Class 7

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

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 1 Integers Ex 1.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 1 Integers Ex 1.2

1. Find the value of:
(a) 32 + 15
(b) 17 + (-18)
(c) (-25) + (21)
(d) (-8) + (-11)
(e) (-13) + (21)
(f) (-19) + (0)
(g) (-85) – (-10)
(h) (15) – (6)
(i) (45) – (-27)
(j) (-62) – (52)
Answer:
(a) 32 + 15 = 47
(b) 17 + (-18) = – 1
(c) (-25) + (21) = -4
(d) (-8) + (- 11) = – 19
(e) (-13) + (21) = 8
(f) (-19) + (0) = – 19
(g) (- 85) – (- 10) = – 85 + 10 = – 75
(h) (15) – (6) = 9
(i) (45) – (- 27) = 45 + 27 = 72
(j) (- 62) – (52) = – 62 – 52 = – 114

2. Solve the following :

Question (a).
(-3) + 7 + (-8)
Answer:
(- 3) + 7 + (- 8) = (- 3) + (- 8) + 7
= – 11 + 7
[∵ (- 3) + (- 8) = – 11] = – 4

Question (b).
(- 2) – (- 1) – (4)
Answer:
(- 2) – (- 1) – (4)= – 2 + 1 – 4
= (- 2) + (- 4) + 1
= – 6 + 1 = – 5

Question (c).
8 + (- 7) – (- 6)
Answer:
8 + (- 7) – (- 6) = 8 + (- 7) + (6)
= (8) + (6) + (- 7)
= 14 + (- 7) = 7

Question (d).
(- 12) – (- 17)+ (- 25)
Answer:
(- 12) – (- 17) + (- 25) = (- 12) + (+ 17) + (- 25)
= (- 12) + (- 25) + (+ 17)
[∵ (- 12) + (- 25) = (- 37)]
= (- 37) + (+ 17)
= – 20

3. Find the value of:

Question (a).
15 – (- 5) + 12 + (- 8) + (- 3)
Answer:
15 – (- 5) + 12 + (- 8) + (- 3)
= 15 + (+ 5) + 12 + (- 8) + (- 3)
= 32 + (- 11)
= 21
[∵ 15 + (+ 5) + 12 = 32 and (- 8) + (- 3) = (- 11)]

Question (b).
(- 32) – (-11) + (- 25) + 27 – 13 + (- 7)
Answer:
(- 32) – (- 11) + (- 25) + 27 – 13 + (- 7)
= (- 32) + (+ 11) + (- 25) + 27 – 13 + (- 7)
= 11 + 27 + (- 32) + (- 25) – 13 + (- 7)
= 38 + (- 77)
= – 39.
[∵ 11 + 27 = 38 and (- 32) + (- 25) – 13 + (- 7) = – 77]

Question (c).
160 + (- 150) + (- 130) – (-100)
Answer:
160 + (- 150) + (- 130) – (- 100)
= 160 + (- 150) + (- 130) + (+ 100)
= 160 + (+ 100) + (- 150) + (- 130)
= 260 + (- 280)
= – 20
[∵ 160 + (+ 100) = 260 and (-150) + (- 130) = – 280)]

Question (d).
25 – (- 15) + (- 12) + 21 – 65 – (- 38)
Answer:
25 – (- 15) + (- 12) + 21 – 65 – (- 38)
= 25 + (+ 15) + (- 12) + 21 – 65 + (+ 38)
= 25 + (+ 15) + 21 + (+ 38) + (- 12) – 65
= 99 + (- 77)
= 22
[∵ 25 + (+ 15) + 21 + (+ 38) = 99 and (- 12) – 65 = – 77]

4. Fill in the blanks using properties of addition and subtraction of integers :

Answer:

5. The difference between two integers is – 10. If first integers is 17, then find the other integer ?
Answer:
Difference = – 10
1^st Integer = 17
2^nd Integer = 1^st integer – Difference
= 17 – (- 10)
= 17 + 10 = 27

6. Write three consecutive odd integers succeeding (- 93).
Answer:
Three consecutive odd integers succeeding (- 93) are – 91, – 89, – 87.

7. At sunrise, the outside temperature was 7° below zero. In the afternoon the temperature rose by 13° and then fell by 8° at night. What was the temperature at the end of the day ?
Answer:
At sunrise, the outside temperature = 0° – 7° = -7°
In the afternoon the temperature = – 7° + 13°
= 6°
At night the temperature
= 6° – 8°
= – 2°
At the end of the day temperature – 2°.

