MCQ Questions for Class 8 Maths Squares And Square Roots Quiz 2

$7928$ is not a perfect square. Give reason.

0%
Because last two digits are $28$
0%
Because last digit of a number is $8$
0%
Because first two digits are not divisible by $8$
0%
Data insufficient

Explanation

We know that numbers ending with

$2, 3, 7$ and

$8$ are not perfect squares.

So, the number

$7928$ is not a perfect square because it ends with

$8$ .

Which of the following is not a perfect square number?

0%
$16$
0%
$25$
0%
$37$
0%
None of the above

Explanation

Here, $16$ can be written as,

$16=4\times4$

$=4^2$

And, $25$ can be written as,

$25=5\times 5$

$=5^2$

But $37$ is a prime number and it cannot be written as a square of any whole number.

Therefore, option $C$ is the correct answer.

What is the value of

$\sqrt4$ ?

0%
$2$
0%
$3$
0%
$1$
0%
None of these

Explanation

$4$ can be written as

$2^2$

$\therefore \sqrt4=\sqrt{2^2}=2$

Hence, option A is correct.

The numbers that end in 2, 3, 7 or 8 are called ______ squared numbers.

0%
perfect
0%
non-perfect
0%
rational
0%
real

Explanation

The numbers end in

$2, 3, 7$ or

$8$ are called non-perfect squared numbers.

If the square number ends with ___, then the number has

$8$ or

$2$ at the units place.

0%
$18$
0%
$6$
0%
$9$
0%
$4$

Explanation

$2 \times 2= 4$ and $8 \times 8 = 6\underline{4}$ (the units place is $4$ )

So, either $8$ or $2$ is necessary in the units place.

$42$ is not a _______ number .

0%
minimum
0%
perfect square
0%
odd
0%
prime

Explanation

As we know that all the perfect square number ends with $1,4,9,6,5,00$ T

Therefore, B is the correct answer.

1 is the square root of

0%
$4$
0%
$1$
0%
$3$
0%
None

Explanation

$1^2=1$ and

$(-1)^2=1$

Therefore, B is the correct answer.

Can we construct sets of Pythagorean Triples with all even numbers?

0%
Real
0%
Yes
0%
No
0%
Negative

Explanation

Yes we can construct Pythagorean Triples with all even number.

An example of such triplet is

$6,8,10.$

where

${ \left( 8 \right) }^{ 2 }{ +(6) }^{ 2 }={ \left( 10 \right) }^{ 2 }$

So option

$B$ is correct.

What is the value of

$\sqrt{625}$ ?

0%
$25$
0%
$-5$
0%
$15$
0%
$-15$

Explanation

The square root of

$625$ is

$25$ as

$5 \times 5 \times 5 \times 5 = 25\times 25 = (25)^2 =625$ .

What is the value of

$\sqrt {441}$ ?

0%
20
0%
21
0%
22
0%
None of these

Explanation

$441$ can be written as

$(21)^2$

So,

$\sqrt{441}=\sqrt{(21)^2}=21$

Hence, option B is correct.

All perfect squares end in 0, 1, 4, 5, 6, __.

0%
3
0%
2
0%
9
0%
7

Explanation

All perfect squares end in

$0, 1, 4, 5, 6, 9$ .
Example:

$484, 169, 625, 100, 676, 121$ are perfect squares.

Identify the Pythagorean triplet.

0%
$\left( 6,8,10 \right)$
0%
$\left( 3,4,7 \right)$
0%
$\left( 5,12,18 \right)$
0%
$\left( 2,3,6 \right)$

Explanation

A pythagorean triplet satisfies

${ c }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }$

In the option

$A,$

Let

$c=10, \ a=6,\ b=8.$

So,

${ c }^{ 2 }={ (10) }^{ 2 }$

$=100$

And,

${ a }^{ 2 }+{ b }^{ 2 }={ 6 }^{ 2 }+{ 8 }^{ 2 }$

$=36+64$

$=100$

$\Rightarrow { c }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }$

$(6,8,10)$ is Pythagorean triplet.

The triplets in the rest of the options do not satisfy the above condition.

How many two-digit numbers satisfy this property: The last digit (unit's digit) of the square of the two-digit number is 8 ?

0%
1
0%
2
0%
3
0%
None of these

Explanation

As we know that all the perfect square number ends with

$1,4,9,6,5,00$

So, a number ending in 8 can never be a perfect square.

Which one of the following can't be the square of natural number?

0%
$30976$
0%
$75625$
0%
$28561$
0%
$143642$
0%
None of these

Explanation

The square of a natural number never ends with

$2$ as the unit's digit.

$\therefore$

$143642$ is not the square of natural number because it has 2 as the unit's digit.

A man plants his orchard with

$5625$ trees, and arranges them so that there are as many rows as there are trees in a row; how many rows are there?

0%
$65$
0%
$75$
0%
$85$
0%
$55$

Explanation

It simply means that number of trees in a row & in a single coloum is same

Let this be

$x$ then it will form a square field with side length

$x$

Hence

$x^2$

$= 5625$

$x = 75$

What will be the units digit of the square of the given number

$7286$ ?

