VIC + averages problem from GMAC paper exam

tonebeeze wrote:If the average (arithmetic mean) of 5 positive temperatures is x degrees Fahrenheit, then the sum of the 3 greatest of these temperatures, in degrees Fahrenheit, could be:

a. 6x
b. 4x
c. 5x/3
d. 3x/2
e. 3x/5

OA = B

First , since the mean of the 5 numbers is x, we know that (A+B+C+D+E)/5 = x
If we take this equation and multiply top and bottom by 5, we get A+B+C+D+E = 5x

Second, let's say that the 5 temperatures are A, B, C, D, and E, and let's say that A<B<C<D<E

We want to know the possible sum of the 3 greatest numbers (i.e., C+D+E)

Since C, D and E are the 3 greatest values, we know that their sum must be greater than 1/2 of the sum of A+B+C+D+E (we can make even stronger conclusions than this, but that isn't necessary)

Since C+D+E is greater than the sum of 1/2(A+B+C+D+E), we know that C+D+E is greater than 1/2(5x) since A+B+C+D+E = 5x

In other words, C+D+E must be greater than 2.5x

This leaves us with answer choices A and B (the other options are less than 2.5x)

Now if A+B+C+D+E = 5x, and if A<B<C<D<E, then C+D+E must be less than 5x.

This leaves us with answer choice B

smackmartine wrote:Hi Brent,
When you say that :

Since C, D and E are the 3 greatest values, we know that their sum must be greater than 1/2 of the sum of A+B+C+D+E (we can make even stronger conclusions than this, but that isn't necessary)

I am just wondering, is there any rule as such?

Sure, we can say that if 0<A<B<C<D<E then C+D+E > (3/5)(A+B+C+D+E)
Similarly, if 0<A<B<C<D<E<F<G then F+G > (2/7)(A+B+C+D+E+F+G)

Etc.

Cheers,
Brent