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19 Aug 2007, 06:35
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N
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Please explain the answer
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19 Aug 2007, 09:32
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I'm getting C, but I am not sure... and takes too long
Set:
L = Ml * Xl + Bl
K = Mk * Xk + Bk
You know that the line intersect at (4,3); thus, you can plug this value in both equations to get:
3 = M*4 + B, and so
Bl = 3 - Ml*4
Bk = 3 - Mk*4
(1) Product of x intercepts are positive. x intercept is when y=0, so find that, you get:
Xl = (-Bl) / (Ml) = (4*Ml - 3) / Ml
Xk = (-Bk) / (Mk) = (4*Mk - 3) / Mk
Product of x intercepts:
((4*Ml - 3) / Ml) * ((4*Mk - 3) / Mk) = positive
For the left or right term to be negative, 0< M < 3/4.
For the left or right term to be positive, M > 3/4 or M < 0
If the product is positive, you can have both negative terms; thus both slopes are positive and between 0 and 3/4. However, if you have both positive terms, you can't tell if the slopes are negative or positive.
INSUFFICIENT.
(2) Product of y intercepts are negative. y intercept when x=0, so
L = Bl
K = Bk
Plug in:
(Bl) * (BK) = negative
(3-Ml*4) * (3-Mk*4) = negative
For the left or right term to be positive, M < 3/4
For the left or right term to be negative, M > 3/4
This doesn't help as the positive term can be negative or positive.
INSUFFICIENT.
Together, we say Ml > 3/4
then Mb < 3/4
then from (1), Mb must be between 0 and 3/4.
This makes the product of slopes always be positive.
SUFFICIENT.
Set:
L = Ml * Xl + Bl
K = Mk * Xk + Bk
You know that the line intersect at (4,3); thus, you can plug this value in both equations to get:
3 = M*4 + B, and so
Bl = 3 - Ml*4
Bk = 3 - Mk*4
(1) Product of x intercepts are positive. x intercept is when y=0, so find that, you get:
Xl = (-Bl) / (Ml) = (4*Ml - 3) / Ml
Xk = (-Bk) / (Mk) = (4*Mk - 3) / Mk
Product of x intercepts:
((4*Ml - 3) / Ml) * ((4*Mk - 3) / Mk) = positive
For the left or right term to be negative, 0< M < 3/4.
For the left or right term to be positive, M > 3/4 or M < 0
If the product is positive, you can have both negative terms; thus both slopes are positive and between 0 and 3/4. However, if you have both positive terms, you can't tell if the slopes are negative or positive.
INSUFFICIENT.
(2) Product of y intercepts are negative. y intercept when x=0, so
L = Bl
K = Bk
Plug in:
(Bl) * (BK) = negative
(3-Ml*4) * (3-Mk*4) = negative
For the left or right term to be positive, M < 3/4
For the left or right term to be negative, M > 3/4
This doesn't help as the positive term can be negative or positive.
INSUFFICIENT.
Together, we say Ml > 3/4
then Mb < 3/4
then from (1), Mb must be between 0 and 3/4.
This makes the product of slopes always be positive.
SUFFICIENT.
19 Aug 2007, 19:45
Kudos
Bookmarks
I don't think any equation is needed. Just need some drawings...
Here's my reasoning:
St1:
Insufficient. We can draw a positive sloping line l passing through (4,3) and a negative sloping line l passing through (4,3) and both cutting the positive side of the x-axis. In this case, product of their gradient = negative. However, we can also draw a negative sloping line l and line k passing through (4,3) and in this case, product of their gradient = positive.
St2:
Insufficient. If the y-intercept of line l is A, and the y-intercept of line k is B, then either (A,B) = (+,-) or (A,B) = (-,+) since we are told A*B = negative.
If (A,B) = (+,-), line l can be negative sloping and line k can be positive sloping and the product of their gradient = negative. But line l and k can also be positive sloping and the product of their gradients = positive.
Using St1 and St2:
We know that line l cannot be a positive sloping line as this will result in a line that intersects the x-axis on the negative side. line k in this case has to be positive sloping and it would intersect the x-axis on the positive side. This violates st1 as we are told the product of their x-intercepts must be positive. Hence, line l has to be negative sloping, line k positive sloping the product of their gradients negative.
Ans C
Here's my reasoning:
St1:
Insufficient. We can draw a positive sloping line l passing through (4,3) and a negative sloping line l passing through (4,3) and both cutting the positive side of the x-axis. In this case, product of their gradient = negative. However, we can also draw a negative sloping line l and line k passing through (4,3) and in this case, product of their gradient = positive.
St2:
Insufficient. If the y-intercept of line l is A, and the y-intercept of line k is B, then either (A,B) = (+,-) or (A,B) = (-,+) since we are told A*B = negative.
If (A,B) = (+,-), line l can be negative sloping and line k can be positive sloping and the product of their gradient = negative. But line l and k can also be positive sloping and the product of their gradients = positive.
Using St1 and St2:
We know that line l cannot be a positive sloping line as this will result in a line that intersects the x-axis on the negative side. line k in this case has to be positive sloping and it would intersect the x-axis on the positive side. This violates st1 as we are told the product of their x-intercepts must be positive. Hence, line l has to be negative sloping, line k positive sloping the product of their gradients negative.
Ans C
