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

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Set A is composed of nine numbers, labeled A1 through A9. Set B is als [#permalink]  10 Nov 2014, 08:39
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Difficulty:

95% (hard)

Question Stats:

37% (03:06) correct 63% (02:13) wrong based on 50 sessions

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.
[Reveal] Spoiler: OA

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Kudos [?]: 10 [0], given: 17

Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als [#permalink]  10 Nov 2014, 10:54

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 = 9-1 = 8
then Set B (2,4,6,8,10,12,14,16,18) - Range = 18-2 = 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) = 26-13 = 13
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Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als [#permalink]  10 Nov 2014, 11:01
I am a bit old fashioned and just wrote it all out on paper and used the plug-in 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= 11-3 = 8 range = 20-4 = 16
mean + range: 15 mean + range: 28

(set B mean + range) - (set A mean + range) = 28 - 15 = 13

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Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als [#permalink]  10 Nov 2014, 19:53
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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 re-organised 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
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Last edited by PareshGmat on 11 Nov 2014, 18:12, edited 2 times in total.
Math Expert
Joined: 02 Sep 2009
Posts: 28781
Followers: 4594

Kudos [?]: 47463 [1] , given: 7123

Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als [#permalink]  11 Nov 2014, 02:49
1
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Expert's post
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 [#permalink]  11 Nov 2014, 06:58
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: B1-1, B2-2, B3-3, B4-4, B5-5....-> -1-2-3-4-5-6-7-8-9 = -45/9 = -5 ( mean of set A)
Range of set A from -1 to -9 = 8

5+8 = 13
-5+8= 4

Hence 13-4= 9

Does it make any sense or have I broken all math rules ? ☺
I am new here.
Thank You !
Math Expert
Joined: 02 Sep 2009
Posts: 28781
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Kudos [?]: 47463 [0], given: 7123

Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als [#permalink]  12 Nov 2014, 03:19
Expert's post
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 re-organised 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 [#permalink]  12 Nov 2014, 04:23

A (A1+.....+A9)
B (B1+.....+B9) = (A1+.....+A9) + 45

A

Mean = (A1+....+A9)/9
Range = A9-A1

B

Mean = (A1+...+A9+45)/9 = (A1+....+A9)/9 + 5
Range = A9+9-(A1+1) = A9-A1+8

Combining

(A1+....+A9)/9 + 5+ A9-A1+8 - ((A1+....+A9)/9+A9-A1) = 5+8 = 13

Answer C - In my view
Math Expert
Joined: 02 Sep 2009
Posts: 28781
Followers: 4594

Kudos [?]: 47463 [0], given: 7123

Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als [#permalink]  12 Nov 2014, 04:26
Expert's post
viniciuszds wrote:

A (A1+.....+A9)
B (B1+.....+B9) = (A1+.....+A9) + 45

A

Mean = (A1+....+A9)/9
Range = A9-A1

B

Mean = (A1+...+A9+45)/9 = (A1+....+A9)/9 + 5
Range = A9+9-(A1+1) = A9-A1+8

Combining

(A1+....+A9)/9 + 5+ A9-A1+8 - ((A1+....+A9)/9+A9-A1) = 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 [#permalink]  12 Nov 2014, 11:31
Bunuel wrote:
viniciuszds wrote:

A (A1+.....+A9)
B (B1+.....+B9) = (A1+.....+A9) + 45

A

Mean = (A1+....+A9)/9
Range = A9-A1

B

Mean = (A1+...+A9+45)/9 = (A1+....+A9)/9 + 5
Range = A9+9-(A1+1) = A9-A1+8

Combining

(A1+....+A9)/9 + 5+ A9-A1+8 - ((A1+....+A9)/9+A9-A1) = 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.
Re: Set A is composed of nine numbers, labeled A1 through A9. Set B is als   [#permalink] 12 Nov 2014, 11:31
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