\frac {24}{7}\)
= 3\(\frac {3}{7}\)

Question (ii).
5 ÷ \(\frac {7}{5}\)
Answer:
5 ÷ \(\frac {7}{5}\)
= 5 × \(\frac {5}{7}\)
= \(\frac {25}{7}\)
= 3\(\frac {4}{7}\)

Question (iii).
4 ÷ \(\frac {8}{3}\)
Answer:
4 ÷ \(\frac {8}{3}\)
= 4 × \(\frac {3}{8}\)
= \(\frac {3}{2}\)
= 1\(\frac {1}{2}\)

Question (iv).
3 ÷ 2\(\frac {3}{5}\)
Answer:
3 ÷ 2\(\frac {3}{5}\)
= 3 ÷ \(\frac {13}{5}\)
= 3 × \(\frac {5}{13}\)
= \(\frac {15}{13}\)
= 1\(\frac {2}{13}\)

Question (v).
5 ÷ 3\(\frac {4}{7}\)
Answer:
5 ÷ 3\(\frac {4}{7}\)
= 5 ÷ 3\(\frac {25}{7}\)
= 5 × \(\frac {7}{25}\)
= \(\frac {7}{25}\)
= 1\(\frac {2}{5}\)

4. Solve the following (Division of a fraction by another fraction)

Question (i).
\(\frac{2}{3} \div \frac{10}{9}\)
Answer:
\(\frac{2}{3} \div \frac{10}{9}\)
= \(\frac{2}{3} \times \frac{9}{10}\)
= \(\frac {3}{5}\)

Question (ii).
\(\frac{4}{9} \div \frac{2}{3}\)
Answer:
\(\frac{4}{9} \div \frac{2}{3}\)
= \(\frac{4}{9} \times \frac{3}{2}\)
= \(\frac {2}{3}\)

Question (iii).
\(2 \frac{1}{2} \div \frac{3}{5}\)
Answer:
\(2 \frac{1}{2} \div \frac{3}{5}\)
= \(\frac{5}{2} \times \frac{5}{3}\)
= \(\frac {25}{6}\)
= 4\(\frac {1}{6}\)

Question (iv).
\(\frac{3}{7} \div 1 \frac{1}{5}\)
Answer:
\(\frac{3}{7} \div 1 \frac{1}{5}\)
= \(\frac{3}{7} \div \frac{6}{5}\)
= \(\frac{3}{7} \times \frac{5}{6}\)
= \(\frac {5}{14}\)

Question (v).
\(5 \frac{1}{2} \div 2 \frac{1}{5}\)
Answer:
\(5 \frac{1}{2} \div 2 \frac{1}{5}\)
= \(\frac{11}{2} \div \frac{11}{5}\)
= \(\frac{11}{2} \times \frac{5}{11}\)
= \(\frac {5}{2}\)
= 2\(\frac {1}{2}\)

Question (vi).
\(3 \frac{1}{5} \div 1 \frac{2}{3}\)
Answer:
\(3 \frac{1}{5} \div 1 \frac{2}{3}\)
\(\frac{16}{5} \div \frac{5}{3}\)
= \(\frac{16}{5} \times \frac{3}{5}\)
= \(\frac {48}{25}\)
= 1\(\frac {23}{25}\)

5. 11 small ropes are cut from 7\(\frac {1}{3}\) m long rope. Find the length of each of the small rope.
Solution:
Length of rope = 7\(\frac {1}{3}\) m = \(\frac {22}{3}\) m,
∴ Length of 11 small ropes = \(\frac {22}{3}\)m
Length of each small rope = \(\frac {22}{3}\)m ÷ 11
= \(\frac{22}{3} \times \frac{1}{11} \mathrm{~m}\)
= \(\frac {2}{3}\)m

6. Multiple choice questions :

Question (i).
Reciprocal of \(\frac {3}{4}\) is :
(a) \(\frac {3}{4}\)
(b) \(\frac {4}{3}\)
(c) 1
(d) none of these
Answer:
(b) \(\frac {4}{3}\)

Question (ii).
\(\frac{5}{7} \div \frac{7}{5}=?\)
(a) 1
(b) \(\frac {49}{25}\)
(c) \(\frac {25}{49}\)
(d) -1
Answer:
(c) \(\frac {25}{49}\)

Question (iii).
\(\frac{5}{7} \div \frac{5}{7}=?\)
(a) 1
(b) \(\frac {49}{25}\)
(c) \(\frac {25}{49}\)
(d) -1
Answer:
(a) 1

7.
Question (i).
The reciprocal of a proper fraction is an improper fraction (True/False)
Answer:
True

Question (ii).
The reciprocal of a whole number is always a whole number (True / False)
Answer:
False.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 1 Integers Ex 1.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3

1. Find the product of :
(i) (-15) × 0
(ii) (-35) × 1
(iii) (-13) × (-12)
(iv) (-20) × 16
(v) (-15) × (-4) × (-5)
(vi) (-8) × (-5) × 9
(vii) (-2) × (-5) × (-4) × (-10)
(viii) (-8) × 0 + [(-5) × (-4)]
Answer:
(i) (-15) × 0 = 0.
(ii) (-35) × 1 = -35.
(iii) (-13) × (-12) = 156.
(iv) (-20) × 16 = -(20 × 16) = -320

(v) (-15) × (-4) × (-5) = (-15 × -4) × (-5)
= 60 × (-5)
= -(60 × 5)
= -300.

