Committee X and Committee Y , which have no common members

[jjykim]

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

The correct answer is C.
I didn't think you could figure out the number of people just based on the mean, but according to the answer choice, I guess you can.
Could someone please explain why that is?

Thanks.

[emma]

Lets say for X:

Number of members = n1
Sum of all the ages = s1
average ages = a1

Now for Y:

Number of members = n2
Sum of all the ages = s2
average ages = a2

And for Z:

Number of members = n1 + n2
Sum of all the ages = s1 + s2
average ages = a3


(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.

Which gives us a1 = 25.7 and a2 = 29.3
Also s1 = n1*25.7
and s2 = n2*29.3

Since we dont have any information on the rest this is INSUFFICIENT

(2) The average (arithmetic mean) age of the members of Committee Z will be 26.6 years

Which gives us a3 = 26.6

Still INSUFFICIENT


TOGETHER:

a3 = (s1+s2) / (n1+n2)

Fill in the values of a3, s1 and s2

26.6 = ((n1*25.7) + (n2*29.3)) / (n1+n2)

Simplify to get .9n1 = 2.7n2

or n1/n2 = 3/1 = 3

That means there are 3 times as many members in X (n1) as in Y (n2)

SUFFICIENT - anwser C

[jjykim]

thanks :)

[RonPurewal]

general comment:

in a WEIGHTED AVERAGE, if you have the 2 distances between the mean and each of the individual component averages, that's good enough to figure out the RATIO of the two components. here's how:

distance between committee A and average = 29.3 - 26.6 = 2.7
distance between committee B and average = 26.6 - 25.7 = 0.9

ratio = 3:1

committee B must be bigger (because the average is closer to committee B's average), so the ratio of committee B : committee A is 3:1 (not 1:3).

that's how it works. the same process applies to any other weighted average - good stuff.

WARNING:
this procedure will NOT allow you to determine the actual SIZE of committee A or committee B, without any additional information; all it will give you is the RATIO of the two sizes. that's plenty sufficient to answer this problem, but, if the problem had asked for the NUMBER of members of any of the committees, the answer would be (e) as you had expected.

--

btw, you don't have to do any of this ratio nonsense to solve this problem; it's sufficient to notice that the overall average is closer to the average for committee B than to that for committee A. if that's the case, then committee B must have a heavier influence on the average, a fact that means committee B has more people.

good stuff.

[jp.jprasanna]

general comment:

in a WEIGHTED AVERAGE, if you have the 2 distances between the mean and each of the individual component averages, that's good enough to figure out the RATIO of the two components. here's how:

distance between committee A and average = 29.3 - 26.6 = 2.7
distance between committee B and average = 26.6 - 25.7 = 0.9

ratio = 3:1

committee B must be bigger (because the average is closer to committee B's average), so the ratio of committee B : committee A is 3:1 (not 1:3).

that's how it works. the same process applies to any other weighted average - good stuff.

Ron - Thanks for the wonderful explanation really helps....

I have one doubt I might probably sound stupid also but

Nx = Na = nos of ppl in committee x (a)
Ny = Nb = nos of ppl in y (b)

Nx / Ny = 29.3-26.6 / 26.6 - 25.7 = 2.7/0.9 = 3/1

Nx/Ny = 3/1 = Na/Nb (i think you have changed this to A and B in ur explanation)

Isn't there more members in Committee X rather Y
i.e Nx = 3/4 (total) and Ny = 1/4(total) also makes sense because avg i.e 26.6 is closer to that of committee X 25.7 than to that of committee Y 29.3

So answer to the question " Does Committee X have more members than Committee Y ? " IS NO RIGHT -

Because your statement " committee B must be bigger (because the average is closer to committee B's average), so the ratio of committee B : committee A is 3:1 (not 1:3). " is saying YES COMMITTEE B IS BIGGER THAN A! Please help me Ron! :-(

[RonPurewal]

jp.jprasanna, i can't follow your explanation, mostly because of the way it's written (i can't make out very many complete sentences, and you aren't using the word "rather" in a way that makes sense).

however, note the following:

Isn't there more members in Committee X rather Y
i.e Nx = 3/4 (total) and Ny = 1/4(total) also makes sense because avg i.e 26.6 is closer to that of committee X 25.7 than to that of committee Y 29.3

i think you are saying that X is bigger. if so, that's correct. ("in X rather Y" is not a meaningful construction; i can't tell which way you are trying to get it to go.)

So answer to the question " Does Committee X have more members than Committee Y ? " IS NO RIGHT -

this is where you've lost me. the lines directly above show that X is bigger, and so then you are answering No to "is X bigger?"
that's incorrect.

perhaps i am misunderstanding you here.

[shubham_sagijain]

general comment:

in a WEIGHTED AVERAGE, if you have the 2 distances between the mean and each of the individual component averages, that's good enough to figure out the RATIO of the two components. here's how:

distance between committee A and average = 29.3 - 26.6 = 2.7
distance between committee B and average = 26.6 - 25.7 = 0.9

ratio = 3:1

committee B must be bigger (because the average is closer to committee B's average), so the ratio of committee B : committee A is 3:1 (not 1:3).

that's how it works. the same process applies to any other weighted average - good stuff.

WARNING:
this procedure will NOT allow you to determine the actual SIZE of committee A or committee B, without any additional information; all it will give you is the RATIO of the two sizes. that's plenty sufficient to answer this problem, but, if the problem had asked for the NUMBER of members of any of the committees, the answer would be (e) as you had expected.

--

btw, you don't have to do any of this ratio nonsense to solve this problem; it's sufficient to notice that the overall average is closer to the average for committee B than to that for committee A. if that's the case, then committee B must have a heavier influence on the average, a fact that means committee B has more people.

good stuff.

The explanation is an absolute beauty by Ron :)..especially the last 7-8 lines of Ron's explanation...

[RonPurewal]

thanks.

[mevicks]

....
btw, you don't have to do any of this ratio nonsense to solve this problem; it's sufficient to notice that the overall average is closer to the average for committee B than to that for committee A. if that's the case, then committee B must have a heavier influence on the average, a fact that means committee B has more people.
...

Hi ron,

If the proportion of x and y was equal this would be the scenario:
A_____________Mid___________B
25.7_________27.5___________29.3

Since the final average is 26.6 (avg of Z) we can answer that A > B

A_____________Mid___________B
25.7___26.6___27.5__________29.3

Correct me if I'm wrong, but here A pulls the average down. Thus A should have more members.

Regards,
Vivek

[RonPurewal]

Looks like your "A" is my "B", and vice versa.

I don't have any idea how I managed to change "X" into "B" and "Y" into "A" near the beginning of the thread... but it happened.