January 19, 2019 January 19, 2019 07:00 AM PST 09:00 AM PST Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. January 20, 2019 January 20, 2019 07:00 AM PST 07:00 AM PST Get personalized insights on how to achieve your Target Quant Score.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 02 Dec 2012
Posts: 177

Seven pieces of rope have an average (arithmetic mean) lengt
[#permalink]
Show Tags
20 Dec 2012, 06:46
Question Stats:
71% (02:28) correct 29% (02:56) wrong based on 2802 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
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 52278

Re: Seven pieces of rope have an average (arithmetic mean) lengt
[#permalink]
Show Tags
20 Dec 2012, 06:56
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Intern
Joined: 16 Aug 2013
Posts: 2
Concentration: Marketing, Strategy

Re: Seven pieces of rope have an average (arithmetic mean) lengt
[#permalink]
Show Tags
08 Sep 2013, 11:28
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




Intern
Joined: 16 Apr 2009
Posts: 14

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



Math Expert
Joined: 02 Sep 2009
Posts: 52278

Re: Seven pieces of rope have an average (arithmetic mean) lengt
[#permalink]
Show Tags
21 Dec 2012, 02:39
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 25 Feb 2013
Posts: 7

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



Math Expert
Joined: 02 Sep 2009
Posts: 52278

Re: Seven pieces of rope have an average (arithmetic mean) lengt
[#permalink]
Show Tags
12 Apr 2013, 01:51
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 11 Jan 2013
Posts: 15
Location: United States

Re: Seven pieces of rope have an average (arithmetic mean) lengt
[#permalink]
Show Tags
07 Jul 2013, 08:24
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



Manager
Joined: 12 Jan 2013
Posts: 152

Re: Seven pieces of rope have an average (arithmetic mean) lengt
[#permalink]
Show Tags
13 Jan 2014, 01:56
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.



Intern
Joined: 03 Jan 2014
Posts: 1

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: 52278

Re: Seven pieces of rope have an average (arithmetic mean) lengt
[#permalink]
Show Tags
28 Apr 2014, 00:15



Intern
Joined: 22 Feb 2014
Posts: 27

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



Math Expert
Joined: 02 Sep 2009
Posts: 52278

Re: Seven pieces of rope have an average (arithmetic mean) lengt
[#permalink]
Show Tags
02 May 2014, 01:06
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?
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Senior Manager
Joined: 07 Apr 2012
Posts: 360

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



Math Expert
Joined: 02 Sep 2009
Posts: 52278

Re: Seven pieces of rope have an average (arithmetic mean) lengt
[#permalink]
Show Tags
14 Jun 2014, 01:31
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Director
Joined: 10 Mar 2013
Posts: 503
Location: Germany
Concentration: Finance, Entrepreneurship
GPA: 3.88
WE: Information Technology (Consulting)

Re: Seven pieces of rope have an average (arithmetic mean) lengt
[#permalink]
Show Tags
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)
_________________
When you’re up, your friends know who you are. When you’re down, you know who your friends are.
Share some Kudos, if my posts help you. Thank you !
800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50 GMAT PREP 670 MGMAT CAT 630 KAPLAN CAT 660



Intern
Joined: 12 Aug 2014
Posts: 17

Re: Seven pieces of rope have an average (arithmetic mean) lengt
[#permalink]
Show Tags
06 Jan 2015, 06:19
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?



Math Expert
Joined: 02 Sep 2009
Posts: 52278

Re: Seven pieces of rope have an average (arithmetic mean) lengt
[#permalink]
Show Tags
06 Jan 2015, 06:50
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?
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 12 Aug 2014
Posts: 17

Re: Seven pieces of rope have an average (arithmetic mean) lengt
[#permalink]
Show Tags
06 Jan 2015, 07:13
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



Math Expert
Joined: 02 Sep 2009
Posts: 52278

Re: Seven pieces of rope have an average (arithmetic mean) lengt
[#permalink]
Show Tags
06 Jan 2015, 07:22




Re: Seven pieces of rope have an average (arithmetic mean) lengt &nbs
[#permalink]
06 Jan 2015, 07:22



Go to page
1 2 3
Next
[ 47 posts ]



