Question with wrong ans I think

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good_dn: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Sun May 05, 2013 6:03 pm

by good_dn » Mon May 06, 2013 5:45 pm

Hello,

Can you please help with this problem?

Tickets to a play cost $10 for children and $25 for adults. If 100 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 $1750

Solution provided in this link

https://www.beatthegmat.com/weighted-ave ... tml#638076

Statement 1: The average revenue per ticket was $18.
The average cost (18) is closer to the cost of an adult ticket (25) than to the cost of a children's ticket (10).
Thus, the number of adult tickets sold must have been greater than the number of children's tickets sold.
SUFFICIENT.

Statement 2: The revenue from ticket sales exceeded $1750.
Plug in the THRESHOLD: 50 of each type of ticket sold.
50(10) + 50(25) = 1750.
The equation above implies the following: for the total revenue to have EXCEEDED 1750, a greater number of the more expensive tickets -- the ADULT tickets -- must have been sold.
SUFFICIENT.

BUT

Is it possible for the average to be 18$???

If that is the case then 25a+10c = 1800 (since total tickets was 100)
and a+c =100 (a=number of adult tickets and c is number of children tickets)

a and c will have non integer values. That is not possible

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Source: Beat The GMAT — Data Sufficiency | Mon May 06, 2013 5:45 pm

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by GMATGuruNY » Mon May 06, 2013 6:46 pm

good_dn wrote:Hello,

Can you please help with this problem?

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

Solution provided in this link

https://www.beatthegmat.com/weighted-ave ... tml#638076

Statement 1: The average revenue per ticket was $18.
The average cost (18) is closer to the cost of an adult ticket (25) than to the cost of a children's ticket (10).
Thus, the number of adult tickets sold must have been greater than the number of children's tickets sold.
SUFFICIENT.

Statement 2: The revenue from ticket sales exceeded $1750.
Plug in the THRESHOLD: 50 of each type of ticket sold.
50(10) + 50(25) = 1750.
The equation above implies the following: for the total revenue to have EXCEEDED 1750, a greater number of the more expensive tickets -- the ADULT tickets -- must have been sold.
SUFFICIENT.

BUT

Is it possible for the average to be 18$???

If that is the case then 25a+10c = 1800 (since total tickets was 100)
and a+c =100 (a=number of adult tickets and c is number of children tickets)

a and c will have non integer values. That is not possible

Good catch: the information in statement 1 contradicts the constraint in the question stem that 100 tickets are sold.
For the average revenue per ticket to be 18 -- as required by statement 1 -- the ratio of adults tickets to children's tickets must be 8:7.
To illustrate:
If 8 adult tickets and 7 children's tickets are sold -- for a total of 15 tickets -- then the average revenue per ticket = (8*25 + 7*10)/15 = 18.
But if adult tickets : children's tickets = 8:7, then the total number of tickets sold must be a multiple of 8+7 = 15.
Not possible, since the total number of tickets sold -- 100 -- is not a multiple of 15.
Since the information in statement 1 contradicts the constraint that 100 tickets are sold, the problem is flawed.

A viable version of the same problem:

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

Here, the following case satisfies both statements and the constraint that 90 tickets are sold:
Adult tickets sold = 48.
Children's tickets sold = 42.
Total revenue = 48*25 + 42*10 = 1620.
Average revenue per ticket = 1620/90 = 18.

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