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by divya23 » Thu Jun 09, 2011 8:30 am
cost of adult ticket is = 5 and child = 2 v have to find d mean of both as in total ticket sold for both
1. ratio of c/a = 3/2
2. total ticket sold = 80
[spoiler]
ans = a[/spoiler]
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by Ashley@VeritasPrep » Thu Jun 09, 2011 8:39 am
Hi there,

You write that "v have to find d mean of both as in total ticket sold for both"

So essentially we have to find the average price per ticket sold, right? But if that's what the problem's asking for, we actually DON'T need to find the total tickets sold, we just need to know how the numbers of tickets sold in each category relate to each other.

If I know that exactly 3 child tickets are sold *every time* exactly 2 adult tickets are sold, I know that each time all of that happens, I've got an average ticket sale price of (3tickets*$2 + 2tickets*$5)/(5tickets), so an average sale price of $3.20. There we have it from just statement 1!

Statement 2 isn't enough, since I have no idea how many of those 80 tickets were the cheap ones versus how many were the expensive ones.

Does that make sense?
Ashley Newman-Owens
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Veritas Prep

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by cans » Tue Jun 14, 2011 2:56 am
A=5;C=2
to find: cost per ticket = [n(A)*5 + n(C)*2] / [n(A) + n(c)]
a)n(C)/n(a)=3/2 -> n(c) = 3n(A)/2
thus cost per ticket = [n(A)*5 + 3n(A)] / [n(A) + 3n(A)/2] = 16/5
sufficient
b)n(A)+n(C)=80 insufficient (1 equation, 2 variables)
IMO A
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