(vi) (-8) × (-5) × 9
= [(-8) × (-5)] × 9
= 40 × 9 = 360.

(vii) (-2) × (-5) × (-4) × (-10)
= [(-2) × (-5)] × [(-4) × (-10)]
= 10 × 40 = 400.

(viii) (-8) × 0 + [(-5) × (-4)]
= [(-8) × 0] + [(-5) × (-4)]
= 0 + 20 = 20.

2.
Question (i).
Verify : 15 × [9 + (-6)] = (15 × 9) + (15 × (-6)]
Answer:
L.H.S. = 15 × [9 + (-6)]
= 15 × [9 – 6]
= 15 × 3 = 45.
R.H.S. = (15 × 9) + [(15 × (-6)]
= 135 + (-90)
= 135 – 90 = 45
∴ L.H.S = R.H.S.

Question (ii).
Verify : 18 × [(-5) + (-4)] = [18 × (-5)] + [18 × (-4)]
Answer:
L.H.S. = 18 × [(-5) + (-4)]
= 18 × (-9)
= -162.
R.H.S. = [(18) × (-5)] + [18 × (-4)] = (-90) + (-72)
= – 162.
∴ L.H.S. = R.H.S.

3. Fill in the blanks :

Answer:
(i) 15 × 0 = 0.
(ii) -25 × -1 = 25.
(iii) (-15) × 18 = 18 × (-15)
(iv) (-10) × [(-15) + (-5)] = (-10) × -15 + (-10) × (-5)
(v) (-6) × [(-5) × (-18)] = (-6) × -5 × -18.

4. Find product using properties :

Question (i).
15 × (-20) + (-20) × (-5)
Answer:
15 × (-20) + (-20) × (-5)
= (-20) × [15 + (-5)]
= (-20) × [15 – 5]
= (-20) × (10)
= -200.

Question (ii).
(15 × 8) × 50
Answer:
(15 × 8) × 50
= 120 × 50
= 6000.

Question (iii).
8 × (40 – 5)
Answer:
8 × (40 – 5)
= 8 × 40 – 8 × 5
= 320 – 40
= 280.

Question (iv).
510 × (-45) + (-510) × 55
Answer:
510 × (-45) + (-510) × 55
= 510 [(-45) + (-55)]
= 510 [-100]
= -51000.

5. In a class test containing 15 questions, 2 marks are awarded for every correct answer and (-1) mark awarded for every incorrect answer and 0 mark for questions not attempted.
(i) Kritika gets 5 correct and 10 incorrect answers. What is her score ?
(ii) Rohan gets 7 correct and 7 incorrect answers out of 14 questions he attempted. What is his score ?
Answer:
(i) Marks given for one correct answer = 2
Marks given for 5 correct answers = 2 × 5 = 10
Marks given for one incorrect answer = -1
Marks given for 10 incorrect answers = -1 × 10 = -10
Total marks Kritika scored in the test = 10 + (-10)
= 10 – 10 = 0.

(ii) Marks given for one correct answer = 2
Marks given for 7 correct answers = 2 × 7 = 14
Marks given for one incorrect answer = -1
Marks given for 7 incorrect answers = – 1 × 7 = -7
Total marks Rohan scored in the test = 14 + (-7)
= 14 – 7 = 7.

6. Multiple Choice Questions :

Question (i).
(-19) – (13) is equal to
(a) -32
(b) 6
(c) -6
(d) None of these.
Answer:
(c) -6

Question (ii).
(-6) × (-5) × 0 is equal to :
(a) 0
(b) -6
(c) -5
(d) 30.
Answer:
(a) 0

(iii) 0 ÷ (-10) is equal to :
(a) 0
(b) -1
(c) -10
(d) None of these.
Answer:
(a) 0

(iv) (-33) × 102 + (-33) × (-2) is equal to :
(a) 3300
(b) -3300
(c) 3432
(d) -3432.
Answer:
(b) -3300

(v) 101 × (-1) + 0 × (-1) is equal to :
(a) -101
(b) 101
(c) -102
(d) 102.
Answer:
(a) -101