x-intercept of n must be

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Mon Oct 17, 2011 1:42 am

Line m has a negative slope and a negative y-intercept. Line n is parallel to line m and has a positive y-intercept. The x-intercept of n must be
(A) negative and greater than the x-intercept of m
(B) negative and less than the x-intercept of m
(C) zero
(D) positive and greater than the x-intercept of m
(E) positive and less than the x-intercept of m

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Source: Beat The GMAT — Problem Solving | Mon Oct 17, 2011 1:42 am

neelgandham: Community Manager; Posts: 1060; Joined: Fri May 13, 2011 6:46 amFollowed by: 52 members; Location: Utrecht, The Netherlands; Thanked: 318 times

by neelgandham » Mon Oct 17, 2011 1:56 am

Let the equation of line m be of the form y = mx + c , where c<o and m<0,
Let the equation of line n be of the form y = mx + c1 , where c1>o and m<0, (m = slope and c, c1 are y intercepts).

To obtain the x intercepts, substitute the value of y = 0 in the equations and find the value of x i.e

Line m : mx + c = 0 => x = -c/m = +ve number / -ve number = -ve number
Line n : mx + c1 = 0 => x = -c1/m = -ve number/ -ve number = +ve number

Hence option D

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shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Mon Oct 17, 2011 1:59 am

Line 'm' will pass through II,III and IV quadrants. -> X intercept -ve
Line 'n' will pass through I,II and IV quadrants -> X intercept +ve

Hence D satisfies this condition

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GmatKiss: Legendary Member; Posts: 2789; Joined: Tue Jul 26, 2011 12:19 amFollowed by: 43 members; Location: Chennai, India; Thanked: 206 times; GMAT Score:640

by GmatKiss » Mon Oct 17, 2011 3:22 am

shankar.ashwin wrote:Line 'm' will pass through II,III and IV quadrants. -> X intercept -ve
Line 'n' will pass through I,II and IV quadrants -> X intercept +ve

Hence D satisfies this condition

Could you please elaborate on this!

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shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Mon Oct 17, 2011 3:38 am

@GmatKiss:

Its derived from y=mx+c.

You know m is the slope and c is the y-intercept.

Case 1: y=x (Slope +ve/ y-intercept 0)
Randomly plot points satisfying this equation (-2,-2)(2,2)
You can see it passes through Quadrants I and III.

Case 2: y=x+2 (Slope +ve/ y-intercept +ve)
Pick values again (0,2)(-2,0)
The line passes through Quadrants I,II and III

Case 3: y=x-2 (Slope +ve/y-intercept -ve)
Random points (0,-2)(2,0)
Now its passes thro Quadrants I,III and IV.

In general,

A line with positive slope passes through the I and III quadrant. If it does not pass through the origin, then it will also pass through the II Quadrant if the y-intercept is +ve. If the y-intercept is -ve, then it will also pass through the IV quadrant.

A line with -ve slope passes through the II and IV quadrants. If it does not pass through the origin, then it will also pass through the I quad if the y-intercept is +ve. If the y-intercept is -ve, then it will pass through the III quad.

Not something I would try to remember; you could always plot points to find out, but it does save time. Hope it helped.[/i]

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