## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 3 Squares and Square Roots Objective Type Questions

**Mental Maths**

Question 1.

Fill in the blanks:

(i) A number ending in ……… is never a perfect square.

(ii) On combining two consecutive triangular number, we get a ………

(iii) If a number has digits ……… in the unit’s place, then its square ends in 1.

(iv) Sum of first 10 odd natural numbers is ………

(v) Number of non-square numbers between 11^{2} and 12^{2} is ………

(vi) Number of zeros in the end of the square of 400 is ………

(vii) Square of any ……… number can be expressed as the sum of two consecutive natural numbers.

(viii)For a natural number m > 1, (2m, m^{2} – 1, m^{2} + 1) is called ………

Solution:

(i) A number ending in 2,3,7 and 8 is never a perfect square.

(ii) On combining two consecutive triangular number,

we get a square number.

(iii) If a number has digits 1 or 9 in the unit’s place,

then its square ends in 1.

(iv) Sum of first 10 odd natural numbers is (10)^{2} = 100.

(v) Number of non-square numbers between

11^{2} and 12^{2} is (12^{2} – 11^{2}) – 1 = 12 + 11 – 1 =22.

(vi) Number of zeros in the end of the square of 400 is 0000 (four zeros).

(vii) Square of any odd number can be expressed as

the sum of two consecutive natural numbers.

(viii) For a natural number m > 1,

(2m, m^{2} – 1, m^{2} + 1) is called Pythagorean triplet.

Question 2.

State whether the following statements are true (T) or false (F):

(i) All natural numbers are not perfect squares.

(ii) A perfect square can never be expressed as the product of pairs of equal prime factors.

(iii) A number having 2,3,7 or 8 at its unit place is never a square number.

(iv) A number having 0, 1, 4, 5, 6 or 9 at its unit place is always a square number.

(v) A number ending in an even number of zeros is always a perfect square.

(vi) Square of an odd number is always an odd number.

(vii) 1, 3, 6, 10, 15, are called triangular numbers.

(viii)There are 2n non-square numbers between the squares of consecutive numbers n and (n + 1).

(ix) (4, 6, 8) is a Pythagorean triplet.

Solution:

(i) All natural numbers are not perfect squares. (True)

(ii) A perfect square can never be expressed as

the product of pairs of equal prime factors. (True)

(iii) A number having 2, 3, 7 or 8 at its unit place is never a square number. (Time)

(iv) A number having 0, 1, 4, 5, 6 or 9 at its unit place is always a square number. (False)

Correct:

As 24, 14, 34, 26, 19, 50, 61, 35, etc are not a perfect square

(v) A number ending in an even number of zeros is always a perfect square. (False)

Correct:

As 200, 500, 8000, etc. are not perfect squares.

(vi) Square of an odd number is always an odd number. (True)

(vii) 1, 3, 6, 10, 15, are called triangular numbers. (True)

(viii) There are 2n non-square numbers between

the squares of consecutive numbers n and (n + 1). (True)

(ix) (4, 6, 8) is a Pythagorean triplet. (False)

Correct:

As 4^{2} + 6^{2} ≠ 8^{2} ⇒ 16 + 36 ≠ 64

⇒ 52 ≠ 64

**Multiple Choice Questions**

**Choose the correct answer from the given four options (3 to 15):**

Question 3.

How many natural numbers lie between 25^{2} and 26^{2}?

(a) 49

(b) 50

(c) 51

(d) 52

Solution:

Natural numbers between 25^{2} and 26^{2}

25 + 26 – 1 = 50 (b)

Question 4.

Square of an even number is always

(a) even

(b) odd

(c) even or odd

(d) none of these

Solution:

Square of an even number is always even. (a)

Question 5.

1+ 3 + 5 + 7 + ……….. up to n terms is equal to

(a) n^{2} – 1

(b) (n + 1)^{2}

(c) n^{2} + 1

(d) n^{2}

Solution:

1 + 3 + 5 + 7 + …….. up to n terms is equal to n^{2} + 1 (c)

Question 6.

\(\sqrt{208+\sqrt{2304}}\) is equal to

(a) 18

(b) 16

(c) 14

(d) 22

Solution:

Question 7.

\(\sqrt{0.0016}\) is equal to

(a) 0.04

(b) 0.004

(c) 0.4

(d) none of these

Solution:

\(\sqrt{0.0016}\) = 0.04

Question 8.

