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

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laxieqv
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There's a set of consecutive integers in which the median [#permalink] New post   21 Jun 2006, 05:58
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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



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Re: There's a set of consecutive integers in which the median [#permalink] New post   21 Jun 2006, 06:21
D

for each of the conditions only the set from 5 to 15 is satisfactory
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Re: There's a set of consecutive integers in which the median [#permalink] New post   21 Jun 2006, 06:44
laxieqv wrote:
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


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...
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Re: There's a set of consecutive integers in which the median [#permalink] New post   21 Jun 2006, 06:52
Agree with Prof and helg. Both conditions can provide the median, largest/smallest number and hence the number of ingeters in the list.
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Re: There's a set of consecutive integers in which the median [#permalink] New post   21 Jun 2006, 10:32
good question.
(D) should suffice as explained byothers.
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Re: There's a set of consecutive integers in which the median [#permalink] New post   21 Jun 2006, 10:42
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.
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Re: There's a set of consecutive integers in which the median [#permalink] New post   21 Jun 2006, 11:08
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
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Re: There's a set of consecutive integers in which the median [#permalink] New post   21 Jun 2006, 11:09
Great way to solve.. Thanks Prof!


Professor wrote:
laxieqv wrote:
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


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...
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laxieqv
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Re: There's a set of consecutive integers in which the median [#permalink] New post   02 Jul 2006, 06:56
Yup, answer is D :)
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Re: There's a set of consecutive integers in which the median [#permalink] New post   02 Jul 2006, 07:41
laxieqv wrote:
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


(D)

With steps,

taking 'A',

1. (max - 5) = ((5 + max) /2)

[range = max - min, median = (max + min)/2 for any progressive set]

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.

=>



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Re: There's a set of consecutive integers in which the median [#permalink]
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