MCQ Questions for Class 8 Maths Squares And Square Roots Quiz 4

A four digit perfect square number whose first two digits and last two digits taken seperately are also perfect squares is:

0%
1,681
0%
1,636
0%
3,664
0%
6,481

Explanation

$1,681$ is the only perfect square from given options.

$1681=41\times 41$

$1681=41^2$

$\therefore$

$1681$ is a square of

$41$ .

First two digits and last two digits of

$1681$ are

$16$ and

$81$

$16=4^2$ and

$81=9^2$

We can see

$16$ and

$81$ are perfect squares of

$4$ and

$9$ .

$\therefore$

$1681$ is a four digit square number whose first two digits taken separately are also perfect square.

Which of the following numbers will not have 1(one) at their unit's place?

0%
$(32)^2$
0%
$(57)^2$
0%
$(69)^2$
0%
$(321)^2$
0%
$(97)^2$
0%
$(265)^2$
0%
$(2091)^2$

Explanation

(A) One's digit of

$(32)^2$ = One's digit of

$2^2$ =

$4$
(B) One's digit of

$(57)^2$ =One's digit of

$7^2$ =

$9$
(C) One's digit of

$(69)^2$ = One's digit of

$9^2$ =

$1$
(D) One's digit of

$(321)^2$ = One's digit of

$1^2$ =

$1$
(E) One's digit of

$(97)^2$ = One's digit of

$7^2$ =

$9$
(F) One's digit of

$(265)^2$ = One's digit of

$5^2$ =

$5$
(G) One's digitof

$(2091)^2$ = One's digit of

$1^2$ =

$1$

Hence, the

$(A),(B),(E)$ and

$(F)$ are the correct options.

The value of

$\sqrt{0.9}$ is (approximately)

0%
0.3
0%
0.03
0%
0.33
0%
0.94

Explanation

$\sqrt{0.9} = {\sqrt{0.90} = \sqrt{\cfrac{90}{100}} = \cfrac{\sqrt{90}}{10}}$

$= \cfrac{9.4}{10} = 0.94$

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

0%
$1936$
0%
$2452$
0%
$1369$
0%
$6561$

Explanation

As we know that all the perfect square number ends with

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

$2452$ is not a perfect square because it ends in 2.

$\text {Also} \quad 1936= (44)^2\\1369=(37)^2\\6561=(81)^2$

Find the square root of 784 by prime factorisation method.

0%
$22$
0%
$28$
0%
$18$
0%
$38$

Explanation

$784=\underline { 2\times2 } \times\underline { 2\times2 } \times\underline { 7\times7 } \\ \Rightarrow \sqrt { 784 } =2\times2\times7=28$

The smallest number by which 396 must be multiplied so that the product becomes a perfect square is:

0%
5
0%
11
0%
3
0%
2

Explanation

$396\, =\, 2\, \times\, 2\, \times\, 3\, \times\, 3\, \times\, 11$
So 396 should be multiplied by 11 to make the product a perfect square.

The least number by which 216 must be divided to make the result a perfect square, is

0%
3
0%
4
0%
6
0%
9

Explanation

$216 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$
Clearly, in order to make it a perfect square, it must be divided by

$2 \times 3$ i.e.,

$6$ .

For the following number, 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.

$252$

0%
$3;42$
0%
$7;42$
0%
$7;30$
0%
$2;42$

Explanation

By prime factorisation we get

$252=\underline{2\times2}\times\underline{3\times3}\times7$

${ \underline { 2\mid 252 } }\\ { \underline { 2\mid 126 } }\\ { \underline { 3\mid 63 } }\\ { \underline { 3\mid 21 } }\\ { \underline { 7\mid 7 } }\\ \; \; \mid 1$
It is clear that in order to get a perfect square, one more

$7$ is required.
So, the given number should be multiplied by

$7$ to make the product a perfect square.

$\therefore\;252\times7=1764$ is a perfect square.
Thus,

$1764=\underline{2\times2}\times\underline{3\times3}\times\underline{7\times7}$

$\therefore\;\sqrt{1764}=2\times3\times7=42$

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

$153$

$257$

$408$

$441$

0%
$153,\,257\;and\;441$ are surely not perfect squares.
0%
$441,\,257\;and\;408$ are surely not perfect squares.
0%
$153,\,441\;and\;408$ are surely not perfect squares.
0%
$153,\,257\;and\;408$ are surely not perfect squares.

A gardener wants to plant

$17956$ trees and arranges them in such a way that there are as many rows as there are trees in a row. What is the number of trees in a row?

0%
$144$
0%
$136$
0%
$154$
0%
$134$

The largest number of

$5$ digits, which is perfect square is

0%
$99999$
0%
$99764$
0%
$99976$
0%
$99856$

Explanation

The largest number of 5 digits

$= 99999$

Required number

$= (99999 - 143) = 99856 = 316^2$

What could be the possible 'one's' digit of the square root of each of the following numbers?

