# Free Squares and Square Roots 02 Practice Test - 8th Grade

### Question 1

Just by looking at the numbers, identify which of the following can not be a perfect square.

382

484

576

529

#### SOLUTION

Solution :A

Numbers ending with 2, 3, 7, or 8 at unit's place are never perfect squares.

Estimation without actually finding roots:

484 can be the perfect square of a number ending with 2 or 8.

576 can be the perfect square of a number ending with 4 or 6.

529 can be the perfect square of a number ending with 3 or 7.

### Question 2

How many non-perfect squares lie in between 752 and 762 ?

75

100

125

150

#### SOLUTION

Solution :D

Between the squares of any two consecutive numbers

nand (n+1), lie2nnon-perfect squares.

Therefore between squares of 75 and 76 (n = 75 and n + 1 = 76), lie (2 x 75) = 150 non-perfect squares.

### Question 3

The lengths 18 cm, 24 cm and 30 cm form the sides of a right-angled triangle.

#### SOLUTION

Solution :A

182+242=324+576=900=302

Since the squares of 18 and 24 add up to the square of 30, the numbers 18, 24 and 30 form a Pythagorean triplet, and hence, can be used to represent the three sides of a right- angled triangle.

### Question 4

Three identical square sheets of paper need to be cut into 4, 5, and 6 stripes of equal size respectively. Find the minimum length of the side of the original square sheet.

45

40

30

25

#### SOLUTION

Solution :C

Since the square sheet needs to be divided equally into 4, 5, and 6 equal parts respectively, its area must be a perfect square divisible by 4, 5, and 6; or the area of the square sheet should be a multiple of 4, 5, and 6.

L. C. M. of 4, 5, and 6 = 60. But 60 is not a perfect square.

60=2×2×3×5=22×3×5.

If we multiply the LCM by 3 and 5 both, it will become a perfect square. Therefore the minimum area of the square sheet =60×3×5=900 sq. units.

Now 900 is a minimum perfect square, which is divisible by 4, 5, and 6. Therefore the side of the square should be the square root of the area of the square (Area of a square = side2 sq. units).

Hence, the minimum length of the side of the square = √(900)=√(22×32×52) = 30 units.

### Question 5

1,024 trees are planted in a grid fashion such that the number of rows is equal to the number of columns. One row and one column of trees are cut such that the number of rows and columns after cutting remain equal. Find the number of trees that were cut?

61

63

65

67

#### SOLUTION

Solution :B

Number of trees = 1024. Since number of rows = number of columns, this means that the number of trees is a perfect square. Therefore number of rows/columns = square root of 1024 (number of trees).

√1024= √2×2×2×2×2×2×2×2×2×2=32

Therefore, the initial number of rows/columns = 32. After cutting 1 row and one column, number of rows/columns = 32-1 = 31.

The remaining number of trees = 312.

Therefore, number of trees that were cut = 322−312 = 1024 - 961 = 63.

### Question 6

Find the square root of 3364 by Long division Method.

#### SOLUTION

Solution :A

Given number: 3364

By grouping the numbers from the left, we get ¯¯¯¯¯¯33 ¯¯¯¯¯¯64.

Now, perform the long division.

5 85¯¯¯¯¯¯33 ¯¯¯¯¯¯6425 ↓108 864 864 0

Thus, the square root of 3364 is 58.

### Question 7

Without adding, find the sum of the following:

i.1+3+5+7+9+11+13+15+17+19+21 =

#### SOLUTION

Solution :Sum of n consecutive odd integers starting from 1 is equal to n2, where 'n' is the number of terms involved in the series.

Therefore,

i. 1+3+5+7+9+11+13+15+17+19+21 = sum of first 11 odd integers = 112 = 121.

### Question 8

It would take

#### SOLUTION

Solution :The square root of a number can be found by repeatedly subtracting odd integers from the number starting from 1.

So in this case,

256- 1-3-5-7-9-11-13-15-17-19-21-23-25-27-29-31 = 0.

(16 steps, hence square root of 256 is 16.)

### Question 9

Use Prime Factorization method to find the square root of the following:-

i. √(1024) = ______

ii. √(2025) = _______

32, 45

36, 35

#### SOLUTION

Solution :A

1024=2×2×2×2×2×2×2×2×2×2

=22×22×22×22×22⇒ √(1024)=√(22×22×22×22×22)=2×2×2×2×2=32

2025=3×3×3×3×5×5

=32×32×52⇒ √(2025)=√(32×32×52)=3×3×5=45

### Question 10

A number "p" exists such that it is the square of a number "q". If "p"> 400, "q" can be

15

19

25

29

#### SOLUTION

Solution :C and D

Given: 'p' is the square of 'q'.

⇒ 'q' is the square root of 'p'

p>400

The square root of 400 is 20.

So if "p">400, its square root 'q' can be any number greater than 20.