Bhagya

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 ਸੰਖਿਆਵਾਂ ਉੱਪਰ ਮੁੱਢਲੀਆਂ ਕਿਰਿਆਵਾਂ Ex 2.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 2 ਸੰਖਿਆਵਾਂ ਉੱਪਰ ਮੁੱਢਲੀਆਂ ਕਿਰਿਆਵਾਂ Ex 2.6

ਪ੍ਰਸ਼ਨ 1.
ਇੱਕ ਕਾਪੀ ਦਾ ਮੁੱਲ ਤੋਂ 15 ਹੈ ।9 ਕਾਪੀਆਂ ਦਾ ਮੁੱਲ ਕਿੰਨਾ ਹੋਵੇਗਾ ?
ਹੱਲ:
ਇੱਕ ਕਾਪੀ ਦਾ ਮੁੱਲ = ₹ 15
9 ਕਾਪੀਆਂ ਦਾ ਮੁੱਲ = ₹ 15 × 9
= ₹ 135.

ਪ੍ਰਸ਼ਨ 2.
ਇੱਕ ਪੈਕਟ ਵਿੱਚ 75 ਪੈਨਸਿਲਾਂ ਆਉਂਦੀਆਂ ਹਨ । ਇਸ ਵਰਗੇ 19 ਪੈਕਟਾਂ ਵਿੱਚ ਕਿੰਨੀਆਂ ਪੈਨਸਿਲਾਂ ਆਉਣਗੀਆਂ ?
ਹੱਲ:
ਇੱਕ ਪੈਕਟ ਵਿੱਚ ਪੈਨਸਿਲਾਂ = 75
19 ਪੈਕਟਾਂ ਵਿੱਚ ਪੈਨਸਿਲਾਂ = 75 × 19
= 1425.

ਪ੍ਰਸ਼ਨ 3.
ਇੱਕ ਮਾਲਾ ਵਿੱਚ 79 ਮੋਤੀ ਹਨ । ਇਸ ਵਰਗੀਆਂ 68 ਮਾਲਾ ਵਿੱਚ ਕਿੰਨੇ ਮੋਤੀ ਹੋਣਗੇ ?
ਹੱਲ:
ਇੱਕ ਮਾਲਾ ਵਿਚ ਮੋਤੀ = 79
68 ਮਾਲਾ ਵਿਚ ਮੋਤੀ = 79 × 68
= 5372.

ਪ੍ਰਸ਼ਨ 4.
ਇੱਕ ਛੋਟੀ ਸਾਇਕਲ ਦਾ ਮੁੱਲ ਤੋਂ 1560 ਹੈ । ਅਜਿਹੀਆਂ 6 ਸਾਇਕਲਾਂ ਦਾ ਮੁੱਲ ਕਿੰਨਾ ਹੋਵੇਗਾ ?
ਹੱਲ:
ਇੱਕ ਛੋਟੀ ਸਾਇਕਲ ਦਾ ਮੁੱਲ = ₹ 1560
ਅਜਿਹੀਆਂ 6 ਸਾਇਕਲਾਂ ਦਾ ਮੁੱਲ = ₹ 1560 × 6 = ₹ 9360.

ਪ੍ਰਸ਼ਨ 5.
ਜੇਕਰ ਕ੍ਰਿਕੇਟ ਦੀ ਇੱਕ ਟੀਮ ਵਿੱਚ 11 ਖਿਡਾਰੀ ਹੋਣ ਤਾਂ 12 ਟੀਮਾਂ ਵਿੱਚ ਕਿੰਨੇ ਖਿਡਾਰੀ ਹੋਣਗੇ ?
ਹੱਲ:
ਇੱਕ ਕ੍ਰਿਕੇਟ ਟੀਮ ਵਿਚ ਖਿਡਾਰੀ = 11
12 ਟੀਮਾਂ ਵਿੱਚ ਖਿਡਾਰੀ = 11 × 12
= 132

ਪ੍ਰਸ਼ਨ 6.
ਇੱਕ ਡੱਬੇ ਵਿਚ 1440 ਸਾਬਣ ਦੀਆਂ ਟਿੱਕੀਆਂ ਹਨ । 6 ਡੱਬਿਆਂ ਵਿੱਚ ਸਾਬਣ ਦੀਆਂ ਕਿੰਨੀਆਂ ਟਿੱਕੀਆਂ ਹੋਣਗੀਆਂ ?
ਹੱਲ:
ਇੱਕ ਡੱਬੇ ਵਿੱਚ ਸਾਬਣ ਦੀਆਂ ਟਿੱਕੀਆਂ = 1440
6 ਡੱਬਿਆਂ ਵਿੱਚ ਸਾਬਣ ਦੀਆਂ ਟਿੱਕੀਆਂ = 1440 × 6 = 8640.

ਪ੍ਰਸ਼ਨ 7.

ਤੁਹਾਡੇ ਮਾਤਾ ਜੀ ਬਜ਼ਾਰ ਗਏ-
(a) ਉਨ੍ਹਾਂ ਨੇ 2 ਕਿਲੋਗ੍ਰਾਮ ਸੇਬ ਅਤੇ 2 ਕਿਲੋਗ੍ਰਾਮ ਅਮਰੂਦ ਖ਼ਰੀਦੇ । ਉਨ੍ਹਾਂ ਨੇ ਕਿੰਨੀ ਰਾਸ਼ੀ ਦੁਕਾਨਦਾਰ ਨੂੰ ਦਿੱਤੀ ਹੈ ?
ਹੱਲ:
2 ਕਿਲੋਗ੍ਰਾਮ ਸੇਬਾਂ ਦਾ ਮੁੱਲ = 120 × 2 = ₹ 240
2 ਕਿਲੋਗ੍ਰਾਮ ਅਮਰੂਦ ਦਾ ਮੁੱਲ = ₹ 35 × 2 = ₹ 70

ਦੁਕਾਨਦਾਰ ਨੂੰ ਕੁੱਲ ਰਾਸ਼ੀ ਦਿੱਤੀ = ₹ 310.

(b) ਜੇਕਰ ਉਹ 3 ਕਿਲੋਗ੍ਰਾਮ ਸੰਤਰੇ ਅਤੇ 2 ਕਿਲੋਗ੍ਰਾਮ ਅਨਾਰ ਖਰੀਦਣ ਤਾਂ ਉਨ੍ਹਾਂ ਨੂੰ ਕਿੰਨੇ ਰੁਪਏ ਦੇਣੇ ਪੈਣਗੇ ?
ਹੱਲ:
3 ਕਿਲੋਗ੍ਰਾਮ ਸੰਤਰੇ ਦਾ ਮੁੱਲ = ₹ 45 × 3 = ₹ 135
2 ਕਿਲੋਗ੍ਰਾਮ ਅਨਾਰ ਦਾ ਮੁੱਲ ₹ 140 × 2 = ₹ 280

ਜਿੰਨੀ ਰਾਸ਼ੀ ਦੇਣੀ ਪਵੇਗੀ = ₹ 415

ਪ੍ਰਸ਼ਨ 8.
ਹੇਠ ਦਿੱਤੇ ਸਾਰੇ ਨੋਟ ਅਤੇ ਸਿੱਕੇ, ਕਰਨ ਨੂੰ ਉਸ ਦੇ ਜਨਮਦਿਨ ‘ਤੇ ਮਿਲੇ । ਤੁਸੀਂ ਦੱਸੋ ਕਿ ਕਰਨ ਕੋਲ ਕਿੰਨੇ ਰੁਪਏ ਹਨ ?

ਹੱਲ:
ਕਰਨ ਕੋਲ ਕੁੱਲ ਰੁਪਏ ਹਨ = ₹ 500 × 5 + ₹ 50 × 3 + ₹ 10 × 7 + ₹ 2 × 3
= ₹ 2500 + ₹ 150 + ₹ 70 + 6
= ₹ 2726.

ਪ੍ਰਸ਼ਨ 9.
ਇੱਕ ਕਾਰ 1 ਲੀਟਰ ਪੈਟਰੋਲ ਨਾਲ 16 ਕਿਲੋਮੀਟਰ ਦੀ ਦੂਰੀ ਤੈਅ ਕਰਦੀ ਹੈ । 28. ਲੀਟਰ ਪੈਟਰੋਲ ਨਾਲ ਉਹ ਕਾਰ ਕਿੰਨੀ ਦੂਰੀ ਤੈਅ ਕਰੇਗੀ ?
ਹੱਲ:
1 ਲੀਟਰ ਪੈਟਰੋਲ ਨਾਲ ਤੈਅ ਕੀਤੀ ਦੂਰੀ = 16 ਕਿਲੋਮੀਟਰ
28 ਲੀਟਰ ਪੈਟਰੋਲ ਨਾਲ ਤੈਅ ਕੀਤੀ ਦੂਰੀ = 16 ਕਿਲੋਮੀਟਰ × 28 = 448 ਕਿਲੋਮੀਟਰ ।

ਪ੍ਰਸ਼ਨ 10.
ਇੱਕ ਫੈਕਟਰੀ ਵਿੱਚ ਇਕ ਘੰਟੇ ਵਿੱਚ 125 ਸਾਬਣ ਦੀਆਂ ਟਿੱਕੀਆਂ ਬਣਦੀਆਂ ਹਨ ? 8 ਘੰਟਿਆਂ ਵਿੱਚ ਕਿੰਨੀਆਂ ਸਾਬਣ ਦੀਆਂ ਟਿੱਕੀਆਂ ਬਣਨਗੀਆਂ ?
ਹੱਲ:
1 ਘੰਟੇ ਵਿੱਚ ਸਾਬਣ ਦੀਆਂ ਟਿੱਕੀਆਂ ਬਣਦੀਆਂ ਹਨ = 125
8 ਘੰਟੇ ਵਿੱਚ ਸਾਬਣ ਦੀਆਂ ਟਿੱਕੀਆਂ ਬਣਦੀਆਂ ਹਨ = 8 × 125
= 1000

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 6 Squares and Square Roots Ex 6.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.3

1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers ?

Question (i).
9801
Solution:
The possible digit at one’s place of the square root of 9801 can be either 1 or 9.
(∵ 1 × 1 = 1 and 9 × 9 = 81)

Question (ii).
99856
Solution:
The possible digit at one’s place of the square root of 99856 can be either 4 or 6.
(∵ 4 × 4= 16 and 6 × 6 = 36)

Question (iii).
998001
Solution:
The possible digit at one’s place of the square root of 998001 can be either 1 or 9.
(∵ 1 × 1 = 1 and 9 × 9 = 81)

Question (iv).
657666025
Solution:
The possible digit at one’s place of the square root of 657666025 can be 5.
(∵ 5 × 5 = 25)

2. Without doing any calculation, find the numbers which are surely not perfect squares:
[Note : The ending digit of perfect square is 0, 1, 4, 5, 6 or 9.
∴ A number having end digit 2, 3, 7 or 8 can never be a perfect square.

Question (i).
153
Solution:
153
Here, the end digit is 3.
∴ 153 cannot be a perfect square.

Question (ii).
257
Solution:
257
Here, the end digit is 7.
∴ 257 cannot be a perfect square.

Question (iii).
408
Solution:
408
Here, the end digit is 8.
∴ 408 cannot be a perfect square.

Question (iv).
441
Solution:
441
Here, the end digit is 1.
∴ 441 can be a perfect square.

