Last visit was: 03 Aug 2024, 06:15 It is currently 03 Aug 2024, 06:15
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# For a certain play performance, adults' tickets were sold for $12 each SORT BY: Tags: Show Tags Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 94776 Own Kudos [?]: 646248 [3] Given Kudos: 86843 GMAT Club Legend Joined: 18 Aug 2017 Status:You learn more from failure than from success. Posts: 8020 Own Kudos [?]: 4262 [0] Given Kudos: 243 Location: India Concentration: Sustainability, Marketing GMAT Focus 1: 545 Q79 V79 DI73 GPA: 4 WE:Marketing (Energy and Utilities) CEO Joined: 07 Mar 2019 Posts: 2636 Own Kudos [?]: 1887 [0] Given Kudos: 763 Location: India WE:Sales (Energy and Utilities) Intern Joined: 21 Feb 2019 Posts: 41 Own Kudos [?]: 14 [0] Given Kudos: 376 Location: United States Schools: Moore '22 (S) GPA: 3.63 Re: For a certain play performance, adults' tickets were sold for$12 each [#permalink]
lnm87 wrote:
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?
Let
Number of adults tickets = A
Number of children tickets = C

(1) The total revenue from the sale of adults’ and children’s tickets for the performance was $5,040. 12A + 8C = 5040 and for A = 2, C = 627 and so on... Since LCM (12,8) = 24, there are 209 combinations possible between number of tickets of adults and children. Hence INSUFFICIENT. (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. $$A = \frac{1}{3}(A + C)$$ 2A = C Since nothing about total amount of tickets price is given, many possibilities exist. Hence INSUFFICIENT. Together 1 and 2 6C + 8C = 5040 C = 360 and A = 180 SUFFICIENT. Answer C Posted from my mobile device Senior Manager Joined: 13 Jul 2022 Posts: 394 Own Kudos [?]: 710 [0] Given Kudos: 267 Location: India Concentration: Finance, Nonprofit GPA: 3.74 WE:Corporate Finance (Non-Profit and Government) Re: For a certain play performance, adults' tickets were sold for$12 each [#permalink]
unraveled wrote:
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?
Let
Number of adults tickets = A
Number of children tickets = C

(1) The total revenue from the sale of adults’ and children’s tickets for the performance was $5,040. 12A + 8C = 5040 and for A = 2, C = 627 and so on... Since LCM (12,8) = 24, there are 209 combinations possible between number of tickets of adults and children. Hence INSUFFICIENT. (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. $$A = \frac{1}{3}(A + C)$$ 2A = C Since nothing about total amount of tickets price is given, many possibilities exist. Hence INSUFFICIENT. Together 1 and 2 6C + 8C = 5040 C = 360 and A = 180 SUFFICIENT. Answer C ­Hi, I did not understand the '209' combinations bit. Can you please help me understand? Re: For a certain play performance, adults' tickets were sold for$12 each [#permalink]
Moderator:
Math Expert
94776 posts