Harley1980 wrote:
The average of a set of 7 integers equal 27. If the smallest number of set equal \(\frac{1}{3}\) of the largest number of set, what is the largest number in the set?
(1) The median of the set equal 23
(2) The range of the set equal 40
1) In order to receive max number we should minimize all other numbers. So before median we have three equal numbers a1, after median we have two numbers equal to median and at the end we have max number:
a1, a1, a1, 23, 23, 23, a7
We know that the smallest number equal \(\frac{1}{3}\) of the largest number so we can write equation:
\(a1 = \frac{1}{3}*a7\)
And as we know average we can find the sum of all numbers: average * number of elements = sum
\(27 * 7 = 189\)
So \(a1 + a1 + a1 + 23 + 23 + 23 + a7 = 189\) and we can rewrite it to
\(3*\frac{1}{3}*a7 + 69 + a7 = 189\) --> \(2a7 = 120\) --> \(a7 = 60\)
Sufficient
2) From this statement we know that max element - min element = 40 and that the smallest number equal \(\frac{1}{3}\) of the largest number, so we can write equation:
\(a1 = \frac{1}{3}*a7\)
By combining this two equations
\(a7 - a1 = 40\) and \(a1 = \frac{1}{3}*a7\)
we receive
\(a7 - \frac{1}{3}*a7=40\) --> \(\frac{2}{3}*a7=40\) --> \(a7 = 60\)
Sufficient
Answer is D
In my opinion, this question cannot be solved sufficiently by (1).
By what you had solved: smallest was 20 and largest was 60.
From what i have solved: smallest is 18 and largest is 54.
So IMO, the answer is B and not D.