Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Five pieces of wood have an average length of 124cm and a [#permalink]

Show Tags

25 Oct 2005, 00:08

5

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

68% (01:02) correct
32% (01:12) wrong based on 285 sessions

HideShow timer Statistics

Five pieces of wood have an average length of 124cm and a median length of 140cm. What is the maximum possible length, in cm, of the shortest piece of wood?

Five pieces of wood have an average (arithmetic mean) length of 124cm and a median length of 140 cm. What is the maximum possible length, in centimetres, of the smallest piece of wood?

think there is something wrong with this Q. If the median is 140cm and AVRG is 124 then the total lenght of the woods is 620cm.Since median is 140cm the values above the median should be > or = to the median.Their minimum value is 140x3=420cm.The max possible lenght of the shortest piece is 200-124=76cm
May be i am wrong

The average of 5 pieces is 124, this means the total length of all pieces=620

We now the median is 140, which gives us that 140 is the middle value of all the fives pieces... which means two pieces are below 140 and two pieces above 140....

To get the maximum length of the smallest piece of wood we must minimize the length of the two largest pieces....

Remember we have the total length of 620... and the middle value 140... all we need is to distribute 380 (i.e. 620-140) in such a way that two of the values would be below 140 and two at least 140... so 100+100+140+140+140=620, the maximum value of the smallest piece is 100, since if we chose 110, two other values would only yield 130 which cannot be the case...

Hope i wrote it in understandable way...

Last edited by SimaQ on 21 Feb 2006, 07:49, edited 1 time in total.

think there is something wrong with this Q. If the median is 140cm and AVRG is 124 then the total lenght of the woods is 620cm.Since median is 140cm the values above the median should be > or = to the median.Their minimum value is 140x3=420cm.The max possible lenght of the shortest piece is 200-124=76cm May be i am wrong

Everything is perfect with your reasoning... you got that 3 values must be at least 140*3=420cm so we need to distribute 200 among the rest pieces (i.e 2) which would equal 100...

The way I solved the question as follows.
Median is 140
Average is 124
Total length is 620
Now based on the available information 2 Lengths <= 140 <= remaining two lengths.

Sum of these 4 lengths has to be 620 - 140 = 480
To maximize the shortest length, I considered that two short lengths are equal is size = x and remaining two lengths equal in size = 140 so now the addition of two short lengths is
2X = 480 - (140 + 140)
2X = 200
X = 100

Five pieces of wood have an average ( arithmetic mean ) length of 124 centimeters and a median length of 140 centimeters. What is the maximum possible length, in centimeters, of the shortest piece of wood?

5 pieces of wood have an average length of 124 cm and a median lenght of 140 cm. What is the maximum possible length, in cm, of the shortest piece of wood?

5 pieces of wood have an average length of 124 cm and a median lenght of 140 cm. What is the maximum possible length, in cm, of the shortest piece of wood?

a) 90 b) 100 c) 110 d) 130 e) 140

b. 100

= (124 x 5 - 140x3)/2 = 200/2 =100

yup. it's B. I forgot some pieces could be the same length

Why should the 2 smallest pieces of wood be 100cm each ?
Why couldn't we have the smallest piece at 71cm and the second smallest at 139cm ??
Since the sum of the 2 smallest should be 200cm.

We could, there are so many possibilities for the two shortest pieces. Remember always to focus on what the question is asking: THE MAXIMUM possible length of the shortest piece. Since the average is below the median, then the lower two pieces shall be of the same length to obtain maximum possible length.

On the other hand, the length of the larger two pieces should be minimzed to get larger smaller pieces and still maintaining the same average and median --> we have now { X , X , 140, 140, 140 }

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...