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The three integers in the set {x, y, z} are all less than 30. How many [#permalink]
reto how did you get possible values of -17, 25, 25? if we add them up, it doesn't =67? Am I missing something?
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Re: The three integers in the set {x, y, z} are all less than 30. How many [#permalink]
SamuelWitwicky wrote:
reto how did you get possible values of -17, 25, 25? if we add them up, it doesn't =67? Am I missing something?


No youre not missing anything. I probably was drunk when writing that. Just ignore the negative sign in my explanation please.

Thanks :)
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Re: The three integers in the set {x, y, z} are all less than 30. How many [#permalink]
reto lol no worries mate.
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The three integers in the set {x, y, z} are all less than 30. How many [#permalink]
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Bunuel wrote:
The three integers in the set {x, y, z} are all less than 30. How many of the integers are positive?

(1) x + y + z = 67
(2) x + y = 40

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x<30
y<30
z<30
x,y,z \(\in\) Integers

Statement 2 is clearly not sufficient. Without the value of z or the total of x+y+z , I can have z > or < 0. Thus giving 2 or 3 as the number of positive integers.

Statement 1, x+y+z =67.

now assume , 1 of these is negative . y =-1 ---> x+z = 68. This is NOT possible as x and z<30 and thus x+z<60 for all cases. Having 2 will make it even more difficult and 3 negatives is not possible as we need a positive value after addition.

Thus all 3 are positive. This statement is sufficient.

A is the correct answer.
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Re: The three integers in the set {x, y, z} are all less than 30. How many [#permalink]
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Bunuel wrote:
The three integers in the set {x, y, z} are all less than 30. How many of the integers are positive?

(1) x + y + z = 67
(2) x + y = 40

Kudos for a correct solution.



Statement 1: x+y+z=67
All three have to be positive.
Let us take x=0, then y and z both have to be greater than or equal to 30. Sufficient
Statement 2: x+y=40. z is not in the equation. Insufficient.
Answer A
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Re: The three integers in the set {x, y, z} are all less than 30. How many [#permalink]
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the answer is A as using max limit we get that all 3 have to be positive to give a sum of 67
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Re: The three integers in the set {x, y, z} are all less than 30. How many [#permalink]
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Bunuel wrote:
The three integers in the set {x, y, z} are all less than 30. How many of the integers are positive?

(1) x + y + z = 67
(2) x + y = 40

Kudos for a correct solution.


800score Official Solution:

(A) Statement (1) alone is sufficient. Since all the numbers are less than 30, all three must be positive for their sum to be 67 or greater because there is no way to get a sum greater than or equal to 67 with just two positive numbers less than 30.

Note: If this is confusing, try plugging in numbers and you will find that it is impossible to have a negative number when 3 numbers less than 30 must sum to a number 67 or greater. Since 29 + 29 = 58, then the third number cannot be negative for the sum of all three to be greater than 67. This provides sufficient information to know how many are positive.

Statement (2) alone is insufficient, because it implies that x and y are positive, but gives no information about z.

Since Statement (1) is sufficient alone, and Statement (2) is not, the correct answer is choice (A).
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Re: The three integers in the set {x, y, z} are all less than 30. How many [#permalink]
From 1
Minimize negative by setting x = -1 thus both y+z can never sum up to 66 .. Hence the 3 integers has to be positive.

B.. Insuff

A


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Re: The three integers in the set {x, y, z} are all less than 30. How many [#permalink]
x,y,z each of them is less than 30. Hence maximum value possible for each of them is 29.
st 1 : x+y+z=67 if y=z=29 then y+z=58. so x=9 All three of them have to be positive. Hence sufficient.
st 2 : x+y = 40 . If maximum possible value of any one of them can be 29.So the other has to be 11. So both of them are positive.But z can be positive,negative or zero.So this statement is insufficient.
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Re: The three integers in the set {x, y, z} are all less than 30. How many [#permalink]
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