Data Sufficiency

clock60 wrote:i saw this query earlier and still can`t understand for what purposes we need point (4.3)
l=y=a1*x+b1
k=y=a2*x+b2, where a1,a2 slopes of linesa l,k we need to estimate their product
(1)if y=0, x=-b1/a1, fro line l and x=-b2/a2 for linek
(-b1/a1)*(-b2/a2)=(b1/a1)*(b2/a2)>0 it is possible if
b1*b2>0 and a1*a2>0. or second case b1*b2<0 and a1*a2<0
we cant find for sure the sign of the product a1*a2
not suficient
(2)b1*b2<0 not suff as the slope is not mentioned
together
if b1*b2<0 then the only case is that a1*a2<0 for the fraction (b1/a1)*(b2/a2) to be +ve
so C seems reasonable answer

Whitney Garner wrote:
Nice work on the math, but I wanted to comment quickly on using the point and "drawing lines". If we quickly sketch the point (4,3), we should only have to draw a few quick graphs (because this is testing Yes/No/

Whitney Garner wrote: Statement (1): If the product of the x-intercepts is positive, then we need both to be positive or both negative. Now, testing Yes/No means to actively look for a sketch that would answer Yes and would answer No. If I make them both negative, then to both cross at (4,3) they would both have to have positive slope = product of the slopes >0.

Whitney Garner wrote: But if I make them both negative { It should be Positive }, I can make them cross with one positive and one negative slope = product of the slopes <0. I can answer the original question either Yes or No. INSUFFICIENT

Whitney Garner wrote: Statement (2): If the product of the y-intercepts is negative, then one must be positive while the other is negative. Again, I need to try to show that I can answer both Yes and No (because if I cannot, then it is Sufficient). If I have one line cross the y-axis at a point higher than (4,3) and the other cross the y-axis at any negative point I find that the slopes have opposite signs = product of the slopes <0. But if I have one line cross the y-axis at a point between 0 and 3 (positive but below the point (4,3)), I see that they are both positively sloped = product of the slopes >0. I can answer the original question either Yes or No. INSUFFICIENT

Statement (1+2) So we need one line to cross the Y above 0, and the other to cross below 0. BUT, we also need both lines to have the same sign on their x-intercepts. If I attempt to draw this I will quickly see that there is no way to draw 2 lines that intersect at (4,3) that have opposite signs on their y-intercepts AND that have negative x-intercepts. Therefore, both must have positive x-intercepts. But if the have positive x-intercepts, the only way I can have them cross at (4,3) is if the line with the positive y-intercept crosses the y-axis above the point and the line with the negative y-axis crosses anywhere below 0. This means that the two lines MUST have slopes with opposite signs = product <0. SUFFICIENT


Whit