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Is √(56x) an integer?

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by M7MBA » Wed May 23, 2018 4:39 am

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Answers

A

B

C

D

E

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Difficulty—

Is √(56x) an integer?

(1) x is a multiple of 14.
(2) 28 is not a factor of x.

The OA is the option E.

Why is E the correct choice? Could anyone explain this DS question to me? Please. Thanks in advance.
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Source: — Data Sufficiency |

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by Jay@ManhattanReview » Thu May 24, 2018 2:08 am
M7MBA wrote:Is √(56x) an integer?

(1) x is a multiple of 14.
(2) 28 is not a factor of x.

The OA is the option E.

Why is E the correct choice? Could anyone explain this DS question to me? Please. Thanks in advance.
For √(56x) to be an integer, 56x must be a perfect square.

Let's factorize √(56x) an integer.

√(56x) = √(2^3*7*x) = 2.√(2*7*x) = 2√(14x)

=> If 14x is a perfect square, the answer is yes, else no.

Let's see each statement one by one.

(1) x is a multiple of 14.

If x = 14, we see that 14x = 14*14 = 14^2, a perfect squre: the answer is yes; however, x = 14*3, we see that 14x = 14*14*3, not a perfect squre: the answer is No. Insufficient.

(2) 28 is not a factor of x.

If x = 14 (not a factor of 28), we see that 14x = 14*14 = 14^2, a perfect squre: the answer is yes; however, x = 14*3 (not a factor of 28), we see that 14x = 14*14*3, not a perfect squre: the answer is No. Insufficient.

(1) and (2) together

Both the examples discussed above are applicable here too. Insufficient.

The correct answer: E

Hope this helps!

-Jay
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by deloitte247 » Sat May 26, 2018 8:33 am
$$\sqrt{56}x=2\sqrt{14}x$$
Statement 1: 'x' is a multiple of 14.
$$56x=8\cdot7\cdot x$$
'x' can be 14 or 28 as a multiple of 14 and we will get different result.
Therefore, statement 1 is NOT SUFFICIENT

Statement 2: 28 is not a factor of 'x'. 'x' can be 14 and 15 and we still get different result. Statement 2 is therefore NOT SUFFICIENT.

For statement 1 and 2: we can say that 'x' is odd and a multiple of 14. if 'x' is 42 or 126. We still get different result, it is not sufficient.

Therefore, Option E is the answer because statement 1 and 2 together are not sufficient.
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