Equivalent to arithmetic ave

[caspermonday]

Another question from EST 31:

Which of the following procedures is always equivalent to adding 5 given numbers and then dividing the sum by 5?
(I) Multiplying the 5 numbers and then finding the 5th root of the product
(II) Adding the five numbers, doubling the sum, and then moving the decimal point one place to left
(III) Ordering the 5 numbers numerically and then selecting the middle number

a) None
b) I only
c) II only
d) III only
e) I and III

OA c
How is that? Is there such property? How can we prove it besides plugging numbers? Thank you!

[Stuart@KaplanGMAT]

Another question from EST 31:

Which of the following procedures is always equivalent to adding 5 given numbers and then dividing the sum by 5?
(I) Multiplying the 5 numbers and then finding the 5th root of the product
(II) Adding the five numbers, doubling the sum, and then moving the decimal point one place to left
(III) Ordering the 5 numbers numerically and then selecting the middle number

a) None
b) I only
c) II only
d) III only
e) I and III

OA c
How is that? Is there such property? How can we prove it besides plugging numbers? Thank you!

Hi.

The first thing to note is that what we're doing is taking the average of the set.

Average = (sum of terms) / (# of terms)

which is exactly what we're doing with our 5 numbers.

So, we either need another formula that's the exact same as the original or another way to calculate average.

(I) Multiplying the 5 numbers and then finding the 5th root of the product

As soon as we see "taking the 5th root", we know that there's no way this will work out the same as the original - eliminate (b) and (e).

(II) Adding the five numbers, doubling the sum, and then moving the decimal point one place to left

The first step is the same: adding the 5 numbers. So, are the next two steps the equivalent of "dividing by 5"?

Doubling is multiplying by 2; moving the decimal point one place to the left is dividing by 10.

Is 1/5 = 2/10? Yes! Therefore, (II) is the exact same as the original process.

The only choice remaining with (II) is (c) - we're done!

If for some reason you still wanted to analyze (III):

(III) Ordering the 5 numbers numerically and then selecting the middle number

This describes the median of the set, not the arithmetic mean, which is what we're looking for.

If you attacked this question by picking numbers (a good idea) and picked overly simplistic/patterned numbers (a bad idea), you may have thought that (III) is correct.

For example, if we pick {1, 2, 3, 4, 5} as our set, (III) would in fact hold true.

However, if we pick a set that's a bit more offbeat, e.g. {1, 2, 3, 5, 100}, we quickly see that the middle term doesn't have to be the average.

[caspermonday]

Awesome! Thank you, Stuart!