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Tickets to a play cost \(\$10\) for children and \(\$25\) for adults. If \(90\) tickets were sold, were more adult ticke

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Tickets to a play cost \(\$10\) for children and \(\$25\) for adults. If \(90\) tickets were sold, were more adult ticke

by Vincen » Wed Oct 28, 2020 8:22 am

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A

B

C

D

E

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Tickets to a play cost \(\$10\) for children and \(\$25\) for adults. If \(90\) tickets were sold, were more adult tickets sold than children's tickets?

(1) The average revenue per ticket was \(\$18.\)

(2) The revenue from ticket sales exceeded \(\$1600.\)

Answer: D

Source: Manhattan GMAT
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Source: — Data Sufficiency |
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Re: Tickets to a play cost \(\$10\) for children and \(\$25\) for adults. If \(90\) tickets were sold, were more adult t

by deloitte247 » Sat Oct 31, 2020 11:33 pm

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Your Answer

A

B

C

D

E

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Let the no. of Children's ticket = x
Let the no. of Adult's ticket = y
x + y = 90 and x = 90 - y
Target question: Were more adult tickets sold than children's tickets?
i.e is y > x?
Statement 1 => The average revenue per ticket was $18
$$\frac{10x+25y}{90}=18$$
$$10x+25y=1620$$ From question stem, x = 90 - y
Therefore, 10 (90 - y) + 25y = 1620
900 - 10y + 25y = 1620
15y = 720
y = 720/15 = 48
and x = 90 - 48 = 42
y > x
Statement 1 is SUFFICIENT.

Statement 2 => The revenue from ticket sales exceeded $1600.
10x + 25y > 1600
From the question stem, x = 90 - y
10 (90 - y) + 25y > 1600
900 - 10y + 25y > 1600
15y > 700
y > 700/15
y > 46
If the least value of y=46, then, x = 90 - 46 = 44. So, at all possible occurrences of y, there will be y>x. So, therefore, statement 2 is SUFFICIENT.

Since each statement alone is SUFFICIENT, the correct answer is Option D.
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