3. Find the square roots of 100 and 169 by the method of repeated subtraction.

Question (i).
100
Solution:
100 – 1 = 99   99 – 3 = 96
96 – 5 = 91   91 – 7 = 84
84 – 9 = 75   75 – 11 = 64
64 – 13 = 51   51 – 15 = 36
36 – 17 = 19   19 – 19 = 0
∴ 100 is a perfect square.
∴ \(\sqrt{100}\) = 10

Question (ii).
169
Solution:
169 – 1 = 168   168 – 3 = 165
165-5 = 160   160 – 7 = 153
153-9 = 144   144 – 11 = 133
133-13 = 120   120 – 15 = 105
105-17 = 88   88 – 19 = 69
69-21 = 48   48 – 23 = 25
25 – 25 = 0
∴ 169 is a perfect square.
∴ \(\sqrt{169}\) = 13

4. Find the square roots of the following numbers by the Prime Factorisation Method:

Question (i).
729
Solution:
729
\(\begin{array}{l|r}
3 & 729 \\
\hline 3 & 243 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
729 = 3 × 3 × 3 × 3 × 3 × 3
= 3² × 3² × 3²
∴ \(\sqrt{729}\) = 3 × 3 × 3
= 27

Question (ii).
400
Solution:
400
\(\begin{array}{l|r}
2 & 400 \\
\hline 2 & 200 \\
\hline 2 & 100 \\
\hline 2 & 50 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
400 = 2 × 2 × 2 × 2 × 5 × 5
= 2² × 2² × 5²
∴ \(\sqrt{400}\) = 2 × 2 × 5
= 20

Question (iii).
1764
Solution:
1764
\(\begin{array}{l|r}
2 & 1764 \\
\hline 2 & 882 \\
\hline 3 & 441 \\
\hline 3 & 147 \\
\hline 7 & 49 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
1764 = 2 × 2 × 3 × 3 × 7 × 7
= 2² × 3² × 7²
∴ \(\sqrt{1764}\) = 2 × 3 × 7
= 42

Question (iv).
4096
Solution:
4096
\(\begin{array}{l|r}
2 & 4096 \\
\hline 2 & 2048 \\
\hline 2 & 1024 \\
\hline 2 & 512 \\
\hline 2 & 256 \\
\hline 2 & 128 \\
\hline 2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)
4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2² × 2² × 2² × 2² × 2² × 2²
∴ \(\sqrt{4096}\) = 2 × 2 × 2 × 2 × 2 × 2
= 64

Question (v).
7744
Solution:
7744
\(\begin{array}{r|r}
2 & 7744 \\
\hline 2 & 3872 \\
\hline 2 & 1936 \\
\hline 2 & 968 \\
\hline 2 & 484 \\
\hline 2 & 242 \\
\hline 11 & 121 \\
\hline 11 & 11 \\
\hline & 1
\end{array}\)
7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11
= 2² × 2² × 2² × 11²
∴ \(\sqrt{7744}\) = 2 × 2 × 2 × 11
= 88

Question (vi).
9604
Solution:
9604
\(\begin{array}{l|r}
2 & 9604 \\
\hline 2 & 4802 \\
\hline 7 & 2401 \\
\hline 7 & 343 \\
\hline 7 & 49 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
9604 = 2 × 2 × 7 × 7 × 7 × 7
= 2² × 7² × 7²
∴ \(\sqrt{9604}\) =2 × 7 × 7
= 98

Question (vii).
5929
Solution:
5929
\(\begin{array}{r|r}
7 & 5929 \\
\hline 7 & 847 \\
\hline 11 & 121 \\
\hline 11 & 11 \\
\hline & 1
\end{array}\)
5929 = 7 × 7 × 11 × 11
= 7² × 11²
∴ \(\sqrt{5929}\) = 7 × 11
= 77

Question (viii).
9216
Solution:
9216
\(\begin{array}{r|r}
2 & 9216 \\
\hline 2 & 4608 \\
\hline 2 & 2304 \\
\hline 2 & 1152 \\
\hline 2 & 576 \\
\hline 2 & 288 \\
\hline 2 & 144 \\
\hline 2 & 72 \\
\hline 2 & 36 \\
\hline 2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
= 2² × 2² × 2² × 2² × 2² × 3²
∴ \(\sqrt{9216}\) = 2 × 2 × 2 × 2 × 2 × 3
= 96

Question (ix).
529
Solution:
529
\(\begin{array}{l|r}
23 & 529 \\
\hline 23 & 23 \\
\hline & 1
\end{array}\)
529 = 23 × 23
= 23²
∴ \(\sqrt{529}\) = 23

Question (x).
8100
Solution:
8100
\(\begin{array}{l|r}
2 & 8100 \\
\hline 2 & 4050 \\
\hline 3 & 2025 \\
\hline 3 & 675 \\
\hline 3 & 225 \\
\hline 3 & 75 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5
= 2² × 3² × 3² × 5²
∴ \(\sqrt{8100}\) = 2 × 3 × 3 × 5
= 90

5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained:

Question (i).
252
Solution:
252
\(\begin{array}{l|r}
2 & 252 \\
\hline 2 & 126 \\
\hline 3 & 63 \\
\hline 3 & 21 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
252 = 2 × 2 × 3 × 3 × 7
The prime factor 7 is unpaired.
∴ [252] × 7 = [2 × 2 × 3 × 3 × 7] × 7
1764 = 2 × 2 × 3 × 3 × 7 × 7
= 2² × 3² × 7²
∴ \(\sqrt{1764}\) = 2 × 3 × 7 = 42
Thus, 252 should be multiplied by smallest whole number 7 to get a perfect square.

Question (ii).
180
Solution:
180
\(\begin{array}{l|r}
2 & 180 \\
\hline 2 & 90 \\
\hline 3 & 45 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
180 = 2 × 2 × 3 × 3 × 5
Here, the prime factor 5 is unpaired.
∴ [180] × 5 = [2 × 2 × 3 × 3 × 5] × 5
∴ 900 = 2 × 2 × 3 × 3 × 5 × 5
= 2² × 3² × 5²
∴ \(\sqrt{900}\) = 2 × 3 × 5 = 30
Thus, 180 should be multiplied by smallest whole number 5 to get a perfect square.

Question (iii).
1008
Solution:
1008
\(\begin{array}{l|r}
2 & 1008 \\
\hline 2 & 504 \\
\hline 2 & 252 \\
\hline 2 & 126 \\
\hline 3 & 63 \\
\hline 3 & 21 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7
Here, the prime factor 7 is unpaired.
∴ [1008] × 7 = [2 × 2 × 2 × 2 × 3 × 3 × 7] × 7
∴ 7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
= 2² × 2² × 3² × 7²
∴ \(\sqrt{7056}\) = 2 × 2 × 3 × 7 = 84
Thus, 1008 should be multiplied by smallest whole number 7 to get a perfect square.

Question (iv).
2028
Solution:
2028
\(\begin{array}{r|r}
2 & 2028 \\
\hline 2 & 1014 \\
\hline 3 & 507 \\
\hline 13 & 169 \\
\hline 13 & 13 \\
\hline & 1
\end{array}\)
2028 = 2 × 2 × 3 × 13 × 13
Here, the prime factor 3 is unpaired.
∴ [2028] × 3 = [2 × 2 × 3 × 13 × 13] × 3
∴ 6084 = 2 × 2 × 3 × 3 × 13 × 13
= 2² × 3² × 13²
∴ \(\sqrt{6084}\) = 2 × 3 × 13 = 78
Thus, 2028 should be multiplied by smallest whole number 3 to get a perfect square.

Question (v).
1458
Solution:
1458
\(\begin{array}{l|r}
2 & 1458 \\
\hline 3 & 729 \\
\hline 3 & 243 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3
Here, the prime factor 2 is unpaired.
∴ [1458] × 2 = [2 × 3 × 3 × 3 × 3 × 3 × 3] × 2
∴ 2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
= 2² × 3² × 3² × 3²
∴ \(\sqrt{2916}\) = 2 × 3 × 3 × 3 = 54
Thus, 1458 should be multiplied by smallest whole number 2 to get a perfect square.

Question (vi).
768
Solution:
768
\(\begin{array}{l|r}
2 & 768 \\
\hline 2 & 384 \\
\hline 2 & 192 \\
\hline 2 & 96 \\
\hline 2 & 48 \\
\hline 2 & 24 \\
\hline 2 & 12 \\
\hline 2 & 6 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
Here, the prime factor 3 is unpaired.
∴ [768] × 3 = [2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3] × 3
∴ 2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
= 2² × 2² × 2² × 2² × 3²
∴ \(\sqrt{2304}\) = 2 × 2 × 2 × 2 × 3 = 48
Thus, 768 should be multiplied by smallest whole number 3 to get a perfect square.

6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained:

Question (i).
252
Solution:
252
\(\begin{array}{r|r}
2 & 252 \\
\hline 2 & 126 \\
\hline 3 & 63 \\
\hline 3 & 21 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
252 = 2 × 2 × 3 × 3 × 7
Here, the prime factor 7 is unpaired. So, given number should be divided by 7.
∴ [252] ÷ 7 = [2 × 2 × 3 × 3 × 7] ÷ 7
∴ 36 = 2 × 2 × 3 × 3
= 2² × 3²
∴ \(\sqrt{36}\) = 2 × 3 = 6
Thus, 252 should be divided by smallest whole number 7 to get a perfect square number.

Question (ii).
2925
Solution:
2925
\(\begin{array}{r|r}
3 & 2925 \\
\hline 3 & 975 \\
\hline 5 & 325 \\
\hline 5 & 65 \\
\hline 13 & 13 \\
\hline & 1
\end{array}\)
2925 = 3 × 3 × 5 × 5 × 13
Here, the prime factor 13 is unpaired. So, given number should be divided by 13.
∴ [2925] ÷ 13 = [3 × 3 × 5 × 5 × 13] ÷ 13
∴ 225 = 3 × 3 × 5 × 5
= 3² × 5²
∴ \(\sqrt{225}\) = 3 × 5 = 15
Thus, 2925 should be divided by smallest whole number 13 to get a perfect square number.

Question (iii).
396
Solution:
396
\(\begin{array}{r|r}
2 & 396 \\
\hline 2 & 198 \\
\hline 3 & 99 \\
\hline 3 & 33 \\
\hline 11 & 11 \\
\hline & 1
\end{array}\)
396 = 2 × 2 × 3 × 3 × 11
Here, the prime factor 11 is unpaired. So, given number should be divided by 11.
∴ [396] ÷ 11 = [2 × 2 × 3 × 3 × 11] ÷ 11
∴ 36 = 2 × 2 × 3 × 3
= 2² × 3²
∴ \(\sqrt{36}\) = 2 × 3 = 6
Thus, 396 should be divided by smallest whole number 11 to get a perfect square number.

Question (iv).
2645
Solution:
2645
\(\begin{array}{r|r}
5 & 2645 \\
\hline 23 & 529 \\
\hline 23 & 23 \\
\hline & 1
\end{array}\)
2645 = 5 × 23 × 23
Here, the prime factor 5 is unpaired. So, given number should be divided by 5.
∴ [2645] ÷ 5 = [5 × 23 × 23] ÷ 5
∴ 529 = 23 × 23 = 23²
∴ \(\sqrt{529}\) = 23
Thus, 2645 should be divided by smallest whole number 5 to get a perfect square number.

