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Committee X and Committee Y , which have no common members [#permalink]
 01 Jul 2010, 05:57
Question Stats:
70% (02:13) correct
29% (00:26) wrong based on 2 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 (arithmetic mean) age of the members of Committee X is 25.7 years and the average age of the members of Committee Y is 29.3 years. (2) The average (arithmetic mean) age of the members of Committee Z will be 26.6 years.
Last edited by Bunuel on 16 Mar 2012, 01:00, edited 1 time in total.
Edited the question
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Re: DS quest from paperbased test31 [#permalink]
01 Jul 2010, 09:39
OK, i replied to this topic 3 times but failed all, this is the last time! Kudo me if I'm right Let X be sums of ages of mem of Com X, n be the number of mems of Com X Let Y be sums of ages of mem of com Y, m be the number of mems of Com Y X/n=25.7 so X=25.7*n Y/m=29.3 so Y=29.3*m (X+Y)/(m+n)= (25.7*n+29.3*m)/(m+n)=26.6 So 25.7*n+29.3*m=26.6*m+26.6*n => 2.7m=0.9n =>3m=n =>n>m C suff
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Re: DS quest from paperbased test31 [#permalink]
01 Jul 2010, 14:16
sunland wrote: 19. 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 (arithmetic mean) age of the
members of Committee X is 25.7 years and the
average age of the members of Committee Y is
29.3 years.
(2) The average (arithmetic mean) age of the
members of Committee Z will be 26.6 years. 1 Statement 1 is insufficient. Knowing the average of the groups gives us no idea about how many members there are 2. Statement 2 is insufficient because knowing the average of the group does not tell you anything about the number in the group Together they are suff because the info together can tell you the relative portion of the two groups. A(25.7) + B(29.3) / A+B = 26.6 is a weighted average problem. It can tell you the relative size of one group to the other. You can also think of the problem another way. Since the avg age of the groups is closer to community X. You know there must be more people in X.
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Re: Committee X and Committee Y , which have no common members, [#permalink]
15 Mar 2012, 13:48
Question: Couldn't both committees have the same number of members but one committee just have older members?
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Re: Committee X and Committee Y , which have no common members, [#permalink]
15 Mar 2012, 14:08
If that were true u would expect mean age of z to b the arithmatic mean, which would be directly betw een x and y . The weighs would be equal for each group in that case Posted from my mobile device [img] http://gmatclub.com/static/mobile.png[/img]
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Re: Committee X and Committee Y , which have no common members, [#permalink]
15 Mar 2012, 15:08
I had to think about this a few times  I forgot to rephrase the question and thats what got me in the end.. Basically the question is asking what is # of X members/# of Y Members 1) Ins 2) Ins 1 + 2 ) Ins to solve for X and Y but based on X as number of comm x members and Y as number of comm y members... you can say: (25.7 * x + 29.3 (y)) / (x + y) = 26.6 then you can say that 25.7 x + 29.3 y = 26.6x + 26.6y thus .9x = 2.7y or x/y = 1/3 I kept banging my head against a wall thinking i cant solve for x or y with one equation..
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Re: Committee X and Committee Y , which have no common members [#permalink]
16 Mar 2012, 01:23
There is actually no need for some variables and formulas here. 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 (arithmetic mean) age of the members of Committee X is 25.7 years and the average age of the members of Committee Y is 29.3 years. Insufficient on its own. (2) The average (arithmetic mean) age of the members of Committee Z will be 26.6 years. Insufficient on its own. (1)+(2) The difference between the averages of Z and X is 26.6-25.7=0.9 and the difference between the averages of Z and Y is 29.3-26.6=2.7. Since the average of Z is closer to the average of X then X must have more members than Y (to draw the overall average closer to its own average). Sufficient. Notice that we cannot find # of members in X or Y, though based on the difference in averages (0.9 and 2.7) we can deduce that the ration of members in X and Y is 3:1. Answer: C.
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Re: Committee X and Committee Y , which have no common members [#permalink]
09 Nov 2012, 01:05
Bunuel wrote: There is actually no need for some variables and formulas here.
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 (arithmetic mean) age of the members of Committee X is 25.7 years and the average age of the members of Committee Y is 29.3 years. Insufficient on its own.
(2) The average (arithmetic mean) age of the members of Committee Z will be 26.6 years. Insufficient on its own.
(1)+(2) The difference between the averages of Z and X is 26.6-25.7=0.9 and the difference between the averages of Z and Y is 29.3-26.6=2.7. Since the average of Z is closer to the average of X then X must have more members than Y (to draw the overall average closer to its own average). Sufficient.
Notice that we cannot find # of members in X or Y, though based on the difference in averages (0.9 and 2.7) we can deduce that the ration of members in X and Y is 3:1.
Answer: C. I agree with this explanation Bunuel, however I have one doubt while solving questions using this approach. What if one committee has say 3 members with age 200(avg 200) and other has say 10 members with age 10 (avg=10) . In this case the average age of group (266) is closer to the committee with fewer members. guess there is a problem in my basics...if u could please throw some light on weighted avg
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Re: Committee X and Committee Y , which have no common members [#permalink]
09 Nov 2012, 05:02
Maverick04308 wrote: Bunuel wrote: There is actually no need for some variables and formulas here.
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 (arithmetic mean) age of the members of Committee X is 25.7 years and the average age of the members of Committee Y is 29.3 years. Insufficient on its own.
(2) The average (arithmetic mean) age of the members of Committee Z will be 26.6 years. Insufficient on its own.
(1)+(2) The difference between the averages of Z and X is 26.6-25.7=0.9 and the difference between the averages of Z and Y is 29.3-26.6=2.7. Since the average of Z is closer to the average of X then X must have more members than Y (to draw the overall average closer to its own average). Sufficient.
Notice that we cannot find # of members in X or Y, though based on the difference in averages (0.9 and 2.7) we can deduce that the ration of members in X and Y is 3:1.
Answer: C. I agree with this explanation Bunuel, however I have one doubt while solving questions using this approach. What if one committee has say 3 members with age 200(avg 200) and other has say 10 members with age 10 (avg=10) . In this case the average age of group (266) is closer to the committee with fewer members. guess there is a problem in my basics...if u could please throw some light on weighted avg The average age of the of the whole group cannot be more than individual averages of the smaller groups, so 266 just cannot be the average age since it's more than both 200 and 10. Generally the weighted average of 2 individual averages (200 and 10) must lie between these individual averages.In this case the average age of the whole group of 13 members is (3*200+10*10)/13=~54, which is closer to 10 than to 200. Hope it's clear.
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Re: Committee X and Committee Y , which have no common members
[#permalink]
09 Nov 2012, 05:02
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