Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 563 reviewsTarget Test Prep

the 8 fruit boxes

Expert replies
Post new topic Post Reply
Subscribe topic
Previous Topic Next Topic
User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members
Location: India
Thanked: 267 times
GMAT Score:760
by sanju09 » Thu May 13, 2010 6:02 am
A box of apples costs $2, a box of pears costs $3, and a box of plums costs $4. If 8 fruit boxes cost together cost $23, what is the largest possible number of the 8 fruit boxes that contain plums?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
The mind is everything. What you think you become. –Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Top
Join the discussion
Source: — Problem Solving |

GMAT/MBA Expert

GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members
Location: Milpitas, CA
Thanked: 447 times

by Rahul@gurome » Thu May 13, 2010 6:30 am
Let there be x boxes of apples, y boxes of pears and z boxes of plums.
Then x + y + z = 8
2x + 3y + 4z = 23
Let us put the values of x , y and z in the above equation, 2x + 3y + 4z = 23,
such that z is maximum and x + y + z = 8.
If x = 4, y =1 and z = 3 then 2(4) + 3(1) + 4(3) = 23
So, the largest possible number of 8 fruit boxes that contain plums is 3.

The correct answer is (C).
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
Join the discussion
Legendary Member
Posts: 610
Joined: Fri Jan 15, 2010 12:33 amFollowed by: 2 members
Thanked: 47 times

by kstv » Thu May 13, 2010 7:15 am
what Rahul@gurome has shown is the std. solution and in 2 mins not many smart solutions will come forth
still
total amount is $ 23
there has to be a box of pears = $ 3 , there cannot be even nos. of pear box
Left with $ 20
Answer options
E) 5 boxes of plums = $ 20 tot 6 boxes and no boxes of apples we need 8
D) 4 boxes of plums = $16 tot. 7 boxes getting close
C) 3 boxes of plums = $ 12 , 4 boxes of apples , tot 8

It is assumed that there should be atleast one box of each fruit.
Top
Join the discussion
Legendary Member
Posts: 759
Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members
Thanked: 85 times

by clock60 » Thu May 13, 2010 12:30 pm
slightly detailed method of Rahul@gurome, let us go a little bit far
x+y+z=8
2x+3y+4z=23

multiply 1 st equation by 2 and substract

2x+2y+2z=16
2x+3y+4z=23

y+2z=7, and we need to max z, the largest value for z so y be an integer is 3, and y-1
1+2*3=7
so C
Top
Join the discussion