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# Tickets for next month's production of The Sea Gull at the local commu

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Re: Tickets for next month's production of The Sea Gull at the local commu [#permalink]
If we go for typical algebra for this question it will take over 6 minutes I feel. After the first few basic algebra steps , we realise that 10, 11 or 12. And that A has to be odd number (which will lead to A being 11). Then on can substitute 6 and 8 in the basic algebra equation to get the answer as 6 . A tough question
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Re: Tickets for next month's production of The Sea Gull at the local commu [#permalink]
HarshaBujji wrote:
My initial thought
Are you sure the question didn't provide any further details, As there are 4 variables and 2 equations.

Later, while commenting got this idea. But is this really an OG question ??
But one solution as per my analysis is A:11, C:6

Between the 2 days the there are 11 tickets & it costed 101 rs more.

Hence these 11 tickets contributed to 101 rs.

Let x be adults whose ticket is A rs,
So 11-x children with C rs

Ax + C(11-x) = 101.

POE

Let see at the options we have for C.

(A-C)x = 101 - 11C. > 0 (As A-C is >0 )

C cannot be 12,11,10.

C has to be either 6 or 8.

Lets take C as 6.

(A-6)x = 35.

A cannot be an even number. So A has to be 11 so x=7,

7 Adults with 11 rs, 4 children with 6 rs. = 77 + 24 =101.

Now let's take C as 8.

(A-8)x = 13. x cannot be 13, & A cannot be 21.

Hence only choice is A:11 , C:6.­

HarshaBujji, there is a concern about your POE: x is the difference of adults between 2 days, x could be positive or negative
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Tickets for next month's production of The Sea Gull at the local commu [#permalink]
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cats wrote:
­Tickets for next month’s production of The Sea Gull at the local community theater went on sale yesterday. The ticket price for a person older than 12 is $A, and the ticket price for a person 12 or under is$C, where A > C. Yesterday, the sale of 33 tickets generated a total revenue of $323, and today the sale of 44 tickets generated a total revenue of$424.

In the table, select a value for A and a value for C that are jointly consistent with the given information. Make only two selections, one in each column.

the sale of 33 tickets generated a total revenue of $323, From 33 tickets revenue is$323 which means that the average revenue per ticket is 9.something.
Hence, cost C should be 6/8 and cost A should be 10/11/12.

Next, to generate a revenue of $323, an odd amount, we need one price to be odd. Except 11, all other 4 options are even. Hence A must be$11.

Next, assume all 33 tickets were for $11. The revenue then would be$363. But the actual revenue is $40 less (it is$323) because some of these tickets were cheaper.
If some of these tickets were for $6 i.e.$5 less than $11, then it would explain the reduced revenue of$40 (8 tickets were cheaper).
But if some of these tickets were for $8 i.e.$3 less than $11, then we cannot reduce the revenue by$40. We will not get an integer number of tickets and hence C cannot be $8. Hence C is$6.

­
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Re: Tickets for next month's production of The Sea Gull at the local commu [#permalink]

KarishmaB wrote:
cats wrote:
­Tickets for next month’s production of The Sea Gull at the local community theater went on sale yesterday. The ticket price for a person older than 12 is $A, and the ticket price for a person 12 or under is$C, where A > C. Yesterday, the sale of 33 tickets generated a total revenue of $323, and today the sale of 44 tickets generated a total revenue of$424.

In the table, select a value for A and a value for C that are jointly consistent with the given information. Make only two selections, one in each column.

the sale of 33 tickets generated a total revenue of $323, From 33 tickets revenue is$323 which means that the average revenue per ticket is 9.something.
Hence, cost C should be 6/8 and cost A should be 10/11/12.

Next, to generate a revenue of $323, an odd amount, we need one price to be odd. Except 11, all other 4 options are even. Hence A must be$11.

Next, assume all 33 tickets were for $11. The revenue then would be$363. But the actual revenue is $40 less (it is$323) because some of these tickets were cheaper.
If some of these tickets were for $6 i.e.$5 less than $11, then it would explain the reduced revenue of$40 (8 tickets were cheaper).
But if some of these tickets were for $8 i.e.$3 less than $11, then we cannot reduce the revenue by$40. We will not get an integer number of tickets and hence C cannot be $8. Hence C is$6.

­

­Hi KarishmaB
I couldn't understand the yellow part in the above explanation. How can we conclude that C can be 6/8 and A can be 10/11/12?
One thing i could figure out while solving was that, ''11'' will be one of the values, because total = 323 is an odd no.
Quote:
Hence, cost C should be 6/8 and cost A should be 10/11/12.''

­
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Re: Tickets for next month's production of The Sea Gull at the local commu [#permalink]
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Quote:
KarishmaB
cats wrote:
cats

Quote:
­Tickets for next month’s production of The Sea Gull at the local community theater went on sale yesterday. The ticket price for a person older than 12 is $A, and the ticket price for a person 12 or under is$C, where A > C. Yesterday, the sale of 33 tickets generated a total revenue of $323, and today the sale of 44 tickets generated a total revenue of$424.

In the table, select a value for A and a value for C that are jointly consistent with the given information. Make only two selections, one in each column.

the sale of 33 tickets generated a total revenue of $323, From 33 tickets revenue is$323 which means that the average revenue per ticket is 9.something.
Hence, cost C should be 6/8 and cost A should be 10/11/12.

Next, to generate a revenue of $323, an odd amount, we need one price to be odd. Except 11, all other 4 options are even. Hence A must be$11.

Next, assume all 33 tickets were for $11. The revenue then would be$363. But the actual revenue is $40 less (it is$323) because some of these tickets were cheaper.
If some of these tickets were for $6 i.e.$5 less than $11, then it would explain the reduced revenue of$40 (8 tickets were cheaper).
But if some of these tickets were for $8 i.e.$3 less than $11, then we cannot reduce the revenue by$40. We will not get an integer number of tickets and hence C cannot be $8. Hence C is$6.

­Hi KarishmaB
I couldn't understand the yellow part in the above explanation. How can we conclude that C can be 6/8 and A can be 10/11/12?
One thing i could figure out while solving was that, ''11'' will be one of the values, because total = 323 is an odd no.Hence, cost C should be 6/8 and cost A should be 10/11/12.''

When you sell 33 tickets and get $323 for it, your average price per ticket is 323/33 = 9.8 (approximately) If your tickets are actually priced at 2 different price points, then the average must be somewhere between those price points. Say you are selling tickets of$8 and $10. The average will be between 8 and 10. Say you are selling tickets of$6 and \$8, the average will be between 6 and 8.

If your average is 9.8, one ticket price will be less than 9.8 and the other will be more than 9.8.
So C will be 6 or 8 (only two options less than 9.8) and A will be 10/11/12 (three options more than 9.8).
­
­
Re: Tickets for next month's production of The Sea Gull at the local commu [#permalink]
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