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

Redeem

Woody

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 90
Joined: Mon Mar 02, 2009 6:06 am
Thanked: 3 times

Woody

by Baldini » Tue Mar 10, 2009 7:13 am
Five pieces of wood have an average (arithmetic mean) length of 124 cm and a median of 140 cm. what is the maximum possible length, in cm, of the shortest piece of wood?

1. 90
2. 100
3. 110
4. 130
5. 140

OA is 100

How does one attain the answer?
Thx
GMAX
Top
Source: — Problem Solving |
Junior | Next Rank: 30 Posts
Posts: 10
Joined: Tue Mar 10, 2009 2:24 am
Thanked: 2 times

by zephir21 » Tue Mar 10, 2009 7:49 am
Answer 2 : 100

a<b<c<d<e the 5 lengths
Asume c=d=e=140

(140-124)*3 = 48

To get the mean equal to 124 you need max(a)=b=124-48/2=100

...
Top
Senior | Next Rank: 100 Posts
Posts: 90
Joined: Mon Mar 02, 2009 6:06 am
Thanked: 3 times

by Baldini » Wed Mar 11, 2009 1:43 am
Thanks for the response. But why does A=B?

When I tried to work this question out, I did A + 140 + 140 + 140 + E = 620 therefore A + E = 200. But I couldn't put a limit to A or E...

Thanks in advance
GMAX
Top
Junior | Next Rank: 30 Posts
Posts: 10
Joined: Tue Mar 10, 2009 2:24 am
Thanked: 2 times

by zephir21 » Wed Mar 11, 2009 2:35 am
Other explanation :
In order to maximize A you need to have the smallest standard deviation for your list of numbers. So you get the mean equal to :
Mean = [max(A+B)+min(C+D+E)]/5
min(C+D+E)= 140+140+140 'cause E>=D>=C=140 (so you choose E=D=C=140)

A=B 'cause you want to maximize A (as B>=A, to maximize A you need A=B)
So max(A+B)=2A

Thus (2A+140*3)/5=124
Do the math and you get A=100

:-) I don't know if it's clear, that's how it work in my head lol
Top
Post new topic Post Reply
• Page 1 of 1