List R contains 5 numbers...CAT 6

[SD828]

List R contains five numbers that have an average value of 55. If the median of the numbers in the list is equal to the mean and the largest number is equal to 20 more than two times the smallest number, what is the smallest possible value in the list?

This problem is poorly written, and there are many like that in the MGMAT... The solution is easy

Note how it says THE LARGEST NUMBER... not its largest number(s), or its largest value, or maximum value. It leads to believe that there is only one number!

If there is only one number as the problem states THE LARGEST NUMBER IS (one), the answer is 30.

There isn't consistency, first sentence refers to numbers and second in reference to the same (group of) numbers, it refers to ONE in particular and uses singular IS. It's ambiguous. Using the word value would make it a bit clearer. There seems to be much controversy around this simple problem, clearly showing that MGMAT should reconsider the wording of this problem.

[RonPurewal]

nope. there's no "ambiguity", nor is there any "inconsistency".

words like greatest/least/biggest/smallest/shortest/longest/etc. DO NOT imply a unique value—and there's no context in which those words can ever be taken to imply such a thing.

if uniqueness is actually a condition of the problem, that's easy enough to specify with additional words ("different numbers", "unique largest value", etc.)—and that's exactly what will be done.

otherwise, "largest value" just means largest value, which MAY OR MAY NOT be unique—in the same way as, say, "rectangle" means something that may or may not be a square.

if you think about it, this is the only sensible way to interpret the terms, anyway, since you have to allow for the possibility of uncertainty.
think about it: what if you don't know the other values in the list?
if your definition were accurate—i.e., if "greatest" could only be used for a single, unique value—then it would actually be IMPOSSIBLE to talk about the biggest number in a list, since we wouldn't actually know whether it was unique. (also, your definition leaves us with no way to talk about the biggest number when there are two or more "copies" of that value, anyway!)

in any case, yeah. your interpretation of the terms is incorrect, and our problem is written exactly according to the conventions that prevail in the rest of the mathematical community.

[RonPurewal]

in fact, if you think our usage of these words is wrong, then you'll have to try to convince GMAC that their usage, too, is wrong.
(:

consider, for instance, problem #196 in the 2016 edition OG, in which there are three different pieces of rope that are all "the shortest piece of rope".

[DhruvA227]

Unless the numbers are equally spaced, you can't just average the first and last terms to get the overall average. You would have to add all five numbers and divide by 5.

Just trying to seek some clarification here for myself: If the average and the median are equal for numbers in a set, does it not imply that the numbers in the set are consecutive?

[Sage Pearce-Higgins]

No, this is a important distinction. If a set is equally spaced, then the mean is equal to the median. But you can't deduce from this that if the mean is equal to the median, then the set is equally spaced. In philosophical logic, you would say that "A therefore B" does not imply "B therefore A". For example, chocolate is something that tastes good, but not all things that taste good are chocolate.

To give a set example, think of the set {4,4,5,5,7}. The mean and the median are equal, but the set is not equally spaced.

[DanielK799]

lets assume the number is arrange as follow:

a, b, 55, c, 2a + 20
a <=b <= 55
55 <= c <= 2a+20

we are being told that average is 55, hence sum of all 5 numbers is 275. The sum of the smallest and the largest numbers is 3a+20.

since we are asked about the minimum a, then we have to find the maximum sum b+c, which is satisfied when b=55 and c=2a+20

we now then have to try 1 by 1 of the solution, in which B+C must suit description above

if a=15, 2a+20 = 50 (impossible since 2a+20 < 55)

if a=20, 2a+20 = 60, b+c = 275-55-80 = 140 (impossible since average of b & c is 70 , greater than 2a+20)

if a=25, 2a+20 = 70, b+c = 275-55-95 = 125 (satisfied when b=55 and c=70)

Therefore the answer is C (a=25)

I was a bit confused until this explanation cleared up maximizing "b" and "c". For whatever reason, I was thinking "c" would also be 55. But after realizing it was 2a + 20 as well, I was able to do the rest algebraically pretty easily.

The 5 sets of numbers are a, 55, 55, 2a +20, 2a + 20

And they all add up to 275

a + 55 + 55 + (2a + 20) + (2a +20)= 275

a + 110 +4a + 40= 275

5a + 150= 275

5a= 125

a= 25

Looks like a lot of algebra, but I was able to do it pretty quickly in order to solve a.

Just another method for those that may have found the "working backwards" part a bit confusing.

[Sage Pearce-Higgins]

Good contribution, thanks for adding that. It seems that the most confusing step is to set up the equation, but you've thought out that if you have five numbers with a fixed average and you want to find the minimum value of the smallest one, then you need to maximize the other four numbers. The first step is probably to lay out the five numbers: a .... 55 .... (20+2a) [where a = smallest value]. Then think: the second number can't be bigger than 55, otherwise the median would change, and the fourth number can't be bigger than the biggest one. So that you get the 5 numbers in algebraic terms as you have them. After that, simplifying the algebra, as you say, isn't so bad.

It's great to try out problems in different ways. This is a good method if you can do it, but remember that working backwards from the answer has the advantage that you can usually eliminate some answers quite quickly so that you're ready to guess and move on if you need to.