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Committee X and Committee Y, which have no common members,

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Senior Manager
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Committee X and Committee Y, which have no common members, [#permalink] New post 24 Oct 2006, 07:12
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 100% (01:17) wrong based on 3 sessions
Committee X and Committee Y, which have no common members, will combine to form Committee Z. Does Committee X have more members than Committee Y?

(1) The average age of members of Committee X is 25.7 yrs and the ave age of the members of Committee Y is 29.3 yrs

(2) The ave age of the members of Committee Z will be 26.6 yrs

My answer is E. Any thoughts on how it's possible to have an OA of C?
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Answer is C [#permalink] New post 24 Oct 2006, 08:31
1) The average age of members of Committee X is 25.7 yrs and the ave age of the members of Committee Y is 29.3 yrs

(2) The ave age of the members of Committee Z will be 26.6 yrs
Assume x has m members and y has n members
Total Age of x = 25.7*m
Total Age of y = 29.3*n
Average age of Z will be (25.7*m+29.3*n)/(m+n) = 26.6 (****)

So u get 25.7* m + 29.3 * n = 26.6*m + 26.6*n
2.7n=.9m
m/n=2.7/.9=3 ==> m=3n

Although in this question if u reach (****) u can choose C as answer
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Re: Answer is C [#permalink] New post 24 Oct 2006, 10:19
Damager wrote:
1) The average age of members of Committee X is 25.7 yrs and the ave age of the members of Committee Y is 29.3 yrs

(2) The ave age of the members of Committee Z will be 26.6 yrs
Assume x has m members and y has n members
Total Age of x = 25.7*m
Total Age of y = 29.3*n
Average age of Z will be (25.7*m+29.3*n)/(m+n) = 26.6 (****)

So u get 25.7* m + 29.3 * n = 26.6*m + 26.6*n
2.7n=.9m
m/n=2.7/.9=3 ==> m=3n

Although in this question if u reach (****) u can choose C as answer


This is very good to know in case we are faced with PS question and need to solve all the way through. For a DS as long as you know this basic concept you do not need to solve all the way and could approach this intuitively.

IF:

25.7+29.3/2 = 27.5 Which is > then the mean age of committee Z. So that means there must be more x members to bring the avg. age down.
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Re: PS: Committee XYZ [#permalink] New post 08 Jun 2011, 07:28
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Hi All,
Was wondering if this question can be taken as a weighted avg question and solved orally than getting into calculations?
we know that weighted avg will be closer to the unit that has higher weights attached.
from st.2 we know that the avg of Z is closer to X than to Y. isnt it enough for us to proceed and mark C?

Thanks for your help

Sudhir
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Re: PS: Committee XYZ [#permalink] New post 08 Jun 2011, 07:53
sudhir18n wrote:
Hi All,
Was wondering if this question can be taken as a weighted avg question and solved orally than getting into calculations?
we know that weighted avg will be closer to the unit that has higher weights attached.
from st.2 we know that the avg of Z is closer to X than to Y. isnt it enough for us to proceed and mark C?

Thanks for your help

Sudhir


I don't see why not!!! Good thinking!!
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Re: PS: Committee XYZ [#permalink] New post 08 Jun 2011, 09:17
sudhir18n wrote:
Hi All,
Was wondering if this question can be taken as a weighted avg question and solved orally than getting into calculations?
we know that weighted avg will be closer to the unit that has higher weights attached.
from st.2 we know that the avg of Z is closer to X than to Y. isnt it enough for us to proceed and mark C?

Thanks for your help

Sudhir


Yes. I solved it in the same way. Weighted average is best method for such questions.
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Re: PS: Committee XYZ   [#permalink] 08 Jun 2011, 09:17
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