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

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Manager
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Positive integers a, b, c, d and e are such that a<b<c<d<e  [#permalink]

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18 Oct 2010, 10:57
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65% (hard)

Question Stats:

66% (02:26) correct 34% (02:51) wrong based on 490 sessions

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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

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18 Oct 2010, 14:25
11
2
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

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Joined: 11 Aug 2010
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18 Oct 2010, 11:14
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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06 Feb 2012, 09:01
2
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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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06 Feb 2012, 10:08
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Re: Positive integers a, b, c, d and e are such that a<b<c<d<e  [#permalink]

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06 Feb 2012, 12:02
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]

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25 Sep 2013, 23:52
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]

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26 Sep 2013, 02:45
saintforlife wrote:
Great set of practice questions from Bunuel. I just wish I knew how to bookmark a particular post or thread.

Attachment:

2013-09-26_1342.png [ 72.76 KiB | Viewed 14828 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]

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30 Sep 2013, 11:24
1
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]

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17 Feb 2014, 07:51
1
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
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Positive integers a, b, c, d and e are such that a<b<c<d<e  [#permalink]

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02 Jul 2014, 04:42
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

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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Re: Positive integers a, b, c, d and e are such that a<b<c<d<e  [#permalink]

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09 Sep 2016, 07:26
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?

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

IMO B

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Re: Positive integers a, b, c, d and e are such that a<b<c<d<e  [#permalink]

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04 Aug 2017, 00:59
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?

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

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Re: Positive integers a, b, c, d and e are such that a<b<c<d<e  [#permalink]

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02 Nov 2018, 06:26
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