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If 0<x<2 and 0<y<2 on the xy-plane, what is the probability that x+y<1

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Math Revolution GMAT Instructor
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If 0<x<2 and 0<y<2 on the xy-plane, what is the probability that x+y<1  [#permalink]

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New post 24 Dec 2018, 00:03
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D
E

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Question Stats:

14% (01:20) correct 86% (01:25) wrong based on 37 sessions

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[Math Revolution GMAT math practice question]

If \(0<x<2\) and \(0<y<2\) on the \(xy\)-plane, what is the probability that \(x+y<1\)?

\(A. \frac{1}{2}\)
\(B. \frac{1}{4}\)
\(C. \frac{1}{6}\)
\(D. \frac{1}{8}\)
\(E. \frac{1}{10}\)

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Re: If 0<x<2 and 0<y<2 on the xy-plane, what is the probability that x+y<1  [#permalink]

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New post 24 Dec 2018, 01:39
1
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If a point (x,y) is ramdomly chosen in the region \(0<x<2\) and \(0<y<2\) of the \(xy\)-plane, what is the probability that \(x+y<1\)?

\(A. \frac{1}{2}\)
\(B. \frac{1}{4}\)
\(C. \frac{1}{6}\)
\(D. \frac{1}{8}\)
\(E. \frac{1}{10}\)


Image


\(? = {{{S_{\Delta OAB}}} \over {{S_{{\rm{square}}}}}}\,\,\,\,\,\,\,\left( {{\rm{geometric}}\,\,{\rm{probability}}} \right)\)


\(A = \left( {0,1} \right)\,\,,\,\,B = \left( {1,0} \right)\)


\(? = {{\,\,{{1 \cdot 1} \over 2}\,\,} \over {2 \cdot 2}} = {1 \over 8}\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: If 0<x<2 and 0<y<2 on the xy-plane, what is the probability that x+y<1  [#permalink]

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New post 24 Dec 2018, 02:32
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If \(0<x<2\) and \(0<y<2\) on the \(xy\)-plane, what is the probability that \(x+y<1\)?

\(A. \frac{1}{2}\)
\(B. \frac{1}{4}\)
\(C. \frac{1}{6}\)
\(D. \frac{1}{8}\)
\(E. \frac{1}{10}\)


Hey MathRevolution guys i LOVE your out of the box questions and approaches! I am more than curious about why this must be 1/8 and not 1/4... please enlighten me!

Best, gota900
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GMAT 1: 760 Q51 V42
GPA: 3.82
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Re: If 0<x<2 and 0<y<2 on the xy-plane, what is the probability that x+y<1  [#permalink]

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New post 26 Dec 2018, 00:45
=>

Attachment:
12.26.png
12.26.png [ 5.03 KiB | Viewed 207 times ]


The outside square contains all points satisfying \(0 < x < 2\) and \(0 < y < 2.\)
The shaded region consists of all points satisfying \(0 < x < 2\) and \(0 < y < 2\) such that \(x + y < 1\). Its area is \(\frac{1}{2}\), and the area of the outside square is \(4\).
Thus, the required probability is \((\frac{1}{2}) / 4 = \frac{1}{8}.\)

Therefore, the answer is D.
Answer: D
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Re: If 0<x<2 and 0<y<2 on the xy-plane, what is the probability that x+y<1 &nbs [#permalink] 26 Dec 2018, 00:45
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