Positive integers a, b, c, d and e are such that a<b<c<d<e

Well it's basic logic. If you want maximize one of the numbers you should minimize all other numbers if you're given the average or sum of the series.
An example would be there are five numbers the sum of the five numbers add to 20. All the numbers are positive what is the largest possible number in the set.
The smallest positive number is 1. 20-1-1-1-1 =16. If other numbers were not minimize then the answer would be less than 16. 20-1-2-1-2 = 14.

For the above question, the smallest positive integer is 1. Therefore A=1. Since they can't be the same number the next smallest integer is 2. B = 2. We know D-B = 3, so 2+3 = 5. D=5. C has to be between 2 and 5. 3 is the smallest number. Finally the last number should be 30-5-3-2-1=19.

Let's say you didn't choose the smallest number a = 3, then b = 4, c = 5, d = 7. E would be only be 11. The range would only be 8.

You can check similar questions to practice:
5-peices-of-wood-have-an-average-length-of-124-inches-and-123513.html
in-a-class-of-30-students-2-students-did-not-borrow-any-101980.html
set-r-contains-five-numbers-that-have-an-average-value-of-102087.html
three-boxes-of-supplies-have-an-average-arithmetic-mean-105819.html
the-average-arithmetic-mean-of-the-5-positive-integers-k-107059.html
if-the-average-of-5-positive-integers-is-40-and-the-127038.html

Hope it helps.

Bunuel, thank you. I will take a look at them.

Great set of practice questions from Bunuel. I just wish I knew how to bookmark a particular post or thread.

To bookmark click a star to the left of the topic name.
To review go to MY bookmarks.

Hope it helps.

a<b<c<d<e

As all these integers are +ve then minimum value of a=1.

Furthermore, to maximize the range we will have to minimize the value of "a"
and maximize the value of "e".

a=1
b=2
c=3
d= 5 (Since d=b+3)
e=19

Range = e - a = 19 -1
Range = 18 !

My way:: let's say that 'a' the smallest integer is 'x' and then we have that b=x+1, c=x+2, d=x+4 and e=x+k. We then have that the sum 5x+7+k = 30, and that k = 23-5x. Since the smallest value of x can be 1. Therefore k = 18 which is also the range is the maximum value. C is the correct answer

Hope this helps
Cheers
J

I have some inference like you, but

a + b + c + d + e = 30
d-b =3 => a + 2b + c + e = 27 => e = 27 - a - 2b - c
range = e - a = 27 - 2a - 2b - c

range is max when a, b, e are min => a=1, b=2, c=3 (as a,b,c are positive integers and a<b<c)

Max range = 27 - 2*1 - 2*2 - 3 = 18 => C

IMO B

a,b,c,d,e are positive integers and a<b<c<d<e
So, let a =1. To make the range largest we need to make e largest possible and other number smallest possible.
So, b = 2
c =3
d = b+3 = 5

a+b+c+d = 1+2+3 +5 =11
also a+b+c+d+e = 6*5 = 30
e = 30-11 =19

Range = 19-1 =18

Answer C

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