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# The average of 25 values is equal to product of these values. If only

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The average of 25 values is equal to product of these values. If only  [#permalink]

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26 Jan 2015, 05:45
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Difficulty:

65% (hard)

Question Stats:

51% (01:18) correct 49% (01:07) wrong based on 299 sessions

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The average of 25 values is equal to product of these values. If only one of the numbers is 0, at most how many numbers can be greater than zero?

A. 0
B. 12
C. 13
D. 23
E. 24

Kudos for a correct solution.

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Re: The average of 25 values is equal to product of these values. If only  [#permalink]

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26 Jan 2015, 12:32
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HI DesiGmat,

This is a great "concept" question - it really tests a person's thoroughness when dealing with "restrictions" (or lack of restrictions).

We're asked for the MAXIMUM number of terms that COULD be greater than 0, given the following restrictions:

1) There are 25 terms in total
2) One of the terms is 0
3) The average of the terms = the product of the terms

Since one of the terms is 0, the product = 0. Since we're told that the average = the product, the average ALSO = 0

Now we have to think about how we can make the average = 0 AND maximize the number of terms that are greater than 0. Notice how the question did NOT say anything about "consecutive" or "duplicates" - so we have a lot of ways to "play around" with this idea.

A 0, Twenty-three 1s and a -23
The product = 0 (because we have a 0 in the group)
The average = 0 + 23(1) +1(-23) = 0/25 = 0

In this situation, we have 23 numbers that are greater than 0. From a concept-standpoint, it can't be 24 though, since we need at least one negative term to offset the positives and bring the average back down to 0.

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Re: The average of 25 values is equal to product of these values. If only  [#permalink]

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Updated on: 26 Jan 2015, 09:38
3
1
ans is D...... 23 is the answer
0 being one number means product is 0 therefore mean should be 0..
so the sum of +ive numbers should be equal to sum of -ive numbers to have total value as 0..
it is possible with different 23 +ive numbers and one -ive number, which is equal to total sum of 23 +ive numbers taken with a -ive sign...
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Originally posted by chetan2u on 26 Jan 2015, 07:41.
Last edited by chetan2u on 26 Jan 2015, 09:38, edited 1 time in total.
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Re: The average of 25 values is equal to product of these values. If only  [#permalink]

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26 Jan 2015, 08:56
Here we go---

Given that avg of 25 values is equal to product of these values.
One of the number is 0.
So avg = 0

Now Surplus above the mean = Deficit below the mean

There can be at most 12 numbers greater than 0 to make the avg 0.

Option B is correct
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Re: The average of 25 values is equal to product of these values. If only  [#permalink]

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26 Jan 2015, 22:42
Ohh...
I completely missed this point....

We can have 23 1s creating the surplus above the mean = 23
and we can have -23 creating deficit below the mean = -23-----> maintaining the avg = 0

Thank you so much.
Kudos to you.
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Re: The average of 25 values is equal to product of these values. If only  [#permalink]

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26 Jan 2015, 23:20

One number is 0 .....So product of all numbers will be 0.....Avg will be 0.....

Now we can have 23 numbers whose sum is equal to one large negative number to make the resultant 0....

Good question
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Re: The average of 25 values is equal to product of these values. If only  [#permalink]

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02 Feb 2015, 02:23
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Bunuel wrote:
The average of 25 values is equal to product of these values. If only one of the numbers is 0, at most how many numbers can be greater than zero?

A. 0
B. 12
C. 13
D. 23
E. 24

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

Since the product of these values is 0, then the average must be 0. In order to then maximize the number of positive values, there are 24 values left. If all 24 were positive, then the average could not be 0, as the sum of the values needs to be 0. But if there were one negative value that was equal to -1 times the sum of the other 23 nonzero values, then the sum of values would be 0 and a total of 23 values could be positive. Therefore, the answer is D.
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Re: The average of 25 values is equal to product of these values. If only  [#permalink]

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03 Feb 2015, 20:56
2

Let the remaining values be from a1 to........... a24

Given that $$\frac{0 + (a1 + a2 + a3 + ............. + a24)}{25} = 0$$

a1 + a2 + a3 .................. + a24 = 0

a1 + a2 + a3 + ............. + a23 = -a24

So all values from a1 to a23 can be positive, whose sum would give a negative a24
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Re: The average of 25 values is equal to product of these values. If only  [#permalink]

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06 Feb 2015, 12:09
Well, since one is 0 we know that the product is zero.

For the mean to be zero, we need their sum to be zero.

Out of the 25 numbers, we know the value of one: 0.
We still have 24 numbers to think about.

Now, we know for example that 10-10=0. So, we need the same number with opposite signs to get a sum of 0.

How is this possible in our case? It is possible if for the remaining 24 numbers, 23 equal to one ammount and the 24th to the same ammount with the opposite sign. L

et's say that 23 numbers have the value of 1. Then their sum would be 23, and we need a -23 to get a zero.
So, we could have the most 23 positive numbers and 1 negative number to end up with 0.
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Re: The average of 25 values is equal to product of these values. If only  [#permalink]

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27 Feb 2016, 23:36
Given The average of 25 values is equal to product of these values.
Since, one of the value is 0, Mean must be equal to 0
Total number of numbers (a1 to a25) = 25, one value is 0
Mean = a1 + a2 + a3 .................. + a24 = 0
The maximum positive values that can exist satisfying the above relation is = 23

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Re: The average of 25 values is equal to product of these values. If only  [#permalink]

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26 Jul 2017, 16:18
1
Bunuel wrote:
The average of 25 values is equal to product of these values. If only one of the numbers is 0, at most how many numbers can be greater than zero?

A. 0
B. 12
C. 13
D. 23
E. 24

Since the average is equal to the product and one value is zero, the product is zero. Thus, in order to have an average of zero, we need the sum to be zero also. We can have 23 numbers that are positive, one that is zero, and one that is negative with its absolute value as the sum of the 23 positive numbers.

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Re: The average of 25 values is equal to product of these values. If only   [#permalink] 26 Jul 2017, 16:18
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