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For a recent play performance, the ticket prices were $25 pe

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For a recent play performance, the ticket prices were $25 pe [#permalink]

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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.

Data Sufficiency
Question: 14
Category: Algebra Simultaneous equations
Page: 154
Difficulty: 650


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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 --> \(25a+15c=10,500\) --> \(25a+15(500-a)=10,500\). We can solve for a. Sufficient.

(2) The average (arithmetic mean) price per ticket sold was $21 --> \(\frac{25a+15c}{500}=21\) --> \(25a+15c=10,500\). The same info as above. Sufficient.

Answer: D.
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Re: For a recent play performance, the ticket prices were $25 pe [#permalink]

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New post 01 Jan 2014, 22:31
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The best answer is D!

After reading the given problem we get this problem:
x +y = 500
25x +15y = (The sum of money spent for tickets)=> This is what we need to solve the problem.

(1) is sufficient. It tells us about revenue which is the sum of all prices of tickets.
(2) is also sufficient. If we have average price and the number of all tickets then we can find the total sum.
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Re: For a recent play performance, the ticket prices were $25 pe [#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 3163 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 pe [#permalink]

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New post 02 Jan 2014, 01:17
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 pe [#permalink]

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New post 05 Jan 2014, 11:54
SOLUTION

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 --> \(25a+15c=$10,500\) --> \(25a+15(500-a)=$10,500\). We can solve for a. Sufficient.

(2) The average (arithmetic mean) price per ticket sold was $21 --> \(\frac{25a+15c}{500}=21\) --> \(25a+15c=$10,500\). The same info as above. Sufficient.

Answer: D.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: For a recent play performance, the ticket prices were $25 pe [#permalink]

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Re: For a recent play performance, the ticket prices were $25 pe   [#permalink] 17 May 2016, 14:36
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