Question (v).
2800
Solution:
2800
\(\begin{array}{l|r}
2 & 2800 \\
\hline 2 & 1400 \\
\hline 2 & 700 \\
\hline 2 & 350 \\
\hline 5 & 175 \\
\hline 5 & 35 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7
Here, the prime number 7 is unpaired. So, given number should be divided by 7.
∴ [2800] ÷ 7 = [2 × 2 × 2 × 2 × 5 × 5 × 7] ÷ 7
∴ 400 = 2 × 2 × 2 × 2 × 5 × 5
= 2² × 2² × 5²
∴ \(\sqrt{400}\) = 2 × 2 × 5 = 20
Thus, 2800 should be divided by smallest whole number 7 to get a perfect square number.

Question (vi).
1620
Solution:
1620
\(\begin{array}{l|r}
2 & 1620 \\
\hline 2 & 810 \\
\hline 3 & 405 \\
\hline 3 & 135 \\
\hline 3 & 45 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5
Here, the prime factor 5 is unpaired. So, given number should be divided by 5.
∴ [1620] ÷ 5 = [2 × 2 × 3 × 3 × 3 × 3 × 5] ÷ 5
∴ 324 = 2 × 2 × 3 × 3 × 3 × 3
= 2² × 3² × 3²
∴ \(\sqrt{324}\) = 2 × 3 × 3 = 18
Thus, 1620 should be divided by 5 to get a perfect square number.

7. The students of Class VIII of a school donated ₹ 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.
Solution:
Let the number of students be x.
Amount each student donated = Number of students in the class.
So, amount donated by each student = ₹ x
Total amount donated by class = ₹ x × x = x²
\(\begin{array}{l|r}
7 & 2401 \\
\hline 7 & 343 \\
\hline 7 & 49 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
∴ x² = 2401
∴ \(\sqrt{x^{2}}=\sqrt{2401}\)
∴ x = \(\sqrt{7 \times 7 \times 7 \times 7}\)
= \(\sqrt{7^{2} \times 7^{2}}\)
∴ x = 7 × 7 = 49
Hence, number of students in the class is 49.

8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Solution:
Let the number of rows be x.
Number of rows = Number of plants in each row
So, number of plants in a row = x
∴ Number of plants to be planted in a garden = x × x = x²
\(\begin{array}{l|r}
3 & 2025 \\
\hline 3 & 675 \\
\hline 3 & 225 \\
\hline 3 & 75 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
∴ x² = 2025
∴ \(\sqrt{x^{2}}=\sqrt{2025}\)
∴ x = \(\sqrt{3 \times 3 \times 3 \times 3 \times 5 \times 5}\)
∴ \(\sqrt{3^{2} \times 3^{2} \times 5^{2}}\)
∴ x = 3 × 3 × 5 = 45
Hence, the number of rows is 45 and the number of plants in each row is 45.

9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Solution :
[Note: LCM is the number, which is divided by all factors of it without leaving remainder. ]
Here, the smallest square number divisible by each one of 4, 9 and 10 is equal to some multiple of the LCM of 4, 9 and 10.
\(\begin{array}{l|ll}
2 & 4, & 9, & 10 \\
\hline 2 & 2, & 9, & 5 \\
\hline 3 & 1, & 9, & 5 \\
\hline 3 & 1, & 3, & 5 \\
\hline 5 & 1, & 1, & 5 \\
\hline & 1, & 1, & 1
\end{array}\)
LCM of 4, 9 and 10 = 2 × 2 × 3 × 3 × 5 = 180
The prime factor 5 is unpaired.
So, 180 must be multiplied by 5.
∴ [180] × 5 = [2 × 2 × 3 × 3 × 5] × 5
∴ 900 = 2² × 3² × 5²
Hence, 900 is the required perfect square number.

10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
Solution:
[Note: LCM is the number, which is divided by all factors of it without leaving remainder.]
Here, the smallest square number divisible by each of 8, 15 and 20 is equal to some multiple of the LCM of 8, 15 and 20.
\(\begin{array}{r|rrr}
2 & 8, & 15, & 20 \\
\hline 2 & 4, & 15, & 10 \\
\hline 2 & 2, & 15, & 5 \\
\hline 3 & 1, & 15, & 5 \\
\hline 5 & 1, & 5, & 5 \\
\hline & 1, & 1, & 1
\end{array}\)
LCM of 8, 15 and 20 = 2 × 2 × 2 × 3 × 5 = 120
The prime factors 2, 3 and 5 are unpaired.
So, 120 should be multiplied by 2 × 3 × 5 = 30.
∴ [120] × 2 × 3 × 5 = [2 × 2 × 2 × 3 × 5] × 2 × 3 × 5
∴ 3600 = 2 × 2 × 2 × 3 × 5 × 2 × 3 × 5
∴ 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
Hence, 3600 is the required perfect square number.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1

1. Identify the terms, their coefficients for each of the following expressions:

(i) 5xyz² – 3zy
(ii) 1 + x + x²
(iii) 4x²y² – 4x²y²z² + z²
(iv) 3 – pq + qr – rp
(v) \(\frac{x}{2}+\frac{y}{2}\) – xy
(vi) 0.3a – 0.6ab + 0.5b
Solution:

Terms	Coefficient of terms
(i) 5 xyz2 -3zy	5 – 3
(ii) 1 x x2 ‘	1 1 1
(iii) 4x2y2 – 4x2y2z2 z2	4 -4 1
(iv) 3 – pq qr – rp	3 – 1 1 – 1
(v) \(\frac {x}{2}\) \(\frac {y}{2}\) – xy	\(\frac {1}{2}\) \(\frac {1}{2}\) – 1
(vi) 0.3 a – 0.6ab 0.5b	0.3 – 0.6 0.5

2. Classify the following polynomials as monomials, binomials, trinomials. which polynomials do not fit in any of these three categories?
x + y, 1000, x + x² + x³ + x⁴, 7 + y + 5x, 2y – 3y², 2y – 3y² + 4y³, 5x – 4y + 3xy, 4z – 15z², ab + bc + cd + da, pqr, p²q + pq², 2p + 2q.
Solution:

Monomials	Binomials	Trinomials
1000 pqr	x + y 2y – 3y2 4z – 15y2 p2q +pq2 2p + 2q	7 + y + 5x 2y – 3y2 + 4y3 5x – 4y + 3xy

Following polynomials do not fit in any catagories:
x + x² + x³ + x⁴ [∵ Polynomial has 4 terms] ab + bc + cd + da [∵ Polynomial has 4 terms]

3. Add the following:

Question (i)
ab – bc, bc – ca, ca – ab
Solution:
To add, let us arrange like terms one below the other.

Question (ii)
a – b + ab, b – c + bc, c – a + ac
Solution:

Question (iii)
2p²q² – 3pq + 4, 5 + 7pq – 3p²q²
Solution:

Question (iv)
l² + m², m² + n², n² + l², 2lm + 2mn + 2nl
Solution:

4.

Question (a)
Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
Solution:

Question (b)
Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
Solution:

Question (c)
Subtract 4p²q – 3pq + 5pq² – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq² + 5p²q
Solution:

Punjab State Board PSEB 6th Class English Book Solutions English Grammar Noun Exercise Questions and Answers, Notes.

PSEB 6th Class English Grammar Noun

A noun is a name of a person, place or thing; as-
किसी व्यक्ति स्थान अथवा वस्तु के नाम को अंग्रेजी में noun कहते हैं।

India. Mohan, taxi, class, toy, boy, table, etc.
Look at these sentences:

1. Geeta went to Patiala.
2. The fox is looking at the grapes.
3. The boys are playing football.

The underlined words in the above sentences are all nouns because they are the names of some person, place, animal or thing.

There are four kinds of noun:

1. Common Noun
2. Proper Noun
3. Abstract Noun
4. Collective Noun.

1. Common Nouns

A Common Noun is the name of every person, place or thing of the same class; as- pen, cow. bird, man, animal, bridge.
Look at these sentences:

1. The boys are playing.
2. These mangoes are pulpy.
3. The birds build nests.

The underlined words in the given sentences are Common Nouns because they are common to every person, place or thing.

Exercises (Solved)

I. Underline the Common Nouns in the following sentences:

1. Keep the books on the table.
2. The shops are closed today.
3. The tiger lives in the forest.
4. The farmer bought a tractor.
5. This building has many offices.
6. There is a dairy near our house.
7. Ail birds do not build their nests.
8. A fish lives in water and not on land.

Hints:
1. books, table
2. shops
3. tiger, forest
4. farmer, tractor building, offices
6. dairy, house
7. birds, nests
8. fish.

II. Add five Common Nouns in each set:

1. birds : parrot, sparrow, peacock, crow, niehtineale. pigeon.
2, colours : red, white, black, green, yellow, orange
3. games : hockey, football, volleyball, cricket, basketball, baseball.
4. animals : dog, camel, cow, sheep, buffalo, goat.
5. vegetables : potato, tomato, cabbage, radish, brinial, pumpkin.
6. fruits : mango, grape, apple, banana, papaya, pear.
7. In a school : library, science room, assembly hall office, staff-room, class room
8. In a house : kitchen, bathroom, dining room, store, bed-room, guest room.

2. Proper Nouns

A Proper Noun is the name of some particular (विशेष) person.
Delhi, Kolkata, Beas, Kama!

Look at these sentences:

1. Moti loves to play.
2. My brother lives in Amritsar.
3. J.C. Bose was a great scientist.
4. The Shan-e-Puniab has left just now.

The underlined w’ords in the above sentences are proper nouns because they are the names of particular persons, places or things.

Note that-
A Proper Noun always begins with a capital letter.

Proper Nouns include (शामिल हैं) the names of people, countries, cities, villages, rivers, ships, streets, buildings, mountains, seas, months of the year, days of the week, festivals, etc.

Exercises (Solved)

I. Underline the Proper Nouns in the following sentences

1. We named the cat Silky.
2. Kabir was a great saint.
3. We visited the. Taj in Agra.
4. Delhi is the capital of India.
5. I have never been to Mumbai.
6. Misha and Monu went to Delhi.
7. Do you know Sunny and Chinkv ?
8. We visited the Golden Temple on Sunday.

Hints:
1. Silky
2. Kabir
3. Taj. Agra
4. Delhi. India
5. Mumbai
6. Misha, Manu, Delhi
7. Sunny, Chinky
8. Golden Temple.

II. Rewrite each Proper Noun correctly in these sentences:

1. Have you visited the taj mahal ?
Have you visited the Taj Mahal ?
2. lam going to ropar on Monday.
3. The amritsar mail goes to kolkata.
4. muslims go to mosques on fridays,
5. black beauty is the story of a horse.
6. Where were the last Olympics held ?
7. bill clinton was the president of ameriea.
Hints:
2. Ropar, Monday
3. Amritsar Mail, Kolkata
4. Muslims, Mosques, Fridays
5. Black Beauty
6. Olympics
7. Bill Clinton, America.

3. Abstract Nouns

An Abstract Noun is the name of a quality, feeling or state (गुण भाव या स्थिति); as-
goodness, hardness, wisdom, love, hatred, theft, boyhood, slavery, freedom.

Look at these sentences:

1. Fire gives us heat.
2. He had pain in his body.
3. He acted upon my advice.
4. Do you know the depth of this well ?

The underlined words in the given sentences are abstract nouns because they refer to some quality, feeling or state.

The following words are all Abstract Nouns:
theft
peace
poverty
kindness
hope
misery
honesty
darkness
truth
greed
courage
weakness
sleep
sorrow
sickness
childhood
death
hunger
patience
treatment

Exercises (Solved)

I. Underline the Abstract Nouns in the following sentences:

1. Please control your anger.
2. Honesty is the best policy.
3. There was silence all around.
4. We get knowledge from books.
5. There was darkness in the room.
6. What is the height of this building?
7. You should have kindness for the poor.
8. Wars always bring death and destruction.
Hints:
1. anger
2. Honesty, best policy
3. silence
4. knowledge
5. darkness
6. Height
7. kindness
8. death, destruction.

