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shrive555
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Let for notation sakes the difference b/w any 2 integers x<y be yx

Here we have 5 differences : ba, cb, dc, ed
Hence we have to maximize (ba+cb +dc+ed) = (ba+3+ed)
a + b + c + d + e = 30

a + (a+ba) + (a+ba+cb) + (a+ba+3) + (a+ba+3+ed) = 30

ba+3+ed = 30 - {5a + 3ba + cb +3}

Min (5a + 3ba + cb +3) = 5+3+1+3 = 12 (Using the properties given in the question)

Hence Max Range = 30-12 = 18
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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.
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Bunuel, thank you. I will take a look at them.
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Great set of practice questions from Bunuel. I just wish I knew how to bookmark a particular post or thread.
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Great set of practice questions from Bunuel. I just wish I knew how to bookmark a particular post or thread.

Attachment:
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2013-09-26_1342.png [ 72.76 KiB | Viewed 33871 times ]

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

Hope it helps.
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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 !
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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
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shrive555
Positive integers a, b, c, d and e are such that a < b < c < d < e. If the average (arithmetic mean) of the five numbers is 6 and d - b = 3, then what is the greatest possible range of the five numbers?

12
17
18
19
20

Range = e-a.
d-b=3 or d=b+3

Average = 6 thus a+b+c+d+e=30

To maximize range, we need to minimize a and maximize e. Minimum a is 1, so we choose that.
Since sum is fixed, we should choose minimum possible values for b,c,d to maximize e.
b>a, minimum choice is 2
c>b, minimum choice is 3
d=b+3 so d=5
So e(max) = 30 - a(min) -b(min) -c(min) - d(min) = 30-1-2-3-5 = 19
Range(max) = e(max)-a(min) = 19-1 =18

Answer is (c)

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
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shrive555
Positive integers a, b, c, d and e are such that a < b < c < d < e. If the average (arithmetic mean) of the five numbers is 6 and d - b = 3, then what is the greatest possible range of the five numbers?

A. 12
B. 17
C. 18
D. 19
E. 20


IMO B

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shrive555
Positive integers a, b, c, d and e are such that a < b < c < d < e. If the average (arithmetic mean) of the five numbers is 6 and d - b = 3, then what is the greatest possible range of the five numbers?

A. 12
B. 17
C. 18
D. 19
E. 20

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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