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For a recent play performance, the ticket prices were $25 pe [#permalink] ### Show Tags 31 Dec 2013, 06:21 2 This post received KUDOS Expert's post 2 This post was BOOKMARKED 00:00 Difficulty: 35% (medium) Question Stats: 66% (02:17) correct 34% (01:29) wrong based on 444 sessions ### HideShow timer Statistics 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 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you! [Reveal] Spoiler: OA _________________ Math Expert Joined: 02 Sep 2009 Posts: 36638 Followers: 7106 Kudos [?]: 93661 [3] , given: 10583 Re: For a recent play performance, the ticket prices were$25 pe [#permalink]

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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.

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01 Jan 2014, 23:57
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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 $21. Refer attachment Attachment: WA.PNG [ 12.23 KiB | Viewed 3621 times ] Now$ 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 _________________ “If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.” Manager Status: GMATting Joined: 21 Mar 2011 Posts: 108 Concentration: Strategy, Technology GMAT 1: 590 Q45 V27 Followers: 1 Kudos [?]: 90 [0], given: 104 Re: For a recent play performance, the ticket prices were$25 pe [#permalink]

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02 Jan 2014, 00:17
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] ### 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. Answer: D. _________________ GMAT Club Legend Joined: 09 Sep 2013 Posts: 13549 Followers: 578 Kudos [?]: 163 [0], given: 0 Re: For a recent play performance, the ticket prices were$25 pe [#permalink]

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