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If a, b, & x are greater than zero, is x > 0

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Source: — Data Sufficiency |
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by [email protected] » Wed May 17, 2017 10:22 am
Hi Mo2men,

I think there's a typo in this question (either in the original source material or in how you transcribed it). The prompt states that X is GREATER THAN ZERO, but the question then asks if X is greater than 0.... The answer would be 'yes' without having to consider the two Facts.

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by Mo2men » Wed May 17, 2017 10:25 am
[email protected] wrote:Hi Mo2men,

I think there's a typo in this question (either in the original source material or in how you transcribed it). The prompt states that X is GREATER THAN ZERO, but the question then asks if X is greater than 0.... The answer would be 'yes' without having to consider the two Facts.

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Rich
Thanks Rich for your observation. It is my problem actually. Problem fixed.
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by GMATGuruNY » Wed May 17, 2017 10:47 am
Mo2men wrote:If a, b, & x are greater than zero, is x > 4?

1) (ax^2 - 9a)/(12xb) = (ax - 3a)/ (bx + 3b)

2)a + b =5 and a - b= 4

Statement 1: (ax² - 9a)/(12xb) = (ax - 3a) / (bx + 3b)

a(x² - 9)/(12xb) = a(x-3) / b(x+3)

(x² - 9)/(12x) = (x-3) / (x+3)

(x+3)(x-3) / (12x) = (x-3) / (x+3)

[(x+3)(x-3) / 12x] - [(x-3) / (x+3)] = 0

(x-3)[ (x+3)/(12x) - 1/(x+3) ] = 0.

Case 1: x-3 = 0
Here, x=3.

Case 2: (x+3)/(12x) - 1/(x+3) = 0
Thus:
(x+3)/(12x) = 1/(x+3)
(x+3)² = 12x
x² + 6x + 9 = 12x
x² - 6x + 9 = 0
(x-3)² = 0
x=3.

Since x=3 in each case, the answer to the question stem is NO.
SUFFICIENT.

Statement 2: a+b = 5 and a-b = 4
No information about x.
INSUFFICIENT.

The correct answer is A.
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by Mo2men » Wed May 17, 2017 11:08 am
GMATGuruNY wrote:
Mo2men wrote:If a, b, & x are greater than zero, is x > 4?

1) (ax^2 - 9a)/(12xb) = (ax - 3a)/ (bx + 3b)

2)a + b =5 and a - b= 4

Statement 1: (ax² - 9a)/(12xb) = (ax - 3a) / (bx + 3b)

a(x² - 9)/(12xb) = a(x-3) / b(x+3)

(x² - 9)/(12x) = (x-3) / (x+3)

(x+3)(x-3) / (12x) = (x-3) / (x+3)

[(x+3)(x-3) / 12x] - [(x-3) / (x+3)] = 0

(x-3)[ (x+3)/(12x) - 1/(x+3) ] = 0.

Case 1: x-3 = 0
Here, x=3.

Case 2: (x+3)/(12x) - 1/(x+3) = 0
Thus:
(x+3)/(12x) = 1/(x+3)
(x+3)² = 12x
x² + 6x + 9 = 12x
x² - 6x + 9 = 0
(x-3)² = 0
x=3.

Since x=3 in each case, the answer to the question stem is NO.
SUFFICIENT.
Dear Mitch,

In third line in green, why did not you cancel out (x-3) from both sides? x is greater zero so we zero won't be answer.
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by GMATGuruNY » Wed May 17, 2017 11:15 am
Mo2men wrote: Dear Mitch,

In third line in green, why did not you cancel out (x-3) from both sides? x is greater zero so we zero won't be answer.
Only NONZERO factors may be canceled out from both sides.
If x=3, then x-3 = 0.
Since it's possible that x-3 = 0, we cannot cancel out x-3.
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by Mo2men » Wed May 17, 2017 2:40 pm
GMATGuruNY wrote:
Mo2men wrote: Dear Mitch,

In third line in green, why did not you cancel out (x-3) from both sides? x is greater zero so we zero won't be answer.
Only NONZERO factors may be canceled out from both sides.
If x=3, then x-3 = 0.
Since it's possible that x-3 = 0, we cannot cancel out x-3.
Suppose in a prompt y > 0 and after factoring out we have following:

(1+y)(..........) = (1+y) (.............)

Can we cancel out (1 +y) as we are sure is always positive? or are their any other restrictions?

Thanks in advance
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by GMATGuruNY » Thu May 18, 2017 3:28 am
Mo2men wrote:Suppose in a prompt y > 0 and after factoring out we have following:

(1+y)(..........) = (1+y) (.............)

Can we cancel out (1 +y) as we are sure is always positive? or are their any other restrictions?

Thanks in advance
In an equation, any nonzero factor can be canceled out from both sides.
If y>0, then 1+y>0.
Since 1+y is nonzero, it can be canceled out from both sides of the equation above.
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