Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 0907:00 AM PDT
-10:00 AM PDT
Meet AdComs and explore top Master’s programs - MiM, MiF, MSc, MSBA and more. - Application Fee Waivers - Free 1-Week of GMAT Club Tests: - Master's Application Toolkit - Grand Prize Giveaway- Sep 09
10:00 AM PDT
-11:00 AM PDT
Three MBA applications. Three rejections. No interview invites. A year later, Aman was admitted to Cambridge Judge, SMU, and IMD. In this episode of MBA Admit Stories, Aman shares how he rebuilt his MBA application... 
Sep 0410:00 AM EDT
-11:59 PM EDT
We’ve worked incredibly hard to build TTP into the best test prep experience possible, and it would mean a lot to us to win Newsweek’s 2026 Readers’ Choice Award for Best Test Prep. If TTP has helped you, we’d be incredibly grateful for your vote.
10 Nov 2022, 09:16
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
56% (02:07) correct
44%
(02:04)
wrong
based on 84
sessions
History
Date
Time
Result
Not Attempted Yet
Joe sells twice as many $10 tickets as Sue and Sue sells three times as many $5 tickets as Joe. How many tickets did Joe sell? (tickets only come in $10 or $5)
(1) Sue sold a total of 35 tickets.
(2) Together Joe and Sue sold 70 tickets for $500.
(1) Sue sold a total of 35 tickets.
(2) Together Joe and Sue sold 70 tickets for $500.
12 Nov 2022, 11:06
Kudos
Bookmarks
Given that Joe sells twice as many $10 tickets as Sue and Sue sells three times as many $5 tickets as Joe and we need to find How many tickets did Joe sell? (tickets only come in $10 or $5)
Let $10 tickets sold by Sue be x
=> $10 tickets sold by Joe = twice as many = 2x
Let $5 tickets sold by Joe = y
=> $5 tickets sold by Sue = three times = 3y
Tickets sold by Joe = 2x + y (we need to find this!)
STAT 1: Sue sold a total of 35 tickets.
=> x + 3y = 35
We can't find the value of x + 2y from this
=> NOT SUFFICIENT
STAT 2: Together Joe and Sue sold 70 tickets for $500.
Total tickets sold = 70
=> 2x + y + x + 3y = 70
=> 3x + 4y = 70 ...(1)
Tickets sold worth $500
=> $5 tickets price + $10 tickets price = $500
=> $5*(y + 3y) + $10*(x+2x) = $500
=> 4y + 6x = 100
=> 2y + 3x = 50
=> 3x + 2y = 50..(2)
We have two equation and two variables to solve for which is SUFFICIENT to answer.
We can mark B as answer and move on.
I am solving to complete the problem
Adding (1) and (2) we get
6x + 6y = 120
=> x + y = 20
=> 2x + 2y = 40 ...(3)
(1) - (3) we get
x + 2y = 30
So, Answer will be B
Hope it helps!
Let $10 tickets sold by Sue be x
=> $10 tickets sold by Joe = twice as many = 2x
Let $5 tickets sold by Joe = y
=> $5 tickets sold by Sue = three times = 3y
Tickets sold by Joe = 2x + y (we need to find this!)
STAT 1: Sue sold a total of 35 tickets.
=> x + 3y = 35
We can't find the value of x + 2y from this
=> NOT SUFFICIENT
STAT 2: Together Joe and Sue sold 70 tickets for $500.
Total tickets sold = 70
=> 2x + y + x + 3y = 70
=> 3x + 4y = 70 ...(1)
Tickets sold worth $500
=> $5 tickets price + $10 tickets price = $500
=> $5*(y + 3y) + $10*(x+2x) = $500
=> 4y + 6x = 100
=> 2y + 3x = 50
=> 3x + 2y = 50..(2)
We have two equation and two variables to solve for which is SUFFICIENT to answer.
We can mark B as answer and move on.
I am solving to complete the problem
Adding (1) and (2) we get
6x + 6y = 120
=> x + y = 20
=> 2x + 2y = 40 ...(3)
(1) - (3) we get
x + 2y = 30
So, Answer will be B
Hope it helps!
Moderators:
