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A set of numbers contains 7 integers and has an average

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Re: A set of numbers contains 7 integers and has an average  [#permalink]

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New post 26 Feb 2019, 07:21
Bunuel wrote:
Joy111 wrote:
A set of numbers contains 7 integers and has an average (arithmetic mean) value as well as a median value of 23. If the largest value is equal to 15 more than 4 times the smallest number, what is the largest possible range for the numbers in the set?

A. 33
B. 35
C. 38
D. 48
E. 75


The average of 7 numbers is 23 --> the sum of these numbers is 7*23=161;
The median of 7 numbers is 23 --> 23 is the middle term: {*, *, *, 23, *, *, *};
The largest value is equal to 15 more than 4 times the smallest number --> say the smallest number is x then the largest number would be 4x+15, so our set is {x, *, *, 23, *, *, 4x+15};

Now, in order to maximize the range we need to make the second and the third numbers equal to x and the fifth and sixth numbers equal to 23, so the set should be {x, x, x, 23, 23, 23, 4x+15}.

Since the sum is 161 then x+x+x+23+23+23+4x+15=161 --> x=11.

The range is (4x+15)-x=3x+15=48.

Answer: D.



How did we consider the 5th and 6th terms as 23 too ? Can someone please elaborate the reason ?
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Re: A set of numbers contains 7 integers and has an average  [#permalink]

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New post 28 Jul 2019, 10:21
If we have to maximize a1 to get the largest range, why aren't we taking the median as a1?
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Re: A set of numbers contains 7 integers and has an average  [#permalink]

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New post 28 Jul 2019, 11:21
Hi sinkingbot,

This prompt places a number of 'restrictions' on us - the big one being that the AVERAGE has to be 23 (meaning that the SUM of the 7 terms is 161). If we made the first term 23, then that would NOT actually maximize the range (since that number - along with all of the other "lower" numbers - would make up too much of the 161 total).

To make the largest number as large as possible, we have to make EVERYTHING ELSE as small as possible. Here's how to do it:

X X X 23 23 23 (4X+15)

Summing all of these values gives us:

7X + 84 = 161
7X = 77
X = 11

So, the smallest number = 11 and the largest number = 4(11)+15 = 59

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Re: A set of numbers contains 7 integers and has an average  [#permalink]

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New post 03 Dec 2019, 06:26
Joy111 wrote:
A set of numbers contains 7 integers and has an average (arithmetic mean) value as well as a median value of 23. If the largest value is equal to 15 more than 4 times the smallest number, what is the largest possible range for the numbers in the set?

A. 33
B. 35
C. 38
D. 48
E. 75


n=7, avg_med=23, sum=7*23=161, tn=15+4a
minimize 'a' to find largest range: {a,a,a,md,md,md,tn}
3a+3md+tn=161…3a+3(23)+15+4a=161
7a=161-15-69;a=77/7=11
range: tn-a=15+4a-a=15+44-11=48

\(Ans (D)\)
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Re: A set of numbers contains 7 integers and has an average   [#permalink] 03 Dec 2019, 06:26

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