DS-Mean & Median

B.

Because St. 1 tells us : Median is 15.
So the five numbers can be 19, (one number between 19-15), 15, (2 numbers less than 15)

St. 2 gives us a range between the largest and the smallest number.
Now visualize the numbers on the numberline, the largest number is 4 greater than the median, and median in 10 greater than the smallest number. Now the Mean will be in the exact middle of this range. Therefore, it has to be less than the Median.

I know this is an instinctive/visual approach, but it works for me.

What is the source of the question?..............

nitya34 wrote:Of the 5 numbers, the largest number is 4 greater than the median. Is the mean greater than the median?
(1) The largest number plus the median is 34.
(2) The median minus the smallest number is 10.

OA is.................... B

keep posing questions like this......................

Wonderful question

Here's an answer:

With the given problem, let the numbers be

a b c d (c+4)

in ascending order. Now, A) states that

c + (c + 4) = 34 -> c = 15.

Does this say much? Nope.

Next, look at B. This says that a = c - 10. We now have:

(c-10) b c d (c + 4)

Now, c < d < c+4 and c-10 < b < c. Now, we need to see if we can say that

[(c-10) + b + c + d + (c+4)]/5 > or < c

To simplify things, let b = c - x and d = c + y. To see the motivation for this, we know that b is less than c by some value (x) and d is greater than c by some value (y). Substituting into the above inequality, we have

y - x > or < 6

Can we state this for sure? We know that x and y are both positive, but y is AT MOST 4. Thus, we can state for sure that y - x < 6, so answer choice B suffices.

Again, great question!

I dont htikn you can assume b<c and d>c. They can all be the same.

@Above reply: I guess answer still remains same .

Well, b <= c <= d. Equality doesn't change the answer.