II. Form Abstract Nouns from the given words:

laugh – laughter
hate – hatred
true – truth
treat – treatment
child – childhood
soft – softness
cruel – cruelty
bright – brightness
brave – bravery
strong – strength
punctual – punctuality
dangerous – danger

III. Use any five Abstract Nouns in sentences of your own:

1. She likes the softness of her skin.
2. Always speak the truth.
3. He was rewarded for his bravery.
4. Punctuality is a great virtue.
5. Her life was in danger.

4. Collective Nouns

A Collective Noun is the name of a group of persons, animals or things of the same kind; as-
flock, cattle, class, army, family, committee.

Look at these sentences:

1. Our army won the battle.
2. I have lost my bunch of keys.
3. The cattle are grazing in the field.

The underlined words in the above sentences are collective nouns because they refer to a collection of persons or things of the same kind.
The word army is a collection of soldiers.
The word cattle is a collection of farm animals.
The word bunch is a collection of things tied together.

Learn the following Collective Nouns:
1. a shoal of fish
2. a hive of bees
3. a pride of lions
4. a herd of cattle
5. a flight of stairs
6. a bunch of keys
7. a flock of sheep
8. a crew of sailors
9. a heap of stones
10. a string of pearls
11. a suite of rooms
12. a basket of fruits
13. a gang of thieves
14. a library of books
15. a bundle of sticks
16. a bench of judges
17. a crowd of people
18. a brood of chickens
19. a band of musicians
20. a wardrobe of clothes
21. a regiment of soldiers
22. a fleet of ships or cars
23. a litter of pups / piglets
24. a pack of cards / wolves

Exercises (Solved)

I. Match the Collective Nouns with the given phrases:

1. A collection of pups	Pack
2. A collection of ships	flock
3. A collection of sheep	fleet
4. A collection of books	suite
5. A collection of rooms	litter
6. A collection of wolves	herd
7. A collection of flowers	library
8. A collection of elephants	bouquet

Hints:
1. litter
2. fleet
3. flock
4. library
5. suite
6. pack
7. bouquet
8. herd.

II. Fill in the blanks with suitable Collective Nouns:

1. A filght of stairs.
2. A ………… of fish.
3. A ………… of lions.
4. A ………… of cows.
5. A ………… of cards.
6. A ………… of fruits.
7. A ………… of pearls.
8. A ………… of judges.
9. A ………… of grapes.
10. A ………… of clothes.
11. A ………… of thieves.
12. A ………… of soldiers.

Hints:
2. shoal
3. pride
4. herd
5. pack
6. basket
7. string
8. bench
9. bunch
10. wardrobe
11. gang
12. regiment.

Miscellaneous Exercises (Solved)

I. What is a Noun ?

II. Name the different kinds of Noun.
Give two examples of each.

III. The italicized words in the following sentences are Nouns. Classify these Nouns (Common / Proper /Abstract / Collective):

1. He won much praise.
2. Nitin lives in Mumbai.
3. I saw a flock of sheep.
4. Silver is a white metal.
5. You cannot cheat God.
6. My sweater is made of wool.
7. I bought some new furniture.
8. The old woman was very happy now.
Hints:
1. praise – abstract
2. Mumbai – roper
3. sheep – common
4. silver – common, metal- common
5. God – proper
6. sweater – common, wool – common
7. furniture – collective
8. woman – common.

IV. Choose suitable Nouns to fill in the blanks:

duty, profit, courage, marriage, need, weight, freedom, childhood

1. Be careful about your weight.
2. We want to live in …………..
3. Her ………….. took place last month.
4. It is our………….. to obey our parents.
5. Seema lost her parents in her …………..
6. We helped him when he was in …………..
7. The soldier was rewarded for his …………..
8. Jatin made good ………….. from his business.
Hints:
2. freedom
3. marriage
4. duty
5. childhood
6. need
7. courage
8. profit.

V. Pick out the Nouns in the following sentences and say whether they are Common, Proper, Collective or Abstract:

1. I love music.
2. Meera studies in sixth class.
3. Ludhiana is an industrial city.
4. He bought a doll for his sister.
5. These tables are made of wood.
6. A drowning man catches at a straw.
7. His father left for London yesterday.
8. Mathematics is my favourite subject.
Hints:
1. music – abstract.
2. Meera – proper; class – collective.
3. Ludhiana – proper, city – common.
4. doll – common, sister – common.
5. tables – common, wood – common.
6. man – common, straw – common.
7. father – common, London – proper.
8. Mathematics – collective, subject – common.

VI. Choose a suitable Abstract Noun to match each phrase:

pride, silence, poverty, courage, strength, greatness, innocence, intelligence

1. A quiet room [silence]
2. A clever boy
3. A great king
4. A strong girl
5. A proud child
6. A poor beggar
7. A brave policeman
8. An innocent woman
Hints:
1. silence
2. intelligence
3. greatness
4. strength
5. pride
6. poverty
7. courage
8. innocence.

The Noun – Number

Singular and Plural Nouns
A noun that refers to one thing is said to be Singular; as-
ball, chair, book, town, cow etc.

A noun that refers to more than one thing is said to be Plural; as-
books, balls, chairs, towns, animals, etc.

Now look at these sentences:

1. Rita has three dolls.
2. Reema has a bag of books.
3. All the babies were crying.
4. Joy got a big ball on his birthday.

The underlined nouns in the above sentences are either singular or plural. They tell whether they refer to one or more than one thing.

Forming Plurals of Nouns

1. As a general rule, the plural of a noun is formed by adding -s to a singular noun.

Singular – Plural
cat – cats
cap – caps
ball – balls
flag – flags
doll – dolls
bird – birds
hare – hares
goat – goats
horse – horses
rat – rats
toy – toys
son – sons
owl – owls
lion – lions
page – pages
table – tables
sister – sisters
orange – oranges

2. Nouns ending in -s, -x, -ch, or -sh form their plurals by adding -es.

Singular – Plural
bunch – bunches
brush – brushes
dish – dishes
church – churches
match – matches
fox – foxes
bush – bushes
dress – dresses
gas – gases
class – classes
loss – losses
box – boxes
glass – glasses
tax – taxes

3. Nouns ending in -y (with a consonant before them) form their plural by changing -y to -ies.

Singular – Plural
city – cities
story – stories
fairy – fairies
lady – ladies
pony – ponies
sky – skies
dairy dairies
baby – babies
family – families
puppy – puppies
butterfly – butterflies
country – countries

4. Nouns ending in -y (with a vowel before them), form their plural by taking an -s only.

Singular – Plural
key – keys
valley – valleys
ray – rays
storey – storeys
day – days
holiday – holidays
boy – boys
journey – journeys
play – plays
monkey – monkeys

5. Nouns ending in -f or -fe form their plural by changing -f or -fe to -ves.

Singular – Plural
calf – calves
loaf – loaves
wolf – wolves
shelf – shelves
life – lives
half – halves
knife – knives
thief – thieves

6. Some nouns ending in -/ or -fe form their plural by taking -s only.

Singular – Plural
roof – roofs
safe – safes
proof – proofs
hoof – hoofs
chief – chiefs
dwarf – dwarfs

7. Nouns ending in -o (with a consonant before them), form their plural by taking -es.

Singular – Plural
echo – echoes
negro – negroes
hero – heroes
mango – mangoes
potato – potatoes
volcano – volcanoes
buffalo – buffaloes
mosquito – mosquitoes

Exceptions : The words photo and piano take -s only to form their plural.

8. Nouns ending in -o (with a vowel before them), form their plural by taking -s only.

Singular – Plural
radio – radios
cuckoo – cuckoos
bamboo – bamboos

9. Some nouns have irregular plurals.

Singular – Plural Singular – Plural
man – men
foot – feet
tooth – teeth
goose – geese
ox – oxen
louse – lice
mouse – mice
child – children

10. A compound noun generally forms its plural by adding -s to the principal word.

daughters-in-law	lookers-on	step-daughters
mothers-in-law	step-sons	maid-servants
fathers-in-law	sons-in-law	passers-by

11. The following Compound Nouns take a double plural.

man-servant – men-servants
woman-teacher – women-teachers
woman-servant – women-servants

Exercises (Solved)

I. Give the plural form of:

fly – flies
box – boxes
hero – heroes
roof – roofs
shoe – shoes
shelf – shelves
dwarf – dwarfs
potato – potatoes
pencil – pencils
mouse – mice
life – lives
fish – fishes
foot – feet
child – children
piano – pianos

II. Give the singular form of:

foxes – fox
teeth – tooth
halves – half
armies – army
watches – watch
gases – gas
ladies – lady
mosquitoes – mosquito
oxen – ox
copies – copy
knives – knife
negroes – negro
chimneys – chimney
shoes – shoe
wolves – wolf

III. Rewrite each sentence using the plural form of Nouns:

1. The monkey was in a cage.
The monkeys were in cages.
2. The knife is on the shelf.
3. He put his foot on the bench.
4. The hero in the film acted well.
5. The policeman chased the thief.
6. The woman told the child a story.
7. Sam plucked a leaf from the tree.
8. The maid washed the glass and the dish.
Answer:
2. The knives are on the shelves.
3. He put his feet on the benches.
4. The heroes in the films acted well.
5. The policemen chased the thieves.
6. The women told the children stories.
7. Sam plucked leaves from the trees.
8. The maids washed the glasses and the dishes.

IV. Rewrite each sentence using the singular form of Nouns

1. The oxen are pulling the carts The ox is pulling the cart.
2. Neha heard the cries of wolves.
3. The women rode on the ponies.
4. The loaves are kept in the boxes.
5. The mice were afraid of the geese.
6. The children were bitten by mosquitoes.
7. These stories are about witches and fairies.
8. The men told the ladies stories of Indian heroes.
Answer:
2. Neha heard the cry of a wolf.
3. The woman rode on a pony.
4. The loaf is kept in the box.
5. The mouse was afraid of the goose.
6. The child was bitten by a mosquito.
7. This story is about a witch and a fairy.
8. The man told the lady a story of an Indian hero.

Remember that-
1. Some Nouns have the same form in the plural and the singular; as-
कुछ Nouns एकवचन तथा बहवचन में एक जैसे होते हैं: जैसे-
deer, sheep, fish, dozen, score, hundred, thousand.

The following Nouns have a plural form but always take the singular verb; as-
निम्नलिखित Nouns बहुवचन में होते हैं, परन्तु उनके साथ हमेशा एकवचन क्रिया (verb) लगती है; जैसे-
news, civics, politics, physics, mathematics, means, gallows.

1. This news is true.
2. Physics is a difficult subject.

3. The following Nouns are always used in the plural form and take the plural verb; as-
निम्नलिखित Nouns का प्रयोग हमेशा बहुवचन में किया जाता है और उनके साथ बहुवचन क्रिया लगती है; जैसे-
thanks, scissors, trousers, pants, alms, wages, spectacles, socks.

1. My thanks are to you all.
2. The scissors were blunt.

4. The following Nouns are used only in the singular form and take the singular verb; as-
नीचे लिखे Nouns केवल एकवचन में प्रयोग किये जाते हैं और उनके साथ एकवचन क्रिया लगती है; जैसे-
furniture, scenery, luggage, machinery, advice, bread, hair, business, mischief.

