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Seven pieces of rope have an average (arithmetic mean) lengt
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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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Re: Seven pieces of rope have an average (arithmetic mean) lengt
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20 Dec 2012, 07:56
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
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08 Sep 2013, 12:28
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  11814=104 C)120  12014 = 106 D)134  13414 = 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




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Re: Seven pieces of rope have an average (arithmetic mean) lengt
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20 Dec 2012, 10:59
Bunuel wrote: Walkabout wrote: Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?
(A) 82 (B) 118 (C) 120 (D) 134 (E) 152 Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g > \(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\). The average length = 68 centimeters > the total length 7*68 centimeters. The median = 84 centimeters > d=84. The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece > \(g=4a+14\). We need to maximize g. Now, to maximize g, we need to minimize all other terms. The minimum value of b and c is a and the minimum value of e and f is median=d. Thus we have that \(a+a+a+84+84+84+(4a+14)=7*68\) > \(a=30\) > \(g_{max}=4a+14=134\). Answer: D. Thanks Bunnel.. Could you please elaborate more about WHY "We need to maximize g. Now, to maximize g, we need to minimize all other terms." I just want to understand how does it work.



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Re: Seven pieces of rope have an average (arithmetic mean) lengt
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21 Dec 2012, 03:39
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
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11 Apr 2013, 17:24
Bunuel wrote: Walkabout wrote: Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?
(A) 82 (B) 118 (C) 120 (D) 134 (E) 152 Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g > \(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\). The average length = 68 centimeters > the total length 7*68 centimeters. The median = 84 centimeters > d=84. The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece > \(g=4a+14\). We need to maximize g. Now, to maximize g, we need to minimize all other terms. The minimum value of b and c is a and the minimum value of e and f is median=d. Thus we have that \(a+a+a+84+84+84+(4a+14)=7*68\) > \(a=30\) > \(g_{max}=4a+14=134\). Answer: D. to minimize y we took 84, we cud have valued all the rope lengths to our minimum=a.



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Re: Seven pieces of rope have an average (arithmetic mean) lengt
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12 Apr 2013, 02:51
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 (\(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\)). Similar questions to practice: gmatdiagnostictestquestion79347.htmlsevenpiecesofropehaveanaveragearithmeticmeanlengt144452.htmlasetof25differentintegershasamedianof50anda129345.htmlthemedianofthelistofpositiveintegersaboveis129639.htmlinacertainsetoffivenumbersthemedianis128514.htmlgivendistinctpositiveintegers1113x2and9whic109801.htmlsetscontainssevendistinctintegersthemedianofsets101331.htmlthreeboxeshaveanaverageweightof7kgandamedianweigh99642.htmlthreestraightmetalrodshaveanaveragearithmeticmean148507.htmlHope it helps.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt
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07 Jul 2013, 09:24
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
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13 Jan 2014, 02:56
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
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27 Apr 2014, 03:44
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
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01 May 2014, 15:30
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
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02 May 2014, 02:06
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
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14 Jun 2014, 02:23
sunshinewhole wrote: Bunuel wrote: Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g > \(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\).
The average length = 68 centimeters > the total length 7*68 centimeters. The median = 84 centimeters > d=84. The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece > \(g=4a+14\).
We need to maximize g. Now, to maximize g, we need to minimize all other terms.
The minimum value of b and c is a and the minimum value of e and f is median=d.
Thus we have that \(a+a+a+84+84+84+(4a+14)=7*68\) > \(a=30\) > \(g_{max}=4a+14=134\).
Answer: D. to minimize y we took 84, we cud have valued all the rope lengths to our minimum=a. It's a little confusing.... Since we have the equation \(g=4a+14\), to maximize g, we also need to maximize a... isn't this the case?



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Re: Seven pieces of rope have an average (arithmetic mean) lengt
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14 Jun 2014, 02:31
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
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05 Jul 2014, 03:38
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
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06 Jan 2015, 07:19
Bunuel wrote: Walkabout wrote: Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?
(A) 82 (B) 118 (C) 120 (D) 134 (E) 152 Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g > \(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\). The average length = 68 centimeters > the total length 7*68 centimeters. The median = 84 centimeters > d=84. The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece > \(g=4a+14\). We need to maximize g. Now, to maximize g, we need to minimize all other terms. The minimum value of b and c is a and the minimum value of e and f is median=d. Thus we have that \(a+a+a+84+84+84+(4a+14)=7*68\) > \(a=30\) > \(g_{max}=4a+14=134\). Answer: D. Im unable to understand that how minimizing all other terms would maximize value of g, can you please explain?



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06 Jan 2015, 07:50
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? 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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Re: Seven pieces of rope have an average (arithmetic mean) lengt
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06 Jan 2015, 08:13
Im unable to understand that how minimizing all other terms would maximize value of g, can you please explain?[/quote] Answer this: the sum of two positive integers is 10. What is the maximum possible value of the largest of the integers?[/quote] I got your point, i don't know how i was thinking before, but i just got it . We must minimize all other values in order to get maximum value for g. Thanks



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