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GMAT Quants Average Question

by sanjay12590 » Sat Aug 16, 2014 5:42 am
Guys need your help with this one

The average of four numbers is 32 and their range is 12. Which of the following options could possibly be a value of the largest of these numbers?

A)40
B)42
C)43
D)44
E)45

Thanks !!
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by Brent@GMATPrepNow » Sat Aug 16, 2014 6:08 am
sanjay12590 wrote:
The average of four numbers is 32 and their range is 12. Which of the following options could possibly be a value of the largest of these numbers?

A)40
B)42
C)43
D)44
E)45
One option is to TEST the answer choices.

First, when it comes to averages, we know that sum of n values = (average value)(n)
Here the average of 4 numbers is 32, so the SUM of the 4 numbers = (32)(4) = 128

Now, let's TEST the answer choices

A)40
If the greatest value is 40, and the range is 12, then the SMALLEST value is 28
The two remaining numbers must lie between 28 and 40 INCLUSIVE
Also, 28 + 40 = 68, and 128 - 68 = 60.
So, the two remaining numbers must add to 60 [so that all 4 numbers add to 128]
So, for example, it's possible that the two remaining numbers are 30 and 30
We get: {28, 30, 30, 40}
Since it's possible for 40 to be the value of the largest number, the correct answer is A

Cheers,
Brent
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by sanjay12590 » Sat Aug 16, 2014 6:13 am
Thanks a lot Brent !!
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by [email protected] » Sat Aug 16, 2014 1:05 pm
Hi sanjay12590,

I'm a big fan of TESTing THE ANSWERS (as Brent showcased). Like most Quant questions on the GMAT, you can answer this question in other ways though. Here's an approach that emphasizes the algebra involved:

We're told that the average of 4 numbers is 32..

Sum/4 = 32
Sum = 128

The prompt does NOT say that the numbers are distinct/unique/different, so we could possibly have duplicate numbers. We're also told that the range of the numbers is 12, so we know that the biggest number and smallest number are relatively close to one another. We're asked for what could be the largest number in the group. This is a "limit" question - to make one value as big as possible, we have to make the others as small as possible.

I'm going to call the first 3 numbers X; with a range of 12, the largest number would be X+12

X, X, X, X+12

We know the sum is 128 so....

X + X + X + X+12 = 128
4X + 12 = 128
4X = 116
X = 29

So one POSSIBLE set of numbers would be 29, 29, 29, and 41. 41 is not in the answers though, so we have to "play" with this group of numbers just a little bit.

If we took two of the 29s and turned them into 30s, then we could turn the other 29 into a 28 and the 41 into 40:

28, 30, 30, 40

Final Answer: A

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by Matt@VeritasPrep » Sun Aug 17, 2014 4:19 pm
Let's call the four integers a, b, c, and d. To maximize the biggest one, we'll minimize the other three, so we'll say a is the biggest and that b = c = d.

Since the average is 32, we have
a + b + c + d = a + d + d + d = a + 3d = 128

Since the range is 12, we have
a - d = 12

The solution to the system is a = 41, d = 29.

Since 41 is the MAXIMUM value of a, any other values must be less than 41, so the answer is 40.
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