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pappueshwar: 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

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Source: Beat The GMAT — Problem Solving | Thu Mar 01, 2012 8:13 am

krusta80: 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

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Abhishek009: 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

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killer1387: 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

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GMATGuruNY: 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.

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