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Tough Coordinate geomentry problem

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Tough Coordinate geomentry problem

by glorydefined » Thu Aug 27, 2009 2:04 pm
In the xy-plane, region R consists of all the points (x, y) such that 2x + 3y <= 6. Is the point (r, s) in region R ?

(1) 3r + 2s = 6

(2) r <= 3 and s <= 2

Also can someone please let me know how to tackle such problems? I am having difficulty in understanding such problems :?
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by pradeepsarathy » Thu Aug 27, 2009 4:42 pm
IMO E

We are asked to figure out whether 2r+3s<=6

Stmt 1:
3r + 2s = 6
This alone gives no information about (r,s).
combining this stmt with the question stem -

3r+2s=6 and 2r+3s=6(taking the max value)
solving for r and s, we get (6/5,6/5).
But this value is not unique as 2r+3s can also be <6.
Say for example 2r+3s=5 and solving for (r,s) with 3r+2s=6, we get (8/5,6/5).
Hence stmt 1 alone is insufficient.

Stmt 2:
r<=3, s<=2
Case 1:
r = 3 and s = 2.
Substituting this pair in the equation 2r+3s, we get 12, which is clearly greater than 6.

Case 2:
r = 0, s = 0.
Substituting this pair in the equation 2r+3s, we get 0, which is clearly less than 6.
Hence stmt 2 alone is insufficient.

Combining both Stmt 1 and Stmt 2:
Still we cannot definitley prove that any given value of (r,s) lies in the region R(i.e, it satisfies the eqn 2r+3s<=6).

Hence this too is insufficient.
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by glorydefined » Fri Aug 28, 2009 1:44 am
Hi Pradeep,

Thanks for the reply, your answer is indeed correct. But am I missing something here. The question asks if 2x+3y<=6 and not 2R+3S<=6. so my question is does "region R consists of all the points (x,y)" mean that x=R and y=s? please help me understand.
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by pradeepsarathy » Fri Aug 28, 2009 5:15 am
@glorydefined,

The question states that,all points (x,y) in region R satisfy the eqn 2x+3y<=6.

We are asked to find whether (r,s) is in region R.


Hence we need to prove that 2r+3s<=6.
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by fruti_yum » Tue Sep 01, 2009 9:13 am
pradeepsarathy wrote:IMO E

We are asked to figure out whether 2r+3s<=6

Stmt 1:
3r + 2s = 6
This alone gives no information about (r,s).
combining this stmt with the question stem -

3r+2s=6 and 2r+3s=6(taking the max value)
solving for r and s, we get (6/5,6/5).
But this value is not unique as 2r+3s can also be <6.
Say for example 2r+3s=5 and solving for (r,s) with 3r+2s=6, we get (8/5,6/5).
Hence stmt 1 alone is insufficient.

Stmt 2:
r<=3, s<=2
Case 1:
r = 3 and s = 2.
Substituting this pair in the equation 2r+3s, we get 12, which is clearly greater than 6.

Case 2:
r = 0, s = 0.
Substituting this pair in the equation 2r+3s, we get 0, which is clearly less than 6.
Hence stmt 2 alone is insufficient.

Combining both Stmt 1 and Stmt 2:
Still we cannot definitley prove that any given value of (r,s) lies in the region R(i.e, it satisfies the eqn 2r+3s<=6).

Hence this too is insufficient.
We are given 2x +3y <= 6 and we're given that it is in region R.

We are asked to find if points r,s are in region R! I do not understand Pradeepsarathy's explanation
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by shanrizvi » Tue Sep 01, 2009 12:15 pm
Well, if you plot the equation 3y<=6-2x on a graph, you will see that the x-intercept is 3, the y-intercept is 2 and anything below the line satisfies the inequality. As long as x is less than 3 and y is less than 2, the point will be in this region which satisfied the inequality.

In my opinion, the answer is B.
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