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Updated on: 05 Jan 2015, 05:04
3
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Difficulty:

55% (hard)

Question Stats:

61% (01:23) correct 39% (01:19) wrong based on 426 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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07 Feb 2010, 17:07
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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Updated on: 08 Feb 2010, 20:07
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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09 Feb 2010, 17:03
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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Re: At a certain theater, the cost of each adult's ticket is $5 and cost [#permalink] ### Show Tags 03 Jan 2015, 12:56 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 _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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18 Oct 2016, 10:26
This is a double Matrix Question :

We know Adult price is $5 and Child Price is$2.

Option A:

The ratio of C:A is 3:2

Number Price Total
Child 3x 2 6x
Total 5x 16x

Average can be found as we have the total and individual components.

Option B: Individual component of Adult number of tickets are given, giving no clue to the number of tickets sold for the children.

Hope it helps.
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08 Dec 2016, 14:50
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.

Statement 1 - let number of children's ticket and adult's ticket sold = 3x and 2x
So average cost = (2x *5 + 3x*2)/3x+2x
= 16x/5x = 16/5 ----sufficient

Statement 2 - Number of adult tickets sold = 80 but no info about number of children's ticket. --not sufficient.

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01 Sep 2018, 09:22
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.

OA: A
Let the Adult's ticket price be $$T_{Adult}$$ and Child's ticket price be $$T_{Child}$$

$$T_{Adult}=5 \qquad T_{Child}=2$$

Let the number of Adults be $$A$$ and the number of children be $$C$$.

We have to find out the average cost of all tickets sold at the theater i.e We have to find $$\frac{A*T_{Adult}+C*T_{Child}}{A+C}......(1)$$

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.

$$\frac{C}{A}=\frac{3}{2}$$ ;$$C = \frac{{3A}}{2}......(2)$$

putting (2) and $$T_{Adult}=5 ; T_{Child}=2$$ in (1), we get

Average cost of all tickets sold $$= \frac{A*T_{Adult}+C*T_{Child}}{A+C} =\frac{A*5+\frac{3A}{2}*2}{A+\frac{3A}{2}} =\frac{8A}{\frac{5A}{2}}=\frac{16}{5}$$

Statement 1 alone is sufficient.

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

We know about the number of Adult's ticket sold but are not aware about the number of children ticket sold.
Statement 2 alone is not sufficient
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