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by vinay1983 » Wed Sep 11, 2013 4:32 am
For a recent play performance, the ticket prices were $25 per adult and S15 per child. A total of 500 tickets were sold for the performance. How many of the tickets sold were for adults?

(1) Revenue from ticket sales for this performance totaled S10,500.

(2) The average (arithmetic mean) price per ticket sold was S21.

Source: OG Q-2-ED OA D

Statement 1

According to me if the the total price of the tickets is 10500$, then

Maximum tickets of $25 sold can be 300 i.e 300*25=7500

remaining 10500-7500=3000 so $ 15*200=3000

Hence sufficient

Statement 2

Mean is 21, this means no.of $ 25 tickets sold are more than $15 tickets, but the question asked for a number, hence insufficient.

What did I do wrong?
You can, for example never foretell what any one man will do, but you can say with precision what an average number will be up to!
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Source: — Data Sufficiency |
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by faraz_jeddah » Wed Sep 11, 2013 6:16 am
vinay1983 wrote:For a recent play performance, the ticket prices were $25 per adult and S15 per child. A total of 500 tickets were sold for the performance. How many of the tickets sold were for adults?

(1) Revenue from ticket sales for this performance totaled S10,500.

(2) The average (arithmetic mean) price per ticket sold was S21.

Source: OG Q-2-ED OA D


Statement 2

Mean is 21, this means no.of $ 25 tickets sold are more than $15 tickets, but the question asked for a number, hence insufficient.

What did I do wrong?
Check out the allegation method that is very popular on BTG.
Here is an example by Mitch - https://www.beatthegmat.com/mixture-and- ... 67444.html

25---21---15

So the ratio of Adult tickets to child tickets = 6:4

6x + 4x = 500
10x = 500
x= 50

Adult tickets = 6x = 6*50 = 300
A good question also deserves a Thanks.

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by Brent@GMATPrepNow » Wed Sep 11, 2013 6:16 am
vinay1983 wrote:For a recent play performance, the ticket prices were $25 per adult and S15 per child. A total of 500 tickets were sold for the performance. How many of the tickets sold were for adults?

(1) Revenue from ticket sales for this performance totaled S10,500.

(2) The average (arithmetic mean) price per ticket sold was S21.
Target question: How many of the tickets sold were for adults?

Given: A total of 500 tickets were sold for the performance
Let C = # of child tickets sold
Let A = # of adult tickets sold
So, C + A = 500

Statement 1: Revenue from ticket sales for this performance totaled S10,500
In other words, 25A + 15C = 10,500
When we add our given equation, C + A = 500, we can see that we have a system of 2 different linear equations with 2 variables.
Since we COULD solve this system for A, we COULD answer the target question with certainty.
So statement 1 is SUFFICIENT

Statement 2: The average (arithmetic mean) price per ticket sold was S21.
We'll use this fact: average of n numbers = (sum of the n numbers)/n
Rearrange to get sum of the n numbers = (average of n numbers)(n)
If 500 tickets were sold and the average ticket price was $21, then the sum of all tickets sold = (21)(500) = $10,500
IMPORTANT: Statement 2 is just another way of telling us that the total revenue from ticket sales was S10,500 (this is exactly what statement 1 told us)
Since statement 1 was SUFFICIENT, statement 2 must also be SUFFICIENT

Answer = D

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Brent
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by [email protected] » Thu Sep 12, 2013 12:22 pm
Hi vinay1983,

Both farad_jeddah and Brent offered methods to solve this problem that are correct.

Here's one other method that you might find useful.

From the prompt, you know there are a total of 500 tickets. Let's say...

A = number of adult tickets
C = number of child tickets

A + C = 500

The question asks us for the value of A. Since we have one equation already, if we get another, unique equation, then we'll have a "system" and we can solve for A.

Fact 2 tells us the average price per ticket. We can create the following equation:

(25A + 15C)/(A + C) = 21

This gives us another unique equation. Combined with the original equation from the prompt, we have a 2V2E (2 variables, 2 unique equations), so we CAN solve for A and there would be just one answer. The cool part is that we don't have to do the math. We KNOW that Fact 2 is SUFFICIENT.

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