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31 Dec 2013, 06:21
3
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Difficulty:

35% (medium)

Question Stats:

70% (01:16) correct 30% (01:24) wrong based on 742 sessions

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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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### Show Tags

01 Jan 2014, 21:31
1

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] ### Show Tags 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 [ 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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### Show Tags

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.

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Re: For a recent play performance, the ticket prices were $25 pe [#permalink] ### Show Tags 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 $25 pe [#permalink] ### Show Tags 10 Dec 2018, 19:29 Got this wrong in a GMATFocus test, but I realise I just made silly mistakes. 25A + 15C = Total$ 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