Problem Solving

Mike@Magoosh wrote:Dear wied81,

I'll do my best to explain this as clearly as possible, but it is a very difficult question requiring some very subtle thinking. I'm happy to give it a whirl.

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?

So, the fact that the scenario is about rope is annoyingly irrelevant. We have a list of seven numbers and, given some constraints, we want to know what the largest value of the maximum is.

First of all, we are given the mean of the seven numbers is 68. Whenever you know the mean and the number of items on a list, you automatically know the sum of the list.

mean = (sum)/(number of items) ----> sum = (mean)*(number of items)

Here, that sum is 7*68 = 448. It's almost always true, in the harder Quantitative questions, that if you are given an average and the number of items on a list, you almost inevitably will have to figure out the sum to figure out what they are asking. Here, we just find it preemptively.

We are told the median = 84; since there are seven numbers, an odd number, we absolutely know that the middle number, the fourth number on the list, has to have the numerical value of 84.

that the longest side is 14 plus 4 times the smallest side. Let the smallest side be a. The max = 4a + 14.

Currently, here is our list in ranked numerical order from smallest to biggest.

{a, b, c, 84, d, e, 4a + 14)

The question is asking: what is the maximum possible length of the longest piece of rope? Because we have a given and fixed average, we know the only way to keep that average fixed is to keep the sum of the seven numbers fixed. This in turn means --- if we increase any one number, we have to decrease another number by the same amount. Our goal is to make the maximum as large as possible. That means, we have to make the other six numbers as small as possible.

a is the minimum, so that's already as low as it can be.

b & c can be bigger than a, or equal to a. If they are both equal to a, then they are as low as they can go, and that does not violate any of the constraints. At this point, we have:

{a, a, a, 84, d, e, 4a + 14)

Now, d & e we can't push all the way down to a, because in order for 84 to be the median, d & e can't be below this median. BUT, d and e could equal 84, and 84 would still be the median --- as long as the fourth largest number equals 84, the median will be 84, and it doesn't matter at all whether other entries also equal 84. Since d & e can be either greater than or equal to 84, the lowest they can be is 84. Now, we have:

{a, a, a, 84, 84, 84, 4a + 14)

Now, we have made the other six numbers as small as they can be, within the given constraints, so the max is as big as it can be.

Now, we will find the sum and set it equal to 448, which we already found preemptively.

a + a + a + 84 + 84 + 84 + 4a + 14 = 448

3a + 3*84 + 4a = 434

3a + 252 + 4a = 434

7a = 182

a = 26

max = 4a + 14 = 4*26 + 14 = 104 + 14 = 118

Answer = B

Here's a DS questions using some of these concepts.
https://gmat.magoosh.com/questions/941
When you submit your answer to that question, the next page will have the complete video explanation.

Does all this make sense? Le me know if you have any further questions.

Mike

wied81 wrote:183. 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

OA: D

The book ( OG 13) posts a pretty confusing way to solve, would love to hear any other recommendations.

Thanks in advance for any guidance.

karthikchandru wrote:Hi Mitch,

"Thus, when 14 is subtracted from the correct answer choice, the result will be a multiple of 4"
There is nothing in the problem which says that this number has to be an integer, right? Then, why do we assume that the result has to be a multiple of 4?

Thanks.

Mike@Magoosh wrote:

GMATGuruNY wrote:GMAT problems about real-world scenarios -- someone's age, the length of a rope -- tend to involve integer values. The combination of factors above virtually guarantees that all of the lengths in the problem will be integer values.

Mitch:
First of all, I want to express tremendous respect for you as a fellow professional. I truly regard you as your screen name suggests, a GMAT Guru.

I want to play Devil's Advocate on this particular point. Yes, if the the numbers in the problem are ages, days until St. Aloysius Day, number of donkeys, etc. then of course they must be restricted to positive integers.

Here, the numbers are lengths of rope. A piece of rope is an excellent visual metaphor for the continuous infinity of the real number line. What inherently about this situation would lead us to know that the GMAT was implicitly restricting the problem to positive integers?

It's true, for answer choices (C) and (E), if we work backwards, we get half-integer values for the smallest length, and since there are seven elements in the set, we would multiply that minimum length by 3, which results in a half-integer value for the sum, which is not correct. BUT, suppose the set consisted of nine elements, in another otherwise similar scenario --- then the smallest length would be multiplied by 4, and even if the smallest length were a half-integer value, the sum would still be an integer. In that case, would there be any a priori reason to dismiss the possibility of half-integer values for the smallest lengths?

In fact, doesn't GMAC take a kid of perverse delight in present problems with only positive integers in the prompt and with only positive integer in the answer choices, but which involve negatives and/or fractions in some of the intermediary steps? Isn't that a trick they have up their sleeve for 700-800 questions?

What do you think? I am keenly interested in your opinion on this.

With great respect,
Mike McGarry

GMATGuruNY wrote:

wied81 wrote:183. 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

OA: D

The book ( OG 13) posts a pretty confusing way to solve, would love to hear any other recommendations.

Thanks in advance for any guidance.

The sum of the lengths = 7*68 = 476.

We can plug in the answer choices, which represent the maximum length of the longest piece.
The longest piece is 14 centimeters more than 4 TIMES the length of the shortest piece. Thus, when 14 is subtracted from the correct answer choice, the result will be a multiple of 4.
Subtracting 14 from each answer choice, we get:
68, 104, 106, 120, 138.
106 and 138 are not multiples of 4.
Eliminate C and E.

Since the question asks for the MAXIMUM possible length, we should start with the GREATEST of the remaining answer choices.
To MAXIMIZE the length of the longest piece, we must MINIMIZE the lengths of all of the other pieces.

Answer choice D: 134
Shortest piece = 120/4 = 30.
The 7 pieces are:
30, b, c, 84, e, f, 134.
The minimum possible value for b and c is 30, implying 3 pieces with a length of 30.
The minimum possible value for e and f is 84, implying 3 pieces with a length of 84.
Sum = 3(30) + 3(84) + 134 = 90+252+134 = 476.
Success!

The correct answer is D.