Download App

Maths M.C.Q. [1 Marks Each]

Square Root and Cube-Cube Root · Jharkhand Board (JAC) STD 8 (Jharkhand - English Medium). 14 questions.

Questions

M.C.Q. [1 Marks Each]

Take a timed test

14 questions · auto-graded multiple-choice test.

MCQ 11 Mark
Which of the following is a square of an even number?
  • ✓
    $144$
  • B
    $169$
  • C
    $441$
  • D
    $625$
Answer
Correct option: A.
$144$
Here, $144 = (12)^2$
Similarly, $169 = (13)^2$
$441 = (21)^2$
$625 = (25)^2$
Thus, $144$ is a square of an even number.
Alternate Method
We know that, square of an even number is always an even number.
Hence, $169, 441$ and $625$ are not even numbers.
So, only $144$ is an even number, which is the square of $12.$
View full question & answer→
MCQ 21 Mark
The sum of first $n$ odd natural numbers is:
  • A
    $2n + 1$
  • ✓
    $n^2$
     
  • C
    $n^2- 1$
     
  • D
    $n^2+ 1$
Answer
Correct option: B.
$n^2$
 
Sum of frist $n$ odd natural numbers
$=\sum(2\text{n}-1)=2\sum\text{n}-\text{n}$
$=\frac{2\times\text{n}(\text{n}+1)}{2}-\text{n}$
$=\text{n}(\text{n}+1)-\text{n}$
$=\text{n}^2+\text{n}-\text{n}$
$=\text{n}^2$
View full question & answer→
MCQ 31 Mark
A perfect square number having n digits where n is even will have square root with:
  • A
    $\text{n}+1\text{ digit}$
  • ✓
    $\frac{\text{n}}{2}\text{ digit}$
  • C
    $\frac{\text{n}}{3}\text{ digit}$
  • D
    $\frac{\text{n}+1}{2}\text{ digit}$
Answer
Correct option: B.
$\frac{\text{n}}{2}\text{ digit}$
A perfect square number having n digits, where n is even, will have square root with $\frac{\text{n}}{2}\text{ digit}.$
View full question & answer→
MCQ 41 Mark
Which of the following cannot be a perfect square?
  • A
    $841$
  • B
    $529$
  • ✓
    $198$
  • D
    All of the above.
Answer
Correct option: C.
$198$
We know that, a number ending with digits $2, 3, 7$ or $8$ can never be a perfect square. So, $198$ cannot be written in the form of a perfect square.
View full question & answer→
MCQ 51 Mark
Which of the following will have $4$ at the units place?
  • A
    $ 14^2 $
     
