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by advita » Tue Jan 11, 2011 2:09 am
Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

78
55 1/7
66 1/7
77 1/5
52

oa-78
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by jaymw » Tue Jan 11, 2011 2:35 am
What's the source of this question?

I backsolved this problem, meaning I plugged in the answer choices.

Before I did so I wrote down the set as: x,>=x,55,>=55,3x+20

Now the range is always MAXvalue - MINvalue, so here we have: 3x+20-x=78(answer choice A)
Solving for X yields 29. That means, for answer choice A, the smallest number of the set must be 29. The biggest number must be 3(29)+20=107.

Now that we have three values, we can calculate the sum of those.

29+55+107=191

In total our set should have a sum of 5*55=275.

That leaves a difference of 84 to be distributed between the two missing numbers in the set.

The second number in the ascending ordered set can now be 29 and the fourth number can be 55. Bingo! A must be correct and there's no need to plug in other answer choices.

Hope this was helpful.
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by Anurag@Gurome » Tue Jan 11, 2011 3:38 am
Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?
Say the numbers are a, b, c, d, and e; where a ≤ b ≤ c ≤ d ≤ e
Now, median = c = 55
and, e = (3a + 20)

Therefore, range = (max - min) = (e - a) = (2a + 20)
Therefore to maximize the range we have to maximize a.

Also, (a + b + c + d + e) = 5*55 = 275
=> (a + b + 55 + d + 3a + 20) = 275
=> (4a + b + d) = 200
=> 4a = (200 - b - d)

Now, a will be maximum when both b and d will be minimum.
Minimum value of b is a and that of d is c, i.e. 55.

Hence, when a is minimum,
  • 4a = (200 - a - 55) => 5a = 145 => a = 29
Hence maximum value of the range is (2*29 + 20) = 78

The correct answer is A.
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by ankur.agrawal » Tue Jan 11, 2011 3:47 am
I have read sumehere that for evenly spaced integers in a set mean & median are same.

Also we can find the mean of an evenly spaced integers in a set by just averaging the first & last terms.

Can we make use of the above two concepts to solve this problem. I tried but failed.

Help!!!!
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by jaymw » Tue Jan 11, 2011 4:15 am
@ankur.agrawal:

Unfortunately, you can't use these concepts for the above problem because it doesn't revolve around a set of evenly spaced integers.

Also, you aren't asked to find the mean or some other average but you're asked to find the range.

Therefore, you'd have to solve it my way or Anrug's, which, admittedly, is more scientific than mine.
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by GMATGuruNY » Tue Jan 11, 2011 4:56 am
advita wrote:Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

78
55 1/7
66 1/7
77 1/5
52

oa-78
To maximize the range, we want to make the smallest number as small as possible and the largest number as large as possible. So the five numbers would be:

x, x, 55, 55, y

The set above will minimize the smallest number (x) because the two smallest values are the same. It will maximize the largest number (y) because the next largest number (55) is the same as the median.

Since the largest number is 20 more than 3 times the smallest number, y = 20+3x. So our list of numbers can be rewritten as:

x, x, 55, 55, 20+3x

Since the average is 55, the sum of the five numbers = 5*55 = 275.
Thus, x + x + 55 + 55 + 20+3x = 275.
5x = 145
x= 29.

Thus, y = 20+3x = 20 + 3*29 = 107.

Thus, the largest possible range is y-x = 107-29 = 78.

The correct answer is A.
Last edited by GMATGuruNY on Fri Oct 21, 2011 10:39 am, edited 1 time in total.
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by ankurmit » Sun Jan 16, 2011 10:11 pm
ankur.agrawal wrote:I have read sumehere that for evenly spaced integers in a set mean & median are same.

Also we can find the mean of an evenly spaced integers in a set by just averaging the first & last terms.

Can we make use of the above two concepts to solve this problem. I tried but failed.

Help!!!!
I tried the same concept and got 65 as answer :(

which is not the provided answer.
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by pesfunk » Thu Jan 27, 2011 6:23 am
Excellent idea.

Basically the highest range can only be achieved when the smallest number is really small and biggest number is really big. In that case....Y need to take the load of right side and both numbers on the left needs to be as small as possible.


GMATGuruNY wrote:
advita wrote:Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

78
55 1/7
66 1/7
77 1/5
52

oa-78
To maximize the range, we want to make the smallest number as small as possible and the largest number as large as possible. So the five numbers would be:

x, x, 55, 55, y

The set above will minimize the smallest number (x) because the two smallest values are the same. It will maximize the largest number (y) because the next largest number (55) is the same as the median.

Since the largest number is 20 more than 3 times the smallest number, y = 20+3x. So our list of numbers can be rewritten as:

x, x, 55, 55, 2x+30

Since the average is 55, the sum of the five numbers is 5*55 = 275.
Thus, x + x + 55 + 55 + 20+3x = 275.
5x = 145
x= 29.

Thus, y = 20+3x = 20 + 3*29 = 107.

Thus, the largest possible range is y-x = 107-29 = 78.

The correct answer is A.
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