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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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New post 31 Dec 2013, 06:21
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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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Re: For a recent play performance, the ticket prices were $25 pe  [#permalink]

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New post 31 Dec 2013, 06:22
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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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Re: For a recent play performance, the ticket prices were $25 pe  [#permalink]

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New post 01 Jan 2014, 21:31
1
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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New post 01 Jan 2014, 23:57
1
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 6193 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, 00:17
1
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, 10:54
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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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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New post 27 Sep 2018, 03:18
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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For a recent play performance, the ticket prices were $25 pe  [#permalink]

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New post 10 Dec 2018, 19:29
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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For a recent play performance, the ticket prices were $25 pe &nbs [#permalink] 10 Dec 2018, 19:29
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