Manhattan CAT Question

[blue.vikasbhardwaj]

Q)
Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

78
77 1/5
66 1/7
55 1/7
52

The correct Answer is 78

Solution :

Please help me point out where is my approach going wrong

Mean = Median in case of an Arithmetic Progression Series

So let the five numbers be a-2d,a-d,a,a+d and a+2d and therefore the Maximum range will be 4d

Given Average(Mean) = 55 , implies that Median = 55 ; a = 55

As per the question a + 2d = 20 + 3(a-2d)
-2a + 8d = 20
4d = 10 + a
Since a = 55 , implies 4d = 65

And hence the Maximum Range is 65 .

[jnelson0612]

Hi blue,
Let me walk you through how I would handle this problem and if you still have questions please ask.

I know that I have five numbers with an average of 55. That means that their total is 275. I also know that mean equals the median, so the median number is 55.

Thus, I have something that looks like this:
____ ____ 55 ____ ____

I know that the largest number is 20 more than 3 times the smallest number. If I call the smallest number x, then the largest number is 3x + 20. Let's add that to our representation:
x ____ 55 ____ 3x + 20

Now, the question is what are the remaining two numbers? Here I need to keep in mind that my goal is to produce the maximum range possible. Thus, I want to make the largest term as big as I possibly can, so I can create the largest possible difference between it and the smallest number, maximizing the range.

So, what should the number between x and 55 be (in place 2)? Since I am not told that these numbers are distinct, let's just call it x. I do that because I want to make every number below the largest one as small as possible, so I have more total value left out of the 275 total to assign to the largest number. x is the smallest possible number for place 2.

Using the same logic, I assign the number in place 4 as 55. Again, I am making that number as small as possible within the constraints that it has to be at least as big as the number in place 3.

My new range looks like this:
x x 55 55 3x + 20

I know this totals 275, so 5x + 130 = 275, or x is 29.

Thus I know the smallest number of my range is 29, and the largest is 3(29)+20, or 107. The difference between the two is 78, and that is the maximum possible range.

Once you grasp the concept of making each of the first four numbers as small as possible so we can make the biggest number as large as possible, maximizing our range, this problem is easy to solve.

I hope this helps.

[Pinkynagpaloberoi]

Does this mean that if Mean and median of a series are equal, it does not necessarily should be a AP series, because if you try solving this question using that logic, the answer comes out to be 65

[tim]

if it's an AP, the mean and median will be equal. the reverse is not automatically true. consider 1 3 4 5 7 as an example; both the mean and median are 4, but this is definitely not an arithmetic progression..