8. Manjeet Singh has a bank balance of -₹430 at the start of the month. What was the bank balance, after he deposited ₹ 250 ?
Answer:
Manjeet Singh’s bank balance in the start of the month = -₹430
Amount deposited in the bank = ₹ 250
The bank balance after depositing = -₹430 + ₹ 250
= -₹(430 + 250)
= -₹180

9. Mount Everest, the highest elevation in Asia, is 29028 feet above the sea level. The Dead Sea is 1312 feet below the sea level. What is the difference between these two elevations ?
Answer:
The elevation of Mount Everest = + 29028 feet
The elevation of the Dead sea = – 1312 feet
Difference between these two elevations = [+ 29028 – (- 1312) feet
= (29098 + 1312) feet
= 30340 feet.

10. In a quiz, Team A scored 70, – 15, 30. Team B scored – 15, 70, 30 and team C score 30, 70, – 15. Which team scored better ? What conclusion do you draw ?

Answer:
Total scores scored by team A = 70 + (- 15) + 30
= 70 + 30 + (- 15)
= 100 – 15 = 85
Total scores scored by team B
= (- 15) + 70 + 30
= (- 15) + 100 = 85
Total scores scored by team C
= 30 + 70 + (- 15)
= 100 + (- 15) = 85
Scores are equal addition of integers is associative Ans.

11. In a competition there are 5 Teams and three rounds. The scores of all the teams are given below in the table. Complete the table and find, the teams at 1st, Ind and IIIrd positions.

Answer:

Ist – A, IInd – C, III – D.

12. Multiple choice questions :

Question (i).
(- 5) + (5) =
(a) -10
(b) 5
(c) 10
(d) 0.
Answer:
(d) 0.

Question (ii).
(- 10) + (- 12) =
(a) -2
(b) 22
(c) -22
(d) 2.
Answer:
(c) -22

Question (iii).
(- 1) – (-1) =
(a) – 2
(b) -1
(c) 2
(d) None of these.
Answer:
(d) None of these.

Question (iv).
Which of the following statements is incorrect ?
(a) Sum of two integers is also an integer.
(b) For all integers a and b, a + b = b + a.
(c) Difference of two integers is also an integer.
(d) Subtraction of integers is commutative.
Answer:
(d) Subtraction of integers is commutative.

Question (v).
Which of the following is correct ?
(a) (- 7) – (3) = 3 – (- 7)
(b) (- 7) + 3 = 3 + (- 7)
(c) (- 1) + [(5) + (- 3)] = [(- 1) + (5)] – (- 3)
(d) None of these.
Answer:
(b) (- 7) + 3 = 3 + (- 7)

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals Ex 2.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

1. Solve the following fractions :

Question (i).
4 + \(\frac {7}{8}\)
Answer:
4 + \(\frac {7}{8}\)
= \(\frac{4 \times 8+7}{8}\)
= \(\frac{32+7}{8}\)
= \(\frac {39}{8}\)
= 4\(\frac {7}{8}\)

Question (ii).
\(\frac{9}{11}-\frac{4}{15}\)
Answer:
\(\frac{9}{11}-\frac{4}{15}\)
= \(\frac{9 \times 15-4 \times 11}{11 \times 15}\)
= \(\frac{135-44}{165}\)
= \(\frac {91}{165}\)

Question (iii).
\(\frac{11}{16}-\frac{2}{5}+\frac{8}{10}\)
Answer:
\(\frac{11}{16}-\frac{2}{5}+\frac{8}{10}\)

LCM of 16, 5 and 10
= 2 × 2 × 2 × 2 × 5
= 80
= \(\frac{11 \times 5-2 \times 16+8 \times 8}{80}\)
= \(\frac{55-32+64}{80}\)
= \(\frac {87}{80}\)
= 1\(\frac {7}{80}\)

Question (iv).
\(2 \frac{1}{5}+6 \frac{1}{2}\)
Answer:
\(2 \frac{1}{5}+6 \frac{1}{2}\)
= \(\frac{11}{5}+\frac{13}{2}\)
= \(\frac{11 \times 2+13 \times 5}{5 \times 2}\)
= \(\frac{22+65}{10}\)
= \(\frac {87}{10}\)
= 8\(\frac {7}{10}\)

