Haryana State Board HBSE 8th Class Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 Textbook Exercise Questions and Answers.

## Haryana Board 8th Class Maths Solutions Chapter 6 Square and Square Roots Exercise 6.3

Question 1.

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

(i) 9801

(ii) 99856

(iii) 998001

(iv) 657666025

Solution:

(i) 9801^{2} = 9801 × 9801

Possible one’s digits =1 [∵ 1 × 1 = 1]

(ii) 99856^{2} = 99856 × 99856

Possible one’s digits =6 [∵ 6 × 6 = 36]

(iii) 998001^{2} = 998001 × 998001

∴ Possible one’s digits = 1

(iv) 657666025^{2} = 657666025 × 657666025

∴ Possible one’s placed digit = 5 [∵ 5 × 5 = 25]

Question 2.

Without doing any calculation, fin#the numbers which are surely not perfect squares.

(i) 153

(ii) 257

(iii) 408

(iv) 441

Solution:

(i) 153, (ii) 257 and (iii) 408 are surely not perfect squares.

[∵ The numbers end with 2, 3, 7, or 8 can never be a perfect square.]

Question 3.

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

Solution:

For \(\sqrt {100}\)

(i) 100 – 1 = 99

(ii) 99 – 3 = 96

(iii) 96 – 5 = 91

(iv) 91 – 7 = 84

(v) 84 – 9 = 75

(vi) 75 – 11 = 64

(vii) 64 – 13 = 51

(viii) 51 – 15 = 36

(ix) 36 – 17 = 19

(x) 19 – 19 = 0

For \(\sqrt {169}\)

(i) 169 – 1 = 168

(ii) 168 – 3 = 165

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(xiii) 23 – 23 = 0

From 100 and 169 we have subtracted successive odd numbers starting from 1 and obtain 0 at 10th and 13th step.

∴ \(\sqrt {100}\) = 10

and \(\sqrt {169}\) = 13

Question 4.

Find the square roots of the follow¬ing numbers by the Prime Factorisation Method.

(i) 729

(ii) 400

(iii) 1764

(iv) 4096

(v) 7744

(vi) 9604

(vii) 5929

(viii) 9216

(ix) 529

(x) 8100.

Solution:

Question 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.

(i) 252

(ii) 180

(iii) 1008

(iv) 2028

(v) 1458

(vi) 768

Solution:

By prime factorisation, we get

252 = \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\) × 7

We see that in prime factorisation of 252, there exists a number 7 which is unpaired.

Hence, we will have to multiply 252 by 7 to make a pair of 7.

∴ 252 × 7 = \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\) × \(\underline{7 \times 7}\)

\(\sqrt {1764}\) = 2 × 3 × 7

= 42.

(ii) By prime factorisation, we get

180 = \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\) × 5

The number 5 is unpaired.

∴ The smallest number by which 180 should be multiplied to make it a perfect square is 5.

Now,

180 × 5 = \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\) × \(\underline{5 \times 5}\)

∴ 900 = \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\) × \(\underline{5 \times 5}\)

∴ \(\sqrt {900}\) = 2 × 3 × 5 = 30.

(iii) 1008 = \(\underline{2 \times 2}\) × \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\) × 7.

Here, 7 is unpaired

∴ We should multiply 1008 by to make it a perfect square.

Now,

1008 × 7 = \(\underline{2 \times 2}\) × \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\) × \(\underline{7 \times 7}\)

⇒ 1056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7

or \(\sqrt {1056}\) = 2 × 2 × 3 × 7

= 84.

(iv) 2028 = \(\underline{2 \times 2}\) × 3 × \(\underline{13 \times 13}\)

We see that the number 3 is unpaired.

∴ We should multiply 2028 by 3 to make it a perfect square.

Now,

2028 × 3 =\(\underline{2 \times 2}\) × \(\underline{3 \times 3}\) × \(\underline{13 \times 13}\)

Thus, 2084 = 2 × 2 × 3 × 3 × 13 × 3

∴ \(\sqrt {2084}\) = 2 × 3 × 13 = 78

(v) 1458 = 2 × \(\underline{3 \times 3}\) × \(\underline{3 \times 3}\) × \(\underline{3 \times 3}\).

Since a number 2 is unpaired.

So, smallest number by which 1458 should be multiplied is 2.

∴ Required number

= 1458 × 2

= 2916

Thus,

2916 = \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\) × \(\underline{3 \times 3}\) × \(\underline{3 \times 3}\)

\(\sqrt {2916}\) = 2 × 3 × 3 × 3 = 54

(vi) By prime factorisation, we get

768 = \(\underline{2 \times 2}\) × \(\underline{2 \times 2}\) × \(\underline{2 \times 2}\) × \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\)

Since a number 3 is unpaired.

∴ We should multiply 768 by 3 to make it a perfect square .

