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What level question is this?

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by bek_gmat » Sat Jun 16, 2012 11:06 am
Can anyone (preferably Experts) tell me what level question is that in terms of GMAT score? And OG explained the answer in a much more complicated way, i'm sure anyone try that in exam would run out of time, so if anybody has a shortcut solution to the problem, feel free to share.

OG 12edn: #148

If x, y, and k are positive numbers such that

(x/x+y)10+ (y/x+y)20= k, if x<y, what could be the value of k?

(A) 10
(B) 12
(C) 15
(D) 18
(E) 30

The answer is D
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by karthikpandian19 » Sat Jun 16, 2012 2:47 pm
One way that i would do this problem is as per the following:

****Use the strategy of checking the answers from bottom. ie. from E*********

I always use this strategy and this works.......


Simply the problem, (10x+20y)/(x+y) = k

now substitute 30 for k, and find out whether the x<y condition satisfies

Start with E (as per the strategy)

10x+20y=30x+30y
20x=-10y
x=-0.5y .........condition is not being satisfied

now move to D,

10x+20y=18x+18y
8x=2y
x=0.25y .........condition is being satisfied

Now you can stop and there is no neccessity to check other choices

and D is your answer


bek_gmat wrote:Can anyone (preferably Experts) tell me what level question is that in terms of GMAT score? And OG explained the answer in a much more complicated way, i'm sure anyone try that in exam would run out of time, so if anybody has a shortcut solution to the problem, feel free to share.

OG 12edn: #148

If x, y, and k are positive numbers such that

(x/x+y)10+ (y/x+y)20= k, if x<y, what could be the value of k?

(A) 10
(B) 12
(C) 15
(D) 18
(E) 30

The answer is D
Regards,
Karthik
The source of the questions that i post from JUNE 2013 is from KNEWTON

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by GMATGuruNY » Sat Jun 16, 2012 3:30 pm
I posted an explanation here:

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by Anurag@Gurome » Sat Jun 16, 2012 11:05 pm
bek_gmat wrote:...so if anybody has a shortcut solution to the problem, feel free to share.
The easiest and least time consuming method to solve this problem is to note that k is nothing but the weighted average of 10 and 20, the weights being x and y respectively. Hence, k must be some number between 10 and 20.

As x < y, the contribution of 10 in the weighted average will be less than that of 20. In other words, the weighted average will be more close to 20 than 10.

Only such option is 18.

The correct answer is D.
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