Gmat Prep - Theater Tickets

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Gmat Prep - Theater Tickets

by dubeystuts » Sun Feb 01, 2009 3:14 am
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?
1) Y''day ratio of # of children's ticket sold to the # of adult's ticketr sold was 3 to 2
2) Y'day 80 adult's tickets were sold at the theater.

I got C but the correct answer is A. Can someone please explain? Many Thanks.

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Re: Gmat Prep - Theater Tickets

by piyush_nitt » Sun Feb 01, 2009 3:35 am
dubeystuts wrote: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?
1) Y''day ratio of # of children's ticket sold to the # of adult's ticketr sold was 3 to 2
2) Y'day 80 adult's tickets were sold at the theater.

I got C but the correct answer is A. Can someone please explain? Many Thanks.
Stmt 1

Ratio given = 3/2

Let X be the common factor therefore

Total Number for tickets = 3x + 2x = 5x

Therefore, Average = 3x(5) + 2x(2)/5x = 3.8

SUFF

stmt 2

INSUFF

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by dubeystuts » Sun Feb 01, 2009 2:47 pm
Many thanks Piyush.

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by ogbeni » Thu Aug 27, 2009 4:40 pm
:shock: :shock: :shock: :( :(

This is such a trick question. I went for C as well. How do you know when you encounter such tricky questions?

Is there a key word or something that just sets off the alarms?

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by lunarpower » Mon Aug 31, 2009 4:29 am
ogbeni wrote::shock: :shock: :shock: :( :(

This is such a trick question. I went for C as well. How do you know when you encounter such tricky questions?

Is there a key word or something that just sets off the alarms?
you should be VERY VERY suspicious of choice (c) on problems like this one.

on this problem, if you have choice (c), you have ALL the information in the problem. IE you know the exact numbers of both types of ticket sold.

this is ALMOST NEVER the way official data sufficiency problems work.**
the hallmark of official data sufficiency problems is to give situations, the vast majority of the time (though not 100% of the time), in which you can find the DESIRED information even though OTHER information remains unknown.

on a problem like this, if both statements together will give ALL possible information, you should not pick (c) unless you single-, double-, and triple-check the statements to make sure that neither of them is sufficient alone.

in any case, this is NOT a "trick question" AT ALL. this is a totally routine averages question, written in the same fashion as 80-90+% of other such data suff questions.

here's another such problem:
https://www.manhattangmat.com/forums/wha ... t5301.html

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**this is one of the many important differentiators between quality prep materials and subpar prep materials. plenty of second-class prep materials have lots and lots of questions in which you have to find everything in the problem to get "sufficient". this is bad, because it reinforces the horrible habit of assuming that you have to find every quantity in the problem to get "sufficient".
Last edited by lunarpower on Mon Aug 31, 2009 4:32 am, edited 2 times in total.
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by lunarpower » Mon Aug 31, 2009 4:29 am
--

update:

i don't think i have EVER seen a ratio or average problem on which you actually had to find all the information to get "sufficient".

so, on problems such as this one, you should be even more suspicious of (c). (and, of course, (e) is impossible.)

the cool thing about this realization is that the suspicion of (c), the impossibility of (e), and the impossibility of (b) and (d) (since the second statement is clearly insufficient) lead automatically to a guess of (a) - the correct answer - should you have to guess.
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by ogbeni » Mon Aug 31, 2009 4:52 am
Thnx for the response. My guard is up and I've learned a lesson from this! :)

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by rbansal » Fri May 13, 2011 6:03 pm
piyush_nitt wrote:
dubeystuts wrote: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?
1) Y''day ratio of # of children's ticket sold to the # of adult's ticketr sold was 3 to 2
2) Y'day 80 adult's tickets were sold at the theater.

I got C but the correct answer is A. Can someone please explain? Many Thanks.
Stmt 1

Ratio given = 3/2



Hello in the question isnt the ratio 3:2, 3 children for every 2 adults

Childrens price is $2, and Adults price is $5

so it would be:

3x(2) + 2x(5)/5x =3.2 roughly is this correct or did I miss something.



Let X be the common factor therefore

Total Number for tickets = 3x + 2x = 5x

Therefore, Average = 3x(5) + 2x(2)/5x = 3.8

SUFF

stmt 2

INSUFF

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by smackmartine » Sat May 14, 2011 12:53 am
One other way to see this question:

A-> # of adult's ticket (cost $5 per ticket)
C-> # of child's ticket (cost $2 per ticket)

Asked: value of (5A+2C) / A+C

St 1: C/A = 3/2

Rearranging the target expression (5A+2C) / A+C

-> dividing numerator and denominator by A ---> (5+2[C/A])/ (1+ [C/A]) (Bammm!!! we know C/A) SUFFICIENT

St 2 : A = 80 , no clue of C so INSUFFICIENT

So, A

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by Scott@TargetTestPrep » Tue Dec 05, 2017 5:51 pm
dubeystuts wrote:At a certain theater, the cost of each adult's ticket is $5 and cost of each child's ticket 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.

Answer: A

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