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

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Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 20 Dec 2012, 06:46
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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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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 20 Dec 2012, 06:56
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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.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 20 Dec 2012, 09:59
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.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 21 Dec 2012, 02:39
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Drik 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.


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.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 23 Dec 2012, 03:26
Again,We can plugin here. Algebraic method is good too.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 11 Apr 2013, 16:24
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.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 12 Apr 2013, 01:51
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sunshinewhole 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.



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}).

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Hope it helps.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 07 Jul 2013, 08:24
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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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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 08 Sep 2013, 11:28
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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

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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 20 Oct 2013, 18:56
Extremely tough question to do in 2 min IMO.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 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?
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 13 Jan 2014, 01:56
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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.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 27 Apr 2014, 02:44
pls help me here! Why cant we just take all the values as 84 itself. :?
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 28 Apr 2014, 00:15
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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.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 01 May 2014, 14:30
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...
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 02 May 2014, 01:06
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drkomal2000 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.


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

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?
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 14 Jun 2014, 01:31
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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?


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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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post 05 Jul 2014, 02:38
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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Re: Seven pieces of rope have an average (arithmetic mean) lengt   [#permalink] 05 Jul 2014, 02:38
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