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Re: Set X consists of eight terms; 5 of them are equal and the remaining 3
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09 Jan 2024, 11:25
Let's say b = the value of the 5 terms that are equal, c = the value of the 3 remaining terms. We're given that b = 3c.
With sets of unknown values, I find that it helps me to write out each term with as much info as possible. Since we don't know whether the values are positive or negative, the set could be ordered (least to greatest) in 2 ways:
- all terms are Positive: \({c, c, c, b, b, b, b, b} = {c, c, c, 3c, 3c, 3c, 3c, 3c}\)
- all terms are Negative: \({b, b, b, b, b, c, c, c} = {3c, 3c, 3c, 3c, 3c, c, c, c}\)
Either way, we can see that b will be the median.
Going through each option:
(A) The lowest possible average of any four terms of the set is 12.
This forces us into the "positive" set, where the first 3 terms are c and 4th term is 3c.
Set up the average formula:
\(\frac{(c + c + c + 3c)}{4} = 12\)
\(6c = 48\)
c = 8, b = 24
Thus, we're able to say the median would be 24.
(B) The range of all the terms in the set is twice the value of the least term.
Positive Scenario:
\(3c - c = 2 * c\)
This doesn't provide any info on the values of b nor c, so we can't determine the median.
Negative Scenario:
\(c - 3c = 2 * (3c)\)
\(-2c = 6c\)
Leads us to a contradiction, so this isn't possible.
(C) The difference between the highest and the lowest term is 16.
Positive Scenario:
\(3c - c = 16\)
\(c = 8, b = 24\)
So we can determine the median is 24.
Negative Scenario:
\(c - 3c = 16\)
c = -8, b = -24
So we can determine the median is -24.