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DS - weighted avg

by karthikpandian19 » Wed Jul 04, 2012 9:25 pm
What percentage of the dogs at a particular dog show are male?

The average weight of a male dog at the show is 15 pounds heavier than the average weight of all of the dogs at the show.
The average weight of a female dog at the show is 32 pounds.
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by eagleeye » Wed Jul 04, 2012 10:07 pm
karthikpandian19 wrote:What percentage of the dogs at a particular dog show are male?

The average weight of a male dog at the show is 15 pounds heavier than the average weight of all of the dogs at the show.
The average weight of a female dog at the show is 32 pounds.
We need to find the percentage of male dogs at the show. In other words, we need to determine the male/female ratio.

Method of balancing works well to answer this question.
A quick observation of the statements reveals that both statement 1 and statement 2 are insufficient taken separately since they only talk about one of the sexes.

Taken together, to find the male/female ratio, we need two values:
1) Difference of average weight of males vs overall average. We are given this statement 1. yippee!
2) Difference of average weight of females vs overall average. Unfortunately, statement 2 only talks about the value of the weight of the female and not the difference that we seek. arrrgh!.

Since we don't have enough data, together they are insufficient as well. Hence, E is the final answer
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by karthikpandian19 » Wed Jul 04, 2012 10:36 pm
I understand this logical and easy way "dont solve" for this problem.

But when i sit in the exams, mind is not opening up for the Logical gateway.....it goes only thru algebraic way.

Is there a solution tht u can tell me in algebraic way?
eagleeye wrote:
karthikpandian19 wrote:What percentage of the dogs at a particular dog show are male?

The average weight of a male dog at the show is 15 pounds heavier than the average weight of all of the dogs at the show.
The average weight of a female dog at the show is 32 pounds.
We need to find the percentage of male dogs at the show. In other words, we need to determine the male/female ratio.

Method of balancing works well to answer this question.
A quick observation of the statements reveals that both statement 1 and statement 2 are insufficient taken separately since they only talk about one of the sexes.

Taken together, to find the male/female ratio, we need two values:
1) Difference of average weight of males vs overall average. We are given this statement 1. yippee!
2) Difference of average weight of females vs overall average. Unfortunately, statement 2 only talks about the value of the weight of the female and not the difference that we seek. arrrgh!.

Since we don't have enough data, together they are insufficient as well. Hence, E is the final answer
Regards,
Karthik
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by eagleeye » Wed Jul 04, 2012 11:06 pm
karthikpandian19 wrote:I understand this logical and easy way "dont solve" for this problem.

But when i sit in the exams, mind is not opening up for the Logical gateway.....it goes only thru algebraic way.

Is there a solution tht u can tell me in algebraic way?
Sure karthik:

Using algebra:
Let the number of male dogs be m, and that of female dogs be f. Then we need to find percentage of male dogs = m*100/(m+f). = 100/(1+m/f). So if we can find m/f ratio, we have the answer.

Let average weight of each male dog be Wm, and that of each female be Wf.
Then weighted average of all dogs Wa = (Wm*m + Wf*f)/(m+f)

=> Wa*m+Wa*f = Wm*m + Wf*f.
=> (Wa-Wm)*m = (Wf-Wa)*f
=> m/f = (Wa-Wm)/(Wf-Wa)

To find m/f, we need to find the ratio of Wa-Wm to Wf-Wa.
Now first statement tells us that:

1. The average weight of a male dog at the show is 15 pounds heavier than the average weight of all of the dogs at the show.
We are told that Wm = 15+ Wa
=> Wm-Wa = 15. Doesn't tell us about Wa-Wf. Insufficient.

2. The average weight of a female dog at the show is 32 pounds.
We are told that Wf=32.
Doesn't tell us anything about Wm or Wa. Insufficient.

Together, We know Wm-Wa, and Wf by itself. We don't know what Wa-Wf is. Insufficient.

Hence final answer is E.
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by GMATGuruNY » Thu Jul 05, 2012 3:26 am
karthikpandian19 wrote:What percentage of the dogs at a particular dog show are male?

The average weight of a male dog at the show is 15 pounds heavier than the average weight of all of the dogs at the show.
The average weight of a female dog at the show is 32 pounds.
Let F = the average weight of the female dogs, M = the average weight of the male dogs, and FM = the average weight of all of the dogs.
Try EXTREME values.

Statements 1 and 2 combined:
If F=32, FM = 33, and M=48, then the average weight of all of the dogs is much closer to the average weight of the FEMALE dogs, implying that MOST of the dogs are FEMALE.
If F=32, FM = 100, and M=115, then the average weight of all of the dogs is much closer to the average weight of the MALE dogs, implying that MOST of the dogs are MALE.
Thus, it is not possible to determine what percentage of the dogs are male.
INSUFFICIENT.

The correct answer is E.
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by karthikpandian19 » Thu Jul 05, 2012 2:33 pm
OA is E
Regards,
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