Weighted Average problem

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Weighted Average problem

by [email protected] » Sat Jul 22, 2017 1:46 am
A charitable association sold an average of 66 raffle tickets per member. Among the female members, the average was 70 raffle tickets. The male to female ratio of the association is 1:2. What was the average number of tickets sold by the male members of the association

A. 50
B. 56
C. 58
D. 62
E. 66

The women's differential multiplied by 2 will cancel with the men's differential multiplied by 1. If the men's differential is m, then:

1*m + 2*(+4) = 0
m + 8 = 0
m = - 8

The men sold an average of 8 fewer tickets than the total average: 66 - 8 = 58.

Hello,

Could you please further explain why the 1*m+2*4 must offset so that we can find m = -8?

Many thanks

Regards

Anna

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by [email protected] » Sat Jul 22, 2017 9:35 am
Hi Anna,

When you're dealing with any type of 'average formula' question, the numbers above the average will always 'cancel out' the numbers below the average - because that's how averages 'work.'

For example, if you have the numbers 1 and 3, then the average is (1+3)/2 = 4/2 = 2.
The larger number (3) is '1 greater' than the average.
The smaller number (1) is '1 less' than the average.
Those differences 'cancel out' and you have the average.

With a 'weighted average', the concept is exactly the same, but you have to account for a different number of terms above the average and below the average.

For example, the average of 1, 1 and 4 is (1+1+4)/3 = 2.
The larger number (4) is '2 greater' than the average.
Each of the two smaller numbers (1 and 1) is '1 less' than the average.
"2 greater' + '1 less' + '1 less' cancels out and you have the average.

That same idea applies to this prompt. We know that the average is 66 and that there are two larger numbers (two 70s) for each smaller number (X).
The two larger numbers (70) are each '4 greater' than the average.
'4 greater' + '4 greater' = '8 greater'
So the smaller number (X) has to be '8 less' than the average to cancel out the two larger numbers.
X = 66 - 8 = 58

Final Answer: C

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by Matt@VeritasPrep » Sun Jul 23, 2017 2:35 pm
Another way to think of this: since there are twice as many women as men, the average should be twice as close to the female average as it is to the male average.

With that in mind, if the difference between the female average and the overall average is 4 (or 70 - 66), the difference between the male average and the overall average should be 8, since 4 is twice as close to the average as 8 is.

From there, the male average is just 66 - 8, or 58, and we're done!

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by Matt@VeritasPrep » Sun Aug 06, 2017 10:49 pm
We could do it algebraically too, though I like the conceptual approach better.

Let's call m the male average and f the female average. We know that the total number of tickets can be represented in two ways:

(Average) * (# of People)

OR

(Female Average) * (# of Women) + (Male Average) * (# of Men)

Since these represent the same number, they must be equal:

66 * (m + w) = x * m + 70 * w

We know that w = 2m, so we can replace w with 2m:

66 * (m + 2m) = x * m + 70 * 2m

198m = xm + 140m

58m = xm

58 = x

and we're done! :)

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by agrineer » Sun Aug 20, 2017 9:23 am
male to female is 1:2
1*m+2*f
1*m+2*70=198~~66is avg for 1, so 3*66=198
m=198-140
m=58