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If m and n are positive integers and

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If m and n are positive integers and  [#permalink]

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New post 26 Jul 2018, 00:44
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

50% (01:33) correct 50% (01:13) wrong based on 48 sessions

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[Math Revolution GMAT math practice question]

If \(m\) and \(n\) are positive integers and \(\sqrt{m}\sqrt{n}=6\), then \(n=?\)

1) Both \(m\) and \(n\) are the squares of integers.
2) \(m > n\)

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If m and n are positive integers and  [#permalink]

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New post 26 Jul 2018, 01:06
1) First statement says that \(\sqrt{m}\) and \(\sqrt{n}\) are both integers. For a product to be 6, we'd need one of the following scenarios:
\(\sqrt{m}\) = 6 and \(\sqrt{n}\) = 1
\(\sqrt{m}\) = 3 and \(\sqrt{n}\) = 2
\(\sqrt{m}\) = 2 and \(\sqrt{n}\) = 3
\(\sqrt{m}\) = 1 and \(\sqrt{n}\) = 6

Insufficient

2) Second statement by itself doesn't provide much information. Insufficient.

1) and 2) Combined eliminates the last two scenarios from the ones exposed above, but we're still left with two possible scenarios:
\(\sqrt{m}\) = 6 and \(\sqrt{n}\) = 1
\(\sqrt{m}\) = 3 and \(\sqrt{n}\) = 2

Therefore the answer is E.
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Re: If m and n are positive integers and  [#permalink]

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New post 26 Jul 2018, 09:47
Given info can be rephrased as m x n = 36 and since m & n are positive integers,
36 = 1 x 36, 2 x 18, 3 x 12, 4 x 9, 6 x 3, 9 x 4, 12 x 3, 18 x 2, 36 x 1 ( there are 9 possible combinations of m & n) and if we can determine one, we can determine the other

(1) Both m and n are the squares of integers: This dramatically reduces the possible combinations to 4. Only 4 x 9, 9 x 4, 1 x 36 or 36 x 1 are possible. But still there are 4 possible values of n and this statement is insufficient

(2) m>n : We still can't determine values of m & n from this. Assuming first number to be m would be wrong and still would give multiple possibilities. This statement is also insufficient.

Combining (1) and (2): There are 4 possible combinations from (1) and (2) doesn't help us to reduce any further for we don't know .

Correct answer is (E)
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Re: If m and n are positive integers and  [#permalink]

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New post 29 Jul 2018, 17:29
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (m and n) and 1 equation, D is most likely to be the answer. So, we should consider each of the conditions on its own first.

Condition 1)
There are at least two possible pairs of solutions. Two of these pairs are: m = 4, n = 9 and m = 9, n = 4.
Since we don’t have a unique solution, condition 1) is not sufficient.

Condition 2)
There are more than two pairs of solutions. Two of these are m = 9, n = 4 and m = 12, n = 3.
Since we don’t have a unique solution, condition 2) is not sufficient.

Conditions 1) and 2)
There are two possible pairs of solutions: m = 36, n = 1 and m = 9, n = 4.
Since we don’t have a unique solution, both conditions together are not sufficient.
Note: This question includes an example of a “hidden 1”.

Therefore, E is the answer.

Answer: E

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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Re: If m and n are positive integers and &nbs [#permalink] 29 Jul 2018, 17:29
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