BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

ps

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Senior | Next Rank: 100 Posts
Posts: 48
Joined: Sun Jul 17, 2011 12:04 am
by shikh » Fri Dec 02, 2011 4:53 am
After installing a powerful pump onto a large gasoline tank, John pumped gasoline out of the tank. Each time John ran the pump, it removed 1/3 of the gas remaining in the tank. Assuming the gas tank was full when John began, what fraction of the total gas in the tank at the beginning was removed if John ran the pump three times?
A) 8/27
B) 1/27
C) 2/3
D) 19/27
E) 27/28
OAD
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

by Anurag@Gurome » Fri Dec 02, 2011 5:02 am
shikh wrote:After installing a powerful pump onto a large gasoline tank, John pumped gasoline out of the tank. Each time John ran the pump, it removed 1/3 of the gas remaining in the tank. Assuming the gas tank was full when John began, what fraction of the total gas in the tank at the beginning was removed if John ran the pump three times?
A) 8/27
B) 1/27
C) 2/3
D) 19/27
E) 27/28
OAD
Let the tank has x units
No. of units removed by 1st pump = (1/3) * x = x/3
Remaining units = x - x/3 = 2x/3
No. of units removed by 2nd pump = (1/3) * (2x/3) = 2x/9 units
Remaining units = 2x/3 - 2x/9 = 4x/9
No. of units removed by 3rd pump = (1/3) * (4x/9) = 4x/27 units
Remaining units = 4x/9 - 4x/27 = 8x/27
Since only 8x/27 units are left, total no. of units removed = x - 8x/27 = 19x/27
Therefore, required fraction = 19/27

The correct answer is D.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Fri Dec 02, 2011 6:57 am
shikh wrote:After installing a powerful pump onto a large gasoline tank, John pumped gasoline out of the tank. Each time John ran the pump, it removed 1/3 of the gas remaining in the tank. Assuming the gas tank was full when John began, what fraction of the total gas in the tank at the beginning was removed if John ran the pump three times?
A) 8/27
B) 1/27
C) 2/3
D) 19/27
E) 27/28
OAD
Let gas = 27.
Amount removed by the first pump = (1/3)27 = 9.
Amount remaining = 27-9 = 18.
Amount removed by the second pump = (1/3)18 = 6.
Amount remaining = 18-6 = 12.
Amount removed by the third pump = (1/3)12 = 4.

Total removed/total gas = (9+6+4)/27 = 19/27.

The correct answer is D.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 349
Joined: Wed Sep 28, 2011 3:38 pmFollowed by: 54 members
Location: Austin, TX
Thanked: 236 times
GMAT Score:770

by GmatMathPro » Fri Dec 02, 2011 8:02 am
Each time he runs the pump, the amount that remains is 2/3 of the previous amount. Thus, if he runs it x times, the fraction that remains is (2/3)^x, and the fraction that was pumped out is 1-(2/3)^x. He runs it three times, so 1-(2/3)^3 = 1-8/27 = 19/27

Ans: D
Pete Ackley
GMAT Math Pro
Free Online Tutoring Trial
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1