$9801$

$99856$

0%
$1\;or\;9\;;\;4\;or\;6$
0%
$2\;or\;8\;;\;4\;or\;6$
0%
$1\;or\;9\;;\;2\;or\;8$
0%
$2\;or\;8\;;\;5\;or\;0$

Explanation

As we know that all the perfect square number ends with

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

Thus, the possible unit digit of square root of 9801 is either 1 or 9 and it will odd square root and the possible unit digit of square root of 99856 is either 4 or 6 and it will have even square root.

Find the square root of the following decimal numbers:

$2.56$
&

$51.84$

0%
$1.4$
&
$7.8$
0%
$1.6$
&
$7.2$
0%
$1.4$
&
$7.2$
0%
$1.6$
&
$7.8$

Explanation

Here, the number of decimal places is already even. So, mark off periods proceed as under

$\;\;\;\;\;\;\;1.6\\ \overline { 1\; \; \mid \;\;\bar { 2 } .\bar { 56 } } \\ { { \; \; \; \; \; \mid -1 } }\\ { \overline { 26\mid\;\; 156 } }\\ \; \; \; { \underline { \; \; \mid -156 } }\\ \; \; \; \; \mid \; \; 0$

$\therefore\;\sqrt{2.56}=1.6$
Here, the number of decimal places are already even. So, mark off periods proceed as under

$\;\;\;\;\;\;\;\;\;7.2\\ \;\;\;\overline { 7\; \; \mid \;\;\bar { 51 } .\bar { 84 } } \\ { { \; \; \; \; \;\;\;\;\mid -49 } }\\ \;\;{ \overline { 142\mid\;\; 284 } }\\ \; \; \; { \underline { \; \; \;\;\;\mid -284 } }\\ \; \; \; \; \;\;\;\mid \; \; 0$

$\therefore\;\sqrt{51.84}=7.2$

Find the square root of each of the following numbers by division method:

$2304$

$4489$

$3481$

0%
$38;67;59$
0%
$48;57;59$
0%
$48;67;69$
0%
$48;67;59$

Explanation

By long division, we have

$\therefore\;\sqrt{2304}=48$

$\;\;\;\;\;\;\;\;\;\;48\\ \overline { 4\;\;\mid \;\;\bar {23 } \bar { 04 } } \\ { { \; \; \;\;\; \mid -16 } }\\ { \overline { 88\mid\;\;\;\;704 } }\\ \; \; \; { \underline { \;\;\mid -704 } }\\ \; \;\;\;\mid \; \; 0$
By long division, we have

$\therefore\;\sqrt{4489}=67$

$\;\;\;\;\;\;\;\;\;\;67\\ \overline { \;\;6\; \mid \;\;\bar {44 } \bar { 89 } } \\ { { \; \; \; \; \; \;\mid -36 } }\\ { \overline { 127\mid\;\; 889 } }\\ \; \; \; { \underline { \; \;\;\mid -889 } }\\ \; \; \; \;\; \mid \; \; 0$
By long division, we have

$\therefore\;\sqrt{3481}=59$

$\;\;\;\;\;\;\;\;\;\;59\\ \overline { \;\;5\; \mid \;\;\bar{34 } \bar{ 81 } }\\ { { \; \; \; \; \; \;\mid -25 } }\\ { \overline { 109\mid\;\; 981 } }\\ \; \; \; { \underline { \; \;\;\mid -981 } }\\ \; \; \; \;\; \mid \; \; 0$

The students of class VII of a school denoted

$Rs. 2401$ in all, for Prime Minister's National Relief Fund. Each student denoted as many rupees as the number of students in the class. Find the number of students in the class.

0%
$47$
0%
$51$
0%
$49$
0%
$53$

Explanation

Let

$x$ be the number of students in the class.
Therefore, total money denoted by students

$=Rs.\;(x\times x)=Rs.\;x^2$ . But it is given as

$Rs.\;2401$ .

$\therefore\;x^2=2401$

$\Rightarrow\;x=\sqrt{2401}$

${ \underline { 7\mid 2401 } }\\ { \underline { 7\mid 343 } }\\ { \underline { 7\mid 49 } }\\ { \underline { 7\mid 7 } }\\ \; \; \mid 1$

$=\sqrt{\underline{7\times7}\times\underline{7\times7}}$

$=7\times7=49$ .

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

0%
$143642$
0%
$27859$
0%
$57896$
0%
$27896$

Explanation

As we know that all the perfect square number ends with

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

$\therefore$ 143642 is not the square of natural number because it ends with 2.

Find the square root of

$10$ , correct to four places of decimal.

0%
$3.4623$
0%
$3.1023$
0%
$3.1693$
0%
$3.1623$

Explanation

.....................................	$3.16227$ ...........................
$3$ $+3$	$10$ $9$
$61$ $+1$	$100$ $61$
$626$ $+6$	$3900$ $3756$
$6322$ $+2$	$14400$ $12644$
$63242$ $+2$ ------------- $632447$	$175600$ $126484$ --------------- $4911600$ $4427129$

$\sqrt{10}=3.16227\simeq 3.1623$

$\therefore$ The square root of

$10$ correct to four places of decimal is

$3.1623$

What least number must be substracted from

$176$ to make it a perfect square?

0%
$16$
0%
$10$
0%
$7$
0%
$4$

Explanation

$176$ is not a perfect square .

$13^{2} = 169 < 176$

$176-169=7$
Therefore,

$7$ need to subtract for make perfect square.

Which of the following is a pythagorean triplet?

0%
$(2,\,3,\,5)$
0%
$(5,\,7,\,9)$
0%
$(6,\,9,\,11)$
0%
$(8,\,15,\,17)$

Explanation

$\textbf{Step1-Check each option which satisfy the condition of Pythagorean triplet}$

$\text{(a,b,c) will be a Pythagorean triplet if its satisfies the condition } a^2+b^2=c^2$

$\text{Here, we will check the option on by one till we get the answer}$

$\text{In option A we have (a,b,c) = (2,3,5)}$

$\Rightarrow 2^2+3^2=5^2$

$\Rightarrow 4+9=25$

$\Rightarrow 13\neq 25$

$\text{Hence option A is not correct}$

$\text{In option B we have (a,b,c) = (5,7,9)}$

$\Rightarrow 5^2+7^2=9^2$

$\Rightarrow 25+49=81$

$\Rightarrow 74 \neq 81$

$\text{Hence option B is not correct}$

$\text{In option C we have (a,b,c) = (6,9,11)}$

$\Rightarrow 6^2+9^2=11^2$

$\Rightarrow 36+81=121$

$\Rightarrow 117\neq 121$

$\text{Hence option C is not correct}$

$\text{In option D we have (a,b,c) = (8,15,17)}$

$\Rightarrow 8^2+15^2=17^2$

$\Rightarrow 64+225=286$

$\Rightarrow 286=286$

$\text{Hence option D is correct}$

$\textbf{Hence option D is correct}$

Check whether

$1152$ is a perfect square. If not, then find the least number by which it should be divided so that, the quotient becomes a perfect square. Also, find the square root of the quotient.

0%
$2; 24$
0%
$3; 24$
0%
$2; 19$
0%
$3; 19$

Explanation

$1152=\underline{2\times2}\times\underline{2\times2}\times\underline{2\times2}\times2\times\underline{3\times3}$
All the prime factors except

$2$ can be paired.

Thus

$1152$ is not a perfect square
Thus, we should divide

$1152$ by

$2$ to get the perfect square.

$1152\div2=576$

$576=\underline{2\times2}\times\underline{2\times2}\times\underline{2\times2}\times\underline{3\times3}$
Thus, the required least number is

$2\;and\;\sqrt{576}=24$ .

Find the square root of each of the following correct to three places of decimal.

$17$

$1.7$

$2.5$

$\displaystyle\frac{7}{8}$

0%
$4.153\;;\;1.304\;;\;1.581\;;\;0.935$
0%
$4.123\;;\;1.304\;;\;1.581\;;\;0.995$
0%
$4.123\;;\;1.304\;;\;1.581\;;\;0.935$
0%
$4.123\;;\;1.304\;;\;1.501\;;\;0.935$

Explanation

$17$

	$4.123$
$4$ $+4$	$17$ $16$
$822$ $+2$	$100$ $81$
$8243$	$1900$ $1644$
	$25600$ $24729$ $871$

$\therefore$ The square root of

$17=4.123$

$1.7$

$0$	$1.303$
$1$ $+1$	$1.7$ $1$
$23$ $+3$	$70$ $69$
$2603$	$10000$ $7809$
	$1191$

$\therefore$ The square root of

$1.7=1.303$

$2.5$

$0$	$1.581$
$1$ $+1$	$2.5$ $1$
$25$ $+5$	$150$ $125$
$308$ $+8$	$2500$ $2464$
$3161$	$3600$ $3161$ $439$

$\therefore$ The square root of

$2.5=1.581$

$\frac{7}{8}$

$8$	$70$ $64$	$0.875$
	$60$ $56$
	$40$ $40$ $0$

	$0.935$
$0$	$0.875$ $0$
$9$ $9$	$87$ $81$
$183$ $+3$	$650$ $549$
$1865$	$10100$ $9325$
	$925$

$\therefore$ The square root of

$\frac{7}{8}=0.935$

Find the least number which must be added to each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained.

