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Set A is composed of nine numbers, labeled A1 through A9. Set B is als
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10 Nov 2014, 09:39
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27% (02:37) correct 73% (02:33) wrong based on 259 sessions
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Tough and Tricky questions: Sets. 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? A. 4 B. 9 C. 13 D. 17 E. cannot be determined Kudos for a correct solution.
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Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als
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Updated on: 11 Nov 2014, 19:12
Answer = E = Cannot be determined
Set A = { a1, a2 , a3 ......... a9}
Nowhere in the question its mentioned that a1 < a2 < a3 < .............. <a9 as we are "assuming" it to be "ideally"
Now, this will NOT make effect for computing mean, however will effect on computing "Range"
For Example:
Set A = {9, 8, 7, 6, 5, 4, 3, 2, 1}
Set B = {10, 10, 10, 10, 10, 10, 10, 10, 10, 10}
To calculate range of set A, it has to be reorganised from {1,2.......... 9}
Range = 8; Mean = 5
Set B
Range = 0; Mean = 10
Difference = 3
We may take multiple examples like this which will give different answers.
Bunuel: Thank you so much for indicating that C was incorrect answer
Originally posted by PareshGmat on 10 Nov 2014, 20:53.
Last edited by PareshGmat on 11 Nov 2014, 19:12, edited 2 times in total.




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Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als
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10 Nov 2014, 11:54
Answer: C  13 Range of set B will always twice that of Set A in value. if the Set A (1,2,3,4,5,6,7,8,9)  Range = 91 = 8 then Set B (2,4,6,8,10,12,14,16,18)  Range = 182 = 16 Mean of Set B will be twice that of Set A if the Set A (1,2,3,4,5,6,7,8,9)  Mean = 45/9 = 5 then Set B (2,4,6,8,10,12,14,16,18)  Mean = 90/9 = 10 Set A (Range + Mean) = 8+5 = 13 Set B (Range + Mean) = 16+10 = 26 Set B (Range + Mean)  Set A (Range + Mean) = 2613 = 13



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Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als
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10 Nov 2014, 12:01
I am a bit old fashioned and just wrote it all out on paper and used the plugin method. At first I chose #s for A and then I used the rule that was given to figure out the values for B. Calculations are below. This took me 2min25sec.
A1=3 B1= 4 A2=4 B2 = 6 A3=5 B3=8 A4=6 B4=10 A5=7 B5=12 A6=8 B6=14 A7=9 B7=16 A8=10 B8=18 A9=11 B9=20 mean= sum/9 = 7 mean= sum/9 = 12 range= 113 = 8 range = 204 = 16 mean + range: 15 mean + range: 28
(set B mean + range)  (set A mean + range) = 28  15 = 13
Answer C!



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Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als
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11 Nov 2014, 03:49
C is not correct. Anyone else?
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Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als
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11 Nov 2014, 07:58
Set A  A1, A2, A3, A4 and so on......
Set B: A1 +1, A2 +2, A3+ 3, A4+4 ,A+5...> 1+2+3+4+5+6+7+8+9 = 45/9 =5 (mean of set B) Range of set B from 1 to 9 = 8
Set A: B11, B22, B33, B44, B55....> 123456789 = 45/9 = 5 ( mean of set A) Range of set A from 1 to 9 = 8
5+8 = 13 5+8= 4
Hence 134= 9
Answer B.
Does it make any sense or have I broken all math rules ? ☺ I am new here. Thank You !