Now,

768 × 3 = \(\underline{2 \times 2}\) × \(\underline{2 \times 2}\) × \(\underline{2 \times 2}\) × \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\)

⇒ \(\sqrt {2304}\) = 2 × 2 × 2 × 2 × 3 = 48.

Question 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.

(i) 252

(ii) 2925

(iii) 396

(iv) 2645

(v) 2800

(vi) 1620

Solution:

(i) We have,

252 = \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\) × 7 (by prime factorisation)

If we divide 252 by 7 then

252 ÷ 7 = 36 = 2 × 2 × 3 × 3

which is a perfect square.

Thus, the required smallest number is 7

and \(\sqrt {36}\) = 2 × 3 = 6

(ii) We have

2925 = \(\underline{5 \times 5}\) × \(\underline{3 \times 3}\) × 13

Since a number 13 is unpaired.

We should multiply 768 by 3 to make it a perfect square .

∴ 2925 ÷ 13 = 225

= \(\underline{5 \times 5}\) × \(\underline{3 \times 3}\)

which is a perfect square.

∴ The required smallest number is 13.

And \(\sqrt {225}\) = \(\sqrt{5 \times 5 \times 3 \times 3}\)

= 5 × 3 = 15

(iii) By prime factorisation, we get

396 = \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\) × 11

If we divide 396 by 11, then

396 ÷ 11 = 36 = 2 × 2 × 3 × 3

which is a perfect square.

∴ The required smallest number is 11.

And, \(\sqrt {36}\) = 2 × 3 = 6

(iv) 2645 = 5 × \(\underline{23 \times 23}\)

Since a number 5 is a unpaired.

So, we should divide 2645 by 5.

2645 ÷ 5 = 529 = 23 × 23

which is a perfect square.

∴ The required smallest number = 5

and \(\sqrt {529}\) = 23.

(v) By prime factorisation, we get

2800 = \(\underline{2 \times 2}\) × \(\underline{2 \times 2}\) × \(\underline{5 \times 5}\) × 7

If we divide 2800 by 7, then

2800 ÷ 7 = 400

= \(\underline{2 \times 2}\) × \(\underline{2 \times 2}\) × \(\underline{5 \times 5}\)

which is a perfect square.

Therefore, the required smallest number 7.

And, \(\sqrt {400}\) = 2 × 2 × 5 = 20

(vi) By prime factorisation, we get

1620 = \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\) × \(\underline{3 \times 3}\) × 5

Since a number 5 is unpaired.

∴ We should divide 1620 by 5.

1620 ÷ 5 = 324

= \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\) × \(\underline{3 \times 3}\)

which is a perfect square.

Thus, \(\sqrt {324}\) = 2 × 3 × 3 = 18.

Question 7.

The students of Class VIII of a school donated Rs. 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 in the class VIII be x.

Then, the rupees donated by each students will be Rs. x.

According to the given condition of the question :

x × x = 2401

⇒ x^{2} = 2401

⇒ x = \(\sqrt{2401}\)

= \(\sqrt{7 \times 7 \times 7 \times 7}\)

= 7 × 7

Hence, the number of students in the class is 49.

Question 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 no. of plants contains in each row be x.

Then, the no. of rows will be also x.

∴ According to the given condition of the question

Thus, x × x = 2025

⇒ x^{2} = 2025

⇒ x^{2} = 5^{2} × 3^{2} × 3^{2}

⇒ x^{2} = (5 × 3 × 3)^{2}

⇒ x^{2} = 452

x = 45.

∴ The number of rows and the number of plants in each row is 45.

Question 9.

Find the smallest square number that is divisible by each of the number 4, 9 and 10.

Solution:

This has to be done in three steps.

Step 1. First of all find the LCM of the numbers 4,9 and 10.

LCM of 4, 9 and 10 = 180

Step 2. Then, find the prime factorisation of the LCM so obtained i.e., 180.

180 = \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\) × 5

Step 3. Now, multiply the number 180 by 5 to make it a perfect square.

⇒ 180 × 5 = \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\) × \(\underline{5 \times 5}\)

⇒ 900 = \(\underline{2 \times 2}\) × \(\underline{3 \times 3}\) × \(\underline{5 \times 5}\)

Now, 900 is obviously a perfect square.

So, the required square number is 900.

Question 10.

Find the smallest square number that is divisible by each of the number 8, 15 and 20.

Solution:

Follow the same procedure as explained in the solution of Question (9)

LCM of 8, 15 and 20 = 120

Prime factorisation of 120 = \(\underline{2 \times 2}\) × 2 × 3 × 5

We see that 2, 3 and 5 are not in pair.

Therefore, 120 should be multiplied by 2 × 3 × 5 i.e., 30 in order to get a perfect square.

Hence, the required square number is

120 × 30 = 3600.