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The average of 6 numbers in a set is equal to 0. What is the

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The average of 6 numbers in a set is equal to 0. What is the [#permalink] New post 14 Dec 2012, 08:22
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The average of 6 numbers in a set is equal to 0. What is the number of positive numbers in the set minus the number of negative numbers in the set?

(1) Each of the positive numbers in the set equals 10.
(2) Each of the negative numbers in the set equals –5.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 14 Dec 2012, 08:26, edited 1 time in total.
Renamed the topic and edited the question.
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Re: The average of 6 numbers in a set is equal to 0. What is the [#permalink] New post 14 Dec 2012, 08:35
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The average of 6 numbers in a set is equal to 0. What is the number of positive numbers in the set minus the number of negative numbers in the set?

Since the average of 6 numbers in the set is equal to 0, then the sum of 6 numbers is also 0. Which means that the sum of the negative numbers must compensate the sum of the positive numbers.

(1) Each of the positive numbers in the set equals 10. Not sufficient.
(2) Each of the negative numbers in the set equals –5. Not sufficient.

(1)+(2) The number of negative numbers (-5) must be twice the number of positive numbers (10). Possible cases are {-5, -5, -5, -5, 10, 10} and {-5, -5, 0, 0, 0, 10}. Thus the number of positive numbers minus the number of negative numbers could be either -2 or -1. Not sufficient.

Answer: E.
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Re: The average of 6 numbers in a set is equal to 0. What is the [#permalink] New post 14 Dec 2012, 08:58
Bunuel wrote:
The average of 6 numbers in a set is equal to 0. What is the number of positive numbers in the set minus the number of negative numbers in the set?

Since the average of 6 numbers in the set is equal to 0, then the sum of 6 numbers is also 0. Which means that the sum of the negative numbers must compensate the sum of the positive numbers.

(1) Each of the positive numbers in the set equals 10. Not sufficient.
(2) Each of the negative numbers in the set equals –5. Not sufficient.

(1)+(2) The number of negative numbers (-5) must be twice the number of positive numbers (10). Possible cases are {-5, -5, -5, -5, 10, 10} and {-5, -5, 0, 0, 0, 10}. Thus the number of positive numbers minus the number of negative numbers could be either -2 or -1. Not sufficient.

Answer: E.


If "Each of the positive numbers in the set equals 10" and "Each of the negative numbers in the set equals -5", don't you think we have gotten our answer?
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The average of 6 numbers in a set is equal to 0. What is the [#permalink] New post 14 Dec 2012, 09:03
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knightofdelta wrote:
Bunuel wrote:
The average of 6 numbers in a set is equal to 0. What is the number of positive numbers in the set minus the number of negative numbers in the set?

Since the average of 6 numbers in the set is equal to 0, then the sum of 6 numbers is also 0. Which means that the sum of the negative numbers must compensate the sum of the positive numbers.

(1) Each of the positive numbers in the set equals 10. Not sufficient.
(2) Each of the negative numbers in the set equals –5. Not sufficient.

(1)+(2) The number of negative numbers (-5) must be twice the number of positive numbers (10). Possible cases are {-5, -5, -5, -5, 10, 10} and {-5, -5, 0, 0, 0, 10}. Thus the number of positive numbers minus the number of negative numbers could be either -2 or -1. Not sufficient.

Answer: E.


If "Each of the positive numbers in the set equals 10" and "Each of the negative numbers in the set equals -5", don't you think we have gotten our answer?


I don't understand your question.

Solution gives two possible sets which give two different answers to the question. Therefore the answer is E.
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Re: The average of 6 numbers in a set is equal to 0. What is the [#permalink] New post 14 Dec 2012, 09:22
Bunuel wrote:
knightofdelta wrote:
Bunuel wrote:
The average of 6 numbers in a set is equal to 0. What is the number of positive numbers in the set minus the number of negative numbers in the set?

Since the average of 6 numbers in the set is equal to 0, then the sum of 6 numbers is also 0. Which means that the sum of the negative numbers must compensate the sum of the positive numbers.

(1) Each of the positive numbers in the set equals 10. Not sufficient.
(2) Each of the negative numbers in the set equals –5. Not sufficient.

(1)+(2) The number of negative numbers (-5) must be twice the number of positive numbers (10). Possible cases are {-5, -5, -5, -5, 10, 10} and {-5, -5, 0, 0, 0, 10}. Thus the number of positive numbers minus the number of negative numbers could be either -2 or -1. Not sufficient.

Answer: E.


