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

boxes

Expert replies
Post new topic Post Reply
Subscribe topic
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 234
Joined: Fri Oct 01, 2010 7:28 pmFollowed by: 4 members
Location: chennai
Thanked: 5 times
by pappueshwar » Thu Mar 01, 2012 8:13 am
Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight, in kilograms, of the lightest box?


A. 3kg B. 4kg C. 5kg D. 6kg E. 7kg
Top
Join the discussion
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 143
Joined: Mon Mar 14, 2011 3:13 amFollowed by: 5 members
Thanked: 34 times

by krusta80 » Thu Mar 01, 2012 8:22 am
pappueshwar wrote:Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight, in kilograms, of the lightest box?


A. 3kg B. 4kg C. 5kg D. 6kg E. 7kg
Since we have an odd number of boxes, we know that the median value has to be equal to the weight of the second box, when placed in non-descending weight order: 9kg.

The total weight of the three boxes equals 3*(mean) = 21kg

This leaves us 21-9 = 12 kgs of "slack" for the remaining two boxes.

In order to get the heaviest possible light box, we need to find the lightest possible heavy box, which in this case is 9kg (same as the median box).

12-3 leaves 3kg for the heavies possible light box. A
Top
Join the discussion
User avatar
Master | Next Rank: 500 Posts
Posts: 359
Joined: Wed Mar 11, 2009 4:37 amFollowed by: 2 members
Location: Kolkata, India
Thanked: 50 times

by Abhishek009 » Thu Mar 01, 2012 9:27 am
pappueshwar wrote:Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight, in kilograms, of the lightest box?


A. 3kg B. 4kg C. 5kg D. 6kg E. 7kg
Let the weight of the 3 boxes be -

a , b and c

Now , a+b+c = 7*3

Or, a + b + c = 21

Further it is given median weight is 9 kgs

Hence -

b = 9

Now it's given to find the maximum possible weight, in kilograms, of the lightest box -

a + b + c = 21

a <= b <= c

b = 9

From a + 9 + c = 21 we can say a + c = 12

Now ,to find the maximum possible weight, of the lightest box let's consider weight of box c = weight of box b

So , a + 18 = 21

Hence a = 3...
Abhishek
Top
Join the discussion
Legendary Member
Posts: 581
Joined: Sun Apr 03, 2011 7:53 amFollowed by: 5 members
Thanked: 52 times

by killer1387 » Thu Mar 01, 2012 11:23 pm
median is 9
hence
x,9,y
y>=9
for x maximum,
y=9
i.e. x=(7*3)-(9+y)=3
hence A
Top
Join the discussion
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 Mar 02, 2012 5:47 am
pappueshwar wrote:Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight, in kilograms, of the lightest box?


A. 3kg B. 4kg C. 5kg D. 6kg E. 7kg
The solutions above are great.
Just want to make sure the approach is clear.
This is a MAX/MIN problem.
In order to MAXIMIZE one value -- the weight of the lightest box -- we must MINIMIZE everything else.

Let L = the lightest box and H = the heaviest box.
The 3 boxes are L, 9, H.
The total weight = number*average = 3*7 = 21.
Thus, L+H = 21-9 = 12.
In order to MAXIMIZE L, we must MINIMIZE H.
The smallest possible value of H is 9.
Thus, the MAXIMUM possible value of L = 12-9 = 3.

The correct answer is A.
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