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In the XY-plane, region X consists of all the points (x, y) such that

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In the XY-plane, region X consists of all the points (x, y) such that  [#permalink]

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New post 12 Nov 2018, 00:19
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  95% (hard)

Question Stats:

36% (02:24) correct 64% (02:07) wrong based on 75 sessions

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In the XY-plane, region X consists of all the points (x, y) such that 3x + 4y ≤ 12. Is the point (r, s) in region X?

(1) 4r + 3s = 12
(2) r ≤ 4 and s ≤ 3

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In the XY-plane, region X consists of all the points (x, y) such that  [#permalink]

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New post 12 Nov 2018, 03:36
Bunuel wrote:
In the XY-plane, region X consists of all the points (x, y) such that 3x + 4y ≤ 12. Is the point (r, s) in region X?

(1) 4r + 3s = 12
(2) r ≤ 4 and s ≤ 3


from given eqn 3x+4y<=12
we can find region of the line which would be :

y=-3/4*x+3
slope is -3/4 and points would be (0,3) and ( 4,0)

from statement 1:
4r + 3s = 12

if r=0 , s=4 , (r,s does not fall in region 3x + 4y ≤ 12)

if =3, s=0 ( r,s falls in region 3x + 4y ≤ 12)

in sufficient

from statement 2:
r ≤ 4 and s ≤ 3 all values will fall under region r ≤ 4 and s ≤ 3
r=4 and s=3, we get 16+9=25 no
and r=1, s=1 , 4+3<12 yes


hence option IMO E is correct.
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In the XY-plane, region X consists of all the points (x, y) such that  [#permalink]

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New post 16 Feb 2019, 04:22
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This problem can be solved easily through graphs.

The question says x/4 + y/3 <= 1. And asks us to find if (r,s) falls within that segment i.e. if (r,s) falls below or on the line x/4 + y/3 = 1.

(1) From this, we can get r/3 + s/4 = 1. Plot the line that passes through (3,0) and (0,4). We see that a portion of the line extends beyond the required region. So, insufficient.

(2) A few values that satisfies r<=4 and s<=3 do not fall within the required region. E.g. (4, 3). Hence insufficient.

(1) + (2), we see that a tiny portion that satisfies both the equation doesnt fall under the required region. The tiny portion is denoted by the region shaded in black.

Hence, Ans: E
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In the XY-plane, region X consists of all the points (x, y) such that   [#permalink] 16 Feb 2019, 04:22
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