1. This furniture is not for sale.
2. Where is my luggage ?

5. The word ‘hair’ is used in the plural when a definite number of hairs are to be mentioned.
जब बालों का निश्चित संख्या में उल्लेख किया जाए, तो hair शब्द का प्रयोग बहुवचन (hairs) में किया जाता है।
1. There were two hairs in my tea.
2. My aunt has four white hairs on her head.

Miscellaneous Exercises (Solved)

I. Give the plural of the following nouns:
1. ox
2. leaf
3. knife
4. chief
5. tooth
6. fox
7. wife
8. child
9. story.
10. mouse.
Answer:
1. oxen
2. leaves
3. knives
4. chiefs
5. teeth
6. foxes
7. wives
8. children
9. stories
10. mice.

II. Rewrite each sentence with a plural subject:

1. A cow eats grass.
2. The child is playing
3. The army was fighting.
4. A crow is sitting in the tree.
5. The ox is grazing in the field.
6. This road is closed for repairs.
Answer:
1. Cows eat grass.
2. The children are playing.
3. The armies were fighting.
4. Crows are sitting in the tree.
5. The oxen are grazing in the field.
6. These roads are closed for repairs.

III. Fill in the blanks with the correct form of the given words:

1. She has white …………..
2. I have lost my …………..
3. This ………….. is not true.
4. The ………….. were crying.
5. The house has two …………..
6. Your ………….. were not new.
Hints:
1. teeth
2. shoes
3. news
4. babies
5. storeys
6. trousers

IV. Correct the following sentences:

1. Her hairs are black.
2. Your scissor is blunt.
3. Where is my trouser ?
4. Please accept my thank.
5. These furnitures are for sale.
6. We saw many wolfs in the zoo.
Answer:
1. Her hair is black
2. Your scissors are blunt.
3. Where are my trousers ?
4. Please accept my thanks.
5. This furniture is for sale.
6. We saw many wolves in the zoo.

The Noun – Gender

Gender means being a male (नर) or a female (मादा).
On the basis of gender, we can classify nouns into four kinds:

1. Masculine Gender
2. Feminine Gender
3. Common Gender
4. Neuter Gender

1. A noun that refers to a male is said to be of the Masculine (पुरुषवाचक) Gender; as-
horse, boy, man, lion, king, dog.

2. A noun that refers to a female is said to be of the Feminine (स्तरीवाचक) Gender; as-
mare, girl, woman, lioness, queen, bitch.

3. A noun that refers to both a male and a female, is said to be of the Common (सामान्य) Gender; as-
child, baby, parent, cousin, friend, student, thief.

4. A noun that refers to neither a male nor a female, is said to be of the Neuter (नपुंसक) Gender; as-
book, pen, toy, house, table, knife, etc.

Genders

1. Masculine (male)
2. Feminine (female)
3. Common (either sex)
4. Neuter (neither sex)

Change of Gender

We can change the gender of a Noun in different ways; as-
1. By using a different word:

Masculine – Feminine
monk – nun
fox – vixen
father – mother
uncle – aunt
boy – girl
son – daughter
man – woman
nephew – niece
bull – cow
cock – hen
king – queen
brother – sister
husband – wife
sir – madam
gentleman – lady
dog – bitch
horse – mare
bachelor – maid

2. By adding ‘-ess’, ‘-ss’ to the masculine and making some change in it:

Masculine – Feminine
god – goddess
prince – princess
lion – lioness
master – mistress
tiger – tigress
emperor – empress

3. By changing a part of the word:

Masculine – Feminine
bride – bridegroom
granduncle – grandaunt
peacock – peahen
he-goat – she-goat
landlord – landlady
headmaster – headmistress
milkman – milkwoman
father-in-law – mother-in-law
grandfather – grandmother
brother-in-law – sister-in-law

Exercises (Solved)

I. Put each word in the column it belongs to:

van
duke
horse
milkmaid
bull
child
flower
governess
box
book
parent
gentleman
nun
baby
servant
hairdresser
aunt
table
duchess
shopkeeper
road
monk
daughter
policeman
duchess
Answer:
Table

II. Change the Gender of the following:

sir – madam
lion lioness
bull – cow
cock – hen
mare – horse
uncle – aunt
tigress – tiger
peacock – peahen
gentleman – lady
grandfather – grandmother

Miscellaneous Exercises (Solved)

I. Give the opposite Gender of the following:

1. sir
2. aunt
3. mare
4. king
5. lady
6. cock
7. horse
8. tiger
9. wife
10. male
11. lioness
12. mother
Answer:
1. madam
2. uncle
3. horse
4. queen
5. gentleman
6. hen
7. mare
8. tigress
9. husband
10. female
11. lion
12. father.

II. Rewrite each sentence, changing the Gender of Nouns and Pronouns:

1. A cruel man killed the fox.
2. Mr. Sharma is a businessman.
3. The Emperor welcomed the Duke.
4. The dog is barking at the servant.
5. Madam, my aunt wants to see you.
6. His nephew went to Shimla with his son.
7. The headmaster punished the naughty boys.
8. The bride touched the feet of her mother-in-law.
Answer:
1. A cruel woman killed the vixen.
2. Mrs. Sharma is a businesswoman.
3. The Empress welcomed the Duchess.
4. The bitch is barking at the maid.
5. Sir, my uncle wants to see you.
6. Her niece went to Shimla with her daughter.
7. The headmistress punished the naughty girls.
8. The bridegroom touched the feet of his father-in-law.

III. Fill in the blanks with the Feminine gender of the words in italics:

1. We pray to gods and …………….
2. The hotel has a waiter and a …………….
3. The actor married an ……………. in Mumbai.
4. The lion and the ……………. are in their den.
5. The witch changed the prince into a …………….
6. The tiger and the ……………. look after their cubs.
7. The emperor and the ……………. of Japan live in Tokyo.
8. The guests were received by the host and the …………….
Hints:
1. goddesses
2. waitress
3. actress
4. lioness
5. wizard, princess
6. tigress
7. empress
8. hostess.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 15 Probability Ex 15.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1

Question 1.
Complete the following statements:

(i) Probability of an event E + Probability of the event ‘not E’ = _________.
Solution:
Probability of an event E +
Probability of the event ‘not E’ = 1

(ii) The probability of an event that cannot happen is ___________. Such an event is called _________.
Solution:
The probability of an event that cannot happen is 0. Such an event is called impossible event.

(iii) The probability of an event that is certain to happen is _________. Such an event is called ________.
Solution:
The probability of an event that is certain to happen is 1. Such event is called sure event.

(iv) The sum of the probabilities of all the elementary events of an experiment is __________.
Solution:
The sum of the probabilities of all the elementary events of an experiment is 1.

(v) The probability of an event is greater than or equal to _________ and less than or equal to _________.
Solution:
The probability of an event is greater than or equal to 0 and less than or equal to 1.

Question 2.
Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.
Solution:
When a dnver attempts to start a car the car starts normally. Only when there is some defects the car does not start. So the outcome is not equally likely.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
Solution:
When a player attempts to shoot a basketball the outcome in this situation is not equally likely because the outcome depends on many factors such as the training of the player, quality of the gun used etc.

(iii) A trial is made to answer a true – false question. The answer is right or wrong.
Solution:
Since for a question there are two possibilities either right or wrong the outcome in this trial of true-false question is either true or false i.e. one out of the two and both have equal chances to happen. Hence, the two outcomes are equally likely.

(iv) A baby is born. It is a boy or a girl.
Solution:
A new baby (i.e. who took birth at a moment) can be either a boy or a girl and both the outcome have equally likely chances.

Question 3.
Why is tossing a coin considered to he a fair way of deciding which team should get the ball at the beginning of a football game?
Solution:
When a coin is tossed there are only two possibilities i.e. Head or tail both are equally likely to happen. Result of the toss of a fair coin is completely unpredictable.

Question 4.
Which of the following cannot be the probability of an event?
(A) \(\frac{2}{3}\)
(B) – 1.5
(C) 15 %
(D) 0.7
Solution:
As we know probability of event cannot be less than O and greater than 1
i.e. 0 ≤ P ≤ 1
∴ (B) – 1.5 is not possible.

Question 5.
If P(E) = 0.05, what is the probability of not E.
Solution. As we know P (E) + P \((\overline{\mathrm{E}})\) = 1
P\((\overline{\mathrm{E}})\) = 1 – P(E)
= 1 – 0.05 = 0.95.

Question 6.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Solution:
(i) Since bag contains only lemon flavoured candies
∴ There is no orange candies
∴ It is impossible event.
∴ Probability of getting orange flavoured = 0.

(ii) Since there are only lemon flavoured candies, it is sure event
∴ Probability, of getting lemon flavoured candy = \(\frac{1}{1}\) = 1.

Question 7.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 studenís have the same birthday?
Solution:
Let A is event that two students have same birthday
∴ \((\overline{\mathrm{A}})\) is event that 2 students not having same birthday is 0.992
∴ P \((\overline{\mathrm{A}})\) = 0.992
∴ P (A) = 1 – P (A) (P (A) + P \((\overline{\mathrm{A}})\) = 1)
= 1 – 0.992 = 0.008
∴ Probability that two students have saine birthday = 0.008.

Question 8.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probabifity that the ball drawn is
(i) red?
(ii) not red?
Solution:
Number of Red balls = 3
Number of Black balls = 5
Total number of balls = 3 + 5 = 8
One hail is drawn at random
(i) Probability of getting Red ball = \(\frac{\text { Number of favourable cases }}{\text { Total number of cases }}\)
P (Red ball) = \(\frac{3}{8}\).

(ii) Probability of getting not red ball = 1 – P (Red ball)
= 1 – \(\frac{3}{8}\) = \(\frac{3}{8}\) [P \((\overline{\mathrm{A}})\) = 1 – P(E)].

Question 9.
A box contaIns 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red ?
(ii) white?
(iii) not green?
Solution:
Number of red marbles = 5
Number of white marbles = 8
Number of green marbles = 4
Total number of marbles = 5 + 8 + 4 = 17
Since, one marble is taken out
(i) There are 5 Red marbles
Probability of drawing Red marble = \(\frac{\text { Number of favourable cases }}{\text { Total number of cases }}\)
= \(\frac{5}{17}\)

(ii) Since there are 8 white marbles
Probability of drawing white marble = \(\frac{\text { Number of favourable cases }}{\text { Total number of cases }}\)
= \(\frac{8}{17}\)

(iii) There are 4 green bails
Probability of drawing green ball = \(\frac{\text { Number of favourable cases }}{\text { Total number of cases }}\)
= \(\frac{4}{17}\)

∴ Probability of not drawing green ball = 1 – Probability of green ball
= 1 – \(\frac{4}{17}\) = \(\frac{17-4}{17}\) = \(\frac{13}{17}\).

Question 10.
A piggy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin
(i) will be a 50 p coin?
(ii) will not be a ₹ 5 coin?
Solution;
Number of 50 coins = 100
Number of ₹ 1 coins = 50
Number of ₹ 2 coins =20
Number of ₹ 5 coins = 10
∴ Total number of coins = 100 + 50 + 20 + 10 = 180

(i) Since there are 100 ; 50’ p coin
Probability of getting 50p coin = \(\frac{\text { Number of favourable cases }}{\text { Total number of outcomes }}\)

= \(\frac{100}{180}\)

P (50 p coins) = \(\frac{5}{9}\).

