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 = 476. 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 = 476
3a + 3*84 + 4a = 462
3a + 252 + 4a = 462
7a = 210
a = 30
max = 4a + 14 = 4*30 + 14 = 120 + 14 = 134
Answer =
D
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
