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Difficulty:   55% (hard)

Question Stats: 61% (01:20) correct 39% (01:21) wrong based on 380 sessions

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At a certain theater, the cost of each adult's ticket is $5 and cost of each child's ticked is$2. What was the average (arithmetic mean) cost of all adult's and children's tickets sold at the theater yesterday.

(1) Yesterday the ratio of the number of children's tickets sold at the theater to the number of adult's tickets sold at the theater was 3 to 2.

(2) Yesterday 80 adult's tickets were sold at the theater.

Originally posted by prashi82 on 21 Aug 2008, 21:21.
Last edited by Bunuel on 05 Jan 2015, 05:04, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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But the ratio provided by A doesn't hold true if you use different values: let's say that 120 children tickets were sold and 80 adults tickets were sold (same ratio as 3 to 2) then we have:

5(120)+2(80)/200 and the average price is not the same.

I choose C for this for that reason. I know it's wrong but can someone explain. Thanks.
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If you take a conceptual approach to this problem, the ratio is all you need to get weighted average of the sold tickets.
X= the number of adult tickets sold
Y=number of the children ticket sold
Average = (5*X+2*Y)/(X+Y)
As you have a ratio of Y/X, you can express Y in terms of X (or vise versa). As a result Average expression will become independent of X and Y.

A is sufficient obviously

Originally posted by alexBLR on 08 Feb 2010, 15:23.
Last edited by alexBLR on 08 Feb 2010, 20:07, edited 1 time in total.
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I picked A because it gave the same average cost per ticket for different numbers i picked.

algebrically when you form the equation it is like :

( 5x+2y ) / x+y , ST 1 says that y = 3/2x , when you substitute you get one value which is 16/5 or 3.2
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: At a certain theater, the cost of each adult's ticket is $5 and cost [#permalink] Show Tags 2 Hi All, When you're dealing with "ratio data" in a DS question, it's actually really easy to prove if a pattern exists or not - just run through a few quick TESTs... Here, we're told: Adult tickets cost$5 each
Children's tickets cost $2 each We're asked for the AVERAGE COST of all tickets sold yesterday. Fact 1: The ratio of Children's tickets to Adult's tickets was 3:2 yesterday. Some Test Takers can clearly see that this ratio IS enough information to say that Fact 1 is SUFFICIENT. Here's how you can quickly prove the consistency... IF... 3 children 2 adults 3(2) + 2(5) = 16/5 =$3.20 average ticket price

6 children
6(2) + 4(5) = 32/10 = $3.20 average ticket price 9 children 6 adults 9(2) + 6(5) = 48/15 =$3.20 average ticket price

With this ratio, the average is ALWAYS $3.20 Fact 1 is SUFFICIENT Fact 2: 80 Adult tickets were sold Here, we don't know the number of Children's tickets, so the average ticket price would change depending on THAT number. Again, here's the proof: 0 children 80 adults 0 + 80(5) = 400/80 =$5 average ticket price

80 children
80(2) + 80(5) = 560/160 = $3.50 average ticket price Fact 2 is INSUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich _________________ Math Expert V Joined: 02 Sep 2009 Posts: 58435 Re: At a certain theater, the cost of each adult's ticket is$5 and cost  [#permalink]

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1
1
At a certain theater, the cost of each adult's ticket is $5 and the cost of each child's ticket is$2. What was the average cost of all the adult's and children's tickets sold at the theater yesterday?

The average cost = (2*C+5*A)/(C+A)

(1) Yesterday ratio of # of children's ticket sold to the # of adult's tickets sold was 3 to 2 --> 3A =2CA = 2C/3 --> the average cost = C(2+5*2/3)/(C(1+2/3)) --> (2+5*2/3)/(1+2/3). Sufficient.

(2) Yesterday 80 adult's tickets were sold at the theater --> A = 80. We know nothing about C. Not sufficient.

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1
prashi82 wrote:
At a certain theater, the cost of each adult's ticket is $5 and cost of each child's ticked is$2. What was the average (arithmetic mean) cost of all adult's and children's tickets sold at the theater yesterday.

(1) Yesterday the ratio of the number of children's tickets sold at the theater to the number of adult's tickets sold at the theater was 3 to 2.

(2) Yesterday 80 adult's tickets were sold at the theater.

We are given that the cost of each adult's ticket is $5 and cost of each child's ticket is$2. We need to determine the average (arithmetic mean) cost of all adults’ and children's tickets sold at the theater yesterday. If we let a = the number of adults’ tickets sold and c = the number of children’s tickets sold, we can create the following average equation:

Average = (5a + 2c)/(a + c)

Statement One Alone:

Yesterday the ratio of the number of children's tickets sold at the theater to the number of adult's tickets sold at the theater was 3 to 2.

Using the information in statement one, we can create the following equation:

c/a = 3/2

2c = 3a

c = 1.5a

Since c = 1.5a, we can substitute 1.5a for c in our average equation and we have:

Average = (5a + 2 x 1.5a)/(a + 1.5a)

Average = (5a + 3a)/2.5a

Average = 8a/2.5a

Average = 3.2

Statement one alone is sufficient to answer the question. We can eliminate answer choices B, C, and E.

Statement Two Alone:

Yesterday 80 adult's tickets were sold at the theater.

Only knowing the number of adults’ tickets sold yesterday is not enough information to determine the average cost of all tickets sold. Statement two alone is not sufficient to answer the question.

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Here is another approach.

Assume that a adults and c children tickets were sold.
The Question can be modified as follows
(5a +2c)/(a+c) = ?

stmt says c/a = 3/2
ie c = 3a/2 ----------(1)

substitute this ie the question.
(5a + 3a)/(a +3a/2)
simplifying,
16a/5a = 16/5 .. a definite answer
So sufficient.

Stmt 2 is not sufficient.

hence A
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_________________ Re: At a certain theater, the cost of each adults ticket is $5 [#permalink] 22 Oct 2019, 08:36 Display posts from previous: Sort by At a certain theater, the cost of each adult's ticket is$5 and cost

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