GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 25 May 2019, 00:13 ### 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
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. ### Request Expert Reply # Seven pieces of rope have an average (arithmetic mean) lengt

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager  Joined: 09 Jan 2013
Posts: 69
Concentration: Entrepreneurship, Sustainability
GMAT 1: 650 Q45 V34 GMAT 2: 740 Q51 V39 GRE 1: Q790 V650 GPA: 3.76
WE: Other (Pharmaceuticals and Biotech)
Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

BrainLab wrote:
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)

BrainLab, this is true only when you have an evenly spaced set such as 1,2,3,4,5,6,7 where mean = median = $$\frac{first No + last No}{2}$$

However, this is not the case with a unevenly spaced set such as 1, 2, 3, 10, 11, 12, 13
Notice that notice that in this case you would have different results
$$mean = \frac{Sum}{Total No's} = \frac{52}{7}$$
median = middle no = 10
$$\frac{first No + last No}{2} = \frac{1+13}{2}=7$$

you can even cross check with the soln that you have.
$$\frac{30+134}{2} = 82$$ and not 84.

Hope this brings clarity.
Intern  Joined: 21 Sep 2015
Posts: 2
Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

BrainLab wrote:
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)

As Bunuel pointed out in an earlier post "The lengths of the pieces of the rope does not form an evenly spaced set to use (mean)=(first+last)/2."
So IMO, your solution does not hold good.

Thanks.
Current Student Joined: 02 Jun 2015
Posts: 78
Location: Brazil
Concentration: Entrepreneurship, General Management
GPA: 3.3
Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

mulhinmjavid wrote:
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?

Because the mean is the same, and the median is the size of the middle rope (from shortest to longest)
Therefore, if you minimize from one site of the median , you are going to maximize from the other.

Actually in this problem, as you have only integers, you can deduce that the size of the shortest has to be also an integer. So you can start working with the answers.

I like to start to work with the number C.
C) 120-14 = 106 (I cannot divide by 4, so its not the answer)
D) 134-14 = 120 (Bingo! I can divide by 4, lets check the last answer)
E) 152-14 = 138 (I cannot divide by 4)

Therefore the answer is letter D.
Intern  Joined: 27 Feb 2015
Posts: 43
Concentration: General Management, Economics
GMAT 1: 630 Q42 V34 WE: Engineering (Transportation)
Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

option A : largest rope cannot be less than median ;so eliminate.
option C and option E : eliminated because of reasons below,
L= 14+4S
S=L-14/4
when you plugin L=118 or 152 you get a decimal value which will also make mean or median,which is clearly an integer as given, into decimal ; we are looking for an integer value of a measurement (cm)

remaining with option B and D :
both are ok but we need greatest value for rope ; so ans is 134 , i.e. option D , here we get smallest rope as 30 cms
check the ans by taking average of 3 ropes of 30cm(smallest one) , 3 ropes of 84cm(median) and 1 last rope of 134 cm(largest one); we get avg=68

incase if you are thinking why 3 rope of 30cms,1 rope of 84cm and 3 rope of 134cms cant be taken, think about the constraint -i.e. the avg which is 68 , if you take 3 ropes of 134cm , avg will be more than 68 cms.
thanks!
Target Test Prep Representative G
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2823
Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

1
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

We need to first recognize that we are working with a maximum problem. This means that of the seven pieces of rope, we must make 6 of those pieces as small as we possibly can, within the confines of the given information, and doing so will maximize the length of the 7th piece.

We are first given that seven pieces of rope have an average (arithmetic mean) length of 68 centimeters. From this we can determine the sum.

average = sum/quantity

sum = average x quantity

sum = 68 x 7 = 476

Next we are given that the median length of a piece of rope is 84 centimeters. Thus when we arrange the pieces of rope from least length to greatest, the middle length (the 4th piece) will have a length of 84 centimeters. We also must keep in mind that we can have pieces of rope of the same length. Let's first label our seven pieces of rope with variables or numbers, starting with the shortest piece and moving to the longest piece. We can let x equal the shortest piece of rope, and m equal the longest piece of rope.

piece 1: x

piece 2: x

piece 3: x

piece 4: 84

piece 5: 84

piece 6: 84

piece 7: m

Notice that the median (the 4th rope) is 84 cm long. Thus, pieces 5 and 6 are either equal to the median, or they are greater than the median. In keeping with our goal of minimizing the length of the first 6 pieces, we will assign 84 to pieces 5 and 6 to make them as short as possible. Similarly, we have assigned a length of x to pieces 1, 2, and 3.