$525$

$1750$

0%
$4,23;14,48$
0%
$4,27;14,42$
0%
$4,23;14,42$
0%
$4,27;14,48$

Explanation

We try to find out the square root of

$525$ .

$\;\;\;\;\;\;\;22\\ \overline { 2\; \; \mid \;\;\bar { 5 }\bar { 25 } } \\ { { \; \; \; \; \; \mid -4 } }\\ { \overline { 42\mid\;\; 125 } }\\ \; \; \; { \underline { \; \; \mid -84 } }\\ \; \; \; \; \mid \; \; 41$
We observe that

$(22)^2\,<\,525\,<\,(23)^2$
The required number to be added

$=(23)^2-525$

$=529-525=4$

$\therefore$ Required perfect square number

$=525+4=529$
Clearly,

$\sqrt{529}=23$ .
We try to find out the square root of

$1750$

$\;\;\;\;\;\;\;41\\ \overline { 4\; \; \mid \;\;\bar { 17 }\bar { 50 } } \\ { { \; \; \; \; \; \mid -16 } }\\ { \overline { 81\mid\;\; 150 } }\\ \; \; \; { \underline { \; \; \mid -81 } }\\ \; \; \; \; \mid \; \; 69$
We observe that

$(41)^2\,<\,1750\,<\,(42)^2$ .
The required number to be added

$=(42)^2-1750$

$=1764-1750$

$=14$

$\therefore$ Required perfect square number

$=(1750+14)=1764$
Clearly,

$\sqrt{1764}=42$ .

When a number ends with a

$6$ in units place, the units place of a square is____.

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

Explanation

As $6 \times 6 = 36$ , so if a number ends with $6$ , then square of that number also ends with $6$

Therefore, option B is the correct answer.

What could be the possible one's digits of the square root of the following numbers?

$9801$

$99856$

$99800$

$657666025$

0%
$1\;or\;9\;;\;4\;or\;6\;;\;0\;;\;5$
0%
$2\;or\;8\;;\;4\;or\;6\;;\;0\;;\;5$
0%
$1\;or\;9\;;\;2\;or\;8\;;\;0\;;\;5$
0%
$1\;or\;9\;;\;4\;or\;6\;;\;5\;;\;0$

Explanation

As we know that all the perfect square number ends with

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

$9801 \Rightarrow$ Possible one's digits

$=1\;or\;9\;\;(\because\;1\times1=1.9\times9=81)$

$99856 \Rightarrow$ Possible one's digits

$=4\;or\;6$

$99800 \Rightarrow$ Possible one's digit

$=0$

$657666025 \Rightarrow$ Possible one's digit

$=5$

Which one of the following cannot be the square of a natural number ?

0%
$30976$
0%
$75625$
0%
$28561$
0%
$143642$

Explanation

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

So, the square of a natural number never ends in $2$

$\therefore 143642$ is not the square of a natural number.

The square root of an odd number

0%
need not be odd
0%
is odd
0%
is even
0%
cannot be determined

Explanation

The square root of an odd number is odd number. We can check this by taking examples as follows:

$\sqrt[2]{25}=5$

Square root of odd number,

$25$ is a odd number

$5$ .

$\sqrt[2]{49}=7$

Square root of odd number,

$49$ is a odd number

$7$ .

Hence, we can say that the square root of odd number is odd. Option C is correct.

Write a Pythagorean triplet whose one member is

$16$ :

0%
$25, 49$
0%
$25, 61$
0%
$63, 65$
0%
None of these

What is the smallest number by which 338 is multiplied or divided to make a perfect square?

0%
13
0%
4
0%
6
0%
2

Explanation

The number

$338$ is not a perfect square itself.

$338 = 2\times 169$
When we divide it by

$2$ , the quotient left is

$169$ which is a square of

$13$ .
Therefore, divide it by

$2$ .
Answer is Option

$D$

Write a Pythagorean triplet whose one member is 14:

0%
36, 64
0%
48, 50
0%
8, 6
0%
17, 20

Explanation

Let us assume 2 m=14 therefore m=7
Now

$\displaystyle m^{2}+1=7^{2}+1=49+1=50$
And

$\displaystyle m^{2}-1=7^{2}-1=49-1=48$
Test :

$\displaystyle14^{2}+48^{2}=196+1304=2500=50^{2}$
Hence the triplet is 14,48 and 50 Answer

Which of the following numbers is not a perfect square?

0%
36
0%
81
0%
37
0%
Both (A) and (B)
0%
Answer required

Explanation

As we know that all the perfect square number ends with

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

So,

$37$ is not a perfect square because it ends with

$7$ .

CBSE Questions for Class 8 Maths Squares And Square Roots Quiz 4 - 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 4 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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