Question (v).
\(8 \frac{1}{2}-3 \frac{5}{8}\)
Answer:
\(8 \frac{1}{2}-3 \frac{5}{8}\)
= \(\frac{17}{2}-\frac{29}{8}\)
= \(\frac{17 \times 4-29}{8}\)
= \(\frac{68-29}{8}\)
= \(\frac {39}{8}\)
= 4\(\frac {7}{8}\)

Question (vi).
\(\frac{9}{10}-\frac{9}{100}+\frac{9}{1000}\)
Answer:
\(\frac{9}{10}-\frac{9}{100}+\frac{9}{1000}\)
= \(\frac{9 \times 100-9 \times 10+9}{1000}\)
= \(\frac{900-90+9}{1000}\)
= \(\frac {810}{1000}\)

2. Arrange the following in ascending order :

Question (i).
\(\frac{2}{17}, \frac{10}{17}, \frac{3}{17}, \frac{16}{17}, \frac{5}{17}, \frac{8}{17}\)
Answer:
Ascending order of \(\frac{2}{17}, \frac{10}{17}, \frac{3}{17}, \frac{16}{17}, \frac{5}{17}, \frac{8}{17}\) is:
\(\frac{2}{17}, \frac{3}{17}, \frac{5}{17}, \frac{8}{17}, \frac{10}{17}, \frac{16}{17}\)

Question (ii).
\(\frac{1}{5}, \frac{3}{7}, \frac{7}{10}\)
Answer:
\(\frac{1}{5}, \frac{3}{7}, \frac{7}{10}\)

L.C.M of 5, 7 , 10 = 2 × 5 × 7
= 70

3. The three sides AB, BC and CA of a triangle ΔABC are \(\frac {5}{6}\)cm, \(\frac {2}{3}\)cm and \(\frac {7}{10}\) cm respectively. Find the perimeter of the triangle.

Answer:
Slides of ΔABC are
AB = \(\frac {5}{6}\) cm,
BC = \(\frac {2}{3}\)
CA = \(\frac {7}{10}\)

L.C.M. (6, 3, 10) = 2 × 3 × 5 = 30
Perimeter of ΔABC = AB + BC + CA

4. Ramesh studies for 5\(\frac {2}{3}\) hours daily. He devotes 2\(\frac {4}{5}\) hours of his time for science devotes for other subjects ?
Answer:
Total daily time for all subjects
= 5\(\frac {2}{3}\) hours = \(\frac {17}{3}\) hours
Time for science and mathematics
= 2\(\frac {4}{5}\) hours = \(\frac {14}{5}\) hours
Time for other subjects

5. Sonia jogs once around the rectangular park of sides 10\(\frac {2}{3}\)m and 12\(\frac {1}{2}\)m. Find the total distance covered by the Sonia.

Answer:
Length of rectangular park
= 12\(\frac {1}{2}\)m = \(\frac {25}{2}\)m
Breadth of rectangular park
= 10\(\frac {2}{3}\)m = \(\frac {32}{3}\)m
Total distance covered by Sonia = 2 [Length + Breadth]
= \(2\left(\frac{32}{3}+\frac{25}{3}\right) \mathrm{m}\)
= \(2\left(\frac{32 \times 2+25 \times 3}{3 \times 2}\right) \mathrm{m}\)
= \(2\left(\frac{65+75}{6}\right) \mathrm{m}\)
= \(\frac {278}{6}\) m
= \(\frac {139}{3}\) m
= 46\(\frac {1}{3}\) m

6. Ritu coloured a picture in \(\frac {7}{12}\) hours. Vaibhav coloured the same picture in \(\frac {3}{4}\) hours. Who worked for a longer time and by what fraction ?
Answer:
Time taken by Ritu to colour
= \(\frac {7}{12}\) hours
Time taken by Vaibhav = \(\frac {3}{4}\) hours
= \(\frac {3}{4}\) × \(\frac {3}{3}\)
= \(\frac {9}{12}\) hours
Since 9 > 7
∴ \(\frac {9}{12}\) > \(\frac {7}{12}\)
∴ Vaibhav worked for more time.
Difference between time taken by
Vaibhav and Ritu = \(\frac{3}{4}-\frac{7}{12}\)
= \(\frac{3 \times 3-7}{12}\)
= \(\frac{9-7}{12}=\frac{2}{12}\)
= \(\frac {1}{6}\) of an hour.

