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:

Seven pieces of rope have an average (arithmetic mean) lengt [#permalink]

Show Tags

20 Dec 2012, 07:46

4

This post received KUDOS

117

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

71% (02:33) correct
29% (02:59) wrong based on 1608 sessions

HideShow timer Statistics

Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

(A) 82 (B) 118 (C) 120 (D) 134 (E) 152

Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g --> \(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\).

The average length = 68 centimeters --> the total length 7*68 centimeters. The median = 84 centimeters --> d=84. The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece --> \(g=4a+14\).

We need to maximize g. Now, to maximize g, we need to minimize all other terms.

The minimum value of b and c is a and the minimum value of e and f is median=d.

Thus we have that \(a+a+a+84+84+84+(4a+14)=7*68\) --> \(a=30\) --> \(g_{max}=4a+14=134\).

Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink]

Show Tags

20 Dec 2012, 10:59

Bunuel wrote:

Walkabout wrote:

Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

(A) 82 (B) 118 (C) 120 (D) 134 (E) 152

Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g --> \(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\).

The average length = 68 centimeters --> the total length 7*68 centimeters. The median = 84 centimeters --> d=84. The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece --> \(g=4a+14\).

We need to maximize g. Now, to maximize g, we need to minimize all other terms.

The minimum value of b and c is a and the minimum value of e and f is median=d.

Thus we have that \(a+a+a+84+84+84+(4a+14)=7*68\) --> \(a=30\) --> \(g_{max}=4a+14=134\).

Answer: D.

Thanks Bunnel.. Could you please elaborate more about WHY "We need to maximize g. Now, to maximize g, we need to minimize all other terms." I just want to understand how does it work.

Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

(A) 82 (B) 118 (C) 120 (D) 134 (E) 152

Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g --> \(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\).

The average length = 68 centimeters --> the total length 7*68 centimeters. The median = 84 centimeters --> d=84. The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece --> \(g=4a+14\).

We need to maximize g. Now, to maximize g, we need to minimize all other terms.

The minimum value of b and c is a and the minimum value of e and f is median=d.

Thus we have that \(a+a+a+84+84+84+(4a+14)=7*68\) --> \(a=30\) --> \(g_{max}=4a+14=134\).

Answer: D.

Thanks Bunnel.. Could you please elaborate more about WHY "We need to maximize g. Now, to maximize g, we need to minimize all other terms." I just want to understand how does it work.

Because the question asks to find the maximum possible length of the longest piece of rope, which we denoted as g.
_________________

Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink]

Show Tags

11 Apr 2013, 17:24

Bunuel wrote:

Walkabout wrote:

Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

(A) 82 (B) 118 (C) 120 (D) 134 (E) 152

Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g --> \(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\).

The average length = 68 centimeters --> the total length 7*68 centimeters. The median = 84 centimeters --> d=84. The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece --> \(g=4a+14\).

We need to maximize g. Now, to maximize g, we need to minimize all other terms.

The minimum value of b and c is a and the minimum value of e and f is median=d.

Thus we have that \(a+a+a+84+84+84+(4a+14)=7*68\) --> \(a=30\) --> \(g_{max}=4a+14=134\).

Answer: D.

to minimize y we took 84, we cud have valued all the rope lengths to our minimum=a.

Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

(A) 82 (B) 118 (C) 120 (D) 134 (E) 152

Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g --> \(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\).

The average length = 68 centimeters --> the total length 7*68 centimeters. The median = 84 centimeters --> d=84. The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece --> \(g=4a+14\).

We need to maximize g. Now, to maximize g, we need to minimize all other terms.

The minimum value of b and c is a and the minimum value of e and f is median=d.

Thus we have that \(a+a+a+84+84+84+(4a+14)=7*68\) --> \(a=30\) --> \(g_{max}=4a+14=134\).

Answer: D.

to minimize y we took 84, we cud have valued all the rope lengths to our minimum=a.

Because e, f, and g cannot be less than the median, which is d=84 (\(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\)).

Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink]

Show Tags

08 Sep 2013, 12:28

17

This post received KUDOS

9

This post was BOOKMARKED

I didn't know how to go about solving this sum. So, I took help of the answer choices. The important cue given in the question is that the the largest number = 4*Smallest Number+ 14. So if you subtract 14 from the answer choices, the number obtained should be divisible by 14. A) 82 - Can be cancelled off as it is less than the median B) 118 - 118-14=104 C)120 - 120-14 = 106 D)134 - 134-14 = 120 E)152 - 152 - 14 = 138 Using the divisibility test for 4, you can strike off the answer choices C and E.

Now you're left with B and D. Since the question asks for the largest possible number, it's best to try solving using choice D first. If 134 is the largest number then 4*30 + 14 gives us the smallest possible value of 30. The median is 84. To maximise the value of the largest number all other values lower than the median should be taken as 30 and higher than it as 84. So we have 30*3 + 84*3 + 134 which is equal to the sum of 68*7. Hence the answer is D.

Whenever you're stuck at a PS question the next best approach is POE

Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink]

Show Tags

13 Jan 2014, 02:56

6

This post received KUDOS

1

This post was BOOKMARKED

Walkabout wrote:

Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

(A) 82 (B) 118 (C) 120 (D) 134 (E) 152

They're telling us that:\(L = 4S + 14\), and they want us to maximize L, thus we plug in options to solve for S such that we maximize S (this maximizes L). So our answer is the option that gives us: \(S* = \frac{(L -14)}{4}\), where * stands for the maximum value of S given the options

The answer needs to be divisible by four after subtraction with 14. The only of the options that are capable of this are B and D (104/4 = 26 and 120/4 = 30). A is obviously not a viable option since the middle value is 84.

Of course, they're asking for "LONGEST possible value" so naturally you take the option that has the highest value of the two: So our answer is D.

pls help me here! Why cant we just take all the values as 84 itself.

We are told that the average length is 64, a median length is 84 centimeters and the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope. How can all the pieces be 84 centimeters from this?

Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink]

Show Tags

01 May 2014, 15:30

1

This post received KUDOS

1

This post was BOOKMARKED

Bunuel wrote:

Walkabout wrote:

Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

(A) 82 (B) 118 (C) 120 (D) 134 (E) 152

Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g --> \(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\).

The average length = 68 centimeters --> the total length 7*68 centimeters. The median = 84 centimeters --> d=84. The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece --> \(g=4a+14\).

We need to maximize g. Now, to maximize g, we need to minimize all other terms.

The minimum value of b and c is a and the minimum value of e and f is median=d.

Thus we have that \(a+a+a+84+84+84+(4a+14)=7*68\) --> \(a=30\) --> \(g_{max}=4a+14=134\).

Answer: D.

I tried to solve this question by this approach but I got wrong answer choice..Can you please explain why I cannot use this approach? why I am wrong here..

Mean = 1st and last term/2 (smallest x + largest y/2) Given Mean = x+y/2 = 68 x+y = 136 now replace y= 4x+14 5x+14= 136 x= 24.5 which gives Ans choice E..... PLS explain...

Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

(A) 82 (B) 118 (C) 120 (D) 134 (E) 152

Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g --> \(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\).

The average length = 68 centimeters --> the total length 7*68 centimeters. The median = 84 centimeters --> d=84. The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece --> \(g=4a+14\).

We need to maximize g. Now, to maximize g, we need to minimize all other terms.

The minimum value of b and c is a and the minimum value of e and f is median=d.

Thus we have that \(a+a+a+84+84+84+(4a+14)=7*68\) --> \(a=30\) --> \(g_{max}=4a+14=134\).

Answer: D.

I tried to solve this question by this approach but I got wrong answer choice..Can you please explain why I cannot use this approach? why I am wrong here..

Mean = 1st and last term/2 (smallest x + largest y/2) Given Mean = x+y/2 = 68 x+y = 136 now replace y= 4x+14 5x+14= 136 x= 24.5 which gives Ans choice E..... PLS explain...

