basso25@ wrote:if q is an odd number and the median of q consecutive integers is 120, what is the largest of these integers?

a) (q-1)/2 + 120
b) q/2 + 119
c) q/2 + 120
d) (q+119)/2
e) (q+120)/2

You haven't posted the options properly.

Easiest method to solve this problem is to plug q = 1.
Hence, there is only one integer : 120.

So the correct option should yield 120 when q = 1.

a) (q-1)/2 + 120 = 120
b) q/2 + 119 = Not integer
c) q/2 + 120 = Not integer
d) (q+119)/2 = 60
e) (q+120)/2 = Not integer

The correct answer is A.

basso25@ wrote:if q is an odd number and the median of q consecutive integers is 120, what is the largest of these integers?

a) (q-1)/2 + 120
b) q/2 + 119
c) q/2 + 120
d) (q+119)/2
e) (q+120)/2

Here's another approach.

If q is odd, then the median of the q integers will be the middle number.
So, of the q integers, the middle number is 120
Of the remaining q-1 integers (not counting 120), half of them are greater than 120 and half are less than 120.
In other words, (q-1)/2 of the integers are greater than 120.
Aside: (q-1)/2 represents half of the remaining q-1 integers.
So, to find the biggest number, we'll take the median (120) and add (q-1)/2 to get ...
120 + (q-1)/2
Answer = A

Cheers,
Brent

vipulgoyal wrote:Brent,
I have a query here
its not necessary that q-1/2 numbers are greater then median
for example 119 119 120 120 120
more over here the geatest is NOT 122
120 + (5-1)/2 = 122
please shed some light

Read the question again.
It said "q consecutive integers"
Your example, '119 119 120 120 120' are not 5 consecutive integers.