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# M60-05

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6815
GMAT 1: 760 Q51 V42
GPA: 3.82

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11 Jun 2018, 01:42
00:00

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) $$xy=21$$

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6815 GMAT 1: 760 Q51 V42 GPA: 3.82 Re M60-05 [#permalink] ### Show Tags 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
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Intern
Joined: 29 Oct 2018
Posts: 1

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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.
Intern
Joined: 01 Nov 2018
Posts: 2

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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.

Intern
Joined: 27 Oct 2018
Posts: 1

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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
Intern
Joined: 16 Dec 2018
Posts: 1

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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
Intern
Joined: 15 Jan 2019
Posts: 3

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

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