N Numbers...

[kamalakarthi]

Can you please help me on the attached problem.

[Brent@GMATPrepNow]

A certain list consists of 21 different numbers. If n is in the list and n is 4 times the
average (arithmetic mean) of the other 20 numbers in the list, then n is what fraction of
the sum of the 21 numbers in the list?
A.1/20
B.1/6
C.1/5
D.4/21
E.5/21

A quick solution here is to plug in some values that meet the given criteria.

Aside: I'm going to ignore the part about the numbers being different, since I have a feeling that this is the author's way of eliminating the possibility that all of the values equal zero (which would ruin the question).

So, the first 20 values (excluding n) could all equal 1, in which case their average (mean) would be 1.
Since n is 4 times the average, n would equal 4.
So, the sum of all 21 values is 24.

Question: n is what fraction of the sum of the 21 numbers in the list?
Answer: 4/24
[spoiler]= 1/6 = B[/spoiler]

IMPORTANT: Keep in mind that I broke the rule about all of the numbers being different. What's important here is that we COULD replace the 1's with 20 different numbers that still have a mean of 1, in which case the sum of the first 20 numbers would still be 20, which means the answer will remain B.

Cheers,
Brent

[Brent@GMATPrepNow]

A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average {mean} of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?
A. 1/20
B. 1/6
C. 1/5
D. 4/21
E. 5/21

Thanks

Here's an algebraic solution.

Let T = the sum of the 20 different numbers (EXCLUDING n)
So, the average (mean) of those 20 numbers is T/20
Since n is 4 times the average, we can see that n = 4(T/20)
Simplify to get n = 4T/20, or (even better) n = T/5

When we add n to the sum of the first 20 numbers we get: T + n
Since n = T/5, we can see that the sum of ALL 21 numbers = T + T/5
When we simplify T + T/5, we get: the sum of ALL 21 numbers = 6T/5

Question: n is what fraction of the sum of the 21 numbers in the list?
Answer: (T/5)/(6T/5)
Simplify to get [spoiler]1/6 = B[/spoiler]

Cheers,
Brent

[[email protected]]

Hi kamalakarthi,

Brent's first explanation took a clever approach with the question. In the event that you don't "see" things the way that Brent explained them, you can still follow the instructions and get the correct answer without too much trouble.

This is a perfect question for TESTing VALUES. We have a list of 21 different numbers, so let's choose...
1 through 20 and the number "N"

Since the first 20 values are consecutive, the sum = 10(21) = 210 and the average will be the "middle" of the group. Here, the average = 10.5

We're told that N = 4 times the average of the other 20 values, so N = 4(10.5) = 42

We're asked to determine the value of N/(sum of the list)

N = 42
Sum of the list = 210 + 42 = 252

42/252 might take a little work to reduce, but we can use the answer choices to our advantage....42(6) = 252, so 42/252 = 1/6

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

[kamalakarthi]

Thank you.

[Matt@VeritasPrep]

Here's one last nifty solution:

Let's call the mean of the other 20 numbers x.

We know that (Sum of set)/(# of terms in set) = Mean, so (Sum)/20 = x. That means that Sum = 20x, so the sum of the other 20 numbers is 20x.

n = 4x, so we have

n/(n + others) = 4x/(4x + 20x) = 4x/24x = 1/6.

[Gurpreet singh]

Let avg of first 20 nos =20
sum =20*20=400
n=4 times the avg=4*20=80

Sum of 21 values =400+80=480

n is fraction of total sum=80/480=1/6

[Matt@VeritasPrep]

Let avg of first 20 nos =20
sum =20*20=400
n=4 times the avg=4*20=80

Sum of 21 values =400+80=480

n is fraction of total sum=80/480=1/6

That works too: it's a specific case of the general solution I gave above (in which x = 20).