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M60-05

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Math Revolution GMAT Instructor
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GMAT 1: 760 Q51 V42
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M60-05  [#permalink]

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New post 11 Jun 2018, 01:42
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

88% (00:28) correct 13% (00:56) wrong based on 8 sessions

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If \(x\) and \(y\) are integers greater than \(1\) and \(x>y\), what is the value of \(x\)?



1) \(x+y=10\)

2) \(xy=21\)

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Math Revolution GMAT Instructor
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Re M60-05  [#permalink]

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New post 11 Jun 2018, 01:42
Official Solution:



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 (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) and 2) together first.

Conditions 1) and 2):

\(y = 10 - x\) and \(xy= x(10-x) = 21\)

⇔ \(-x^2 + 10x = 21\)

⇔ \(x2 - 10x + 21 = 0\)

⇔ \((x-3)(x-7) = 0\)

⇔ \(x = 3\) and \(y = 7\), or \(x = 7\) and \(y = 3\).

Since \(x > y\), we must have \(x = 7\) and \(y = 3\).

Thus \(x = 7\), and conditions 1) and 2) are sufficient, when taken together.



Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT 4(A).



Condition 1)

There are two possible solutions: \(x = 6\) and \(y = 4\), and \(x = 7\) and \(y = 3\).

Since the solution is not unique, condition 1) is not sufficient.



Condition 2):

Either \(x = 3\) and \(y = 7\), or \(x = 7\) and \(y = 3\).

Since \(x > y\), \(x = 7\) and \(y = 3\).

Thus, we have the unique solution, \(x = 7\).

Therefore, condition 2) is sufficient.



Therefore, B is the answer.

Answer: B
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Re: M60-05  [#permalink]

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New post 29 Oct 2018, 20:25
Condition 2 is incorrect, as it results in 2 possible solutions: (a) x=7, y=3, and (b) x=21, y=1. So answer should be C.
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Re: M60-05  [#permalink]

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New post 01 Nov 2018, 20:00
Why is x=8 and y=2 not listed as a possible solution for condition 1?



MathRevolution wrote:
Official Solution:



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 (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) and 2) together first.

Conditions 1) and 2):

\(y = 10 - x\) and \(xy= x(10-x) = 21\)

⇔ \(-x^2 + 10x = 21\)

⇔ \(x2 - 10x + 21 = 0\)

⇔ \((x-3)(x-7) = 0\)

⇔ \(x = 3\) and \(y = 7\), or \(x = 7\) and \(y = 3\).

Since \(x > y\), we must have \(x = 7\) and \(y = 3\).

Thus \(x = 7\), and conditions 1) and 2) are sufficient, when taken together.



Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT 4(A).



Condition 1)

There are two possible solutions: \(x = 6\) and \(y = 4\), and \(x = 7\) and \(y = 3\).

Since the solution is not unique, condition 1) is not sufficient.



Condition 2):

Either \(x = 3\) and \(y = 7\), or \(x = 7\) and \(y = 3\).

Since \(x > y\), \(x = 7\) and \(y = 3\).

Thus, we have the unique solution, \(x = 7\).

Therefore, condition 2) is sufficient.



Therefore, B is the answer.

Answer: B
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Re: M60-05  [#permalink]

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New post 27 Nov 2018, 02:28
jasmeet1089 wrote:
Condition 2 is incorrect, as it results in 2 possible solutions: (a) x=7, y=3, and (b) x=21, y=1. So answer should be C.


I've made the same mistake. But the question says x and y are integers greater than 1
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Re M60-05  [#permalink]

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New post 01 Jan 2019, 05:37
I think this is a poor-quality question and I don't agree with the explanation. value of x,y can be 21,1. so this question is incorrect
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Re: M60-05  [#permalink]

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New post 15 Jan 2019, 10:52
integers x and y should be greater than 1
hence 21 1 is not a possibility :cry: :cool: :thumbup:
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Re: M60-05 &nbs [#permalink] 15 Jan 2019, 10:52
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