Manhattan Challenge Problem

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Manhattan Challenge Problem

by akhpad » Wed Jun 09, 2010 3:09 am
Source: Manhattan Challenge Problem

In the xy-plane, region Q consists of all points (x, y) such that x^2 + y^2 ≤ 100. Is the point (a, b) in region Q?

(1) a + b = 14

(2) a > b


IMO: E; Solution and OA will come in next weak.

My Solution
--------------

Statement 1
a + b = 14

a=10, b=4, a^2 + b^2 = 116; NO
a=8, b=6, a^2 + b^2 = 100; YES
a=7.1, b=6.9, a^2 + b^2 = 98.02; YES
a = b = 7, a^2 + b^2 = 98; YES

Insufficient

Statement 2
a > b

a=10, b=4, a^2 + b^2 = 116; NO
a=8, b=6, a^2 + b^2 = 100; YES
a=7.1, b=6.9, a^2 + b^2 = 98.02; YES

Insufficient

Statement 1 and 2
a=10, b=4, a^2 + b^2 = 116; NO
a=8, b=6, a^2 + b^2 = 100; YES
a=7.1, b=6.9, a^2 + b^2 = 98.02; YES

Insufficient

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by liferocks » Wed Jun 09, 2010 3:34 am
I will go with E
x^2 + y^2<= 100 is all the area enclosed in a circle with radius 10

so any point with abscissa or ordinate with more than 10 will not be in that circular region.

From 1
a+b=14
a and b can be any thing like a=b=7..its in x^2 + y^2<= 100
but a=14 and b=0 its not in x^2 + y^2<= 100..insufficient

From 2
7<=a<=14 for some values a,b will be in x^2 + y^2<= 100 for others it will not be..in sufficient.
"If you don't know where you are going, any road will get you there."
Lewis Carroll

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by kstv » Wed Jun 09, 2010 7:36 am
x²+y²<=100
all points within the the circle with radius 10 and origin (0,0) is the region Q
1) a+b = 14
the point (8,6) - 64+36 or (6,8) - 36+64are on the circle.
(7,7) - 49+49 < 100 is inside the circle, but (9,5)- 81 + 25 > 100 is outside Q.
2) a > b
Not sufficient.
Combining (1) & (2) & assuming a and b are integers, then it will be option C, otherwise E.

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