We can plug these variables into our sum equation:

x + x + x + 84 + 84 + 84 + m = 476

3x + 252 + m = 476

3x + m = 224

We also given that the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope. So we can say:

m = 14 + 4x

We can now plug 14 + 4x in for m into the equation 3x + m = 224. So we have:

3x + 14 + 4x = 224

7x = 210

x = 30

Thus, the longest piece of rope is 4(30) + 14 = 134 centimeters.

Answer is D.
_________________

# Jeffrey Miller

Head of GMAT Instruction

Jeff@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Intern  Joined: 12 Apr 2014
Posts: 6
Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

the important thing to remember here is that it is a rope and not numbers otherwise...to obtain maximum rope length that is g.....we have to minimize all other lengths....a is smallest.....keep b and c equal to a keeping in mind inequality and e and f equal to median keeping in mind the equality...insert the values and use average formula to get the right answer
CEO  V
Joined: 12 Sep 2015
Posts: 3727
Location: Canada
Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

1
Top Contributor
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

So, we have 7 rope lengths.
If the median length is 84, then the lengths (arranged in ascending order) look like this: {_, _, _, 84, _, _, _}

The length of the longest piece of rope is 14 cm more than 4 times the length of the shortest piece of rope.
Let x = length of shortest piece.
This means that 4x+14 = length of longest piece.
So, we now have: {x, _, _, 84, _, _, 4x+14}

Our task is the maximize the length of the longest piece.
To do this, we need to minimize the other lengths.
So, we'll make the 2nd and 3rd lengths have length x as well (since x is the shortest possible length)
We get: {x, x, x, 84, _, _, 4x+14}

Since 84 is the middle-most length, the 2 remaining lengths must be greater than or equal to 84.
So, the shortest lengths there are 84.
So, we get: {x, x, x, 84, 84, 84, 4x+14}

Now what?

At this point, we can use the fact that the average length is 68 cm.
There's a nice rule (that applies to MANY statistics questions) that says:
the sum of n numbers = (n)(mean of the numbers)
So, if the mean of the 7 numbers is 68, then the sum of the 7 numbers = (7)(68) = 476

So, we now now that x+x+x+84+84+84+(4x+14) = 476
Simplify to get: 7x + 266 = 476
7x = 210
x=30

If x=30, then 4x+14 = 134
So, the longest piece will be 134 cm long.

Answer =

RELATED VIDEOS

_________________
Manager  B
Joined: 12 Oct 2012
Posts: 111
WE: General Management (Other)
Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

GMATPrepNow 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

So, we have 7 rope lengths.
If the median length is 84, then the lengths (arranged in ascending order) look like this: {_, _, _, 84, _, _, _}

The length of the longest piece of rope is 14 cm more than 4 times the length of the shortest piece of rope.
Let x = length of shortest piece.
This means that 4x+14 = length of longest piece.
So, we now have: {x, _, _, 84, _, _, 4x+14}

Our task is the maximize the length of the longest piece.
To do this, we need to minimize the other lengths.
So, we'll make the 2nd and 3rd lengths have length x as well (since x is the shortest possible length)
We get: {x, x, x, 84, _, _, 4x+14}

Since 84 is the middle-most length, the 2 remaining lengths must be greater than or equal to 84.
So, the shortest lengths there are 84.
So, we get: {x, x, x, 84, 84, 84, 4x+14}

Now what?

At this point, we can use the fact that the average length is 68 cm.
There's a nice rule (that applies to MANY statistics questions) that says:
the sum of n numbers = (n)(mean of the numbers)
So, if the mean of the 7 numbers is 68, then the sum of the 7 numbers = (7)(68) = 476

So, we now now that x+x+x+84+84+84+(4x+14) = 476
Simplify to get: 7x + 266 = 476
7x = 210
x=30

If x=30, then 4x+14 = 134
So, the longest piece will be 134 cm long.

Answer =

RELATED VIDEOS

Hi Brent,

I just have one question :

Do we not have to keep the shortest term unique?? "Shortest".

