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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

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]

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15 Mar 2012, 13:08

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

There is actually no need for 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 ratio of members in X and Y is 3:1.

Re: Committee X and Committee Y , which have no common members [#permalink]

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09 Nov 2012, 00:05

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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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.

Re: Committee X and Committee Y , which have no common members [#permalink]

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17 Jun 2013, 23:01

sunland wrote:

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.

here is my take of the question:

s1: gives u two variable terms which can't be equated to any so its insufficent s2: gives a single variable term : not sufficient s1 and s2 together then we get a two viable equation which can be used ot find out the relation between x and y so c is our picK

Committee X and Committee Y , which have no common members [#permalink]

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16 Jan 2014, 05:31

They totally got me on this one, I thought too the following. "Can't X just have older members?" But it is actually the ratio that matters. Even if X has older member the ratio, that is the weight of each of the components will determine the quantity of members in each of both groups, very tricky question indeed totally fell for it

Re: Committee X and Committee Y , which have no common members [#permalink]

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29 Nov 2017, 14:28

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