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16 Jul 2007, 17:16
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Set A is composed of nine numbers, labeled A1 through A9. Set B is also composed of nine numbers, labeled B1 through B9.
Set B is defined as follows: B1 = 1 + A1; B2 = 2 + A2; and so on, including B9 = 9 + A9. How much larger is the sum of
set B's mean and range than the sum of set A's mean and range?
4
9
13
17
Cannot be determined
Set B is defined as follows: B1 = 1 + A1; B2 = 2 + A2; and so on, including B9 = 9 + A9. How much larger is the sum of
set B's mean and range than the sum of set A's mean and range?
4
9
13
17
Cannot be determined
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16 Jul 2007, 17:45
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Call Set A = {a, b, c, d, e, f, g, h, i}
So Set B = {(a+1), (b+2), (c+3), ... ,(i +9)}
I am assuming that the sets are ordered.
RangeA = (i - a)
RangeB = (i+9) - (a+1) = (i - a) + 8
= RangeA + 8
MeanA = (a+b+...+i)/9
MeanB = ((a+1) + (b+2) +...(i+9))/9
= (a+b+..+i) + (1+2+3+...+9))/9
= (a+b+..+i)/9 + (1+2+3+..+9)/9
Use Sum of n consecutive integers = n * (n+1)/2
= MeanA + ((9*10)/2)/9
= MeanA + 5
RangeB + MeanB = (RangeA + 8) + (MeanA + 5)
= (RangeA + MeanA) + 13
So the answer is 13. If we cannot assume the sets to be ordered, I think the answer cannot be determined.
So Set B = {(a+1), (b+2), (c+3), ... ,(i +9)}
I am assuming that the sets are ordered.
RangeA = (i - a)
RangeB = (i+9) - (a+1) = (i - a) + 8
= RangeA + 8
MeanA = (a+b+...+i)/9
MeanB = ((a+1) + (b+2) +...(i+9))/9
= (a+b+..+i) + (1+2+3+...+9))/9
= (a+b+..+i)/9 + (1+2+3+..+9)/9
Use Sum of n consecutive integers = n * (n+1)/2
= MeanA + ((9*10)/2)/9
= MeanA + 5
RangeB + MeanB = (RangeA + 8) + (MeanA + 5)
= (RangeA + MeanA) + 13
So the answer is 13. If we cannot assume the sets to be ordered, I think the answer cannot be determined.
16 Jul 2007, 17:54
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iamba
I get E b/c the prob stem doesn't tell us anything about the #s. I think we could figure this out if we knew whether the #s in the first set were evenly spaced.
