Average of 4 distinct positive integers is 60. How many of the integers are less than 50?
1. One of the integers is 200
2. Median of the 4 integers is 50
The OA is D but I disagree
I know we can solve it with stem 1
However, I disagree with B being able to tell us anything
Am I not correct that when there is an even number of terms, the median is the average of the middle 2 terms?
Here is the reason I disagree with stem two being an answer
We could have numbers that look like this and still have a median of 50 and an average of 60
50 50 50 90
That would leave 0 numbers below 50
We could also have 30 48 52 110
In that instance the median is 50 but we have two numbers below 50
So with stem 2..we can't tell how many numbers are below 50 ..
Am I way out in left field?
S1: obviously suff.
S/4=60 S=240. lets say the four integers are w,x,y,z. The median is 50. So x+y/2=50 So x+y=100. Thus w+z has to be 140.
Since X,Y must be the middle numbers here. W,Z must be the outer numbers.
Ok so we know that W<X<Y<Z. So we can have 48+49+51+92.
Lets try other numbers: 2+3+97+138.
From this it can be seen that w<50 or the numbers wont work. x must also be less than 50 b/c the numbers are distinct.
Work it out u can't have w or x greater than 50 or this doesnt work.
So B is Sufficient.
Answer is D