A list of number of 25 integers has a median of 50 and a range of 50.

Consider 25 numbers in ascending order to be \(x_1\), \(x_2\), \(x_3\), ..., \(x_{25}\).

The median of a set with odd number of elements is the middle number (when arranged in ascending or descending order), so the median of given set is \(x_{13}=50\);

The range of a set is the difference between the largest and the smallest numbers of a set, so the range of given set is \(50=x_{25}-x_{1}\) --> \(x_{25}=50+x_{1}\);

We want to maximize \(x_{25}\), hence we need to maximize \(x_{1}\). The maximum value of \(x_{1}\) is the median, so 50, hence the maximum value of \(x_{25}\) is \(x_{25}=50+50=100\).

The set could be {50, 50, 50, ..., 50, 100}

Answer: E.

Now, in order the answer to be 88, as given in the OA, the question should state that "A set of 25 different integers has a median of 50..."

In this case since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median-12=50-12=38\) and thus the maximum value of \(x_{25}\) is \(x_{25}=38+50=88\).

The set could be {38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 88}

Answer: D.

Hope it's clear.
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Shouldn't the question actually be "25 different positive integers" then?

That question is here: a-set-of-25-different-integers-has-a-median-of-50-and-a-129345.html
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The range 50 stops us from taking numbers non-consecutive numbers. Right?
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Not necessarily. One possible answer could have been 38,38,38... (9 more times),50,50,50,... (9 more times), 88. This way your set has a median value of 50 and range is 50 as well. In this case the elements of the set are not consecutive integers. No where in the question is it stated that the integers have to be different!

for questions like these, it is always better to use the options provided as follows:

You know that the median of 25 integers (or the 13th term) has to be 50 and that the range has to be 50.

Now start with option (E). Can the max. be 100? Lets check. If the max. is 100, then the minimum has to be 50 (for range to be 50) and we can then have a set as : 50,50...(22 more times), 100. Yes, this is a possible answer and thus is the correct answer. You dont have to spend any more time on this. But I agree with Bunuel that for the OA to be 88, there needs to additional information that can negate 100 as the maximum possible integer value.
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Set of 25 integers
median = 50
Range = 50

So, Mid number i.e. 13 th number is 50
{x1, x2, ......., x12, 50, x14, x15, ...... x25}
For the integer to be greatest and range to be 50 , we will have to maximize the lowest integer.

This is possible by keeping all the numbers on the left hand side of x13 = 50 as 50.
So set becomes {50,50,......., 50,50, x14, x15, ...... x25}

Now greatest integer can go upto 50+50 = 100.

Answer E

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