  • ✓
    $ 62^2 $
     
  • C
    $ 27^2 $
     
  • D
    $ 35^2 $
Answer
Correct option: B.
$ 62^2 $
 
The unit place of the square of $14 = 4^2= 16 = 6$
The unit place of the square of $62 - 2^2= 4 [\because2^2=4]$
The unit place of square of $27 = 7^2= 49 = 9$
The unit place of the square of $35 = 5^2= 52 = 5$
Clearly, $ 62^2 $ has $4$ at the unit's place.
View full question & answer→
MCQ 61 Mark
Which of the following numbers is a perfect cube?
  • A
    $243$
  • ✓
    $216$
  • C
    $392$
  • D
    $8640$
Answer
Correct option: B.
$216$
$a.$ Resolving $243$ into prime factors, we have
$243 = 3 \times 3 \times 3 \times 3 \times 3$
Grouping the factors in triplets of equal factors, we get,
$243 = (3 \times 3 \times 3) \times 3 \times 3$
Clearly, in grouping, the factors in triplets of equal factors, we are left with two factors $3 \times 3.$
Therefore, $243$ is not a perfect cube.
$b.$ For option $(b)$ We have, $216$ Resolving $216$ into prime factors, we have
$216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3$
Grouping the factors in triplets of equal factors, we get $216 = (2 \times 2 \times 2) \times (3 \times 3 \times 3)$
Clearly, in grouping, the factors of triplets of equal factors, no factor is left over.
So, $216$ is a perfect cube.
$c.$ or option $(c)$ We have, $392$
Resolving $392$ into prime factors, we get
$392 = 2 \times 2 \times 2 \times 7 \times 7$
Grouping the factors in triplets of equal factors, we get
$392 = (2 \times 2 \times 2) \times 7 \times 7$
Clearly, in grouping, the factors in triplets of equal factors, we are left with two factors $7 \times 7.$
Therefore, $392$ is not a perfect cube.
$d.$ For option $(d)$ We have, $8640$
Resolving $8640$ into prime factors, we get,
$8640 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5$
Grouping the factors in triplets of equal factors, we get, $8640 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3) \times 5$
Clearly, in grouping, the factors in triplets of equal factors, we are left with one factor $5.$
Therefore, $8640$ in not a perfect cube.
After solving, it is clear that option $(b)$ is correct.
View full question & answer→
MCQ 71 Mark
If $m$ is the square of a natural number $n,$ then $n$ is:
  • A
    The square of $m.$
  • B
    Greater than $m.$
  • C
    Equal to $m$
  • ✓
    $\sqrt{\text{m}}.$
Answer
Correct option: D.
$\sqrt{\text{m}}.$
Given, $m$ is the square of $n, i,. m = n^2$
Taking square root both sides, we get
$\text{n}=\sqrt{\text{m}}$
View full question & answer→
MCQ 81 Mark
Which of the following numbers is not a perfect cube?
  • A
    $216$
  • ✓
    $567$
  • C
    $125$
  • D
    $343$
Answer
Correct option: B.
$567$
$216 = 6 \times 6 \times 6, 567 = 3 \times 3 \times 3 \times 3 \times 7$
$125 = 5 \times 5 \times 5, 343 = 7 \times 7 \times 7$
Clearly, $567$ is not a perfect cube, because in grouping, the factors in triplets of equal factors, we are left with two factors $3 \times 7.$
View full question & answer→
MCQ 91 Mark
$\sqrt[3]{1000}$ is equal to:
  • ✓
    $10$
  • B
    $100$
  • C
    $1$
  • D
    None of these.
Answer
Correct option: A.
$10$
We have, $\sqrt[3]{1000}=\sqrt[3]{10\times10\times10}$
$=\sqrt[3]{(10)^3}=(10)^\frac{3}{3}=10$
View full question & answer→
MCQ 101 Mark
Given that $\sqrt{4096}=64,$ the value of $\sqrt{4096}+\sqrt{40.96}$ is:
  • A
    $74$
  • B
    $60.4$
  • C
    $64.4$
  • ✓
    $70.4$
Answer
Correct option: D.
$70.4$
Given, $\sqrt{4096}=64$
So, $\sqrt{4096}+\sqrt{40.96}$
$=64+\sqrt{4096\times10^{-2}}$
$=64+\sqrt{4096}\sqrt{10^{-2}}$
$=64+64\times10^{-2}$
$=64+6.4=70.4$
View full question & answer→
MCQ 111 Mark
A number ending in $9$ will have the units place of its square as:
  • A
    $3$
  • B
    $9$
  • ✓
    $1$
  • D
    $6$
Answer
Correct option: C.
$1$
We know that, it a number is ending in $1$ or $9$ in the unit's place, then its square ends in $1.$
The number ending in $9$, will have the unit's place of its square as $1$. $[\because 9 \times 9 = 81]$
View full question & answer→
MCQ 121 Mark
How many natural numbers lie between $52$ and $6^2$?
  • A
    $9$
  • ✓
    $10$
  • C
    $11$
  • D
    $12$
Answer
Correct option: B.
$10$
The natural numbers lying between $5^2$ and $6^2$,
i.e. between $25$ and $36$ are $26, 27, 28, 29, 30, 31, 32, 33, 34$ and $35.$
Hence, $10$ natural numbers lie between $5^2$ and $6^2$.
View full question & answer→
MCQ 131 Mark
The sum of successive odd numbers $1, 3, 5, 7, 9, 11, 13$ and $15$ is:
 
  • A
    $81$
  • ✓
    $64$
  • C
    $49$
  • D
    $36$
Answer
Correct option: B.
$64$
We know that, the sum of first $n$ odd natural numbers is $n^2$.
Given odd numbers are $1, 3, 5, 7, 9, 11, 13$ and $15.$
So, number of odd numbers, $n = 8$
The sum of given odd numbers $= n^2= (8)^2= 64$
View full question & answer→
MCQ 141 Mark
The value of $\sqrt{248+\sqrt{52+\sqrt{144}}}$ is:
  • A
    $14$
  • B
    $12$
  • ✓
    $16$
  • D
    $13$
Answer
Correct option: C.
$16$
We have, $\sqrt{248+\sqrt{52+\sqrt{144}}}$
$=\sqrt{248+\sqrt{52+12}}$ $[\because$ square root of 144 = 12$]$
$=\sqrt{248+\sqrt{64}}$
$=\sqrt{248+8}$ $[\because$ square root of 64 = 8$]$
$=\sqrt{256}=16$ $[\because$ square root of 256 = 16$]$
View full question & answer→
Generate Paper FreeAll Square Root and Cube-Cube Root topics