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09 Mar 2011, 23:43
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
25% (02:54) correct
75%
(02:51)
wrong
based on 427
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How many integral values of k are possible, if the lines 3x+4ky+6 = 0, and kx-3y+9 = 0 intersect in the 2nd quadrant.
A. 5
B. 4
C. 3
D. 6
E. 2
A. 5
B. 4
C. 3
D. 6
E. 2
Updated on: 10 Mar 2011, 00:49
Originally posted by beyondgmatscore on 10 Mar 2011, 00:46.
Last edited by beyondgmatscore on 10 Mar 2011, 00:49, edited 1 time in total.
Last edited by beyondgmatscore on 10 Mar 2011, 00:49, edited 1 time in total.
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vjsharma25
We are given
\(3x+4ky = -6\) .... (1) and
\(kx-3y = -9\) ........(2)
Lets calculate their intersection point in terms of k
Multiply (1) by k and (2) by 3 and then subtract (2) from (1) to solve for y, we will get
\((4k^2+9)*y = 27-6k\) or \(y = (27-6k)/(4k^2+9)\).... (3)
Now, multiply (2) by 4/3k and then add (1) and (2) to solve for x, we will get
\(x = -3*(12k+6)/(4k^2+9)\).... (4)
We know that these intersection points lie in 2nd quadrant, so y has be positive and x has to be negative
y is positive when \(27-6k > 0\)or \(k < 4.5\)
Similarly, x is negative when \(12+6k > 0\) or \(k > -0.5\)
So, we have -0.5<k<4.5, so it can take integer values of 0,1,2,3,4. Hence, five values.
This is tedious approach and takes at least 2.5 to 3 minutes.
I don't know of a faster way to do so
10 Mar 2011, 00:32
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solving for values of x and y from original equations. And knowing x < 0 and y > 0 in II quadrant.
x = -18 - 36k / (4k^2 + 9)
y = (108 - 24k) / 4 (4k^2 + 9)
Hence -18 - 36 k < 0
So k > -0.5
And 108 - 24k >0
k < 108 / 24
k < 18 / 4
k < 9/2
k < 4.5
-0.5 < k < 4.5
Phew !
x = -18 - 36k / (4k^2 + 9)
y = (108 - 24k) / 4 (4k^2 + 9)
Hence -18 - 36 k < 0
So k > -0.5
And 108 - 24k >0
k < 108 / 24
k < 18 / 4
k < 9/2
k < 4.5
-0.5 < k < 4.5
Phew !
.
