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28 May 2026, 19:00
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
50% (02:52) correct
50%
(02:10)
wrong
based on 36
sessions
History
Date
Time
Result
Not Attempted Yet
A small theater sold only adult tickets and student tickets for one afternoon show. Each adult ticket cost $18, and each student ticket cost $10. If the theater sold 45 tickets for the show, how many adult tickets were sold?
(1) The average (arithmetic mean) ticket price was greater than $14.50 but less than $15.00.
(2) The total revenue from ticket sales was greater than $655 but less than $663.
(1) The average (arithmetic mean) ticket price was greater than $14.50 but less than $15.00.
(2) The total revenue from ticket sales was greater than $655 but less than $663.
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BongBideshini
Joined: 25 May 2021
Last visit: 04 Jun 2026
Posts: 290
Given Kudos: 92
Location: India
Concentration: Entrepreneurship, Marketing
Schools: HBS '27 Stanford '27 Wharton '27 Sloan '27
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28 May 2026, 22:45
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Let:
Statement (1)
Average ticket price:
Revenue / 45
14.5 < Revenue/45 < 15
652.5 < Revenue < 675
652.5 < 8A + 450 < 675
202.5 < 8A < 225
25.3125 < A < 28.125
Insufficient.
Statement (2)
655 < 8A + 450 < 663
205 < 8A < 213
25.625 < A < 26.625
A = 26
Sufficient.
Answer: B
- Adult tickets = A
- Student tickets = 45 − A
Statement (1)
Average ticket price:
Revenue / 45
14.5 < Revenue/45 < 15
652.5 < Revenue < 675
652.5 < 8A + 450 < 675
202.5 < 8A < 225
25.3125 < A < 28.125
Insufficient.
Statement (2)
655 < 8A + 450 < 663
205 < 8A < 213
25.625 < A < 26.625
A = 26
Sufficient.
Answer: B
Bunuel
03 Jun 2026, 11:42
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This is a fun problem to solve by using weighted averages to minimize calculations.
Statement 1 tells us the average ticket price was between $14.50 and $15.00 (non-inclusive). We also know, from the setup, that adult tickets were $18 and student tickets were $10. If exactly the same number of adult and student tickets were sold (impossible, given the odd numbered total tickets sold, but bear with me), the average ticket price would be the exact midpoint of the adult and student ticket prices: $14.
Since the average ticket price was slightly higher than $14, and therefore closer to the price of adult tickets, we know that more adult tickets were sold. But is this enough to say exactly how many?
Since adult tickets cost $8 more, every additional adult ticket sold raises the average ticket price per person by $8.00 / 45. Estimate by rounding the denominator to 50. 8/50 → 16/100 —> $.16. So every additional adult ticket sold raises the average by about $.16.
Since $14.51 to $14.99 gives us a range of $.48 for the average ticket price, it’s not specific enough to know exactly how many adult tickets were sold — within that range, the number could fluctuate by 2 - 3 tickets.
Insufficient.
Statement 2 tells us the range of total ticket sales. If all 45 tickets sold were student tickets, the total revenue from tickets would be $450. If all 45 were adult tickets, the total revenue would be 45 * 18 = 45 * 10 + 45 * 2 * 4 = 450 + 90*4 = 450 + 360 = $810. If the ticket sales were divided evenly (again, not possible, but worth considering hypothetically), the total would split the difference. The difference between $810 - $450 = $360. Half of that is $180. Add that to $450 and you get $630. So if sales were evenly balanced, the total revenue would be $630.
Since sales were greater than that, we know that more adult tickets were sold. Again, from our work on Statement 1, we know that adult tickets add an extra $8 per ticket to the revenue. Since the range in total revenue sales is less than $8, this should be precise enough to determine exactly how many adult tickets were sold.
Sufficient. The answer is B.
If you're looking to practice more weighted averages problems, check out ManhattanPrep's [url=Free GMAT Starter Kit]Free GMAT Starter Kit[/url]Free GMAT Starter Kit, which has problems and explanations written by our teachers!
Best,
Ally Bell
ManhattanPrep GMAT Instructor
Statement 1 tells us the average ticket price was between $14.50 and $15.00 (non-inclusive). We also know, from the setup, that adult tickets were $18 and student tickets were $10. If exactly the same number of adult and student tickets were sold (impossible, given the odd numbered total tickets sold, but bear with me), the average ticket price would be the exact midpoint of the adult and student ticket prices: $14.
Since the average ticket price was slightly higher than $14, and therefore closer to the price of adult tickets, we know that more adult tickets were sold. But is this enough to say exactly how many?
Since adult tickets cost $8 more, every additional adult ticket sold raises the average ticket price per person by $8.00 / 45. Estimate by rounding the denominator to 50. 8/50 → 16/100 —> $.16. So every additional adult ticket sold raises the average by about $.16.
Since $14.51 to $14.99 gives us a range of $.48 for the average ticket price, it’s not specific enough to know exactly how many adult tickets were sold — within that range, the number could fluctuate by 2 - 3 tickets.
Insufficient.
Statement 2 tells us the range of total ticket sales. If all 45 tickets sold were student tickets, the total revenue from tickets would be $450. If all 45 were adult tickets, the total revenue would be 45 * 18 = 45 * 10 + 45 * 2 * 4 = 450 + 90*4 = 450 + 360 = $810. If the ticket sales were divided evenly (again, not possible, but worth considering hypothetically), the total would split the difference. The difference between $810 - $450 = $360. Half of that is $180. Add that to $450 and you get $630. So if sales were evenly balanced, the total revenue would be $630.
Since sales were greater than that, we know that more adult tickets were sold. Again, from our work on Statement 1, we know that adult tickets add an extra $8 per ticket to the revenue. Since the range in total revenue sales is less than $8, this should be precise enough to determine exactly how many adult tickets were sold.
Sufficient. The answer is B.
If you're looking to practice more weighted averages problems, check out ManhattanPrep's [url=Free GMAT Starter Kit]Free GMAT Starter Kit[/url]Free GMAT Starter Kit, which has problems and explanations written by our teachers!
Best,
Ally Bell
ManhattanPrep GMAT Instructor