0%
$0$
0%
$1$
0%
$5$
0%
$6$

Explanation

Unit digit of multiplication of

$2$ numbers is obtained by taking the last digit of the multiplication of two numbers i.e. the digits at the unit place

Units digit of

$7286\times 7286$ is obtained by taking the last digit of

$6\times 6=36$

$\Rightarrow$ Unit digit of square of

$7286$ is

$6$ .

What will be the units digit of the square of the given number

$39$ ?

0%
$9$
0%
$3$
0%
$1$
0%
$0$

Explanation

Unit digit of multiplication of 2 numbers is obtained by taking the last digit of the multiplication of two numbers i.e. the digits in the unit place

Units digit of

$39×39$ is obtained by taking the last digit of

$9×9=81$

$\Rightarrow$ Unit digit of square of

$39$ is

$1$ .

Which of the following is not a perfect square?

0%
$16384$
0%
$23857$
0%
$18496$
0%
$11025$

Explanation

$16384= 128\times 128$ (perfect square)

$18496=136\times 136$ (perfect square)

$11025=105\times 105$ (perfect square)

$23857 = 1\times 23857$ (prime no. so not a perfect square)

Hence, option B is correct.

State true or false:

The number

$14641$ is a perfect square of

$121$ .

0%
True
0%
False

Explanation

By prime factorization we have

$14641=11\times 11 \times 11 \times 11 = {(11 \times 11)}^2 = (121)^2$

Hence it is a perfect square.

State true or false:

The number

$1156$ is a perfect square of

$34$ .

0%
True
0%
False

Explanation

$(34)^2=1156$

So, given statement is true.

State true or false:

The number

$2025$ is a perfect square of

$35$ .

0%
True
0%
False

Explanation

$(35)^2=1225$
So, given statement is false.

State true or false:

The number

$4762$ is a perfect square of

$69$ .

0%
True
0%
False

Explanation

$69^2=4761$
So, given statement is false.

Which one of the following numbers is not a perfect square from

$4900, 90000, 810000, 81000$ ?

0%
$4900$
0%
$90000$
0%
$810000$
0%
$81000$

Explanation

$\sqrt { 4900 } =70$ perfect square

$\sqrt { 90000 } =300$ perfect square

$\sqrt { 810000 } =900$ perfect square

$\sqrt { 81000 }$ is not an integer

$\sqrt { 81000 }$ is not a perfect square

Find the value of

$5^2$ .

0%
20
0%
25
0%
22
0%
5

Explanation

$\textbf{Step-1: Finding }$ $5^2$

${5^2= }$ $5\times 5 =$ $25$

$\textbf{Hence,}$ $5\times 5 =$ $25$

Find the square root correct upto

$2$ decimals

$60.92$ .

0%
$7.92$
0%
$7.81$
0%
$7.27$
0%
$7.56$

Explanation

$\sqrt{6092}$

$=\sqrt{\dfrac{6092}{100}}$

$=\dfrac{78.05}{10}$

$=7.81$

The smallest four digit number which is perfect square is ____.

0%
$1000$
0%
$1016$
0%
$1024$
0%
$1036$

Explanation

Consider,

$31\times 31=961$

$32\times 32=1024$

$\therefore$ Smallest four-digit number which is perfect square is

$1024$ .

Which of the following is a square of an even number?

0%
144
0%
169
0%
441
0%
625

Explanation

The unit place of the square of any even number is always an even number, so

$144$ will be the square of an even number.

Square of an even number,

$144 = 12 \times 12 = (12)^2$

Hence, A will be the correct answer.

State whether the statement is true (T) or false (F):

$1000$ is a perfect square.

0%
True
0%
False

Explanation

$\therefore 1000 =2 \times 2 \times 2 \times 5 \times 5 \times 5 = 2^3 \times 5^3$

Since, the powers are not even,

$1000$ is not a perfect square

1649413_1792082_ans_2c784c3c2a9b4140b2958bee63196e79.png

State true or false:

Pythagorean triplet, whose one member is

$14$ , is

$14, 48, 50$

0%
True
0%
False

Explanation

$2k$ ,

$(k^2 + 1)$ and

$(k^2 - 1)$ form a Pythagorean triplet for any number,

$k > 1$

Let

$2k = 14$
Hence,

$k = 7$

and,

$k^2 + 1 = 7^2 + 1= 49 + 1 = 50$
and,

$k^2 - 1 = 7^2 - 1 = 49 - 1 = 48$

Check the answers:

$14^2 + 48^2 = 196 + 2304 = 2500 = 50^2$

Hence, the triplet is

$14, 48,50$

Pythagorean triplet of given number

$6$ is

0%
$6, 8, 10$
0%
$6, 9, 10$
0%
$6, 8, 11$
0%
$6, 7, 10$

Explanation

$2k$ ,

$(k^2 + 1)$ and

$(k^2 - 1)$ form a Pythagorean triplet for any number, k > 1

Let

$2k = 6$
Hence,

$k = 3$

and,

$k^2 + 1 = 3^2 + 1= 9 + 1 = 10$
and,

$k^2 - 1 = 3^2 - 1 = 9 - 1 = 8$

Check the answers:

$6^2 + 8^2 = 36 + 64 = 100 = 10^2$

Hence, the triplet is

$6, 8,10$

CBSE Questions for Class 8 Maths Squares And Square Roots Quiz 2 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Squares And Square Roots Quiz 2 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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