arithmetic mean problem

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arithmetic mean problem

by kishokbabu » Wed Dec 28, 2011 8:13 am
For a recent play performance , the ticket prices were $25 per adult and $15 per child . A total of 500 tickets were sold for the performance . How many of the tickets sold were for adults?

A) Revenue from ticket sales for this performance totaled $10,500
B) The average (arithmetic mean) price per ticket sold was $21.
Source: — Data Sufficiency |

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by GMATGuruNY » Wed Dec 28, 2011 9:43 am
kishokbabu wrote:For a recent play performance , the ticket prices were $25 per adult and $15 per child . A total of 500 tickets were sold for the performance . How many of the tickets sold were for adults?

A) Revenue from ticket sales for this performance totaled $10,500
B) The average (arithmetic mean) price per ticket sold was $21.
Let a = the number of adult tickets sold and c = the number of children's tickets sold.
Thus, a + c = 500.

Statement 1: Total revenue = 10,500.
25a + 15c = 10,500.
Given this equation and the equation in the question stem, we can solve for a.
SUFFICIENT.

Statement 2: Average price per ticket sold = 21.
Average price per ticket = (total revenue)/(total number of tickets sold) = (25a + 15c)/(a+c).
Thus, (25a + 15c)/(a+c) = 21.
Given this equation and the equation in the question stem, we can solve for a.
SUFFICIENT.

The correct answer is D.
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by golu23 » Fri Dec 30, 2011 7:48 am
HI Mitch could you please elaborate on this question

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by ArunangsuSahu » Fri Dec 30, 2011 8:39 am
NO NEED TO SOLVE

Given Info:

Adult(A) =$25
Child(C) = $15

ASSUMPTION:
No of Adults = x
No of Children =y

So,
x+y =500

Statement 1 :

25x+15 y =10,500----(i)
Also given x+y = 500----(ii)

So 2 variables and 2 different equations...Unique values

SUFFICIENT


STATEMENT 2:

A.M=$21

Therefore 25x+15y = 500*$21-----(i)

Also given x+y = 500----(ii)

So 2 variables and 2 different equations...Unique values

SUFFICIENT

(D) is the Choice