Quantitative Aptitude
SQUARE ROOT AND CUBE ROOT MCQs
Square Roots, Cube Roots, Squares And Square Roots
Total Questions : 547
| Page 10 of 55 pages
Answer: Option B. -> 5r-18
Answer: Option A. -> p+q-r
Answer: Option A. -> 0
Answer: Option C. -> 8
Answer: Option A. -> 382
:
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.
:
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.
Answer: Option A. -> 27
:
A
The square root of 729 should have 3 or 7 in its unit place. Because square of 3 and 7 has 9 in it's units place.Only 27 satisfies this.
Thus, Aman's cousin gets 27 coins as a birthday gift.
Also, we can find the square root using prime factorization as
729=3×3×3×3×3×3
729=36
Square root of 729 is equal to
√729=√36=33=27
:
A
The square root of 729 should have 3 or 7 in its unit place. Because square of 3 and 7 has 9 in it's units place.Only 27 satisfies this.
Thus, Aman's cousin gets 27 coins as a birthday gift.
Also, we can find the square root using prime factorization as
729=3×3×3×3×3×3
729=36
Square root of 729 is equal to
√729=√36=33=27
Answer: Option A. -> 2
:
A
If a number ends with n zeroes,its square will end with 2n zeroes.
Here, 60 ends with one zero, so its square will endwith 2 zeroes.
It is important to note thatthis stands true only for natural numbers (not decimals).
:
A
If a number ends with n zeroes,its square will end with 2n zeroes.
Here, 60 ends with one zero, so its square will endwith 2 zeroes.
It is important to note thatthis stands true only for natural numbers (not decimals).
Answer: Option A. -> XX1
:
A
The numbers 7, 2 and 3 do not appear in the unit place ofany perfect square. So, ABC2 and PQR7 can not be perfect squares and XX1 is the only number which could be a perfect square.
:
A
The numbers 7, 2 and 3 do not appear in the unit place ofany perfect square. So, ABC2 and PQR7 can not be perfect squares and XX1 is the only number which could be a perfect square.
Answer: Option A. -> True
:
A
By division method,
323¯¯¯¯¯¯10¯¯¯¯¯¯249↓621241240
∴ The square root of 1024 is 32.
:
A
By division method,
323¯¯¯¯¯¯10¯¯¯¯¯¯249↓621241240
∴ The square root of 1024 is 32.
Answer: Option A. -> 1681
:
A
Perfect squares cannot have 2, 3, 7 or 8in their unit's place.
In the given options only 1681 does not endwith a number other than these.
The perfect squares of numbers ending in 1 and 9 have 1 at their unit's place.
Thus, 1681 could be a perfect square of an integer ending with either 1 or 9.
:
A
Perfect squares cannot have 2, 3, 7 or 8in their unit's place.
In the given options only 1681 does not endwith a number other than these.
The perfect squares of numbers ending in 1 and 9 have 1 at their unit's place.
Thus, 1681 could be a perfect square of an integer ending with either 1 or 9.
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