I have got some questions wrong before because I kept the first few terms identical to maximise the largest and the explanation provided was that since the term "shortest" is mentioned, we cannot take same numbers.

I cannot trace the same question to illustrate further.

Can you tell me how do we identify?

Thanks.
Board of Directors V
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3631
Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

aditi2013 wrote:
GMATPrepNow 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

So, we have 7 rope lengths.
If the median length is 84, then the lengths (arranged in ascending order) look like this: {_, _, _, 84, _, _, _}

The length of the longest piece of rope is 14 cm more than 4 times the length of the shortest piece of rope.
Let x = length of shortest piece.
This means that 4x+14 = length of longest piece.
So, we now have: {x, _, _, 84, _, _, 4x+14}

Our task is the maximize the length of the longest piece.
To do this, we need to minimize the other lengths.
So, we'll make the 2nd and 3rd lengths have length x as well (since x is the shortest possible length)
We get: {x, x, x, 84, _, _, 4x+14}

Since 84 is the middle-most length, the 2 remaining lengths must be greater than or equal to 84.
So, the shortest lengths there are 84.
So, we get: {x, x, x, 84, 84, 84, 4x+14}

Now what?

At this point, we can use the fact that the average length is 68 cm.
There's a nice rule (that applies to MANY statistics questions) that says:
the sum of n numbers = (n)(mean of the numbers)
So, if the mean of the 7 numbers is 68, then the sum of the 7 numbers = (7)(68) = 476

So, we now now that x+x+x+84+84+84+(4x+14) = 476
Simplify to get: 7x + 266 = 476
7x = 210
x=30

If x=30, then 4x+14 = 134
So, the longest piece will be 134 cm long.

Answer =

RELATED VIDEOS

Hi Brent,

I just have one question :

Do we not have to keep the shortest term unique?? "Shortest".

I have got some questions wrong before because I kept the first few terms identical to maximise the largest and the explanation provided was that since the term "shortest" is mentioned, we cannot take same numbers.

I cannot trace the same question to illustrate further.

Can you tell me how do we identify?

Thanks.

Even if we have 3 terms with the value of x and the others with the value greater than x, we will always call x the shortest.

e.g. Can you find out which is the shortest value among the below numbers?

1,1,1,2,3,4,5,6.

Answer to above question will clear your doubt.
_________________
My GMAT Story: From V21 to V40
My MBA Journey: My 10 years long MBA Dream
My Secret Hacks: Best way to use GMATClub | Importance of an Error Log!
Verbal Resources: All SC Resources at one place | All CR Resources at one place

GMAT Club Inbuilt Error Log Functionality - View More.
New Visa Forum - Ask all your Visa Related Questions - here.
New! Best Reply Functionality on GMAT Club!
Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free
Check our new About Us Page here.
CEO  V
Joined: 12 Sep 2015
Posts: 3727
Location: Canada
Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

Top Contributor
aditi2013 wrote:

Hi Brent,

I just have one question :

Do we not have to keep the shortest term unique?? "Shortest".

I have got some questions wrong before because I kept the first few terms identical to maximise the largest and the explanation provided was that since the term "shortest" is mentioned, we cannot take same numbers.

I cannot trace the same question to illustrate further.

Can you tell me how do we identify?

Thanks.

No, we need not keep the shortest term unique.

Cheers,
Brent
_________________
Manager  B
Joined: 24 May 2014
Posts: 89
Location: India
GMAT 1: 590 Q39 V32 GRE 1: Q159 V151 GRE 2: Q159 V153 GPA: 2.9
Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

Hey everyone,

I can see the great explanations provided, but I just directly jumped to the POE method. Only in option 'D', i can solve the equation: 134 = 14+4x, x=30, whereas in all other options, 'x' doesn't yield an integer. Can such methods give correct solutions most of the time.?
Intern  Joined: 25 Aug 2016
Posts: 10
Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

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.

What i do not understand. How are you getting to a = 30? I understand everything, however not this part Could please explain this to me?
Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

SamsterZ wrote:
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.

What i do not understand. How are you getting to a = 30? I understand everything, however not this part Could please explain this to me?

