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

Question

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

Show Answer

Official Answer

E

A. 4 
B. 9 
C. 13 
D. 17 
E. cannot be determined

PareshGmat · Accepted Answer

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

DaVagabond · Answer

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 = 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

DangerPenguin · Answer

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

Answer C!

Bunuel · Answer

C is not correct. Anyone else?

member1 · Answer

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

Answer B.

Does it make any sense or have I broken all math rules ? ☺
I am new here.
Thank You !

Bunuel · Answer

Yes, the correct answer is E.

viniciuszds · Answer

Answer C = 13.

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

Bunuel · Answer

Note that the Official Answer is E, not C.

viniciuszds · Answer

Tks! i saw my mistake, i assumed sets A and B as sequence, i can`t make this inference.

GMATGuruNY · Answer

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 = 13-0 = 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 = 10-13 = -3

Since the two cases yield different results, the difference between B's sum and A's sum cannot be determined.

E

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.

Kinshook · Answer

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

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