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range question

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by runzun » Thu Aug 04, 2011 11:37 am
a set of numbers contains seven integers and has an average and arithmatic mean of 23. if the largest value is 15 more than 4 times the smallest no, what is the largest possible range in no of set..
a.33
b.35
c.38
d.48
e.75
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Source: — Problem Solving |

by goalevan » Thu Aug 04, 2011 5:12 pm
We are given that the arithmetic mean is 23 in a set of 7 integers, so the sum of these seven integers is 7 * 23 = 161.

The range will be (4x + 15) - x = 3x + 15, where x is the smallest integer in the set. In order to maximize this quantity (the range), we need to maximize the smallest integer in the set because the quantity 3x + 15 will increase by 3 for every unit increase in x.

We must also hold to the constraints that x is the smallest integer, that the integers add to 161, and that the largest integer is 4x + 15.

In order to maximize the smallest integer, we need to minimize all the other integers while holding them greater than or equal to x. We can set them equal to x.

The sum then becomes x + x + x + x + x + x + (4x + 15) = 6x + 4x + 15 = 10x + 15

Since the sum of this quantity equals 161, 10x + 15 = 161, or 10x = 146, and x = 14.6

With non-integers, we could have the smallest number as 14.6 and the largest as 4 * 14.6 + 15 = 73.4 with a range of (4 * 14.6 + 15) - 14.6 = 58.8.

In order to hold to the integer constraint, we must move x down to the next integer, 14. 4x + 15 becomes 71 and the range is 71 - 14 = 57. An example where this works is the set {14,15,15,15,15,16,71}.

However this is not an answer choice above, perhaps 48 is the correct answer? What is the OA and source?

3x + 15 (the range) cannot be 75, since the smallest number would then calculate as (75 - 15)/3 = 20, and 20 + 95 + 20 * 5 > 161.
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