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# Set B has three positive integers with a median of 9. if

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Manager
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Set B has three positive integers with a median of 9. if [#permalink]  07 Jan 2005, 12:41
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Set B has three positive integers with a median of 9. if the largest possible range of the three numbers is 19, given a certain mean, what is that mean?

(A) 22

(B) 10

(C) 9.6

(D) 9

(E) It can not be determined from the information given.
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[#permalink]  07 Jan 2005, 12:49
one of those rare, it cannot be determined...know the range but have no idea what the numbers are... (E) is the answer
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[#permalink]  07 Jan 2005, 13:06
Yeah I got E as well.

so if the three numbers are x,y,z (in consecutive order) and the median is 9 then y = 9

that would mean you have x,9,z

then they give you the range = 19

that means x-z = 19

to find the average you need x+z that is why it is insufficient.
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[#permalink]  07 Jan 2005, 13:23
E
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[#permalink]  07 Jan 2005, 13:40
If option E had not been there, I would have picked B...
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[#permalink]  07 Jan 2005, 13:59
Banerjeea, I agree with your calculation which is exactly why I said I would have picked B. However, there are other sets of numbers that will satisfy the condition too but with a different mean.

1) {1,9,20} => Mean = 10 (choice B), Range = 19
2) {4,9,23} => Mean = 12, Range = 19
3) {7,9,26} => Mean = 14, Range = 19

Given that there are other sets why do you pick B as opposed to E?
Or Am I thinking like I would for a Data Sufficiency problem, i.e, if there are more than one solution, it is insufficient?
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Re: PS- Median, Range & Mean [#permalink]  07 Jan 2005, 21:53
mbassmbass04 wrote:
Set B has three positive integers with a median of 9. if the largest possible range of the three numbers is 19, given a certain mean, what is that mean?

(A) 22

(B) 10

(C) 9.6

(D) 9

(E) It can not be determined from the information given.

The range of mean is 10 to 44/3. the value of the mean in the answer choices in between the range is only 10. therefore the OA is 10. However, E underestimates the acceptability of B as OA.
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[#permalink]  08 Jan 2005, 06:15
I choose Answer E,

But in fact the OA is B.

and this is what the answer mentions:

Plug in your answers here. Don't start with C, because 9.6 is an annoying number to calculate with. Start with (B) instead. If the mean is 10 and the median is 9, what would the largest possible range of the three integers be? To find that, our three integers must fit into the equation (a+b+c)/3 = 10. The median, b, equals 9, so a+c=21. The range is defined as c-a, to make c-a as large as possible, given that a+c=21, we can set a=1 and c=20. That does give us a range of 19, so (B) is the correct answer.
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[#permalink]  09 Jan 2005, 09:12
mbassmbass04 wrote:
I choose Answer E,

But in fact the OA is B.

and this is what the answer mentions:

Plug in your answers here. Don't start with C, because 9.6 is an annoying number to calculate with. Start with (B) instead. If the mean is 10 and the median is 9, what would the largest possible range of the three integers be? To find that, our three integers must fit into the equation (a+b+c)/3 = 10. The median, b, equals 9, so a+c=21. The range is defined as c-a, to make c-a as large as possible, given that a+c=21, we can set a=1 and c=20. That does give us a range of 19, so (B) is the correct answer.

but there are other sets which satisfies the requirement. why B then?
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[#permalink]  09 Jan 2005, 22:08
i am still confused why not E. since there are other sets of number which can give similar results??
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[#permalink]  10 Jan 2005, 23:19
would you guys say that this is a good problem for number picking??
Manager
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[#permalink]  29 May 2006, 14:44

Select any other mean in the list of answers A = 22 or D = 9

Try pluggin in and you will get that

for A)

a+9+c/3 = 22 for A => a+c = 66 -9 =57

You need to come with two numbers such that c-a, which is the range = 19, c+a totals 57 with a not exceeding 9. Not possible

for D)

a+9+c/3 = 9 for D => a+c = 27-9 = 18

With the above result you can not get two values a and c such that a+c =18 and c-a, which is the range = 19 (as both c and a are +ve integers)

for B)

a+9+c/3 = 10 for B => a+c = 21

With the above result you can get two values a and c such that a+c = 21 and c -a =19, for Ex: 1 and 20, 2 and 19, 2 and 18 etc.

With the above mean, you can 3 positive numbers, which satisfy the range condition as well as the median.

Hence B.
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[#permalink]  30 May 2006, 15:43
This is a BAD question.

From the info given, we know mean=(x+9+x+19)/3=(2x+28)/3, where x is the smallest of the three integers.
It's easy to see that when x takes different values mean would be different. So E would definitely be the answer.

However, if the question stem changes to "what could be the mean", then B would be the answer. To solve this we would have to plug in the smallest possible x (x=1) and the biggest x (x=8) to see what the mean is. We then know that the range of the mean is from 10 to 44/3, so only B fits the bill.
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Re: PS- Median, Range & Mean [#permalink]  30 May 2006, 17:16
mbassmbass04 wrote:
Set B has three positive integers with a median of 9. if the largest possible range of the three numbers is 19, given a certain mean, what is that mean?

(A) 22
(B) 10
(C) 9.6
(D) 9
(E) It can not be determined from the information given.

yes, its an ambigious question..

but great discussionssssss............
Re: PS- Median, Range & Mean   [#permalink] 30 May 2006, 17:16
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# Set B has three positive integers with a median of 9. if

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