(ii) Number of ₹ 5 coins = 10
∴ Probability of getting ₹ 5 coin = \(\frac{\text { Number of favourable cases }}{\text { Total number of cases }}\)

P (₹ 5 coins) = \(\frac{10}{180}\) = \(\frac{1}{18}\)
Probability of getting not ₹ 5 coin = 1 – P (₹ 5 coins)
= 1 – \(\frac{1}{18}\)
= \(\frac{18-1}{18}\) = \(\frac{17}{18}\).

Question 11.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

Solution:
Number of male fish = 5
Number of female fish = 8
Total number of fish in the tank = 5 + 8 = 13
Probability of getting a male fish = \(\frac{\text { Number of favourable cases }}{\text { Total number of cases }}\)
P(Male fish) = \(\frac{5}{13}\)

Question 12.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

Solution:
(i) Total number of outcomes = {1, 2, 3, 4, 5, 6, 7, 8)
Probability of getting ‘8’ = \(\frac{1}{8}\)

(ii)Odd numbers are = {1, 3, 5, 7)
Probability of getting odd number = \(\frac{4}{8}=\frac{1}{2}\)

(iii) Numbers greater than 2 are {3, 4, 5, 6, 7, 8)
∴ Probability of getting number greater than 2 = \(\frac{6}{8}=\frac{3}{4}\)
P (number greater than 2) = \(\frac{3}{4}\).

(iv) Numbers less than 9 are: {1, 2, 3, 4, 5, 6, 7, 8)
∴ Probability of getting number less than 9 = \(\frac{8}{8}\)
P(a numher less than 9) = 1.

Question 13.
A die is thrown once. Find the probability of getting
(i) a prime number,
(ii) a number lying between 2 and 6;
(iii) an odd number.
Solution:
When dice is thrown number of possible outcomes
S = {1, 2, 3, 4, 5, 6)
(i) Prime numbers are {2, 3, 5)
∴ Probability of getting prime number = \(\frac{3}{6}=\frac{1}{2}\)

(ii) Numbers lying between 2 and 6 = {3, 4, 5}
Probability of getting number between 2 and 6 = \(\frac{3}{6}=\frac{1}{2}\).

(iii) The odd numbers are = {1, 3, 5}
Probability of getting an odd number = \(\frac{3}{6}=\frac{1}{2}\)
P (odd number) = \(\frac{1}{2}\).

Question 14.
One card is drawn from a well. shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds.
Solution:
There are 52 cards in a pack
(i) There are two red kings i.e. king of heart and king of diamond
Probability of getting red king = \(\frac{2}{52}=\frac{1}{26}\)
P(Red king) = \(\frac{1}{26}\)

(ii) There are 12 face cards
i.e. 4 Jack, 4 Queens and 4 kings
Probability of getting face card = \(\frac{12}{52}\)
∴ P (A face card) = \(\frac{3}{13}\).

(iii) Since there are 6 Red face cards i.e 2 Jacks; 2 Queens and 2 Kings
∴ Probability of getting 6 Red face cards = \(\frac{6}{52}\)
P (Red face card) = \(\frac{3}{26}\).

(iv) There is only one Jack of Heart
∴ Probability of getting Jack of Heart = \(\frac{1}{52}\)
P (A Jack card) = \(\frac{1}{52}\)

(v) Since there are 13 spade cards
∴ Probability of getting a spade card = \(\frac{13}{52}\)
P (A spade card) = \(\frac{1}{4}\).

(vi) Since there is only one queen of diamonds
∴ Probability of getting queen of spade card = \(\frac{1}{52}\)
P (A queen of spade) = \(\frac{1}{52}\).

Question 15.
Five cards – the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) if the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
Five cards are ten, jack, queen, king and ace
(i) Probability of getting queen = \(\frac{1}{5}\)
∴ P (A queen) = \(\frac{1}{5}\).

(ii) If the queen is drawn and put aside then there are 4 cards left – Ten, a Jack, a king and an ace.
(a) Probability of getting an ace = \(\frac{1}{4}\)
P (An Ace) = \(\frac{1}{4}\).
There’s no queen left

(b) Probability of getting a queen = \(\frac{0}{4}\) = 0
P (a queen) = 0.

Question 16.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Number of defective pens = 12
Number of good pens = 132
∴ Total number of pens = 12 + 132 = 144
Probability of getting good pen = \(\frac{132}{144}=\frac{11}{12}\)
P (a good pen) = \(\frac{11}{12}\).

Question 17.
(i) A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in
(i) is not defective and is not replaced. Now one bulb is defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
(i) Number of defective bulbs 4
Number of good bulbs (Not defective) = 16
Total number of bulbs = 4 + 16 = 20
Probability of getüng defective bulb = \(\frac{4}{20}=\frac{1}{5}\).

(ii) When a defective bulb drawn is not being replaced, we are left with 19 bulbs
Now probability of getting not defective bulb = \(\frac{15}{19}\)
∴ P (Not defective bulb) = \(\frac{15}{19}\)

Question 18.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Solution:
From 1 to 90 there are 90 numbers in all and 81 two – digit numbers from 10 to 90
(i) Probability of getting two digit number = \(\frac{81}{90}\)
∴ P (two digit number) = \(\frac{81}{90}=\frac{9}{10}\).

(ii) Perfect square numbers are (1, 4, 9, 16, 25, 36, 49, 64, 81 } there are 9 perfect square numbers between 1 to 90
Probability of getting perfect square = \(\frac{9}{90}=\frac{1}{10}\)
∴ P (Perfect square) = \(\frac{1}{10}\)

(iii) Numbers divisible by 5 are (5, 10, 15, 20, 25, 30, 35, 40, 45. 50, 55. 60, 65, 70, 75, 80, 85, 90}
There are 18 numbers divisible by 5
∴ Probability of number getting divisible by 5 = \(\frac{18}{90}=\frac{1}{5}\)
∴ Required probability = \(\frac{1}{5}\).

Question 19.
A child has a die whose six faces show the letters as given below:

The die is thrown. What is the probability of getting
(i) A ?
(ii) D?
Solution:
Number of faces of a die = 6
S = {A, B, C, D, E, A}
n(S) = 6
(i) Since there are two A’s
∴ Probability of getting A = \(\frac{2}{6}=\frac{1}{3}\)
P(A) = \(\frac{1}{3}\)

(ii) Since there is only one face with D
Probability of getting D = \(\frac{1}{6}\)
∴ P(D) = \(\frac{1}{6}\)

Question 20.
Suppose you drop a die at random on the rectangular region shown in Fig. What is the probability that it will land inside the circle with diameter 1 m?

Solution:
Length of rectangle (l) = 3 m
Width of rectangle (b) = 2 m
∴ Area of rectangle = 3 m × 2 m = 6m²
Diameter of circle = 1 m
Radius of circle (R) = \(\frac{1}{2}\) m
∴ Area of circle = πR² = π(\(\frac{1}{6}\))²
= \(\frac{\pi}{4}\) m².

Probability of die to land on a circle = \(\frac{\text { Area of circle }}{\text { Area of rectangle }}\)
= \(\frac{\frac{\pi}{4} \mathrm{~m}^{2}}{6 \mathrm{~m}^{2}}=\frac{\pi}{24}\)
∴ Required Probability = \(\frac{\pi}{24}\).

Question 21.
A lot consists of 144 ball pens of which 20 are défective and the others are good. Nun will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it?
(ii) She will not buy it?
Solution:
Total number of Pens in lot = 144
Number of defective Pens = 20
∴ Number of good Pens = 144 – 20 = 124

(i) Let ‘A’ is event showing she buy the pen
∴ Probability that she buy a Pen = \(\frac{124}{144}\)
P(A) = \(\frac{31}{36}\)

(ii) \(\bar{A}\) is event showing that she will not buy the pen
P \((\bar{A})\) = 1 – P(A)
= 1 – \(\frac{31}{36}\) = \(\frac{36-31}{36}\)
∴ P (Not buy the pen) = \(\frac{5}{36}\).

Question 22.
Two dice, one blue and one grey are thrown at the same time. Write down all the possible outcomes
(i) Complete the following table:

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9 10, 11 and 12. Therefore, each of them has a
probability \(\frac{1}{11}\) Do you agree with this argument ? Justify your answer.
Solution:
When two dices are thrown total number of possible outcomes
{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
S (4, 1) (4, 2) (4, 3) (4, 4) (4,5) (4,6)
(5, 1) (5, 2) (5.3) 5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)}
n(S) = 36
Let A is event of getting sum as 3
∴ A = {(1,2) (2, 1)}
n(A) = 2
∴ Probability of getting sum as 3 = \(\frac{2}{36}=\frac{1}{18}\)
P(A) = \(\frac{1}{18}\)

Let B is event of getting sum as 4 B = ((1, 3), (3, 1), (2, 2))
n(B) = 3
∴ P(B) = \(\frac{3}{36}=\frac{1}{12}\)

Let C is event of getting sum as 5.
C = {(1, 4) (4, 1) (2, 3) (3, 2)}
n(C) = 4
P(C) = \(\frac{4}{36}=\frac{1}{9}\)

Let D is event of getting sum as 6
D = {(1, 5) (5, 1)(2, 4) (4,2) (3, 3)}, n (D) = 5
∴ P(6) = \(\frac{5}{36}\)

Let E is event of getting sum as 7
E = {(1, 6) (6, 1) (2, 5) (5,2) (4, 3) (3, 4)}
∴ P (E) = P (Sum as 7) = \(\frac{6}{36}=\frac{1}{6}\)

Let F is event of getting sum as 8
F = {(2, 6) (6, 2) (3, 5) (4, 4) (5, 3)}
∴ n(F) = 5
P(F) = P(sum as 8) = \(\frac{5}{36}\)

Let G is event of getting sum as 9 when two dices are thrown
G = {(4, 5) (5, 4) (3, 6) (6, 3))
n(G) = 4
∴ P (G) P (Sum as 8) = \(\frac{4}{36}=\frac{1}{9}\)

Let H is event of getting sum as 10
H= {(6, 4) (4, 6) (5, 5)}
n(H) = 3
∴ P (H) = P (sum as 10) = \(\frac{3}{36}=\frac{1}{12}\)

Let I is event of getting sum as 11
I = ((5,6) (6, 5))
n(I) = 2
∴ P(D) = \(\frac{2}{36}=\frac{1}{18}\)

Let J is event of,getting sum as 12
J = {(6,6)}; n(J) = 1
∴ P(J) = \(\frac{1}{36}\)

(ii) No, here all 11 possible outcomes are not equally likely
∴ Three probabilites are different.

Question 23.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e. three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
When a coin tossed three times, then possible out comes are
S = {HHH, HHT HTH, THH, HTF, THT, TTH, TTT)
n(S) = 8
Let A is event of getting all the three same results i.e., {HHH, TTT}
∴ P(A) = \(\frac{2}{8}=\frac{1}{4}\)
Probability of lossing the game = 1 – P (A)
P \((\bar{A})\) = 1 – \(\frac{1}{4}\)
= \(\frac{4-1}{4}\) = \(\frac{3}{4}\)
∴ Probability of losing the game = \(\frac{3}{4}\).

Question 24.
A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
Solution:
When a die is thrown twice all possible outcomes are
S = {(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) 5) (4, 6) (5, 1) (5, 2) (5, 3) (5,4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)}
n(S) = 36
Ler A is event that 5 will come up either time
A = {(1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5) (5, 1) (5, 2) (5, 3) (5, 4) (5, 6)}
n(A) = 11
∴ \((\bar{A})\) is event that 5 will not come up either time.
n\((\bar{A})\) = 36 – 11 = 25.

