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Bunuel
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Let the smallest be Y, The largest hence becomes Y+40
From this we can infer that Y should be greater or equal to 10 and maximum of Y is 40 (If y becomes less than 10, then median will not be 50 and if Y become more than 40, the median of lowest three will be less than 40)

Let the numbers be:
Y, 40,50, Unknown, Y+40

Option 1: Zero: This is possible when Y=10
Option 2: 20: This is possible when Y=30
Option 3: 40, Not possible, Lets give Y its maximum value which is 40, then largest becomes 80 and range of largest three will be 80-50=30
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I thought this way.
_ 40 50 _ _ This is given
smallest number can not be less than 10, otherwise Range of 5 numbers will become more than 40
Hence,
10 40 50 50 50 -- Here Range of 5 is 40 and Range of last 3 is 0
Largest number can not be more than 80, otherwise Range of 5 numbers will become more than 40
Hence,
40 40 50 80 80 -- here Range of 5 is 40 and Range of last 3 is 30
Hence Range of Last 3 numbers will be --> 0 ≤ Range ≤ 30

So Answer will be C
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We have 5 numbers with 50 as median (ordered from greatest to smallest): \(a,\: b,\: 50,\: d,\: e\:\)

Range: \(a-e=40 \) ⭢ \(a=40+e\)

Median of the three smallest numbers is 40 ⭢ \(d=40\)

Therefore, our set is: \(40+e,\: b,\: 50,\: 40,\: e\:\)

We have to find the range of the three gratest numbers, which is \(a-50\).


(I) If \(a-50=0\) ⭢ \(a=50\).
Set will be: \(50,\: 50,\: 50,\: 40,\: 10\)
Possible


(II) \(If a-50=20\) ⭢ \(a=70\)
Set will be: \(70,\: b,\: 50,\: 40,\: 30\:\), with \(b \) that can be any number between 50 and 70.
Possible

(III) \(If a-50=40\) ⭢ \(a=90\)
Set will be: \(90,\: b,\: 50,\: 40,\: 50\:\) ⭢ \(e \) cannot be 50 because it has to be the smallest number.
Not possible


Answer: C
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