Probability
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Hi,
I can't draw a figure and upload it but I will try to explain it. I will be easy to visualize if we can a draw a figure.
x^2 + y^2 = 4 is circle with center at origin and radius '2' units. It intersects X-axis at (-2,0) & (2,0) as well as Y-axis at (0,2) and (0.-2)
Consider the lines y=x+2. It passes through (-2,0) and (0,2). the region above this line is y>x+2
Consider point (m,n) on the circle.
n>m+2 is the set of points on the circle, over the line y = x+2.
It is one quarter of the circumference of the circle.
Si, probability than n>m+2 is 1/4.
I can't draw a figure and upload it but I will try to explain it. I will be easy to visualize if we can a draw a figure.
x^2 + y^2 = 4 is circle with center at origin and radius '2' units. It intersects X-axis at (-2,0) & (2,0) as well as Y-axis at (0,2) and (0.-2)
Consider the lines y=x+2. It passes through (-2,0) and (0,2). the region above this line is y>x+2
Consider point (m,n) on the circle.
n>m+2 is the set of points on the circle, over the line y = x+2.
It is one quarter of the circumference of the circle.
Si, probability than n>m+2 is 1/4.
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
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I'd place this one on the out-of-scope category.sana.noor wrote:is it a gmat type question?
To begin, you aren't expected to know that x^2 + y^2 = 4 describes a circle with radius 2.
Cheers,
Brent
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Will respectfully disagree with Brent here - while I can't remember any such questions on the actual exam, I know I've seen practice questions (from other reputable GMAT prep companies, not just ours) that involve the equation of a circle centered on the origin ... so SOMEBODY must've seen them (at least, you'd suppose).
Anyway, let's do this algebraically.
If m² + n² = 4, and n > m + 2, then
n² = 4 - m²
n = sqrt(4-m²)
Let's substitute this in our equality
sqrt(4 - m²) > m + 2
4 - m² > m² + 4m + 4
0 > 2m² + 4m
0 > 2m(m + 2)
0 > m(m+2)
So m is negative and (m+2) is positive, in which case
0 > m and
m + 2 > 0, or m > -2, so
0 > m > -2
So this works for 0 > m > -2. If 0 > m > -2 and n > m + 2, then 2 > n > 0.
0 > m > -2 and 2 > n > 0 accounts for 1/4 of the circumference of the circle (the portion in Quadrant II), so that's our answer.
Great question!
Anyway, let's do this algebraically.
If m² + n² = 4, and n > m + 2, then
n² = 4 - m²
n = sqrt(4-m²)
Let's substitute this in our equality
sqrt(4 - m²) > m + 2
4 - m² > m² + 4m + 4
0 > 2m² + 4m
0 > 2m(m + 2)
0 > m(m+2)
So m is negative and (m+2) is positive, in which case
0 > m and
m + 2 > 0, or m > -2, so
0 > m > -2
So this works for 0 > m > -2. If 0 > m > -2 and n > m + 2, then 2 > n > 0.
0 > m > -2 and 2 > n > 0 accounts for 1/4 of the circumference of the circle (the portion in Quadrant II), so that's our answer.
Great question!
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In the Math Review section of the Official Guide, the only conic equation mentioned is that for the parabola. If circle equations were part of the curriculum, I'd hope that some instruction would appear in the Official Guide, since this is sometimes the only resource that students use to prepare for the GMAT.
Having said that, I guess it's conceivable that the test-makers might assume that students will see something "Pythagorean-ish" in the equation x^2 + y^2 = 4 and reach the conclusion that this equation represents a circle with radius 2. However, even if that were the case, the difficulty level of the question is certainly flirting in the 800 (or even 800+ ) range. I'm still inclined to say that students will not see this kind of question on test day (unless this topic is first added to the Official Guide).
Cheers,
Brent
Having said that, I guess it's conceivable that the test-makers might assume that students will see something "Pythagorean-ish" in the equation x^2 + y^2 = 4 and reach the conclusion that this equation represents a circle with radius 2. However, even if that were the case, the difficulty level of the question is certainly flirting in the 800 (or even 800+ ) range. I'm still inclined to say that students will not see this kind of question on test day (unless this topic is first added to the Official Guide).
Cheers,
Brent
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This problem seems beyond the scope of the GMAT, but there is still a good take-away that can be gleaned here:
Many geometry problems are best solved by DRAWING.
x² + y² = r² is the equation of a circle centered at the origin with a radius of length r.
Thus, x² + y² = 4 is the equation of a circle centered at the origin with a radius of 2.
The problem asks for the probability that a point on this circle will satisfy the constraint that y > x + 2.
y > x + 2 is the region ABOVE the line y = x + 2.
DRAW the circle and the line y = x + 2:
Only the blue portion of the circle lies above the line y = x+2.
Since the blue portion is equal to 1/4 of the circle, the probability that a point on the circle will satisfy the constraint that y>x+2 is 1/4.
Similar problems that are a bit more GMAT-friendly:
https://www.beatthegmat.com/probability-t90694.html
https://www.beatthegmat.com/coordinate-g ... 22193.html
Many geometry problems are best solved by DRAWING.
x² + y² = r² is the equation of a circle centered at the origin with a radius of length r.
Thus, x² + y² = 4 is the equation of a circle centered at the origin with a radius of 2.
The problem asks for the probability that a point on this circle will satisfy the constraint that y > x + 2.
y > x + 2 is the region ABOVE the line y = x + 2.
DRAW the circle and the line y = x + 2:
Only the blue portion of the circle lies above the line y = x+2.
Since the blue portion is equal to 1/4 of the circle, the probability that a point on the circle will satisfy the constraint that y>x+2 is 1/4.
Similar problems that are a bit more GMAT-friendly:
https://www.beatthegmat.com/probability-t90694.html
https://www.beatthegmat.com/coordinate-g ... 22193.html
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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