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A set of 25 different integers has a median of 50 and . . .

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A set of 25 different integers has a median of 50 and . . .

by M7MBA » Sat Dec 30, 2017 7:25 am
A set of 25 different integers has a median of 50 and a range of 50. What is the greatest possible integer that could be in this set?

(A) 62
(B) 68
(C) 75
(D) 88
(E) 100

The OA is the option D.

I don't know how to solve this PS question. Experts, may you give me some help here? Please. I'd be thankful.
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by [email protected] » Sat Dec 30, 2017 11:40 am
Hi MY7MBA,

We're told that a set of 25 DIFFERENT integers has a MEDIAN of 50 and a RANGE of 50. We're asked for the greatest possible integer that could be in this set.

When it comes to maximizing or minimizing a value in a group of numbers, you have to think about what the other numbers would need to be to accomplish your goal. Here, we have a group of 15 DIFFERENT integers with a median of 50 and a RANGE of 50. That range will dictate how large the largest value can be.

With a median of 50, we know that 12 numbers are LESS than 50 and 12 numbers are GREATER than 50:

_ _ _ _ _ _ _ _ _ _ _ _ 50 _ _ _ _ _ _ _ _ _ _ _ _

To maximize the largest value, we need to maximize the smallest value. Here's how we can do it:

38 39 40 41 42 43 44 45 46 47 48 49 50 _ _ _ _ _ _ _ _ _ _ _ _

With 38 as the smallest value, and a range of 50, the largest value would be 88.

Final Answer: D

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Rich
Contact Rich at [email protected]
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by Scott@TargetTestPrep » Sun Aug 18, 2019 6:31 pm
M7MBA wrote:A set of 25 different integers has a median of 50 and a range of 50. What is the greatest possible integer that could be in this set?

(A) 62
(B) 68
(C) 75
(D) 88
(E) 100

The OA is the option D.

I don't know how to solve this PS question. Experts, may you give me some help here? Please. I'd be thankful.
Since the median (or the 13th number after the set is ordered) is 50 and the integers are different, the maximum value of the smallest possible integer 50 - 12 = 38. Therefore, the maximum value of the largest possible integer is 38 + 50 = 88 (since the range is 50).

Anwer: D

Scott Woodbury-Stewart
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by swerve » Mon Aug 19, 2019 10:02 am
Consider first number \(= x\)
Last number \(= x + 25\)
And we have median (8th number)\(=25\)
Series: \(x,\cdots, 6\) numbers,\(\cdots, 25,\cdots 6\) numbers,\(\cdots, (x+25)\)

In order to have the "last number" \((x+25)\) as the greatest possible, we have to maximize the "first number" \((x)\) (under 25).

To maximize \(x\), identify Integers under 25 such that they are consecutive in descending order.
i.e. 18, 19, 20, 21, 22, 23, 24, 25

Thus first number \(x=18\) and largest number \(x+25=18+25=43\Rightarrow\) __D__
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