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

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Re: DS quest from paperbased test31 [#permalink]
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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]
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]
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]
https://gmatclub.com/static/mobile.png[/img]
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Re: Committee X and Committee Y , which have no common members, [#permalink]
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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]
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.

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]
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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.

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

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

Using Scale Method

25.7----------26.6--------29.3
-X------------Z------------ Y
-------0.3-----:---- 2.7
-------3-------:-----27
-------1--------:----- 9

Therefore
W1/W2 = 9:1
Therefore members in X > Y
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Committee X and Committee Y , which have no common members [#permalink]
Bunuel wrote:
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.

Bunuel
But, the average of x and y is 27.5 {(25.7+29.3)/2}. May I know why did you deduct 25.7 from the average of z (26.6)?
Thanks__
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Re: Committee X and Committee Y , which have no common members [#permalink]
Bunuel wrote:
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.

Can't they have same number of members?

For example, say X has two members of ages: 20 and 66. (Average= 43)
Similarly, say Y also has two members of ages: 20 and 20. (Average= 20)

Committee Z: 20+66+20+20/4= 32 (Average of Z)

(All of these twenty year olds are three different individuals, as the two set don't share any member.)

Now, as Bunuel did: 43-32=11; 32-20=12. So the two groups, X and Y, would be in the ratio 12:11.

As we can see, if we go by the logic above, then X should have more members than Y as the ratio suggests that. But in reality, X and Y have two members each.

What am I missing?
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Re: Committee X and Committee Y , which have no common members [#permalink]
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Re: Committee X and Committee Y , which have no common members [#permalink]
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