The lengths of the pieces of the rope does not form an evenly spaced set to use (mean)=(first+last)/2.

Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink]

Show Tags

14 Jun 2014, 02:23

sunshinewhole wrote:

Bunuel wrote:

Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g --> \(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\).

The average length = 68 centimeters --> the total length 7*68 centimeters. The median = 84 centimeters --> d=84. The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece --> \(g=4a+14\).

We need to maximize g. Now, to maximize g, we need to minimize all other terms.

The minimum value of b and c is a and the minimum value of e and f is median=d.

Thus we have that \(a+a+a+84+84+84+(4a+14)=7*68\) --> \(a=30\) --> \(g_{max}=4a+14=134\).

Answer: D.

to minimize y we took 84, we cud have valued all the rope lengths to our minimum=a.

It's a little confusing.... Since we have the equation \(g=4a+14\), to maximize g, we also need to maximize a... isn't this the case?

Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g --> \(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\).

The average length = 68 centimeters --> the total length 7*68 centimeters. The median = 84 centimeters --> d=84. The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece --> \(g=4a+14\).

We need to maximize g. Now, to maximize g, we need to minimize all other terms.

The minimum value of b and c is a and the minimum value of e and f is median=d.

Thus we have that \(a+a+a+84+84+84+(4a+14)=7*68\) --> \(a=30\) --> \(g_{max}=4a+14=134\).

Answer: D.

to minimize y we took 84, we cud have valued all the rope lengths to our minimum=a.

It's a little confusing.... Since we have the equation \(g=4a+14\), to maximize g, we also need to maximize a... isn't this the case?

No. Because we have fixed total length of the rope: 7*68 centimeters. If you increase a, you'd be decreasing g.
_________________

Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink]

Show Tags

05 Jul 2014, 03:38

7

This post received KUDOS

1

This post was BOOKMARKED

For the solution we don't even need an arithmetic mean. The median has a property that (first number + last number)/2 = median (in case of odd number of values) --> we have smallest number=x and largest number = 4x + 14

(4x+14+x)/2=84 --> x ≈ 30 --> 4x+14=134 (D)
_________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

Share some Kudos, if my posts help you. Thank you !

Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink]

Show Tags

06 Jan 2015, 07:19

Bunuel wrote:

Walkabout wrote:

Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

(A) 82 (B) 118 (C) 120 (D) 134 (E) 152

Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g --> \(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\).

The average length = 68 centimeters --> the total length 7*68 centimeters. The median = 84 centimeters --> d=84. The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece --> \(g=4a+14\).

We need to maximize g. Now, to maximize g, we need to minimize all other terms.

The minimum value of b and c is a and the minimum value of e and f is median=d.

Thus we have that \(a+a+a+84+84+84+(4a+14)=7*68\) --> \(a=30\) --> \(g_{max}=4a+14=134\).

Answer: D.

Im unable to understand that how minimizing all other terms would maximize value of g, can you please explain?

Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

(A) 82 (B) 118 (C) 120 (D) 134 (E) 152

Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g --> \(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\).

The average length = 68 centimeters --> the total length 7*68 centimeters. The median = 84 centimeters --> d=84. The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece --> \(g=4a+14\).

We need to maximize g. Now, to maximize g, we need to minimize all other terms.

The minimum value of b and c is a and the minimum value of e and f is median=d.

Thus we have that \(a+a+a+84+84+84+(4a+14)=7*68\) --> \(a=30\) --> \(g_{max}=4a+14=134\).

Answer: D.

Im unable to understand that how minimizing all other terms would maximize value of g, can you please explain?

Answer this: the sum of two positive integers is 10. What is the maximum possible value of the largest of the integers?
_________________

Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink]

Show Tags

06 Jan 2015, 08:13

Im unable to understand that how minimizing all other terms would maximize value of g, can you please explain?[/quote]

Answer this: the sum of two positive integers is 10. What is the maximum possible value of the largest of the integers?[/quote]

I got your point, i don't know how i was thinking before, but i just got it . We must minimize all other values in order to get maximum value for g. Thanks

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

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...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...