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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7465
GMAT 1: 760 Q51 V42 GPA: 3.82

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Difficulty:   25% (medium)

Question Stats: 80% (00:30) correct 20% (00:48) wrong based on 10 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 V
Joined: 16 Aug 2015
Posts: 7465
GMAT 1: 760 Q51 V42 GPA: 3.82

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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.

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Intern  Joined: 29 Oct 2018
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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.
Intern  Joined: 01 Nov 2018
Posts: 2

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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.

Intern  Joined: 27 Oct 2018
Posts: 1

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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
Intern  Joined: 16 Dec 2018
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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
Intern  Joined: 15 Jan 2019
Posts: 3

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integers x and y should be greater than 1
hence 21 1 is not a possibility    Re: M60-05   [#permalink] 15 Jan 2019, 11:52
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