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Martha obtained an average score of y in a total of x mandat

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Joined: 09 Mar 2013
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Location: India
Martha obtained an average score of y in a total of x mandat [#permalink]

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New post Updated on: 01 May 2013, 02:48
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Difficulty:

  75% (hard)

Question Stats:

56% (02:07) correct 44% (01:54) wrong based on 87 sessions

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Martha obtained an average score of y in a total of x mandatory papers. She also obtained a score of z in an additional optional paper. Does Martha’s average score on all the x + 1 papers exceed her average score on the x mandatory papers by more than 50%?

(1) 3x = y
(2) 2z – 3y = xy

[Reveal] Spoiler:
My trial :

Avg Score is Y
No of Papers is X
Total Score is X.Y

New Avg Score = (X.Y + Z) / X+1

As per question 1/2 of new Ayg = Old Avg

Use this i am getting A alone is insufficient and for B i am not sure if that is correct. I think my approach is wrong for this question.

Can you please help.

Regards
Rohan
[Reveal] Spoiler: OA

Originally posted by rohanchowdhary on 30 Apr 2013, 23:53.
Last edited by Bunuel on 01 May 2013, 02:48, edited 2 times in total.
Renamed the topic and edited the question.
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Concentration: Strategy, Operations
Schools: Booth '17 (M)
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Re: Weighted Avg DS question [#permalink]

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New post 01 May 2013, 00:58
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Statement 1 is clearly insuficient as it has no mention of Z

Statement 2 states that
2z – 3y = xy
=>z = (xy+3y)/2
=> z = y(x+3)/2

So even if x would be equal to 1, which will be the min. value of x, then z=2y therefore it will increase the average of X+1 papers to more than y

Therefore B would be the right answer
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Re: Weighted Avg DS question [#permalink]

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New post 01 May 2013, 01:16
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rohanchowdhary wrote:
Hi All

I am stuck in this DS question.

Martha obtained an average score of y in a total of x mandatory papers. She also obtained a score of z in an additional optional paper. Does Martha’s average score on all the x+ 1 papers exceed her average score on the x mandatory papers by more than 50%?
(1) 3x = y
(2) 2z – 3y = xy

Average of x mandatory papers is y and average of x+1 papers is \(\frac{xy+z}{x+1}\)

The question is whether -
\(\frac{\frac{xy+z}{x+1} - y}{y} > 0.5\) ==> \(\frac{xy+z}{x+1} - y > 0.5y\) ==> \(\frac{xy+z}{x+1} > \frac{3}{2}y\) ==> 2xy + 2z > 3xy +3y ==> 2z - 3y > xy

1) There is no information on z. Insufficient.

2) 2z – 3y = xy
==> 2z – 3y is not greater than xy
==> Martha’s average score on all the x+ 1 papers does not exceed her average score on the x mandatory papers by more than 50%
This statement is sufficient as we have a definite answer.

Answer is B.
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Re: Weighted Avg DS question [#permalink]

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New post 01 May 2013, 01:39
Thanks aceacharya & doe007 for a quick help.
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Re: Martha obtained an average score of y in a total of x mandat [#permalink]

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New post 05 Apr 2014, 09:31
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The question is asking the following. Is xy+z/x+1 > 3/2(y)?

We have that 2xy + 2z>3y(x+1), solving we have that 2z-3y>xy?

Statement 1 is clearly not sufficient, but statement 2 gives as the answer straight away.

Hence B

Hope this clarifies
Cheers
J 8-)
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Re: Martha obtained an average score of y in a total of x mandat [#permalink]

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New post 04 Jan 2018, 03:46
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Re: Martha obtained an average score of y in a total of x mandat   [#permalink] 04 Jan 2018, 03:46
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