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Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als
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12 Nov 2014, 04:19
PareshGmat wrote: Answer = E = Cannot be determined
Set A = { a1, a2 , a3 ......... a9}
Nowhere in the question its mentioned that a1 < a2 < a3 < .............. <a9 as we are "assuming" it to be "ideally"
Now, this will NOT make effect for computing mean, however will effect on computing "Range"
For Example:
Set A = {9, 8, 7, 6, 5, 4, 3, 2, 1}
Set B = {10, 10, 10, 10, 10, 10, 10, 10, 10, 10}
To calculate range of set A, it has to be reorganised from {1,2.......... 9}
Range = 8; Mean = 5
Set B
Range = 0; Mean = 10
Difference = 3
We may take multiple examples like this which will give different answers.
Bunuel: Thank you so much for indicating that C was incorrect answer Yes, the correct answer is E.
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Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als
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12 Nov 2014, 05:23
Answer C = 13.
A (A1+.....+A9) B (B1+.....+B9) = (A1+.....+A9) + 45
A
Mean = (A1+....+A9)/9 Range = A9A1
B
Mean = (A1+...+A9+45)/9 = (A1+....+A9)/9 + 5 Range = A9+9(A1+1) = A9A1+8
Combining
(A1+....+A9)/9 + 5+ A9A1+8  ((A1+....+A9)/9+A9A1) = 5+8 = 13
Answer C  In my view



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Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als
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12 Nov 2014, 05:26
viniciuszds wrote: Answer C = 13.
A (A1+.....+A9) B (B1+.....+B9) = (A1+.....+A9) + 45
A
Mean = (A1+....+A9)/9 Range = A9A1
B
Mean = (A1+...+A9+45)/9 = (A1+....+A9)/9 + 5 Range = A9+9(A1+1) = A9A1+8
Combining
(A1+....+A9)/9 + 5+ A9A1+8  ((A1+....+A9)/9+A9A1) = 5+8 = 13
Answer C  In my view Note that the Official Answer is E, not C.
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Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als
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12 Nov 2014, 12:31
Bunuel wrote: viniciuszds wrote: Answer C = 13.
A (A1+.....+A9) B (B1+.....+B9) = (A1+.....+A9) + 45
A
Mean = (A1+....+A9)/9 Range = A9A1
B
Mean = (A1+...+A9+45)/9 = (A1+....+A9)/9 + 5 Range = A9+9(A1+1) = A9A1+8
Combining
(A1+....+A9)/9 + 5+ A9A1+8  ((A1+....+A9)/9+A9A1) = 5+8 = 13
Answer C  In my view Note that the Official Answer is E, not C. Tks! i saw my mistake, i assumed sets A and B as sequence, i can`t make this inference.



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Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als
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08 Sep 2018, 08:00
Bunuel wrote: Tough and Tricky questions: Sets. 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? A. 4 B. 9 C. 13 D. 17 E. cannot be determined Case 1: A = 0, 0, 0, 0, 0, 0, 0, 0, 0 > range = 0, mean = 0, range + mean = 0 B = 1, 2, 3, 4, 5, 6, 7, 8, 9 > range = 8, mean = 5, range + mean = 13 B's sum  A's sum = 130 = 13 Case 2: A = 9, 8, 7, 6, 5, 4, 3, 2, 1 > range = 8, mean = 5, range + mean = 13 B = 10, 10, 10, 10, 10, 10, 10, 10, 10 > range = 0, mean = 10, range + mean = 10 B's sum  A's sum = 1013 = 3 Since the two cases yield different results, the difference between B's sum and A's sum cannot be determined. The use of the word larger in the question stem incorrectly implies that B's sum must be larger than A's sum. To avoid this miscommunication, the question stem should simply ask for the difference between B's sum and A's sum.
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Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als
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09 Sep 2019, 04:14
Bunuel wrote: Tough and Tricky questions: Sets. 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? A. 4 B. 9 C. 13 D. 17 E. cannot be determined Kudos for a correct solution.Given: 1. Set A is composed of nine numbers, labeled A1 through A9. 2. Set B is also composed of nine numbers, labeled B1 through B9. 3. Set B is defined as follows: B1 = 1 + A1; B2 = 2 + A2; and so on, including B9 = 9 + A9. Asked: How much larger is the sum of set B's mean and range than the sum of set A's mean and range? A = {A1,A2, ,,, A9} B = {B1,B2,,,,B9} = {1+A1,2+A2,,,,,9+A9} Mean of B = \(\frac{A1+A2+.... + A9 + 45}{9} = \frac{A1+A2 +... + A9}{9} + 5\) = Mean of A + 5 Since we don't know minimum and maximum of numbers {A1,A2,,, A9}, Range can not be determined. IMO E




Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als
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