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.
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In any evenly spaced set the arithmetic mean (average) is equal to the median. For example for a set {1, 3, 5, 7, 9} mean=median=5.

But the reverse of this property is not always true. For example a set {0, 1, 1, 1, 2} has the median as well as mean equal to 1, but this set is not evenly spaced.

Hope it's clear.
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for a set of 7 numbers if the median is 23 then we can have 2 cases

1) all numbers are 23 { 23,23,23,23,23,23,23}, in this case too average is 23

2) {# <= 23, # <= 23, # <= 23, 23( center digit equal to 23) , # >= 23, >= 23, >= 23}, median of 7 numbers is 23 and average is 23, means that half of the numbers are below or equal to the median and half of the numbers are equal to or above the median .

Now the given condition in the question is that the maximum = 4X + 15 , where x is the smallest one .So we know that all numbers are not equal .

now in order to maximize the range we have to keep the other 6 as small as possible .

{ x,x,x,23,23,23,4X+15}

now if the smallest is x then largest is 4x+15, this is given, so we cannot change these two.
the center has to be 23 as this is the median.

so the smallest value for the 2, and 3, digit has to be equal to the smallest one which is equal to x , we cannot take anything larger then x as that could make the range smaller , but want to find the largest range , so we have to take the smallest numbers.

similarly:

the smallest value for 4th and 5th digit cannot be smaller than 23, as 23 is the median , and cannot be larger the 23 as again we are looking for the largest range, so we have the keep the 4th and 5th digits as small as possible , and the smallest we can get for the 4th and 5th digit is 23.

hence Bunuel's Statement

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}.

Hope it is clear to you now .Please ask if any doubt remains .

Hi

Remember that the total has to 161 , and we need one variable to sort this out , so by taking x and y , it would be difficult to find the range.

To find the answer the most efficient way is to take smallest one as x and largest one as 4x+15 and , and keep the others as small as possible .

Technically { 11, 11, 12, 23, 23, 24, 59 }this set is also possible , but to arrive at 11 and 59 , we have to proceed as shown by bunuel, one cannot solve the sum
assuming the set to be { x, x, x+1, 23, y, y+1,4x+15}.

solving this way will be complicated and time consuming . Hope it helps .

Isn't the average of first and last term supposed to equal the average hence (4x+15+x)/2 should equal 23 but then we get some other result for x??

I have a question here: now in order to maximize the range we have to keep the other 6 as small as possible

Range is to do with first and last term in a series, how do we assume the other terms. Am I missing something?

hey really nice question! I must admit couldn't fig out a way ahead for over 3mins! i totally missed the point that median has been provided and hence all 6 nos can be 23!

I had a similar problem while solving the question.

A way to look at this is that we need to maximize 3x+15, which is the range.
Hence we need to maximize x.

While maximize x, the set ( x , x+1 , x+ 3 , 23 , y , y+ 1 , 4x + 15) will do worse than ( x, x, x , 23 , 23 , 23 , 4x+15) when the total is fixed.(161) .

Makes sense?

Please check the solutions above.
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