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by bacali » Tue Nov 25, 2008 8:00 am
For a certain play performance, adults’ tickets were sold for $12 each and children’s tickets were sold for $8 each. How many children’s tickets were sold for the performance?

(1) The total revenue from the sale of adults’ and children’s tickets for the performance was $5,040.
(2) The number of adults’ tickets sold for the performance was 1/3 the total number of adults’ and children’s tickets sold for the performance.


OA:

C
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Source: — Data Sufficiency |
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by rohangupta83 » Tue Nov 25, 2008 8:26 am
Number of adult tickets sold = x
number of children tickets sold = y

1). 12x + 8y = 5040

Two unknowns and 1 eq - NOT SUFFICIENT

2). Number of adult tickets = 1/3rd the sum of adult and children tickets

therefore,

x = (x+y)/3
or
3x = x+y
2x = y--------NOT SUFFICIENT

Using both equations

Plugging in equation 1

12x = 6*2*x = 6y

So,
6y + 8y = 5040
y = 5040/14

Hence C
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by jimmiejaz » Tue Nov 25, 2008 8:30 am
totally agree with C.
I would suggest an approach for these questions.

We are asked

How many children’s tickets were sold for the performance?


From 1st stmt, we got total revenue but we dont have any number of tickets sold. hence insuff.

From 2nd stmt, we got the ratio for number for tickets sold. But from this stmt alone, we cant get the total revenue generated. hence insuff.

Now combining we have the total revenue and the ratio of tickets sold.
hence suff.
Ans is C.
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