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
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The answer to Q is E.
I calculated C (13).
Here is my logic:
A = a1, a2,.......a9.
B = a1+1, a2+2,......a9+9.
Now range of A series = a9 - a1
Range of B series = b9 - b1 = a9 + 9 - a1 - 1 = 8.
I tried some range of A such as, -5, -4.....3 and 2,3,....10. In each case the diference is 8.
The same holds tru for Average also.
Can anyone please add your thougts to this questions.....
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- RonPurewal
- Students
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- Joined: Tue Aug 14, 2007 8:23 am
Re: Rangy Mean - Q from MGMAT CAT Maths
careful with those assumptions!
to wit:
not necessarily true. you are assuming, without justification, that a1 and a9 are the smallest and largest, respectively, of the numbers in set a (and therefore likewise for b1 and b9 in set b).
if this isn't true, and the numbers are ordered in some other way, then the ranges won't act the way you think they will.
as a rather extreme counterexample, but one that should certainly get the point across rather plainly, consider the set in which a1 = 9, a2 = 8, a3 = 7, ..., a9 = 1. following the directions given, every single one of b1, b2, ..., b9 will be 10. therefore the range of the b's will be 0.
that's enough to make the answer (e).
--
btw, you're right about the averages: the average of the b's must be 5 greater than the average of the a's, because the sum of the b's is 45 (= 1 + 2 + ... + 9) greater than the sum of the a's.
to wit:
minerr Wrote:Now range of A series = a9 - a1
Range of B series = b9 - b1 = a9 + 9 - a1 - 1 = 8.
not necessarily true. you are assuming, without justification, that a1 and a9 are the smallest and largest, respectively, of the numbers in set a (and therefore likewise for b1 and b9 in set b).
if this isn't true, and the numbers are ordered in some other way, then the ranges won't act the way you think they will.
as a rather extreme counterexample, but one that should certainly get the point across rather plainly, consider the set in which a1 = 9, a2 = 8, a3 = 7, ..., a9 = 1. following the directions given, every single one of b1, b2, ..., b9 will be 10. therefore the range of the b's will be 0.
that's enough to make the answer (e).
--
btw, you're right about the averages: the average of the b's must be 5 greater than the average of the a's, because the sum of the b's is 45 (= 1 + 2 + ... + 9) greater than the sum of the a's.
- Guest
Hello Ron,
Thank you for help.
For the range, in your example:
A series = 9,8,7,6,5,4,3,2,1,
B series = 10,10,...............10.
Range of A series = 8 and Range of B series = 0.
Difference of range = A-B= 8.
My point is that whatever sequence one will select, the defference in range will always be 8 because there is constant (phase) difference between the variables.
I am not abel to find a solution set in which the difference in range is not 8. Request you to please provide me a appropriate sent to A series.
I have my test in 2 days, kindly help me.
Thank you for help.
For the range, in your example:
A series = 9,8,7,6,5,4,3,2,1,
B series = 10,10,...............10.
Range of A series = 8 and Range of B series = 0.
Difference of range = A-B= 8.
My point is that whatever sequence one will select, the defference in range will always be 8 because there is constant (phase) difference between the variables.
I am not abel to find a solution set in which the difference in range is not 8. Request you to please provide me a appropriate sent to A series.
I have my test in 2 days, kindly help me.
- StaceyKoprince
- ManhattanGMAT Staff
- Posts: 9360
- Joined: Wed Oct 19, 2005 9:05 am
- Location: Montreal
5 posts Page 1 of 1