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

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Intern
Joined: 09 Mar 2013
Posts: 10

Kudos [?]: 16 [0], given: 8

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

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30 Apr 2013, 22:53
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Difficulty:

75% (hard)

Question Stats:

56% (02:09) correct 44% (01:48) wrong based on 84 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.

Regards
Rohan
[Reveal] Spoiler: OA

Last edited by Bunuel on 01 May 2013, 01:48, edited 2 times in total.
Renamed the topic and edited the question.

Kudos [?]: 16 [0], given: 8

Current Student
Joined: 04 Mar 2013
Posts: 68

Kudos [?]: 58 [1], given: 27

Location: India
Concentration: Strategy, Operations
Schools: Booth '17 (M)
GMAT 1: 770 Q50 V44
GPA: 3.66
WE: Operations (Manufacturing)
Re: Weighted Avg DS question [#permalink]

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30 Apr 2013, 23:58
1
KUDOS
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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When you feel like giving up, remember why you held on for so long in the first place.

Kudos [?]: 58 [1], given: 27

Senior Manager
Joined: 16 Dec 2011
Posts: 419

Kudos [?]: 248 [1], given: 70

Re: Weighted Avg DS question [#permalink]

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01 May 2013, 00:16
1
KUDOS
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.

Kudos [?]: 248 [1], given: 70

Intern
Joined: 09 Mar 2013
Posts: 10

Kudos [?]: 16 [0], given: 8

Location: India
Re: Weighted Avg DS question [#permalink]

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01 May 2013, 00:39
Thanks aceacharya & doe007 for a quick help.

Kudos [?]: 16 [0], given: 8

Current Student
Joined: 06 Sep 2013
Posts: 1957

Kudos [?]: 770 [1], given: 355

Concentration: Finance
Re: Martha obtained an average score of y in a total of x mandat [#permalink]

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05 Apr 2014, 08:31
1
KUDOS
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

Kudos [?]: 770 [1], given: 355

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Re: Martha obtained an average score of y in a total of x mandat [#permalink]

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04 Jan 2018, 02:46
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Re: Martha obtained an average score of y in a total of x mandat   [#permalink] 04 Jan 2018, 02:46
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