There's a set of consecutive integers in which the median

Question

There's a set of consecutive integers in which the median equals the range. How many integers are there in the group?
 1), the smallest number is 5 
 2), the biggest number is 15

helg · Answer

D

for each of the conditions only the set from 5 to 15 is satisfactory

Professor · Answer

m = (l+s) / 2 
r = l - s

m = r 
(l+s)/2 = l - s
l = 3s

if we put the values for s or l, each st is suff. 

great stuff laxi...

giddi77 · Answer

Agree with Prof and helg. Both conditions can provide the median, largest/smallest number and hence the number of ingeters in the list.

sgrover · Answer

good question.
(D) should suffice as explained byothers.

shampoo · Answer

I comeup with D aswell. Prof your approach is awsome. So, for all cases the largest # L has to be 3*s (s=smallest #) for median and range to be equal.

haas_mba07 · Answer

Let l = largest s = smallest

S1: range = l-s = l-5
 median = 5+(l+1)/2

 Given two equations we can solve for l.
 S1 is sufficient.

S2: Largest is 15 
 Similar to method in S1, S2 is also sufficient

Answer : D

haas_mba07 · Answer

Great way to solve.. Thanks Prof!

laxieqv · Answer

Yup, answer is D

necromonger · Answer

(D)

With steps,

taking 'A',

1. (max - 5) = ((5 + max) /2)

 

solving (1) , 2max - 10 = 5 + max => max = 15. SUFFICIENT

taking 'B'

15 - min = (15 + min)/2 
=> 30 - 2min = 15 + min => 3min = 15 
=> min = 5. SUFFICIENT

Therefore, D.

=>

There's a set of consecutive integers in which the median

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