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M18-13

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M18-13  [#permalink]

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New post 16 Sep 2014, 01:03
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

56% (01:22) correct 44% (01:40) wrong based on 39 sessions

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Re M18-13  [#permalink]

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New post 16 Sep 2014, 01:03
Official Solution:


The x-intercept of a line is the value of \(x\) when \(y=0\). So, the x-intercept of line \(ax+by+c=0\) is \(x=-\frac{c}{a}\). Question basically asks: is \(-\frac{c}{a} \lt 0\)? Or: is \(\frac{c}{a} \gt 0\)? Or: do \(c\) and \(a\) have the same sign?

(1) \(ba \lt 0\). Not sufficient as we cannot answer whether \(c\) and \(a\) have the same sign.

(2) \(ac \gt 0\). From this statement it follows that \(c\) and \(a\) have the same sign. Sufficient.


Answer: B
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Re: M18-13  [#permalink]

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New post 18 Mar 2017, 17:18
How do we know that x=-c/a?
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Re: M18-13  [#permalink]

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New post 19 Mar 2017, 04:25
Cez005 wrote:
How do we know that x=-c/a?


The x-intercept of a line is the value of \(x\) when \(y=0\). Substitute y = 0 into \(ax+by+c=0\) to get \(ax + c=0\) --> \(x=-\frac{c}{a}\).

Check for more here: https://gmatclub.com/forum/math-coordin ... 87652.html
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Re: M18-13  [#permalink]

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New post 15 Dec 2017, 02:18
But what if the original equation is modified to:
y= \(\frac{-a}{b}\)x - c ?

Then when we substitute y=0, x = -\(\frac{b}{a}\)c. For this, we need to know the signs of a,b and c! Which means, how can statement 2 alone be sufficient?
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New post 15 Dec 2017, 02:22
helloksin wrote:
But what if the original equation is modified to:
y= \(\frac{-a}{b}\)x - c ?

Then when we substitute y=0, x = -\(\frac{b}{a}\)c. For this, we need to know the signs of a,b and c! Which means, how can statement 2 alone be sufficient?


If y = 0, the value of b does not play any role.
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M18-13  [#permalink]

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New post 15 Dec 2017, 03:01
helloksin wrote:
But what if the original equation is modified to:
y= \(\frac{-a}{b}\)x - c ?

Then when we substitute y=0, x = -\(\frac{b}{a}\)c. For this, we need to know the signs of a,b and c! Which means, how can statement 2 alone be sufficient?


The equation would become
y= \(\frac{-a}{b}x - \frac{c}{b}\)

So even in that case, for x intercept, y=0,
the equation boils down to x= \(\frac{-c}{a}\)

You missed the "b" in constant term.
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Re M18-13  [#permalink]

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New post 22 Jun 2018, 11:54
I think this is a high-quality question and I agree with explanation.
Re M18-13 &nbs [#permalink] 22 Jun 2018, 11:54
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