[Math Revolution GMAT math practice question]
In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5. If p and q are integers, how many possibilities are there for the point (p,q)?
A. 2
B. 4
C. 8
D. 12
E. 16
In the x-y coordinate plane, the distance between (p,q) and
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- Max@Math Revolution
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To answer this question quickly, sketch a circle of radius 5, the center of which is located at point (1, 1).Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5. If p and q are integers, how many possibilities are there for the point (p,q)?
A. 2
B. 4
C. 8
D. 12
E. 16
On the circumference of the circle, the four points (1, 6), (1, -4), (-4, 1), and (6, 1) - which are above, below, left, and right of point (1, 1), respectively - are each 5 units from point (1, 1).
To quickly find the remaining points, consider a right triangle with a hypotenuse of 5, such that the triangle's hypotenuse coincides with the circle's radius.
If p and q are integers, then the legs of this right triangle have integer lengths. Only the 3-4-5 Pythagorean Triple satisfies this requirement, so the legs of this triangle have lengths of 3 and 4.
Consider the possible locations of point (p, q) in Quadrant I. Point (p, q) will be either 3 units to the right and 4 units above point (1, 1), or it will be 4 units to the right and 3 units above point (1, 1) -- points (4, 5) and (5, 4), respectively.
Similarly, there are two possible locations of point (p, q) in Quadrant II, two in Quadrant III, and two in Quadrant IV.
Thus, the total number of possible locations for point (p, q) is 12, and the correct answer is choice D.
- Max@Math Revolution
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=>
(p-1)^2 + (q-1)^2 = 5^2
If p - 1 = ±3, and q - 1 = ±4, then p = 1 ± 3, and q = 1 ± 4. There are four possible points: ( p, q ) = ( 4, 5 ), ( 4, -3 ), ( -2, 5 ), ( -2, -3 ).
If p - 1 = ±4, and q - 1 = ±3, then p = 1 ±4, and q = 1 ± 3. There are four possible points: ( p, q ) = ( 5, 4 ), ( 5, -2 ), ( -3, 4 ), ( -3, -2 ).
If p - 1 = 0, and q - 1 = ±5, then p = 1, and q = 1 ±5. There are two possible points: ( p, q ) = ( 1, 6 ), ( 1, -4 ).
If p - 1 = ±5, and q - 1 = 0, then p - 1 = ±5, and q = 1. There are two possible points: ( p, q ) = ( 6, 1 ), ( -4, 1 ).
There are a total of 4 + 4 + 2 + 2 = 12 possibilities for the point (p,q).
Therefore, the answer is D.
Answer: D
(p-1)^2 + (q-1)^2 = 5^2
If p - 1 = ±3, and q - 1 = ±4, then p = 1 ± 3, and q = 1 ± 4. There are four possible points: ( p, q ) = ( 4, 5 ), ( 4, -3 ), ( -2, 5 ), ( -2, -3 ).
If p - 1 = ±4, and q - 1 = ±3, then p = 1 ±4, and q = 1 ± 3. There are four possible points: ( p, q ) = ( 5, 4 ), ( 5, -2 ), ( -3, 4 ), ( -3, -2 ).
If p - 1 = 0, and q - 1 = ±5, then p = 1, and q = 1 ±5. There are two possible points: ( p, q ) = ( 1, 6 ), ( 1, -4 ).
If p - 1 = ±5, and q - 1 = 0, then p - 1 = ±5, and q = 1. There are two possible points: ( p, q ) = ( 6, 1 ), ( -4, 1 ).
There are a total of 4 + 4 + 2 + 2 = 12 possibilities for the point (p,q).
Therefore, the answer is D.
Answer: D
Math Revolution
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Score an excellent Q49-51 just like 70% of our students.
[Free] Full on-demand course (7 days) - 100 hours of video lessons, 490 lesson topics, and 2,000 questions.
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