Problem Solving

Which of the following CANNOT be the median of the three positive integers x , y and Z?

A. x
b. z
c. x+z
d. x+z/2
e.x+z/3












I sifter out choices a b and d. I was left with c and e. BY not testing numbers thoroughly enough I chose e. QA is C. How can I pick better numbers to avoid this problem?

can u tell me the sequence for the which u got e as the answer......for all the numbers i tried i got c as the answer.

Enginpasa1 wrote:Which of the following CANNOT be the median of the three positive integers x , y and Z?

A. x
b. z
c. x+z
d. x+z/2
e.x+z/3

I sifter out choices a b and d. I was left with c and e. BY not testing numbers thoroughly enough I chose e. QA is C. How can I pick better numbers to avoid this problem?

First, you need to read the question carefully.

Based on the stem, we're only allowed to pick positive integers. However, nothing says that x, y and/or z have to be different numbers.

We have an odd number of terms, so the median will simply be the middle term in the set.

We can quickly eliminate (a) and (b), since we can arrange our terms in any order.

(d) says that the median is the average of x and z. Could y be right in the middle? Sure, we can pick {2, 4, 6} as our set.

(e) doesn't really apply any specific rule, so let's just try picking numbers. {2, 4, 10} (x, y and z respectively) follows the rule, since (2+10)/3 = 4.

(c) sets the median as the sum of two of the terms. Since we know that the terms are all positive, it's impossible for the median to be bigger than TWO of the individual terms (think Golidlocks and the 3 bears - one term is too small, one is too big, one is just right): choose (c).

excellent now i get it. It was a simple error of not reading the question carefully. Stuart, I have a DS question and having trouble picking numbers b/c it makes it very tough. can you help.. its in my posts og ds 151. thanks for everything