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

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Positive integers a, b, c, d and e are such that a<b<c<d<e [#permalink] New post 18 Oct 2010, 09:57
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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
[Reveal] Spoiler: OA

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Re: Range [#permalink] New post 18 Oct 2010, 10:14
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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Re: Range [#permalink] New post 18 Oct 2010, 13:25
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shrive555 wrote:
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)
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Re: Range [#permalink] New post 06 Feb 2012, 08:01
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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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Re: Range [#permalink] New post 06 Feb 2012, 09:08
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nonameee wrote:
kys123, thanks a lot.


You can check similar questions to practice:
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Hope it helps.
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Re: Positive integers a, b, c, d and e are such that a<b<c<d<e [#permalink] New post 06 Feb 2012, 11:02
Bunuel, thank you. I will take a look at them.
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Re: Positive integers a, b, c, d and e are such that a<b<c<d<e [#permalink] New post 18 Sep 2013, 06:31
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Re: Positive integers a, b, c, d and e are such that a<b<c<d<e [#permalink] New post 25 Sep 2013, 22:52
Great set of practice questions from Bunuel. I just wish I knew how to bookmark a particular post or thread.
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Re: Positive integers a, b, c, d and e are such that a<b<c<d<e [#permalink] New post 26 Sep 2013, 01:45
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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 1330 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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Re: Positive integers a, b, c, d and e are such that a<b<c<d<e [#permalink] New post 30 Sep 2013, 10:24
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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Re: Positive integers a, b, c, d and e are such that a<b<c<d<e [#permalink] New post 17 Feb 2014, 06:51
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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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Positive integers a, b, c, d and e are such that a<b<c<d<e [#permalink] New post 02 Jul 2014, 03:42
shrouded1 wrote:
shrive555 wrote:
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
Positive integers a, b, c, d and e are such that a<b<c<d<e   [#permalink] 02 Jul 2014, 03:42
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