boxes

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 234
Joined: Fri Oct 01, 2010 7:28 pm
Location: chennai
Thanked: 5 times
Followed by:4 members

boxes

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
Source: — Problem Solving |

User avatar
Master | Next Rank: 500 Posts
Posts: 143
Joined: Mon Mar 14, 2011 3:13 am
Thanked: 34 times
Followed by:5 members

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

User avatar
Master | Next Rank: 500 Posts
Posts: 359
Joined: Wed Mar 11, 2009 4:37 am
Location: Kolkata, India
Thanked: 50 times
Followed by:2 members

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

Legendary Member
Posts: 581
Joined: Sun Apr 03, 2011 7:53 am
Thanked: 52 times
Followed by:5 members

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

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
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