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605-655 Level|   Min-Max Problems|   Statistics and Sets Problems|                              
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Seven pieces of rope have an average (arithmetic mean) lengt 20 Dec 2012, 07:46
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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
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D
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Seven pieces of rope have an average (arithmetic mean) lengt 20 Dec 2012, 07:56
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Walkabout
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
Show more

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.
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Seven pieces of rope have an average (arithmetic mean) lengt 08 Sep 2013, 12:28
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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. :-D

Whenever you're stuck at a PS question the next best approach is POE :)
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Seven pieces of rope have an average (arithmetic mean) lengt 12 Apr 2013, 02:51
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sunshinewhole
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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.
Show more

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.
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Seven pieces of rope have an average (arithmetic mean) lengt 13 Jan 2014, 02:56
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Walkabout
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
Show more

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.
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Seven pieces of rope have an average (arithmetic mean) lengt 20 Dec 2012, 10:59
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Bunuel
Walkabout
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.
Show more

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.
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Seven pieces of rope have an average (arithmetic mean) lengt 21 Dec 2012, 03:39
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Drik
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Walkabout
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.
Show more

Because the question asks to find the maximum possible length of the longest piece of rope, which we denoted as g.
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Seven pieces of rope have an average (arithmetic mean) lengt 07 Jul 2013, 09:24
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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
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Seven pieces of rope have an average (arithmetic mean) lengt 27 Apr 2014, 03:44
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pls help me here! Why cant we just take all the values as 84 itself. :?
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Seven pieces of rope have an average (arithmetic mean) lengt 28 Apr 2014, 01:15
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krish1chaitu
pls help me here! Why cant we just take all the values as 84 itself. :?
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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.
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Seven pieces of rope have an average (arithmetic mean) lengt 01 May 2014, 15:30
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Bunuel
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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.
Show more

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...
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Seven pieces of rope have an average (arithmetic mean) lengt 02 May 2014, 02:06
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Bunuel
Walkabout
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...
Show more

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?
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Seven pieces of rope have an average (arithmetic mean) lengt 14 Jun 2014, 02:23
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sunshinewhole
Bunuel

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.
Show more
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?
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Seven pieces of rope have an average (arithmetic mean) lengt 14 Jun 2014, 02:31
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sunshinewhole
Bunuel

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?
Show more

No. Because we have fixed total length of the rope: 7*68 centimeters. If you increase a, you'd be decreasing g.
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Seven pieces of rope have an average (arithmetic mean) lengt 05 Jul 2014, 03:38
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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)
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Seven pieces of rope have an average (arithmetic mean) lengt 06 Jan 2015, 07:19
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Bunuel
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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.
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Im unable to understand that how minimizing all other terms would maximize value of g, can you please explain?
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Seven pieces of rope have an average (arithmetic mean) lengt 06 Jan 2015, 07:50
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Bunuel
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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?
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Answer this: the sum of two positive integers is 10. What is the maximum possible value of the largest of the integers?
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JeffTargetTestPrep
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Seven pieces of rope have an average (arithmetic mean) lengt 19 May 2016, 05:10
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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
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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.
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Seven pieces of rope have an average (arithmetic mean) lengt Updated on: 26 Aug 2020, 10:43
Originally posted by BrentGMATPrepNow on 26 Jul 2016, 07:31.
Last edited by BrentGMATPrepNow on 26 Aug 2020, 10:43, edited 1 time in total.
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Walkabout
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
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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 = D

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Seven pieces of rope have an average (arithmetic mean) lengt 09 Nov 2017, 03:57
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Where could I find this concept of Maximum and Minimum? I never came across this in my studies so far.
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