Hey Tony_Stark - Let me break it down for you!!

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.

Given : (T+J+S)/3 = 120,000 ==> T+J+S = 360,000
To find : Unique Value of Median
Question Type : Value

(1) T-110,000. Therefore J+S = 360,000-110,000 = 250,000
- This is deemed insufficient, but I don't fully understand why.

You are pretty much on track!!
why this answer is insufficient - There can be multiple values of median that can satisfy this option.

J+S = 360,000-110,000 = 250,000 =250K

Example 1
T=110K ,J= 150K , S= 100K -> Median = 110K (middle value after placing the numbers in increasing order)
Example 2
T=110K ,J =120K, S=130K -> Median = 120K

Hence Insufficient!

(2)The price of Jane's house was $120,000.

Total-120,000. Therefore J+S = 360,000-120,000 = 240,000 =240K

Now there can be two cases for this equation to hold true :

Case 1 :

J+S = 360,000-120,000 = 240,000 =240K

When J and S are different.

Irrespective of what value you plug in, the second highest value will be 120K because for sum of two different numbers to be 240K one will be greater than 120K and other less than 120K.

Median -> 120k as one number is 120k and other two numbers are either >120K or <120K


Case 2 :
When J and S are same.

J+S = 360,000-120,000 = 240,000 =240K

Note here , the Max value of J & S will be when they both are same i.e. 120K each
So in this case also T=J=S= 120K -> Median =120K



Hence B!!!!

Let me know if you still have doubts.

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.

In essence, I need to look at these questions and essentially do some arithmetic to disprove the stated median (e.g. 110,000).

Yep mate!!

Plus a tip :-

"Read both choices (1) and (2) and start with the one that seems easy between two" that way you can narrow down your answer choices.

Like in above case I personally found 2nd option more easy than 1st. So after solving 2nd i was pretty sure answer will be either A or D.

This technique not only saves time but also increases your chances to hit the right option while solving DS questions.