By solving $$a+a+a+84+84+84+(4a+14)=7*68$$.
_________________
Current Student D
Joined: 12 Aug 2015
Posts: 2617
Schools: Boston U '20 (M)
GRE 1: Q169 V154 Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

Excellent Question from the official Guide set.
Here is my solution to this one =>
Let the Rope pieces be ->
W1
W2
W3
W4
W5
W6
W7

In increasing order

Mean = 68

$$Mean = \frac{Sum}{#}$$

Hence Sum(7)=68*7 = 476 cms.

Now median = 84
#=7=> odd

Hence Median => Fourth term = W4

So W4=84

Now W7=14+4W1

We have to maximise the largest term i.e W7
For the we must minimise all other terms.

W2=W3=W1 each
W4=W5=W6=84each

Hence => 3W1+3*84+14+4W=476

=> 7W1=210
W1=30

Hence W7=14+4*30=134cms

Hence D

_________________
Manager  B
Joined: 30 Apr 2013
Posts: 78
Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

Where could I find this concept of Maximum and Minimum? I never came across this in my studies so far.
Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

santro789 wrote:
Where could I find this concept of Maximum and Minimum? I never came across this in my studies so far.

14. Min/Max Problems

_________________
Manager  B
Joined: 10 Sep 2014
Posts: 80
Location: Bangladesh
GPA: 3.5
WE: Project Management (Manufacturing)
Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

Bunuel Do you have similar problem's links to practice ?
Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

sadikabid27 wrote:
Bunuel Do you have similar problem's links to practice ?

Please check the post just above yours: https://gmatclub.com/forum/seven-pieces ... l#p1959061
_________________
Intern  B
Joined: 13 Mar 2017
Posts: 4
Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

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 $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$.

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?

Take Bunuel's analogy. x+y=10, to have max y e.g. 9 we have to minimise x to 1. Or seeing it another way y=10-x. So minimising x maximises y.
Doing the same thing in this example rearranging x+a+b+84+c+d+4x+14=7*68 gives us 4x+14=7*68-(x+a+b+84+c+d). So the more you minimise (x+a+b+84+c+d), the more you maximise 4x+14.
VP  D
Joined: 09 Mar 2016
Posts: 1283
Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

### Show Tags

JeffTargetTestPrep 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

We need to first recognize that we are working with a maximum problem. This means that of the seven pieces of rope, we must make 6 of those pieces as small as we possibly can, within the confines of the given information, and doing so will maximize the length of the 7th piece.

We are first given that seven pieces of rope have an average (arithmetic mean) length of 68 centimeters. From this we can determine the sum.

average = sum/quantity

sum = average x quantity

sum = 68 x 7 = 476

Next we are given that the median length of a piece of rope is 84 centimeters. Thus when we arrange the pieces of rope from least length to greatest, the middle length (the 4th piece) will have a length of 84 centimeters. We also must keep in mind that we can have pieces of rope of the same length. Let's first label our seven pieces of rope with variables or numbers, starting with the shortest piece and moving to the longest piece. We can let x equal the shortest piece of rope, and m equal the longest piece of rope.

piece 1: x

piece 2: x

piece 3: x

piece 4: 84

piece 5: 84

piece 6: 84

piece 7: m

Notice that the median (the 4th rope) is 84 cm long. Thus, pieces 5 and 6 are either equal to the median, or they are greater than the median. In keeping with our goal of minimizing the length of the first 6 pieces, we will assign 84 to pieces 5 and 6 to make them as short as possible. Similarly, we have assigned a length of x to pieces 1, 2, and 3.

We can plug these variables into our sum equation:

x + x + x + 84 + 84 + 84 + m = 476

3x + 252 + m = 476

3x + m = 224

We also given that the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope. So we can say:

m = 14 + 4x

We can now plug 14 + 4x in for m into the equation 3x + m = 224. So we have:

3x + 14 + 4x = 224

7x = 210

x = 30

Thus, the longest piece of rope is 4(30) + 14 = 134 centimeters.

Answer is D.

Hello one question, why did you denote the shortest as X as well 2, and 3 as x ? Shouldnt we need to give the shortest piece the unique name to diffirentitate from 2 and 3 ? if you denote the longest as m and m is the unique name
Thanks!  Seven pieces of rope have an average (arithmetic mean) lengt   [#permalink] 04 Jan 2018, 08:46

Go to page   Previous    1   2   3    Next  [ 49 posts ]

Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.

#### MBA Resources  