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Manager  Status: Never ever give up on yourself.Period.
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The average of 6 numbers in a set is equal to 0. What is the  [#permalink]

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10 00:00

Difficulty:   95% (hard)

Question Stats: 42% (01:46) correct 58% (01:40) wrong based on 171 sessions

### HideShow timer Statistics 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.

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Originally posted by daviesj on 14 Dec 2012, 09:22.
Last edited by Bunuel on 14 Dec 2012, 09:26, edited 1 time in total.
Renamed the topic and edited the question.
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5
2
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.

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Re: The average of 6 numbers in a set is equal to 0. What is the  [#permalink]

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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.

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]

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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.

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?

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]

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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.

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?

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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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?

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]

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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.

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]

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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.

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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Re: The average of 6 numbers in a set is equal to 0. What is the  [#permalink]

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Since the average of the 6 numbers in a set is 0, the sum of the 6 numbers is 0. We also know there could be positive numbers and negative numbers in the set. Zero is not mentioned, but this does not rule it out. In order for the sum of the numbers in the set to be 0, either all the terms are 0, or there are some positives and some negatives.

(1) INSUFFICIENT: Statement 1 tells us that the set has at least one positive number, and that each positive term is 10. We should try to prove insufficiency. For instance, the set could be {–2, –2, –2, –2, –2, 10}, and the number of positive terms minus the number of negative terms would be 1 – 5 = –4. Alternatively, the set could be {–20, –20, 10, 10, 10, 10}, and the answer would be 4 – 2 = 2.

(2) INSUFFICIENT: Statement 2 tells us that the set has at least one negative number, and that each negative term is –5. Again, we should try to prove insufficiency. The set could be {–5, 1, 1, 1, 1, 1}, and the number of positive terms minus the number of negative terms would be 5 – 1 = 4. The set could be {–5, –5, –5, 2, 5, 8}, and the answer would be 3 – 3 = 0.

(1) & (2) INSUFFICIENT: The statements together suggest that the set has twice as many –5 terms as 10 terms, in order to maintain a sum of 0. If every term is negative or positive, then the set would have to be {–5, –5, –5, –5, 10, 10} and the definitive answer would be 2 – 4 = –2. However, zero terms are possible, so the set could be {–5, –5, 0, 0, 0, 10} and an alternative answer would be 1 – 2 = –1.

Hence E.
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