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Need expert help - weighted average

by voodoo_child » Tue May 01, 2012 5:31 am
Three groups of people go to the gym. Group A has an average weight of 2; Group B has an average weight of 3; Group C has an average weight of 5. The average weight of all groups was 27/8. What proportion of people from Group A participated in the gym?

a) 3
b) 2
c) 1
d) 4
e) 5

OA - a Any thoughts - how to solve this?
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by aneesh.kg » Tue May 01, 2012 5:46 am
Proportion refers to equality between ratios. So, I think that the phrase 'proportion of people from Group A' is wrong.

What can be found here is the fraction of people from each group.

Let the number of people in group A, B and C be a,b and c respectively.
(2a + 3b + 5c) / (a + b + c) = 27/8
16a + 24b + 40c = 27a + 27b + 27c
13c = 3b + 11a
a = 1, b = 5, c = 2 satisfies the equation.

fraction of people from group A = 1/(1 + 5 + 2) = 1/8
fraction of people from group B = 5/8
fraction of people from group C = 2/8 = 1/4

I'm sorry but I am unable to figure out what he is really asking.
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by mathewmithun » Tue May 01, 2012 7:52 pm
I think question is to find the number of people participating in the gym.

there are 2 possibilities for this question:

A=3 B=2 and C=3

OR

A=1 B=5 and C=2.

So it could be a or c. someone could help how to eliminate one possibility to get unique answer.

@aneesh.kg- pls check the possibility A=3 B=2 and C=3
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by aneesh.kg » Tue May 01, 2012 10:27 pm
Hi Mithun!

If you notice, all the integral multiples of the two solutions shown by you will satisfy the equation. Which means that
A = 3x, B = 2x, C = 3x
A = 1y, B = 5y, C = 2y
(and many more)
are its solutions.
where x and y are integers.

So, there are an infinite number of solutions (which includes all the options - A, B, C, D, E - given here).

Lets not force an answer when there is an ambiguity with the language of the problem.
Aneesh Bangia
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