(i) ∴ Probability of not getting 5 up either time = \(\frac{25}{36}\)
P \((\bar{A})\) = \(\frac{25}{36}\)
Probability that 5 will come up at least once = \(\frac{11}{36}\)
∴ P(A) = \(\frac{11}{36}\).

Question 25.
Which of the following arguments are correct ? Give reasons for your answer:
(i) 1f two coins are tossed simultaneously there are three possible outcomes – two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac{1}{3}\):
(ii) If a die is thrown, there are two possible outcomes – an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac{1}{2}\).
Solution:
(i) When two coins are tossed the possible outcomes are S = {HH, HT, TH, TT}
Probability of getting 2 Heads = \(\frac{1}{4}\)
P(HH) = \(\frac{1}{4}\)
Probability of getting two tails = \(\frac{1}{4}\)
P(TT) = \(\frac{1}{4}\)
Probability of getting one head and one tail = \(\frac{2}{4}=\frac{1}{2}\)
∴ (i) argument is incorrect.

(ii) When a die is thrown possible outcomes are S = (1, 2, 3, 4, 5, 6)
n(S) = 6
Odd numbers are 1, 3, 5
∴ Probability of getting odd number = \(\frac{3}{6}=\frac{1}{2}\)
Even numbers are 2, 4, 6
∴ Probability of getting even number = \(\frac{3}{6}=\frac{1}{2}\)
(ii) argument is correct.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 11 Mensuration Ex 11.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4

1. Given a cylindrical tank, in which situation will you find surface area and in which situation volume:

Question (a)
To find how much it can hold.
Solution:
To find how much a cylindrical tank can hold, we will find its volume.

Question (b)
Number of cement bags required to plaster it.
Solution:
To find number of cement bags required to plaster a cylindrical tank, we will find its surface area.

Question (c)
To find the number of smaller tanks that can be filled with water from it.
Solution:
To find the number of smaller tanks that can be filled with water from it, we will find volume of the tank.

2. Diameter of cylinder A is 7 cm and the height is 14 cm.

Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?
Solution:
Radius of cylinder B is double them that of cylinder A.
∴ Volume of cylinder B should be more than that of cylinder A.

For cylinder A:
radius (r) = \(\frac{diameter}{2}\) = \(\frac{7}{2}\)
height (h) = 14 cm
Volume of cylinder A = πr²h
= \(\frac{22}{7}\) × (\(\frac{22}{7}\))² × 14
= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 14
= 11 × 7 × 7
= 593 cm³

For cylinder B:

radius (r) = \(\frac{diameter}{2}\) = \(\frac{14}{2}\) = 7cm and
height (h) = 7 cm
Volume of cylinder B = πr²h
= \(\frac{22}{7}\) × 7² × 7
= \(\frac{22}{7}\) × 7 × 7 × 7
= 22 × 7 × 7
= 1078 cm³

Total surface area:
For cylinder A:

radius (r) = \(\frac{7}{2}\) cm
height (h) = 14 cm
Total surface area of cylinder A
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × \(\frac{7}{2}\)(\(\frac{7}{2}\) +14)
= 22(\(\frac{35}{2}\))
= 11 × 35
= 385 cm²

For cylinder B :
radius (r) = 7 cm, height (h) = 7 cm
Total surface area of cylinder B
= 2πr (r + h)
= 2 × \(\frac {22}{7}\) × 7(7 + 7)
= 44(14) = 616 cm²
So the surface area of cylinder B is greater than that of cylinder A. Hence, the cylinder with greater volume also has greater surface area.

3. Find the height of a cuboid whose base area is 180 cm² and volume is 900 cm³?
Solution:
Let the height of cuboid be h cm.
Now, = Area of base × Height
∴ 900 = 180 × h
∴ h = \(\frac {900}{180}\) = 5cm
Hence, height of cuboid is 5 cm.

4. A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?
Solution:
Volume of a cuboid = 60 × 54 × 30 cm³
Volume of a cube = 6³ = 6 × 6 × 6 cm³
∴ Number of small cubes
= \(\frac{\text { Volume of cuboid }}{\text { Volume of one cube}}\)
= \(\frac{60 \times 54 \times 30}{6 \times 6 \times 6}\)
= 10 × 9 × 5 = 450
Hence, 450 cubes can be placed in the given cuboid.

5. Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm?
Solution:
For given cylinder:
Volume = 1.54 m³
Radius(r) = \(\frac {diameter}{2}\) = \(\frac {140}{2}\) = 70cm = 0.7 m
Volume of cylinder = πr²h
∴ 1.54 = \(\frac {22}{7}\) × 0.7 × 0.7 × h
∴ h = \(\frac{1.54 \times 7}{22 \times 0.7 \times 0.7}\) = 1m
Hence, height of the cylinder is 1 m.

6. A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank?
Solution:
For given cylindrical milk tank:
Radius (r) = 1.5 m = \(\frac {15}{10}\) m
Height (h) = 7m
Volume of cylindrical milk tank
= πr²h
= \(\frac {22}{7}\) × (\(\frac {15}{10}\))² × 7
= \(\frac {22}{7}\) × \(\frac {15}{10}\) × \(\frac {15}{10}\) × 7
= \(\frac{11 \times 3 \times 3}{2}=\frac{99}{2}\) = 49.5m³
= 49.5 m³
1 m³ = 1000 litres
∴ 49.5 m³ = 49.5 × 1000 = 49500 litres
Hence, 49,500 litres of milk can be stored in the tank.

7. If each edge of a cube is doubled:

Question (i)
How many times will its surface area increase?
Solution:
Let the edge of the original cube be x.
Then, its new edge (by doubling) = 2x
Original surface area of the cube
= 6 (side)²
= 6 (x)²
= 6x²
New surface area of the cube
= 6 (side)²
= 6 (2x)²
= 6 (2x × 2x)
= 6 (4x²) = 24X²
= \(\frac{\text { New surface area of the cube }}{\text { Original surface area of the cube }}\) = \(\frac{24 x^{2}}{6 x^{2}}\) = 4
Hence, surface area of a cube will increase 4 times.

Question (ii)
How many times will its volume increase ?
Solution:
Original volume of the cube
= (side)³
= (x)³
= x³
New volume of the cube = (side)³
= (2x)³
= (2x × 2x × 2x)
= 8x³
Now,
\(\frac{\text { New volume of the cube }}{\text { Original volume of the cube }}\) = \(\frac{8 x^{3}}{x^{3}}\) = 8
Hence, volume of the cube will increase 8 times.

8. Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m³, find the number of hours it will take to fill the reservoir.
Solution:
Volume of the cuboidal reservoir =108 m³ 1 m³ = 1000 litres
∴ 108 m³ = 108 × 1000 litres
= 1,08,000 litres
Water poured in a minute = 60 litres
∴ Water poured in an hour
(∵ 1 hour = 60 minutes) = 60 × 60
= 3600 litres
Time taken to fill reservoir = \(\frac{\text { volume of reservoir }}{\text { water poured in an hour }}\) = \(\frac {108000}{3600}\)
= 30 hours
Hence, 30 hours it will take to fill the reservoir.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 6 Squares and Square Roots Ex 6.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2

1. Find the square of the following numbers:

Question (i).
32
Solution:
Let us use: (a + b)² = a² + 2ab + b²
(32)² = (30 + 2)²
= (30)² + 2 (30) (2) + (2)²
= 900 + 120 + 4
= 1024

Question (ii).
35
Solution:
(35)² = (30 + 5)²
= (30)² + 2 (30) (5) + (5)²
= 900 + 300 + 25
= 1200 + 25
= 1225
Second method :
[Note : The unit digit of 35 is 5.]
(35)² = 3 × (3 + 1) × 100 + 25
= 3 × 4 × 100 + 25
= 1200 + 25
= 1225

Question (iii).
86
Solution:
(86)² = (80 + 6)²
= (80)² + 2 (80) (6) + (6)²
= 6400 + 960 + 36
= 7396

Question (iv).
93
Solution:
(93)² = (90 + 3)²
= (90)² + 2 (90) (3) + (3)²
= 8100 + 540 + 9
= 8649

Question (v).
71
Solution:
(71 )² = (70 + 1)²
= (70)² + 2 (70) (1) + (1)²
= 4900 + 140 + 1
= 5041

Question (vi).
46
Solution:
(46)² = (40 + 6)²
= (40)² + 2 (40) (6) + (6)²
= 1600 + 480 + 36
= 2116

2. Write a Pythagorean triplet whose one member is:

Question (i).
6
Solution:
Here, 2n = 6
∴ n = 3

Now, n² – 1 = 3² – 1
= 9 – 1
= 8
and n² + 1 = 3² + 1
= 9 + 1 = 10
Thus, the required Pythagorean triplet is 6, 8, 10.

Question (ii).
14
Solution:
Here, 2n = 14
∴ n = 7
Now, n² – 1 = 7² – 1 = 49 – 1 = 48
and n² + 1 = 7² + 1 = 49 + 1 = 50
Thus, the required Pythagorean triplet is 14, 48, 50.

Question (iii).
16
Solution:
Here, 2n = 16
∴ n = 8
Now, n² – 1 = 8² – 1 = 64 – 1 = 63
and n² + 1 = 8² + 1 = 64 + 1 = 65
Thus, the required Pythagorean triplet is 16, 63, 65.

Question (iv).
18
Solution:
Here, 2n = 18
∴ n = 9
Now n² – 1 = 9² – 1 = 81 – 1 = 80
and n² + 1 = 92 + 1 = 81 + 1 = 82
Thus, the required Pythagorean triplet is 18, 80, 82.

Punjab State Board PSEB 6th Class English Book Solutions English Grammar Conjunction Exercise Questions and Answers, Notes.

PSEB 6th Class English Grammar Conjunction

A Conjunction is a word which joins two words, phrases or sentences; as-
Conjunction वह शब्द है जो दो शब्दों, शब्द-समूहों अथवा वाक्यों को आपस में जोडता है: जैसे-

1. I know that Raju is a good boy.
2. Ram and Sham are friends.
3. You will fail if you don’t work hard.
4. He wants coffee but I want milk.

Some other Conjunctions of common use are-
because, therefore, yet, for, though, otherwise, so, or

Now look at these sentences:

1. Deepa went to the shop.
She bought a book.
We can combine these two sentences by using ‘and’:
= Deepa went to the shop and bought a book.

2. Anil is thin.
His sister is fat.
= Anil is thin but his sister is fat.

3. He did not go for a walk today.
He is unwell.
= He did not go for a walk today because he is unwell.

4. They played well.
They could not win the match.
= Though they played well, they could not win the match.

Exercises (Solved)

1. Point out the Conjunction in each sentence:

1. He is slow but steady.
2. Nisha and Meera are friends.
3. We went out and had an ice-cream.
4. I missed the train because I was late.
5. Yesterday it rained but today it is sunny.
6. I remember his name but not his address.
7. He bought a radio because he loves music.
8. They went to the market and did some shopping.
Hints:
2. and
3. and
4. because
5. but
6. but.
7. because
8. and.

II. Join each pair of sentences using ‘but’:

1. Seema is tall.
Her brother is short.

2. A bird can fly.
A fish can’t fly.

3. Shimla is cold.
Jaipur is warm.

4. Varun worked hard.
He failed the test.

5. The girls saw a lion.
The lion did not see them.

6. A car has four wheels.
A cycle has two wheels.
Answer:
1. Seema is tall but her brother is short.
2. A bird can fly but a fish cannot.
3. Shimla is cold but Jaipur is warm.
4. Though Varun worked hard, he failed the test.
5. The girls saw a lion but the lion did not see them.
6. A car has four wheels but a cycle has two (wheels).

