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Bunuel
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If y is an integer and (48y)^(1/2) is an integer, which of the followi 22 Jul 2020, 05:39
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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88% (00:38) correct 12% (00:45) wrong based on 78 sessions

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If y is an integer and \(\sqrt{48y}\) is an integer, which of the following is the lowest possible value of y?

A. 1

B. 2

C. 3

D. 4

E. 6
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C
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If y is an integer and (48y)^(1/2) is an integer, which of the followi 22 Jul 2020, 05:42
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A. 1 , not integer value

B. 2 , not integer value

C. 3 , integer value

D. 4 , not integer value

E. 6 , not integer value

Hence, OA is (C).
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If y is an integer and (48y)^(1/2) is an integer, which of the followi 22 Jul 2020, 05:50
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48 = 4² * 3

So, the lowest value of y for √48y to be an integer is 3

Option : C

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If y is an integer and (48y)^(1/2) is an integer, which of the followi 22 Jul 2020, 06:16
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In this problem we need to find out the lowest possible value of y which makes \(\sqrt{48y}\) an integer.

There are two ways of solving this problem

Method 1: Algebra

Now \(\sqrt{48y}\) will be integer only when the value inside the square root, which is 48y, is a perfect square.
So, lets start factorizing 48y
48y = 16*3y = \(4^2\)* 3y
Now, in \(4^2\)* 3y we know that \(4^2\) is a perfect square so to make \(4^2\)* 3y a perfect square we need to make 3y also a perfect square.
So, y should be minimum 3 which will make
\(4^2\)* 3y = \(4^2 * 3^2\)

So, Answer will be C

Method 2: Substitution:

Put each option choice value as value of y in \(\sqrt{48y}\) and see which one gives you an integer

A. 1 => \(\sqrt{48y}\) = \(\sqrt{48*1}\)= \(\sqrt{48}\)= \(4\sqrt{3}\) ≠ INTEGER

B. 2 => \(\sqrt{48y}\) = \(\sqrt{48*2}\)= \(\sqrt{96}\)= \(4\sqrt{6}\) ≠ INTEGER

C. 3 => \(\sqrt{48y}\) = \(\sqrt{48*3}\)= \(\sqrt{144}\)= 12 = INTEGER => ANSWER

We don't need to check further but solving just to complete the solution

D. 4 => \(\sqrt{48y}\) = \(\sqrt{48*4}\)= \(\sqrt{192}\)= \(8\sqrt{3}\) ≠ INTEGER

E. 5 => \(\sqrt{48y}\) = \(\sqrt{48*5}\)= \(\sqrt{240}\)= \(4\sqrt{15}\) ≠ INTEGER

So, Answer will be C

Hope it helps!
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