Seven pieces of rope have an average (arithmetic mean) lengt : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 22 Feb 2017, 14:13

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 02 Dec 2012
Posts: 178
Followers: 5

Kudos [?]: 2430 [2] , given: 0

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

### Show Tags

20 Dec 2012, 06:46
2
KUDOS
83
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

68% (03:35) correct 32% (02:58) wrong based on 1526 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?

(A) 82
(B) 118
(C) 120
(D) 134
(E) 152
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96332 [29] , given: 10737

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

### Show Tags

20 Dec 2012, 06:56
29
KUDOS
Expert's post
46
This post was
BOOKMARKED
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 $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$.

_________________
Intern
Joined: 16 Apr 2009
Posts: 16
Followers: 0

Kudos [?]: 12 [0], given: 5

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

### Show Tags

20 Dec 2012, 09:59
Bunuel 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 $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$.

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.
Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96332 [0], given: 10737

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

### Show Tags

21 Dec 2012, 02:39
Drik wrote:
Bunuel 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 $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$.

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.
_________________
Senior Manager
Status: Prevent and prepare. Not repent and repair!!
Joined: 13 Feb 2010
Posts: 274
Location: India
Concentration: Technology, General Management
GPA: 3.75
WE: Sales (Telecommunications)
Followers: 9

Kudos [?]: 90 [0], given: 282

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

### Show Tags

23 Dec 2012, 03:26
Again,We can plugin here. Algebraic method is good too.
_________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan
Kudos drives a person to better himself every single time. So Pls give it generously
Wont give up till i hit a 700+

Intern
Joined: 25 Feb 2013
Posts: 8
Followers: 0

Kudos [?]: 0 [0], given: 3

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

### Show Tags

11 Apr 2013, 16:24
Bunuel 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 $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$.

to minimize y we took 84, we cud have valued all the rope lengths to our minimum=a.
Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96332 [1] , given: 10737

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

### Show Tags

12 Apr 2013, 01:51
1
KUDOS
Expert's post
13
This post was
BOOKMARKED
sunshinewhole wrote:
Bunuel 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 $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$.

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 ($$a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}$$).

Similar questions to practice:
gmat-diagnostic-test-question-79347.html
seven-pieces-of-rope-have-an-average-arithmetic-mean-lengt-144452.html
a-set-of-25-different-integers-has-a-median-of-50-and-a-129345.html
the-median-of-the-list-of-positive-integers-above-is-129639.html
in-a-certain-set-of-five-numbers-the-median-is-128514.html
given-distinct-positive-integers-1-11-3-x-2-and-9-whic-109801.html
set-s-contains-seven-distinct-integers-the-median-of-set-s-101331.html
three-boxes-have-an-average-weight-of-7kg-and-a-median-weigh-99642.html
three-straight-metal-rods-have-an-average-arithmetic-mean-148507.html

Hope it helps.
_________________
Intern
Joined: 11 Jan 2013
Posts: 16
Location: United States
Followers: 0

Kudos [?]: 31 [1] , given: 12

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

### Show Tags

07 Jul 2013, 08:24
1
KUDOS
We can save a little bit of time by cancelling out the "7" before multiplying the mean by it:

(a+a+a+84+84+84+(14+4a))/7=68

=> (7a + 266)/7=68
=> a + 38 = 68
=> a=30
=> g=4(30)+14=134
Intern
Joined: 16 Aug 2013
Posts: 2
Concentration: Marketing, Strategy
Followers: 0

Kudos [?]: 19 [12] , given: 3

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

### Show Tags

08 Sep 2013, 11:28
12
KUDOS
6
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
Intern
Joined: 09 Sep 2013
Posts: 19
Followers: 1

Kudos [?]: 1 [0], given: 7

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

### Show Tags

20 Oct 2013, 18:56
Extremely tough question to do in 2 min IMO.
Senior Manager
Joined: 07 Apr 2012
Posts: 464
Followers: 2

Kudos [?]: 54 [0], given: 58

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

### Show Tags

21 Oct 2013, 13:12
Bunuel wrote:
Because the question asks to find the maximum possible length of the longest piece of rope, which we denoted as g.

But from how I see it, since the longest piece is 4a+14, to maximize it, we need to maximize a. no?
Manager
Joined: 12 Jan 2013
Posts: 244
Followers: 4

Kudos [?]: 70 [6] , given: 47

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

### Show Tags

13 Jan 2014, 01:56
6
KUDOS
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.
Intern
Joined: 03 Jan 2014
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 21

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

### Show Tags

27 Apr 2014, 02:44
pls help me here! Why cant we just take all the values as 84 itself.
Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96332 [0], given: 10737

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

### Show Tags

28 Apr 2014, 00:15
krish1chaitu wrote:
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?

Complete solution is here: seven-pieces-of-rope-have-an-average-arithmetic-mean-lengt-144452.html#p1159013

Also, check similar questions here: seven-pieces-of-rope-have-an-average-arithmetic-mean-lengt-144452.html#p1211023

Hope this helps.
_________________
Intern
Joined: 22 Feb 2014
Posts: 30
Followers: 0

Kudos [?]: 10 [0], given: 14

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

### Show Tags

01 May 2014, 14:30
Bunuel 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 $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$.

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...
Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96332 [0], given: 10737

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

### Show Tags

02 May 2014, 01:06
drkomal2000 wrote:
Bunuel 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 $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$.

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.

Does this make sense?
_________________
Senior Manager
Joined: 07 Apr 2012
Posts: 464
Followers: 2

Kudos [?]: 54 [0], given: 58

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

### Show Tags

14 Jun 2014, 01:23
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 $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$.

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?
Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96332 [1] , given: 10737

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

### Show Tags

14 Jun 2014, 01:31
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
ronr34 wrote:
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 $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$.

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.
_________________
Director
Joined: 10 Mar 2013
Posts: 608
Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
Followers: 15

Kudos [?]: 288 [5] , given: 200

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

### Show Tags

05 Jul 2014, 02:38
5
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.

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660

Intern
Joined: 12 Aug 2014
Posts: 19
Followers: 0

Kudos [?]: 0 [0], given: 72

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

### Show Tags

06 Jan 2015, 06:19
Bunuel 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 $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$.

Im unable to understand that how minimizing all other terms would maximize value of g, can you please explain?
Re: Seven pieces of rope have an average (arithmetic mean) lengt   [#permalink] 06 Jan 2015, 06:19

Go to page    1   2   3    Next  [ 44 posts ]

Similar topics Replies Last post
Similar
Topics:
3 Seven pieces of wood have an average length of 100 cm and a median len 4 21 Dec 2015, 10:48
2 The average (arithmetic mean) of seven numbers is 12.2 5 28 Apr 2015, 19:36
8 Three straight metal rods have an average (arithmetic mean) 9 27 Feb 2013, 21:23
Three sisters have an average (arithmetic mean) age of 25 ye 4 14 Jan 2013, 05:16
37 Three boxes of supplies have an average (arithmetic mean) 11 05 Dec 2010, 13:17
Display posts from previous: Sort by