III. Fill in the blanks with a suitable Conjunction:

1. She bought nothing because she had no money.
2. He was sad ………….. he had failed.
3. June is warm ………….. January is cold.
4. Arun has a car ………….. he can’t drive.
5. Tom …………. Bob always play together.
6. Neha went to Agra …………. saw the Taj.
7. An elephant is big …………. an ant is small.
8. Go to the garden …………. get some flowers.
9. I went to see him …………. he was not at home.
10. I couldn’t make any tea ……….. there was no milk.
Hints:
2. because
3. but
4. but
5. and
6. and
7. but
8. and
9. but
10. because.

IV. Match the columns to make meaningful sentences:

1. He worked hard	but not tea.
2. Kapil likes coffee	yet he failed.
3. I went to see my mother	because she was ill.
4. No one answered the bell	and rested for a while.
5. Tanu and Manu came home	because everyone was out.

Answer:
1. He worked hard yet he failed.
2. Kapil likes coffee but not tea.
3. I went to see my mother because she was ill.
4. No one answered he bell because everyone was out.
5. Tanu and Manu came home and rested for a while.

V. Join each pair of sentences using a suitable Conjunction. You can use one from the box below:

so, if, but, though, therefore, yet, and, because, otherwise, or

1. You will win. You run fast.
2. He was poor. He was honest.
3. I helped him. He is my friend.
4. She was ill. She did not come.
5. He did not work hard. He failed.
6. Anu came early. Bonny came late.
7. Tell me the truth. I will punish you.
8. Is that story true ? Is that story false ?
9. Ravi ran fast. He couldn’t win the race.
10. Amit can read English. Amit can write English.

Answer:
1. You will win if you run fast.
2. He was poor yet he was honest.
3. I helped him because he is my friend.
4. She was ill, therefore, she did not come.
5. She did not work hard, so she failed.
6. Anu came early but Bonny came late.
7. Tell me the truth otherwise I will punish you.
8. Is that story true or false ?
9. Though Ravi ran fast, he could not win the race.
10. Amit can read and write English.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 6 Squares and Square Roots Ex 6.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1

1. What will be the unit digit of the squares of the following numbers?

Question (i).
81
Solution:
81
Here, the ending digit is 1.
1 × 1 = 1
∴ The unit digit of (81)² will be 1.

Question (ii).
272
Solution:
272
Here, the ending digit is 2.
2 × 2 = 4
∴ The unit digit of (272)² will be 4.

Question (iii).
799
Solution:
799
Here, the ending digit is 9.
9 × 9 = 81
∴ The unit digit of (799)² will be 1.

Question (iv).
3853
Solution:
3853
Here, the ending digit is 3.
3 × 3 = 9
∴ The unit digit of (3853)² will be 9.

Question (v).
1234
Solution:
1234
Here, the ending digit is 4.
4 × 4 = 16
∴ The unit digit of (1234)² will be 6.

Question (vi).
26387
Solution:
26387
Here, the ending digit is 7.
7 × 7 = 49
∴ The unit digit of (26387)² will be 9.

Question (vii).
52698
Solution:
52698
Here, the ending digit is 8.
8 × 8 = 64
∴ The unit digit of (52698)² will be 4.

Question (viii).
99880
Solution:
99880
Here, the ending digit is 0.
0 × 0 = 0
∴ The unit digit of (99880)² will be 0.

Question (ix).
12796
Solution:
12796
Here, the ending digit is 6.
6 × 6 = 36
∴ The unit digit of (12796)² will be 6.

Question (x).
55555
Solution:
55555
Here, the ending digit is 5.
5 × 5 = 25
∴ The unit digit of (55555)² will be 5.

2. The following numbers are obviously not perfect squares. Give reason:

Question (i).
1057
Solution:
1057
Here, the ending digit is 7, which is not one of the digits 0, 1, 4, 5, 6 or 9.
∴ 1057 is not a perfect square.

Question (ii).
23453
Solution:
23453
Here, the ending digit is 3, which is not one of the digits 0, 1, 4, 5, 6 or 9.
∴ 23453 is not a perfect square.

Question (iii).
7928
Solution:
7928
Here, the ending digit is 8, which is not one of the digits 0, 1, 4, 5, 6 or 9.
∴ 7928 is not a perfect square.

Question (iv).
222222
Solution:
222222
Here, the ending digit is 2, which is not one of the digits 0, 1, 4, 5, 6 or 9.
∴ 222222 is not a perfect square.

Question (v).
64000
Solution:
64000
Here, the number of zeros at the end is odd.
∴ 64000 is not a perfect square.

Question (vi).
89722
Solution:
89722
Here, the ending digit is 2, which is not one of the digits 0, 1, 4, 5, 6 or 9.
∴ 89722 is not a perfect square.

Question (vii).
222000
Solution:
222000
Here, the number of zeros at the end is odd.
∴ 222000 is not a perfect square.

Question (viii).
505050
Solution:
505050
Here, the ending digit is 0. (Number of odd zero.)
∴ 505050 is not a perfect square.

3. The squares of which of the following would be odd numbers ?

(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Solution:
[Note: The square of an odd natural number is odd and that of an even number is an even number.]
(i) 431
This is an odd number.
∴ (431)² is an odd number.

(iii) 7779
This is an odd number.
∴ (7779)² is an odd number.

4. Observe the following pattern and find the missing digits:

Solution :
Observing the above pattern, we can find the missing digits as follows :

5. Observe the following pattern and supply the missing numbers :

Solution:
Observing the above pattern, we can find the missing numbers as follows :

6. Using the given pattern, find the missing numbers:
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + …….² = 21²
5² + ……² + 30² = 31²
6² + 7² + ……² = ……²
[To find pattern:
Third number is related to first and second number. How?
Fourth number is related to third number. How?]
Solution:
(a)² + (a + 1)² + {a (a + 1 )}²
= {a (a + 1) + 1}²
The missing numbers are :
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + 20² = 21²
5² + 6² + 30² = 31²
6² + 7² + 42² = 43²

7. Without adding, find the sum:

Question (i).
1 + 3 + 5 + 7 + 9
Solution:
The sum of first five odd numbers = 5²
= 5 × 5
= 25

Question (ii).
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
Solution:
The sum of first ten odd numbers = 10²
= 10 × 10
= 100

Question (iii).
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
The sum of first twelve odd numbers = 12²
= 12 × 12
= 144

8.

Question (i).
Express 49 as the sum of 7 odd numbers.
Solution:
49 = 7² = Sum of first seven odd numbers
∴ 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13

Question (ii).
Express 121 as the sum of 11 odd numbers.
Solution :
121 = 11² = Sum of first eleven odd numbers
∴ 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

9. How many numbers lie between squares of the following numbers?
[Note : Between n² and (n + 1)², there are 2n non-square numbers.]

Question (i).
12 and 13
Solution:
Natural numbers between 12² and 13²
= 2 × 12
= 24. (2n, n = 12)

Question (ii).
25 and 26
Solution:
Natural numbers between 25² and 26²
= 2 × 25
= 50. (2n, n = 25)

Question (iii).
99 and 100
Solution:
Natural numbers between 99² and 100²
= 2 × 99
= 198. (2n, n = 99)

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 5 Data Handling Ex 5.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.3

1. List the outcomes you can see in these experiments:

Question (a).
Spinning a wheel

Solution :
Outcomes of spinning the given wheel are :
A, B, C or D.

Question (b).
Tossing two coins together
Solution:
Outcomes in tossing two coins together are : HH, HT, TT, TH
(H = Head, T = Tail)

2. When a die is thrown, list the outcomes of an event of getting

Question (i).
(a) a prime number.
(b) not a prime number.
Solution:
Possible outcomes are: 1, 2, 3, 4, 5 or 6
Prime numbers = 2, 3, 5
(a) Outcomes of getting a prime number are: 2, 3 or 5.
(b) Outcomes of getting not a prime number are : 1, 4 or 6.

(ii) (a) a number greater than 5.
(b) a number not greater than 5.
Solution:
(a) Outcomes of getting a number greater than 5 is 6.
(b) Outcomes of getting a number not greater than 5 are: 1, 2, 3, 4 or 5.

3. Find the:

Question (a).
Probability of the pointer stopping on D in (Question 1 – (a)) ?
Solution:
There are 5 sectors containing A, B, C and D.
Since, there is only 1 sector containing D.
∴ Favourable outcomes = 1
Number of total outcomes = 5
∴ Probability of the pointer stopping on D = \(\frac {1}{5}\)

Question (b).
Probability of getting an ace from a well shuffled deck of 52 playing cards ?
Solution:
Number of possible outcomes = 52
(Total number of cards in a deck = 52)
Since, there are 4 aces in a pack of 52 cards.
∴ Favourable outcomes = 4
∴ Probability of getting an ace = \(\frac{4}{52}=\frac{1}{13}\)

Question (c).
Probability of getting a red apple. (See figure below)

Solution:
Total number of apples = 7
∴ Possible number of ways = 7
Since, there are 4 red apples.
∴ Favourable outcomes = 4
∴ Probability of getting a red apple = \(\frac {4}{7}\)

4. Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking into it. What is the probability of.

Question (i).
getting a number 6 ?
Solution :
We can get a slip written the number ‘6’ only once.
∴ Number of favourable outcome = 1
Probability of getting the number 6 = \(\frac {1}{10}\)

Question (ii).
getting a number less than 6 ?
Solution:
Numbers less than 6 are 1, 2, 3, 4 and 5. These are five numbers.
∴ Favourable outcomes = 5
Probability of getting the number less than 6 = \(\frac {5}{10}\) = \(\frac {1}{2}\)

Question (iii).
getting a number greater than 6 ?
Solution :
Numbers greater than 6 are 7, 8, 9 and 10. These are four numbers.
∴ Favourable outcomes = 4
Probability of getting the number greater than 6 = \(\frac {4}{10}\) = \(\frac {2}{5}\)

Question (iv).
getting a 1-digit number ?
Solution :
1-digit numbers are 1, 2, 3, 4, 5, 6, 7, 8 and 9. These are nine numbers.
∴ Favourable outcomes = 9
Probability of getting a 1-digit number = \(\frac {9}{10}\)

5. If you have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector, what is the probability of getting a green sector? What is the probability of getting a non-blue sector?
Solution:
There are 5 sectors in all.
(3 + 1 + 1 = 5)
∴ Number of total possible outcomes = 3
Probability of getting a green colour sector = \(\frac {3}{5}\)
There are 4 non-blue sectors. (5 – 1 = 4)
∴ Number of favourable outcomes = 4
Probability of getting a non-blue sector = \(\frac {4}{5}\)

6. Find the probabilities of the events given in Question 2.
Solution:
There are 6 outcomes.
(i) 2, 3 and 5 are prime numbers. These are three numbers.
∴ Favourable outcomes = 3
Probability of getting a prime number = \(\frac {3}{6}\) = \(\frac {1}{2}\)

(ii) 1, 4 and 6 are non-prime numbers. These are three numbers.
∴ Favourable outcomes = 3
Probability of getting a non-prime number = \(\frac {3}{6}\) = \(\frac {1}{2}\)

(iii) 6 is greater than 5. This is only one number.
∴ Favourable outcome = 1
Probability of a number greater than 5 = \(\frac {1}{6}\)

(iv) 1, 2, 3, 4 and 5 are not greater than 5. There are five numbers.
∴ Favourable outcomes = 5
Probability of a number which is not greater than 5 = \(\frac {5}{6}\)