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:

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 }

2X + 3(140) / 5 = 124 --> X = 100

ANSWER: B

I don't quite understand how the average being below the median allows us to determine that the lower two pieces have to be of the same length? Can you provide some clarity around this?

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,

5 x 124= 620

piece.1 -- piece.2 -- 140 -- piece.4 -- piece.5

since we are looking for a max possible length of the smallest piece we need to take least possible values for the larger pieces and since 140 is the median, the smallest possible value of each of the larger pieces is 140.

140 x 3 =420, so we have 200 left for the 2 smaller pieces.

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 }

2X + 3(140) / 5 = 124 --> X = 100

ANSWER: B

I don't quite understand how the average being below the median allows us to determine that the lower two pieces have to be of the same length? Can you provide some clarity around this?

Because increasing length of one piece means decreasing legnth of the second piece, thus automaticly the length of the shortest piece decreases.

100 + 100 = 200---------> smallest is 100

50 + 150= 200----------> smallest is 50

99 + 101= 200----------> smallest is 99

we need biggest length of the shortest piece, it is 100.

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

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

A) 90 B) 100 C) 110 D) 130 E) 140

ok. i will take a shot on it/try to explain.

(a+b+c+d+e)/5=124 that's is the average arithmetic mean. then the sum of their length is 124*5=620

then median let's say C=140 then let's assume a is the shortest of all 5 in order for a to be max short wood c=d=e=140 and a=b so 140*3=420

what is left for a and be is 620-420=200 so the max it can be is 100 as a=b=100

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

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

Need help. Thanks

Five pieces of wood in order of increasing length: A,B,C,D,E A+B+C+D+E = 124*5 = 620 C is the median and equals to 140 Because Maximum (A+B) occurs when D and E at their minimum; and D and E cannot be lower than 140, so min. D and E = 140 So max. A+B = 620 - 3*140 = 200 A+B = 200 and B > A So, max A = 99 when B = 101 if B=A is acceptable, max. A = 100

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

in order to have the maximum possible height for the shortest wood is to have tallest 3 woods the same as the median..and the shortest 2 woods to have the same height..

avg=sum/n 124*5 =620

620 is the total sum of the heights..now lets the tallest 3 woods=140 each..sum of 3 tallest =3*140 =420.

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

Need help. Thanks

For length to be maximum, two largest pieces other than median has to be 140 which means

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

A) 90 B) 100 C) 110 D) 130 E) 140

Because 140 is the mean, we know the 3rd (piece in the middle) is going to be 140. It might be easier to number the pieces to keep them straight. #1, #2, #3, #4, #5.

#3 = 140 or it wouldn't be the median.

The average is 124 so we know that 124 * 5 pieces = 620 for total length of all boards. If the median piece is 140, then 620 - 140 is the total for the 4 other pieces, or 480.

I decided to see if 90 would work. #1 = 90 #2 = ? #3 = 140 #4 = ? >= 140 #5 = ? >= 140

We can see if #5 is also 140, then 3 * 140 = 420 + 90 for #1 = 510, so we'd have 110 for #2 also. It would go (90, 110, 140, 140, 140). We have to remember that we're looking for the maxium length for the smallest number. I see that if we have 90 & 110 for the 2 smallest numbers, we could also have 100, 100 and it wouldn't change the median or the mean. If we were to go any larger than 100, 100, we would have to increase both, but if we take any from the #3, #4, or #5, that will change the median, even if the average stays the same. It looks like 100 (Answer B) is the maxium length of the shortest piece of the 5. _________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Actually, you need to maximise only the shortest piece. But, if you maximise, say, x1, then x2 will become the shortest piece and vice versa. The only way to maximise one and it still remains the shortest piece, is when x1=x2. _________________

average of 5 pieces = 124 => sum of lengths = 124*5=620 median = 140 so we have a,b,140,c,d we have to find "maximum possible" length for shortest piece i.e a. for this c,d should have minimum possible lengths i.e 140 so we have a,b,140,140,140

so a+b=200 (given that all sum up to 620) plug in the answer choices, the "max possible" length of a can be 100.

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...