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How many integral values of k are possible, if the lines 3x+

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How many integral values of k are possible, if the lines 3x+  [#permalink]

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New post 09 Mar 2011, 23:43
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  95% (hard)

Question Stats:

32% (03:08) correct 68% (02:45) wrong based on 218 sessions

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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
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Re: How many integral values of k are possible  [#permalink]

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New post Updated on: 10 Mar 2011, 00:49
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vjsharma25 wrote:
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

I was able to do it but I am not convinced by my approach.
Please explain!!


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 :cry:

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.
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Re: How many integral values of k are possible  [#permalink]

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New post Updated on: 10 Mar 2011, 00:26
I got -0.5 < k < 4.5

hence the integral values are 0,1,2,3,4. should be B. Please check. I am not a fan of algebra, sorry :o

vjsharma25 wrote:
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

I was able to do it but I am not convinced by my approach.
Please explain!!

Originally posted by gmat1220 on 10 Mar 2011, 00:15.
Last edited by gmat1220 on 10 Mar 2011, 00:26, edited 1 time in total.
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Re: How many integral values of k are possible  [#permalink]

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New post 10 Mar 2011, 00:24
gmat1220 wrote:
I got -0.5 < k < 3.375

hence the integral values are 0,1,2,3. should be B

vjsharma25 wrote:
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

I was able to do it but I am not convinced by my approach.
Please explain!!

Can you explain how you got this range?
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Re: How many integral values of k are possible  [#permalink]

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New post 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 !
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Re: How many integral values of k are possible  [#permalink]

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New post 10 Mar 2011, 00:44
@gmat1220
But now you got the correct range and correct answer :-).

I was not taking values of x and y as independent. Thats why i got confused.
Thanks for the explanation.
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Re: How many integral values of k are possible  [#permalink]

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New post 10 Mar 2011, 00:48
1
Man this was monster ! especially my equations go completely haywire when I look at algebra.
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Re: How many integral values of k are possible, if the lines 3x+  [#permalink]

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New post 29 Jun 2015, 22:49
Can someone pls explain how to arrive at these 2 equations mentioned above.
x = -18 - 36k / (4k^2 + 9)
y = (108 - 24k) / 4 (4k^2 + 9)
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Re: How many integral values of k are possible, if the lines 3x+  [#permalink]

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New post 29 Jun 2015, 23:34
2
jayanthjanardhan wrote:
Can someone pls explain how to arrive at these 2 equations mentioned above.
x = -18 - 36k / (4k^2 + 9)
y = (108 - 24k) / 4 (4k^2 + 9)



The given equations are

3x + 4ky = -6 ------1
kx - 3y = -9 ------2

Try solving the two equations:

1. Multiply eq 1 by k, 3kx + 4k^2y = -6k ---------3
2. Multiply eq 2 by 3, 3kx - 9y = -27 ---------4

Solving eq, 3 and 4, we can get the value of y.

where y = (27-6k)/(4k^2 + 9)..

Since y should be > 0, 27-6k > 0, solving k<4.5

Similarly, solving for x gives or sub y in eq 2 , x = -6(6k+3) / 4k^2 + 9

since x < 0, 6k+3 should be > 0 => 6k+3 > 0, solving k < -0.5

Hence k has 5 Integral values 0, 1, 2, 3 and 4.
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Re: How many integral values of k are possible, if the lines 3x+  [#permalink]

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Re: How many integral values of k are possible, if the lines 3x+ &nbs [#permalink] 26 Aug 2017, 21:51
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