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Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16}

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motion2020
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Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16} [#permalink] New post   Updated on: 29 Jul 2021, 11:28
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Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16}
Set B includes numbers the same as in the set A with 3 subtracted from each number and then multipled by 1.5



Quantity A
Quantity B
7
Arithmetic average of the set B


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
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A

Originally posted by motion2020 on 29 Jul 2021, 10:04.
Last edited by GeminiHeat on 29 Jul 2021, 11:28, edited 2 times in total.
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Carcass
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Re: Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16} [#permalink] New post   29 Jul 2021, 10:13
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See here how to create the two quantities. Quite easy https://gre.myprepclub.com/forum/how-to-po ... 12752.html

Also add always the 4 answer choices

Thank you
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motion2020
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Re: Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16} [#permalink] New post   29 Jul 2021, 10:39
Carcass wrote:
See here how to create the two quantities. Quite easy https://gre.myprepclub.com/forum/how-to-po ... 12752.html

Also add always the 4 answer choices

Thank you

thanks!
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motion2020
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Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16} [#permalink] New post   02 Aug 2021, 02:43
This question has not been answered yet..
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Re: Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16} [#permalink] New post   02 Aug 2021, 06:10
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motion2020 wrote:
Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16}
Set B includes numbers the same as in the set A with 3 subtracted from each number and then multiplied by 1.5



Quantity A
Quantity B
7
Arithmetic average of the set B


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


Average of set A \(= \frac{4+5+5+5+5+5+5+5+6+7+11+12+16}{13} = 7\)

If we subtract 3 from each value, then the new average (arithmetic mean) = 7 - 3 = 4

If we now multiply each value by 1,5, the new average = (4)(1.5) = 6

So, the average of Set B = 6

So we get:
QUANTITY A: 7
QUANTITY B: 6

Answer: A
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motion2020
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Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16} [#permalink] New post   02 Aug 2021, 09:33
and morale of this question is that adding and multiplying or dividing the mean (average) by constant should change the mean (new mean is obtained) by that constant's addition, multiplication or division
GreenlightTestPrep wrote:
motion2020 wrote:
Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16}
Set B includes numbers the same as in the set A with 3 subtracted from each number and then multiplied by 1.5



Quantity A
Quantity B
7
Arithmetic average of the set B


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


Average of set A \(= \frac{4+5+5+5+5+5+5+5+6+7+11+12+16}{13} = 7\)

If we subtract 3 from each value, then the new average (arithmetic mean) = 7 - 3 = 4

If we now multiply each value by 1,5, the new average = (4)(1.5) = 6

So, the average of Set B = 6

So we get:
QUANTITY A: 7
QUANTITY B: 6

Answer: A
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Set A is given as A = {4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 11, 12, 16} [#permalink]
02 Aug 2021, 09:33
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