Is the x-intercept of the non-horizontal line L negative?

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Mon Feb 18, 2019 8:54 am

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GMATH practice exercise (Quant Class 20)

Is the x-intercept of the non-horizontal line L negative?

(1) The slope of line L is equal to the y-intercept of line L.
(2) The slope of line L is positive.

Answer: [spoiler]____(A)__[/spoiler]

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pmFollowed by: 24 members; Location: New Delhi, India; Thanked: 205 times

by GMATinsight » Mon Feb 18, 2019 10:54 pm

fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 20)

Is the x-intercept of the non-horizontal line L negative?

(1) The slope of line L is equal to the y-intercept of line L.
(2) The slope of line L is positive.

Answer: [spoiler]____(A)__[/spoiler]

Question : Is the x-intercept of the non-horizontal line L negative?

CONCEPT: A line with negative slope is bound to pass through II and IV quadrants
CONCEPT: A line with negative slope has both intercepts (x and y intercepts) +ve and both intercepts -ve

Statement 1: The slope of line L is equal to the y-intercept of line L.
Since the line may have both slope and Y intercepts +ve as well as -ve but in each of these cases x-intercept will be -ve hence SUFFICIENT

Statement 2: The slope of line L is positive.
a +ve sloping line may have +ve as well as -ve x-intercept hence

NOT SUFFICIENT

Answer: Option A

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Tue Feb 19, 2019 11:06 am

fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 20)

Is the x-intercept of the non-horizontal line L negative?

(1) The slope of line L is equal to the y-intercept of line L.
(2) The slope of line L is positive.

$$c = {\left( {x - {\rm{intercept}}} \right)_{\rm{L}}}\,\,\mathop {\rm{ < }}\limits^? \,\,{\rm{0}}$$
$$L\,\,\,:\,\,\,{\rm{vertical}}\,\,{\rm{or}}\,\,{\rm{non - vertical}}\,\,{\rm{with}}\,\,m = {\rm{slop}}{{\rm{e}}_{\rm{L}}} \ne 0$$

$$\left( 1 \right)\,\,\,L\,\,{\rm{non - vertical}}\,\,{\rm{and}}\,\,L\,\,:\,\,\,y = mx + m = m\left( {x + 1} \right)$$
$$?\,\,\,:\,\,\,0 = m\left( {c + 1} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{m\, \ne \,\,0} \,\,\,\,c + 1 = 0\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$

$$\left( 2 \right)\,\,\,{\rm{INSUFF}}.\,\,\,\,\left( {{\rm{trivial}}\,\,{\rm{geometric}}\,\,{\rm{bifurcation}}} \right)$$

The correct answer is (A).

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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