tony_stark wrote:I've been reading about this problem on various forums but I can't seem to grasp the overall concept being tested.
Tom, Jane, and Sue each purchased a new house. The average (arithmetic mean) price of the three houses was $120,000. What was the median price of the three houses?
(1)The price of Tom's house was $110,000.
(2)The price of Jane's house was $120,000.
OA: B
I understand that we're looking for the median. How I've solved this so far is:
(T+J+S)/3 = 120,000 ==> T+J+S = 360,000
(1) T-110,000. Therefore J+S = 360,000-110,000 = 250,000
- This is deemed insufficient, but I don't fully understand why.
(2) J=120,000. Since Jane's house is the same as the mean, I assumed there was some trick here that I didn't capture. The only thing I could think of was that in evenly spaced sets the mean = median, but I don't think that applies in this question.
Can someone please explain in detail why (1) is insufficient and why (2) is sufficient and the theory I'm missing in answering this question?
Appreciate any help in advance,
Tony_Stark
Let T = Tom's house, J = Jane's house, S = Sue's house.
Sum = number * average = 3*120,000 = 360,000.
Thus, T+J+S = 360,000.
Statement 1: T = 110,000
Thus, J + S = 360,000 - 110,000 = 250,000.
Case 1: J = 10,000, T = 110,000, S = 240,000.
Median = 110,000.
Case 2: T = 110,000, J = 120,000, S = 130,000.
Median = 120,000.
Since the median can be different values, INSUFFICIENT.
Statement 2: J = 120,000
Thus, T + S = 360,000 - 120,000 = 240,000.
Case 1: T = 120,000, J = 120,000, S = 120,000.
Median = 120,000.
Only the values of T and S can be changed.
Case 2: If the value of T increases above 120,000, then the value of S must decrease below 120,000, so that T+S = 240,000.
J will still be in the middle, so the median will remain 120,000.
Case 3: If the value of S increases above 120,000, then the value of T must decrease below 120,000, so that T+S = 240,000.
J will still be in the middle, so the median will remain 120,000.
Since in every case median = 120,000, SUFFICIENT.
The correct answer is
B.
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