GRE Prep 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 28Live Online Class with Top GRE Expert Mike Butville
08:00 PM EDT
-10:00 PM EDT
Top scores are possible when you enroll in a powerful GRE course, taught live online + 6 months access to TTP GRE OnDemand video courses included! Class starts Mon/Wed Sept. 28, 2026 - Dec. 9, 2026, 8:00pm-10:00pm EST
$1,600 worth of $20 bills stacked up that reach 0.35 inches high OR $1
[#permalink]
31 Jan 2023, 07:36
31 Jan 2023, 07:36
Expert Reply
00:00
Question Stats:
89% (02:40) correct
10% (01:39) wrong based on 19 sessions
Hide Show timer Statistics
$1,600 worth of $20 bills stacked up that reach 0.35 inches high OR $1,050 worth of $10 bills are also stacked up (assume all denominations are the same thickness)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The percent by which the height of the stack of $10 bills is greater than that of the stack of $20 bills |
33.5% |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: $1,600 worth of $20 bills stacked up that reach 0.35 inches high OR $1
[#permalink]
18 Feb 2023, 06:39
18 Feb 2023, 06:39
Expert Reply
OE
Because all bills have the same height, you can compare the number of bills in each stack directly to determine the percent increase in height. The number of $20 bills in a stack with a value of $1,600 is: 1600/20 = 80 The number of $10 bills in a stack with a value of $1,050 is 1,050/10 = 105
Plug these values into the percent change formula to evaluate Quantity A:
A=5/16
5/16<33.5
B is the answer
Because all bills have the same height, you can compare the number of bills in each stack directly to determine the percent increase in height. The number of $20 bills in a stack with a value of $1,600 is: 1600/20 = 80 The number of $10 bills in a stack with a value of $1,050 is 1,050/10 = 105
Plug these values into the percent change formula to evaluate Quantity A:
A=5/16
5/16<33.5
B is the answer
gmatclubot
Re: $1,600 worth of $20 bills stacked up that reach 0.35 inches high OR $1 [#permalink]
18 Feb 2023, 06:39
Moderators:
