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I get B but I am not sure. OA pls? Following is the reasoning: stmt 1: we just have two points on the x-axis. we can draw many lines via these two points and having different slopes. Insuff.

stmt2: we have the intersection point in the first quadrant. Now, draw a line perpendicular from (7,2) to the x-axis. This will be a vertical line. Now select two points on the axis just to right and left of (7,0) on x-axis. Let these points be the x-intercepts of the lines. Now the slope of line with a greater x-intercept say, (7.0001, 0 ) will always be -ve [in order to intersect (7,2)] and the slope of the line with a smaller x-intercept say (6.9999, 0 ) will always be +ve [in order to intersect (7,2)]

Approach of economist is nice. Lets do it more mathematically. 1)insuff. No need to explain. 2)insuff. No need to explain.

together line k=> y=mx+n line j=> y=tx+y we know from 1 that n is greater than y. And we know from 2 that they are intersecting at point (7,2). That makes 2=7m+n 2=7t+y 7m+n=7t+y 7m+n<7t+n 7m<7t m<t

In y = mx + b equation, m = slope and b is the Y intercept. From your represenation, how can we say that

maliyeci wrote:

we know from 1 that n is greater than y.

when variables n and y are the Y intercepts. Not X intercepts.

Can you please explain how you made this Inequality based on n > y(assuming y intercepts of L > y intercept of K). I am confused.

maliyeci wrote:

7m+n<7t+n 7m<7t m<t

I beg your pardon for the intercepts. Yes you are rigth. I gave the Y intercept. The x intercepts are respectively; -n/m and -y/t But they are sufficient to find an answer. That is the solution. 7m+n=2 n=2-7m n/m=2/m-7 same as for the other y/t=2/t-7 since -n/m is greater than -y/t; y/t is greater than n/m so 2/m-7>2/t-7 2/m>2/t so we can find the solution C is suff.

Sorry to disappoint you guys, but seems that the answer should be E.

First of all to solve this question think that the best way is graphic approach:

We all agree that (1) and (2) alone are not sufficient.

Lets combine --> draw only X-axis. k intercept is greater than j intercept --> k intercept (k1) point right to j intercept point (j1) --> Interception of lines is above X-axis (point 7;2) --> it will give us only three possible ways to draw the lines: A. j has negative slope, k has negative slope --> slope of k>slope of j, (because j is closer to vertical line); B. j has positive slope, k has negative slope --> slope of k<slope of j; C. j has positive slope, k has positive slope --> slope of k>slope of j, (because k is closer to vertical line);

So (1)+(2) not sufficient. Answer E.

BUT the way you were solving also should get E:

f(k)=mx+n f(j)=tx+p (1) The x-intercept of line k is greater than the x-intercept of line j --> 0=mx+n; 0=tx+p --> -n/m>-p/t or n/m<p/t --> not sufficient; (2) Lines k and j intersect at (7, 2) 2=7m+n; 2=7t+p --> 7m+n=7t+p -->not sufficient;

Combining the way you did: n=2-7m --> n/m=2/m-7 p=2-7t --> p/t=2/t-7 n/m<p/t --> 2/m-7<2/t-7 --> 1/m<1/t --> (t-m)/mt<0 And here is the catch, from above statement you can not determine whether m>t or not. t=1<m=3 statement is true and t=1>m=-3 statement is also true. Answer E. _________________

Sorry to disappoint you guys, but seems that the answer should be E.

First of all to solve this question think that the best way is graphic approach:

We all agree that (1) and (2) alone are not sufficient.

Lets combine --> draw only X-axis. k intercept is greater than j intercept --> k intercept (k1) point right to j intercept point (j1) --> Interception of lines is above X-axis (point 7;2) --> it will give us only three possible ways to draw the lines: A. j has negative slope, k has negative slope --> slope of k>slope of j, (because j is closer to vertical line); B. j has positive slope, k has negative slope --> slope of k<slope of j; C. j has positive slope, k has positive slope --> slope of k>slope of j, (because k is closer to vertical line);

So (1)+(2) not sufficient. Answer E.

BUT the way you were solving also should get E:

f(k)=mx+n f(j)=tx+p (1) The x-intercept of line k is greater than the x-intercept of line j --> 0=mx+n; 0=tx+p --> -n/m>-p/t or n/m<p/t --> not sufficient; (2) Lines k and j intersect at (7, 2) 2=7m+n; 2=7t+p --> 7m+n=7t+p -->not sufficient;

Combining the way you did: n=2-7m --> n/m=2/m-7 p=2-7t --> p/t=2/t-7 n/m<p/t --> 2/m-7<2/t-7 --> 1/m<1/t --> (t-m)/mt<0 And here is the catch, from above statement you can not determine whether m>t or not. t=1<m=3 statement is true and t=1>m=-3 statement is also true. Answer E.

This exactly why I said IMO E . Thanks for explanation Bunuel!

Lets combine --> draw only X-axis. k intercept is greater than j intercept --> k intercept (k1) point right to j intercept point (j1) --> Interception of lines is above X-axis (point 7;2) --> it will give us only three possible ways to draw the lines: A. j has negative slope, k has negative slope --> slope of k>slope of j, (because j is closer to vertical line); B. j has positive slope, k has negative slope --> slope of k<slope of j; C. j has positive slope, k has positive slope --> slope of k>slope of j, (because k is closer to vertical line);

Hi Bunuel,

Request you to please look at the colored portion again and confirm if its correct. Isn't in Option A->Slope of j > Slope of K. If not, please provide your reasoning, may be I'm lacking some concept.

Lets combine --> draw only X-axis. k intercept is greater than j intercept --> k intercept (k1) point right to j intercept point (j1) --> Interception of lines is above X-axis (point 7;2) --> it will give us only three possible ways to draw the lines: A. j has negative slope, k has negative slope --> slope of k>slope of j, (because j is closer to vertical line); B. j has positive slope, k has negative slope --> slope of k<slope of j; C. j has positive slope, k has positive slope --> slope of k>slope of j, (because k is closer to vertical line);

Hi Bunuel,

Request you to please look at the colored portion again and confirm if its correct. Isn't in Option A->Slope of j > Slope of K. If not, please provide your reasoning, may be I'm lacking some concept.

thanks

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