If "Each of the positive numbers in the set equals 10" and "Each of the negative numbers in the set equals -5", don't you think we have gotten our answer?


I don't understand your question.

Solution gives two possible sets which gives two different answers to the question. Therefore the answer is E.


It seems like combining (1) and (2) will only provide the first set in your solution i.e. {-5, -5, -5, -5, 10, 10}. Where did you get the zeros in {-5, -5, 0, 0, 0, 10} when each of the negative number is -5 and each of the positive number is 10?
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Re: The average of 6 numbers in a set is equal to 0. What is the [#permalink] New post 14 Dec 2012, 09:26
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knightofdelta wrote:
Bunuel wrote:
knightofdelta wrote:

If "Each of the positive numbers in the set equals 10" and "Each of the negative numbers in the set equals -5", don't you think we have gotten our answer?


I don't understand your question.

Solution gives two possible sets which gives two different answers to the question. Therefore the answer is E.


It seems like combining (1) and (2) will only provide the first set in your solution i.e. {-5, -5, -5, -5, 10, 10}. Where did you get the zeros in {-5, -5, 0, 0, 0, 10} when each of the negative number is -5 and each of the positive number is 10?


Zero is neither negative nor positive number.

Now consider {-5, -5, 0, 0, 0, 10}: each of the positive numbers in the set equals 10 and each of the negative numbers in the set equals –5.

Hope it's clear.
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Re: The average of 6 numbers in a set is equal to 0. What is the [#permalink] New post 02 Sep 2014, 13:16
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Re: The average of 6 numbers in a set is equal to 0. What is the [#permalink] New post 29 Jul 2015, 15:08
Bunuel wrote:
The average of 6 numbers in a set is equal to 0. What is the number of positive numbers in the set minus the number of negative numbers in the set?

Since the average of 6 numbers in the set is equal to 0, then the sum of 6 numbers is also 0. Which means that the sum of the negative numbers must compensate the sum of the positive numbers.

(1) Each of the positive numbers in the set equals 10. Not sufficient.
(2) Each of the negative numbers in the set equals –5. Not sufficient.

(1)+(2) The number of negative numbers (-5) must be twice the number of positive numbers (10). Possible cases are {-5, -5, -5, -5, 10, 10} and {-5, -5, 0, 0, 0, 10}. Thus the number of positive numbers minus the number of negative numbers could be either -2 or -1. Not sufficient.

Answer: E.


Hi Bunnuel,

I got C. Here is my explanation

(Sum of + numbers)+(Sum of - numbers) = 6

0 = [(Sum of + numbers)+(Sum of 0 numbers)][/6]
0 = (Sum of + numbers)+(Sum of - numbers)

From 1. (Sum of + numbers) = 10*(Quantity of + numbers)

From 2. (Sum of - numbers) = 10*(Quantity of - numbers)

Can you explain where i made my mistake? Thanks.
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Re: The average of 6 numbers in a set is equal to 0. What is the [#permalink] New post 30 Jul 2015, 00:08
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josegf1987 wrote:
Bunuel wrote:
The average of 6 numbers in a set is equal to 0. What is the number of positive numbers in the set minus the number of negative numbers in the set?

Since the average of 6 numbers in the set is equal to 0, then the sum of 6 numbers is also 0. Which means that the sum of the negative numbers must compensate the sum of the positive numbers.

(1) Each of the positive numbers in the set equals 10. Not sufficient.
(2) Each of the negative numbers in the set equals –5. Not sufficient.

(1)+(2) The number of negative numbers (-5) must be twice the number of positive numbers (10). Possible cases are {-5, -5, -5, -5, 10, 10} and {-5, -5, 0, 0, 0, 10}. Thus the number of positive numbers minus the number of negative numbers could be either -2 or -1. Not sufficient.

Answer: E.


Hi Bunnuel,

I got C. Here is my explanation

(Sum of + numbers)+(Sum of - numbers) = 6

0 = [(Sum of + numbers)+(Sum of 0 numbers)][/6]
0 = (Sum of + numbers)+(Sum of - numbers)

From 1. (Sum of + numbers) = 10*(Quantity of + numbers)

From 2. (Sum of - numbers) = 10*(Quantity of - numbers)

Can you explain where i made my mistake? Thanks.


Not sure that I completely understand your approach but I think that you missed that there could be some number of zeros in the set. Check the highlighted part for possible sets, proving E.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: The average of 6 numbers in a set is equal to 0. What is the   [#permalink] 30 Jul 2015, 00:08
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