7. Multiple Choice Questions :

Question (i).
Fraction \(\frac {2}{5}\), \(\frac {7}{5}\) are :
(a) Like fractions
(b) Unlike fractions
(c) Equivalent fractions
(d) None of these
Answer:
(a) Like fractions

Question (ii).
What fraction do 8 hours of a day represents ?
(a) \(\frac {1}{2}\)
(b) \(\frac {1}{3}\)
(c) \(\frac {8}{60}\)
(d) \(\frac {2}{3}\)
Answer:
(b) \(\frac {1}{3}\)

Question (iii).
Equivalent fraction of \(\frac {3}{5}\) is :
(a) \(\frac {13}{15}\)
(b) \(\frac {5}{3}\)
(c) \(\frac {9}{15}\)
(d) \(\frac {5}{13}\)
Answer:
(c) \(\frac {9}{15}\)

Question (iv).
Shaded area of given triangle represents the fractions:
(a) \(\frac {1}{3}\)
(b) \(\frac {3}{4}\)
(c) \(\frac {1}{4}\)
(d) \(\frac {2}{3}\)

Answer:
(b) \(\frac {3}{4}\)

Question (v).
Sum of fractions \(\frac {2}{7}\), \(\frac {3}{4}\) is equal to :
(a) \(\frac {5}{28}\)
(b) \(\frac {1}{3}\)
(c) \(\frac {5}{11}\)
(d) \(\frac {29}{28}\)
Answer:
(d) \(\frac {29}{28}\)

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 1 Integers MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 1 Integers MCQ Questions

Multiple Choice Questions

Question 1.
The value of -| – 21| is :
(a) 21
(b) -21
(c) 1
(d) None of these.
Answer:
(b) -21

Question 2.
17 + (-18) =
(a) 35
(b) 1
(c) -1
(d) -35.
Answer:
(c) -1

Question 3.
(-15) × 0 is equal to :
(a) 0
(b) -15
(c) 15
(d) 1.
Answer:
(a) 0

Question 4.
The product of 3 × -1 is :
(a) 3
(b) – 3
(c) 1
(d) -1.
Answer:
(b) – 3

Question 5.
(-8) ÷ (-1) is equal to :
(a) 8
(b) 1
(c) -8
(d) -1.
Answer:
(a) 8

Fill in the blanks

Question 1.
0 is greater than every …………….. integer
Answer:
Negative

Question 2.
25 – 10 = -10 + ……..
Answer:
25

Question 3.
15 × ……… = 0
Answer:
0

Question 4.
369 ÷ ……… = 369
Answer:
1

Question 5.
20 ÷ ……… = -2.
Answer:
-10

Write True or False

Question 1.
Sum of two integeres is also integer.
Answer:
True

Question 2.
(-7) + 3 = 3 + (-7) (True/False)
Answer:
True

Question 3.
-2 + 2 = 0 (True/False)
Answer:
True

Question 4.
1 ÷ a = 1 (True/False)
Answer:
False

Question 5.
a ÷ 1 = 0. (True/False)
Answer:
True

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 1 Integers Ex 1.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4

1. Evaluate each of the following :
(i) 76 ÷ 19
(ii) (-156) ÷ (-12)
(iii) (-125) ÷ (-1)
(iv) (125) ÷ (-25)
(v) 0 ÷ (-5)
(vi) (-15) ÷ (15)
Answer:

2. Write all even integers between -18 and 0.
Answer:
All even integers between – 18 and 0 are :
-16, -14, -12, -10, -8, -6, -4, -2.

3. Write all odd integers between -9 and 9.
Answer:
All odd integers between -9 and 9 are :
-7, -5, -3, -1, 1, 3, 5, 7.

4. By what number should (-240) be divided to obtain 16.
Answer:
Let the required number be x
∴ -240 ÷ x = 16

Hence, the required number is -15

5. Find the value of :

Question (i).
125 ÷ [5 ÷ (-1)]
Answer:
125 ÷ [5 ÷ (-1)] = 125 ÷ (-5)
= -25 Ans.

Question (ii).
[169 ÷ 13] ÷ [26 ÷ 2]
Answer:
[169 ÷ 13] ÷ [26 ÷ 2]
= [13] ÷ [13] = 1 Ans.

Question (iii).
[(-105) ÷ 3] ÷ 7
Answer:
[(-105) ÷ 3] ÷ 7
= [-35] ÷ 7
= -5.