The smallest number by which 75 should be divided to make it a perfect square is

(a) 1

(b) 2

(c) 3

(d) 4

Solution:

75 = 3 × 5 × 5

Factor 3 is unpaired

∴ By dividing 75 by 3, we get a perfect square of 5.

Question 9.

\(\sqrt{3 \frac{6}{25}}\) is equal to

Solution:

Question 10.

The smallest number by which 162 should be multiplied to make it a perfect square is

(a) 4

(b) 3

(c) 2

(d) 1

Solution:

162 = 2 × 3 × 3 × 3 × 3

For 2 is left unpairs. So, by multiplying 162 by 2,

we get a perfect square.

∴ Required least number to be multiplied = 2 (c)

Question 11.

If the area of a square field is 961 unit^{2}, then the length of its side is

(a) 29 units

(b) 41 units

(c) 31 untis

(d) 39 units

Solution:

Area of a square = 961 unit^{2}

∴ It’s side = \(\sqrt{961}\) unit = 31 unit (c)

Question 12.

The smallest number that should be subtracted from 300 to make it a perfect square is

(a) 11

(b) 12

(c) 13

(d) 14

Solution:

300

Taking the square root of 300,

we see that 11 is left unpaired.

∴ 11 be subtracted. (a)

Question 13.

If one number of Pythagoream triplet is 6, then the triplet is

(a) (4, 5, 6)

(b) (5, 6, 7)

(c) (6, 7, 8)

(d) (6, 8, 10)

Solution:

One number of a Pythagorean triplet is 6

Let 2n = 6 ⇒ n = 3

n^{2} – 1 = 3^{2} – 1 = 8 and n^{2} + 1 = 3^{2} + 1 = 10

∴ Triplet is (6, 8, 10) (d)

Question 14.

nth triangular number is

Solution:

nth triangular number = \(\frac{n(n+1)}{2}\) (a)

Question 15.

Given that \(\sqrt{1521}\) = 39, the value of \(\sqrt{0.1521}+\sqrt{15.21}\) is

(a) 42.9

(b) 4.29

(c) 3.51

(d) 35.1

Solution:

\(\sqrt{1521}\) = 39, then value of \(\sqrt{0.1521}+\sqrt{15.21}\)

= 0.39 + 3.9 = 4.29 (b)

**Value Based Questions**

Question 1.

In a school, students of class VIII collected ₹9216 to give a donation to an NGO working for the education of poor children. If each student donated as many rupees as the number of students in class VIII. Find the number of students in class VIII.

Why should we donate money for the education of poor children? What values are being promoted?

Solution:

Amount collected by students = ₹ 9216

and each donated amount equal to the number of students

Number of students = \(\sqrt{9216}\) = ₹ 96

Donation to an NGO, who is working for the education

of poor children is a noble cause.

Question 2.

A person wants to plant 2704 medicinal plants with a board depicting the diseases in which that can be used. He planted these in the form of rows. If each row contains as many plants as the number of rows, then find the number of rows.

Why should we plant medicinal plants? What values are being promoted?

Solution:

Total number of plants = 2704

These are planted in such a way that

Number of rows = number of plants in each row

∴ Number of rows = \(\sqrt{2704}\)

Medicinal plants = 52

These plants give us many kinds of medicines

which are useful for different diseases.

So, it is a noble cause for society.

**Higher Order Thinking Skills (Hots)**

Question 1.

A square field is to be ploughed. Ramu get it ploughed in ₹34560 at the rate of ₹15 per sq. m. Find the length of side of square field.

Solution:

Total expenditure for ploughing the field = ₹34560

The rate of ploughing = ₹15 per sq. m

∴ Total area of the square field = \(\frac{34560}{15}\)

= 2304 m^{2}

Side of the square field = \(\sqrt{2304}\) m

= 48 m

Question 2.

Lalit has some chocolates. He distributed these chocolates among 13 children in such a way that he gave one chocolate to first child, 3 chocolates to second child, 5 chocolates to third and so on. Find the number of chocolates Lalit had.

Solution:

Number of children = 13

Lalit gave chocolates to the children in such a way that

he gives one chocolate to first child, 3 chocolates to the second child

5 chocolates to a third child and so on

∴ 1 + 3 + 5 + up to 13 terms

(∵ 1 + 3 + 5 + 7 + ……… n terms = n^{2})

= 169 chocolates