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MathRevolution
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You didn't specify that the numbers are integers!!!
With current wording I would answer E.
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MathRevolution
Set S has six numbers and their average (arithmetic mean) is 32. What is the median of the numbers?

1) The six numbers are greater than or equal to 31
2) There is “37” in set S

the stem does not specify these numbers are integers and, in your calculation, you only use integer numbers!

Please check!
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Can anyone help

I am concerned about the word SET S

The first Instant My understanding was SET doesn't contain repeated elements

Hence
(i) a,b,c,d,e,f (Numbers but not just Integers)]
a+b+c+d+e+f = 6*32 => sum(a,b,c,d,e,f) = 192

Case 1: 6*32 --> [32,32,32,32,32,32] -- Not possible since it is a Set (32 is repeating)
[31,32,32,32,32,33] -- Not Possible since it is a Set (32 is repeating)
[31,32,33,34,35,36] -- average not equals 32
Case2: since it is a set and all numbers are >=31
a can be 31,31.5,32,32.5 (considering a as lowest number)
if a = 31
(6/2)*[31+L] = 192
(3)*[31+L] = 192
[31+L] = 64
L = 33 ----> limitation this is considering series in AP (but condition of AP among numbers is missing)
hence first option is insufficient
(ii) set has 37
sum = 192
element in set = 37
rest sum = 192 - 37 => 155
[31,31,31,31,31,37]
[30,31,32,31,31,37]
[29,30,32,33,31, 37] -- median = 63/2
[28,29,33,34,31,37] -- median = 64/2
Not sufficient

Combining Both
[31,31,31,31,31,37] -- But S is set
Hence Answer is E
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You didn't specify that the numbers are integers!!!
With current wording I would answer E.
­You don't need to assume integers, even if you take the elements as numbers you'll get the same answer by combining the statements 31.
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How on earth can a set have repeated (non-unique) numbers ? The answer should be E.
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