6. Simplify : 12 – [8 + 27 ÷ (2 × 8 – 7)]
Answer:
12 – [8 + 27 + (2 × 8 – 7)]
= 12 – [8 + 27 ÷ (16 – 7)]
= 12 – [8 + 27 ÷ (9)]
= 12 – [8 + 3] = 12- 11
= 1

7. Simplify : 10 – [8 – {11 + 30 ÷ (4 + 2)}]
Answer:
10 – [8 – {11 + 30 (4 + 2)}]
= 10 – [8 – {11 + 30 ÷ 6}]
= 10 – [8 – (11 + 5)]
= 10 – [8 – 16]
= 10 – [-8]
= 10 + 8 = 18

8. Multiple Choice Questions :

Question (i).
(-8) ÷ 2 =
(a) -16
(b) -4
(c) 4
(d) -8.
Answer:
(b) -4

Question (ii).
(-7) ÷ (-7) =
(a) -1
(b) 49
(c) -49
(d) None of these.
Answer:
(d) None of these.

Question (iii).
0 ÷ 2 =
(a) 1
(b) 2
(c) -2
(d) 0.
Answer:
(d) 0.

9. The quotient of two integers is always an integer. (True/False)
Answer:
False.

10. If a and b are two unequal non-zero integers then a ÷ b = b ÷ a. (True/False)
Answer:
False.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 1 Integers Ex 1.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1

1. Use the appropriate symbol >, <, = to fill in the blanks

Answer:
(i) -3  >  -5
(ii) – 2  <  5-4
(iii) 8-4  >  – 3
(iv) – 6  <  5-0
(v) 5  =  8-3
(vi) 0  >  – 3.

2. Arrange the following integers in ascending order.

Question (i)
– 2, 12, – 43, 31, 7, – 35, – 10
Answer:
Given positive integers are 12, 31, 7
Ascending order is 7 < 12 < 31
Given negative integers are – 2, – 43, – 35, -10
Ascending order is – 43 < – 35 < – 10 < -2.
Hence, all given integers in ascending order are :
– 43 < – 35 < -10 < – 2 < 7 < 12 < 31.
i.e. – 43, – 35, – 10, – 2, 7, 12, 31.

Question (ii)
– 20, 13, 4, 0, – 5, 5
Answer:
Given positive integers are 13, 4, 5
Ascending order is 4 < 5 < 13
Given negative integers are – 20, – 5
Ascending order is -20 < – 5.
Hence all given integers in ascending order are :
– 20 < – 5 < 0 < 4 < 5 < 13
i.e. – 20, – 5, 0, 4, 5, 13

3. Arrange the following integers in descending order.

Question (i)
0, – 7, 19, – 23, – 3, 8, 46
Answer:
Given positive integers are 19, 8, 46
Descending order is 46 > 19 > 8
Given negative integers are – 7, – 23, – 3
Descending order is – 3 > – 1 > – 23
Hence, all given integers in descending order are :
46 > 19 > 8 > 0 > – 3 > – 7 > – 23
i.e. 46, 19, 8, 0, – 3, – 7, – 23

Question (ii)
30, – 2, 0, – 6, – 20, 8.
Answer:
Given positive integers are 30, 8
Descending order is 30 > 8
Given negative integers are – 2, – 6, – 20
Descending order is-2 > -6 > -20
Hence all given integers in descending order are :
30 > 8 > 0 > -2 > – 6 > – 20
i.e. 30, 8, 0, – 2, – 6, – 20

4. Evaluate :

Question (i)
30 – | -21 |
Answer:
30 – | -21 | = 30 – 21
[∵ | – 21 | = 21]
= 9

Question (ii)
| -25 | – | -18 |
Answer:
| -25 | – | – 18 | = 25 – 18
[∵ | -25 | = 25 and | – 18 | = 18]
= 7

Question (iii)
6 – | -4 |
Answer:
6 – | -4 | = 6 – 4
[∵ | – 4 | = 4]
= 2

Question (iv)
| – 125 | + | 110 |
Answer:
| – 125 | + | 110 | = 125 + 110
[∵ | -125 | = 125 and | 110 | = 110]
= 235

5. Fill in the blanks :

Question (i)
0 is greater than every …………… integer.
Answer:
Negative

Question (ii)
Modulus of a negative integer is always ……………
Answer:
Positive

Question (iii)
The smallest positive integer is ……………
Answer:
1

Question (iv)
The largest negative integer is ……………
Answer:
-1

Question (v)
Every negative integer is less than every …………… integer.
Answer:
Positive