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Re: For a recent play performance, the ticket prices were $25 per adult an [#permalink]
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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?

(1) Revenue from ticket sales for this performance totaled $10,500.
(2) The average (arithmetic mean) price per ticket sold was $21.

Sol: Let A = Total no. of Adult tickets
C: Total no. of Child Tickets
Given A+C=500, we need to find A ?
Price of Adult Ticket: $25
Price of Child Ticket : $15


From St 1, we have 25*A+15*C = 10500
We also know A+C= 500
We have 2 variables and 2 equations and therefore we can solve for A. We can leave it that.
So B C and E ruled out

From St 2 we have Average price is $ 21. Refer attachment

Attachment:
WA.PNG
WA.PNG [ 12.23 KiB | Viewed 20071 times ]


Now $ 21 is 6 $ more than Child Ticket price and $4 Less than Adult ticket price. So by Weighted Average principle.
Total difference between Adult and Child Ticket price is $ 10

Number of Adult Tickets will be: 6/10 *500 = 300 -----> A

Just for clarity purpose we can calculate St 1 as well

5A+3C= 2100-------Eq 1
A+C=500-----> 3A+3C= 1500-------> Eq 2

Subtracting 2 from1, we get 2A=600 or A =300.

Ans D
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Re: For a recent play performance, the ticket prices were $25 per adult an [#permalink]
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Since the question asks for the number of tickets sold for adults, let us assume x to be the number of tickets sold to adults.

Average price/ticket = (no. of adult tickets) * (Price/adult ticket) + (no. of child tickets) * (Price/child ticket)
Av. price = 25 * A + 15 * C
Since A + C = 500; C = 500 - A

Av. price = 25A + 15(500 - A) = 25A + 7500 - 15A = 10A + 7500;
So, if we know the av.price/ticket, we can A;

1) Av.price = 10500/500 = 21; Sufficient
2) Av.price is given as 21; Sufficient.

Hence (D).
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Re: For a recent play performance, the ticket prices were $25 per adult an [#permalink]
nutshell wrote:
Since the question asks for the number of tickets sold for adults, let us assume x to be the number of tickets sold to adults.

Average price/ticket = (no. of adult tickets) * (Price/adult ticket) + (no. of child tickets) * (Price/child ticket)
Av. price = 25 * A + 15 * C
Since A + C = 500; C = 500 - A

Av. price = 25A + 15(500 - A) = 25A + 7500 - 15A = 10A + 7500;
So, if we know the av.price/ticket, we can A;

1) Av.price = 10500/500 = 21; Sufficient
2) Av.price is given as 21; Sufficient.

Hence (D).

This is more clean
D
x: the tickets sold were for adults

1) 25x + (500-x)*15 = 10500
=> sufficient
2) The average (arithmetic mean) price per ticket sold was $21.
=> Revenue from ticket sales 500x21 = 10500
=> sufficient because it' has the same condition with 1)
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Re: For a recent play performance, the ticket prices were $25 per adult an [#permalink]
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Got this wrong in a GMATFocus test, but I realise I just made silly mistakes.

25A + 15C = Total $ However, we are told the # tickets sold, so lets include this now
25A + 15(500-A) = Total$

1. Total$ = 10,500
25A + 15(500-A) = 10,500.

Sufficient to solve for A

2. [25A + 15(500-A)]/500 = 21
25A + 15(500-A) = 21*500
25A + 15(500-A) = 10,500 (Sufficient to solve for A

D
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Re: For a recent play performance, the ticket prices were $25 per adult an [#permalink]
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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?

(1) Revenue from ticket sales for this performance totaled $10,500.
(2) The average (arithmetic mean) price per ticket sold was $21.


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 $10,500
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 $21.
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 $10,500 (this is exactly what statement 1 told us)
Since statement 1 was SUFFICIENT, statement 2 must also be SUFFICIENT

Answer: D

Cheers,
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Re: For a recent play performance, the ticket prices were $25 per adult an [#permalink]
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Bunuel 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?

(1) Revenue from ticket sales for this performance totaled $10,500.
(2) The average (arithmetic mean) price per ticket sold was $21.

Solution:

Question Stem Analysis:

We need to determine the number of adult tickets sold in a recent play performance, given that an adult ticket is $25 and a child ticket is $15. If we let x = the number of adult tickets sold and y = the number of child tickets sold, we can create the equation x + y = 500 for the total number of tickets sold. We need to determine the value of x.

Statement One Alone:

Since the total revenue raised from the ticket sales was $10,500, we can create another equation: 25x + 15y = 10,500. Along with the equation x + y = 500, we see that we can solve for the value of x (and y). Statement one alone is sufficient.

Statement Two Alone:

Since the average price per ticket sold is $21, we can create another question (25x + 15y) / 500 = 21. Along with the equation x + y = 500, we see that we can solve for the value of x (and y). Statement two alone is sufficient.

Answer: D
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