Class 8 Squares and Square Roots Maths Ex 6.3

 

NCERT Solutions For Class 8 Maths Chapter 6 Ex 6.3

ncert solutions for class 8 maths chapter 6, Exercise 6.3 involve complete  answers for each question in the exercise 6.3. The solutions provide students a  strategic methods  to prepare for their exam. Class 8 Maths Chapter 8 Squares and Square Roots Exercise 6.3 questions and answers helps students  to perform better in exam and it will  clear doubts definitely. Students will find it extremely easy to understand the questions and learn solving the problems. NCERT Solutions for Class 8 Maths Chapter 8 Squares and Square Roots Exercise 6.3 prepared by our subject matter experts in very delicate, easy and creative way. 

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

Answer:

(i) If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 9801 is either 1 or 9.

(ii) If the number ends with 6, then the one’s digit of the square root of that number may be 4 or 6. Therefore, one’s digit of the square root of 99856 is either 4 or 6.

(iii) If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 998001 is either 1 or 9.

(iv) If the number ends with 5, then the one’s digit of the square root of that number will be 5. Therefore, the one’s digit of the square root of 657666025 is 5.

Question 2:

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

(i) 153 (ii) 257 (iii) 408 (iv)441

Answer:

The perfect squares of a number can end with any of the digits 0, 1, 4, 5, 6, or 9 at unit’s place. Also, a perfect square will end with even number of zeroes, if any.

(i) Since the number 153 has its unit’s place digit as 3, it is not a perfect square.

(ii) Since the number 257 has its unit’s place digit as 7, it is not a perfect square.

(iii) Since the number 408 has its unit’s place digit as 8, it is not a perfect square.

(iv) Since the number 441 has its unit’s place digit as 1, it is a perfect square.

Question 3:

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

Answer:

We know that the sum of the first n odd natural numbers is n².

Consider √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

We have subtracted successive odd numbers starting from 1 to 100, and obtained 0 at 10^th step.

Therefore, √100=10

The square root of 169 can be obtained by the method of repeated subtraction as follows.

(i) 169 − 1 = 168

(ii) 168 − 3 = 165

(iii) 165 − 5 = 160

(iv) 160 − 7 = 153

(v) 153 − 9 = 144

(vi) 144 − 11 = 133

(vii) 133 − 13 = 120

(viii) 120 − 15 = 105

(ix) 105 − 17 = 88

(x) 88 − 19 = 69

(xi) 69 − 21 = 48

(xii) 48 − 23 = 25

(xiii)25 − 25 = 0

We have subtracted successive odd numbers starting from 1 to 169, and obtained 0 at 13th step.

Therefore, √169=13

Question 4:

Find the square roots of the following 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

Answer:

(i) 729 can be factorised as follows.

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

Answer:

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 number. Also find the square root of the square number so obtained.

(i) 252 (ii) 2925 (iii) 396 (iv) 2645 (v) 2800 (vi) 1620

Answer:

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.

Answer:

It is given that each student donated as many rupees as the number of students of

the class. Number of students in the class will be the square root of the amount

donated by the students of the class.

The total amount of donation is Rs 2401.

Number of students in the class = √2401

∴ √2401=7×7×7×7

√2401=7×7 =49

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.

Answer:

It is given that in the garden, each row contains as many plants as the number of rows.

Number of rows = Number of plants in each row

Total number of plants = Number of rows × Number of plants in each row

Number of rows × Number of plants in each row = 2025

(Number of rows)² = 2025

∴ √2025=3×3×3×3×5×5

√2025=3×3×5 =45

Thus, 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 numbers 4, 9, and 10.
Answer:

Question 10:

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

Answer: