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.