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In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5

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In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5  [#permalink]

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New post 10 Aug 2018, 02:36
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

35% (01:54) correct 65% (01:42) wrong based on 31 sessions

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[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 \((p,q)\)?

\(A. 2\)
\(B. 4\)
\(C. 8\)
\(D. 12\)
\(E. 16\)

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Re: In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5  [#permalink]

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New post 10 Aug 2018, 03:25
MathRevolution 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 \((p,q)\)?

\(A. 2\)
\(B. 4\)
\(C. 8\)
\(D. 12\)
\(E. 16\)


The distance between \((p,q)\) and \((1,1)\) is \(5\)
Or, \((p-1)^2+(q-1)^2=5^2\)

1) \((p-1)^2+(q-1)^2=5^2+0^2\); so\((p-1)^2=0\) and \((q-1)^2=5^2\). Possible pairs of (p,q):-(0,5), (0,-5),(5,0), (-5,0) ----(4 pairs)
2)\((p-1)^2+(q-1)^2=5^2=4^2+3^2\);so \((p-1)^2=4^2\) and \((q-1)^2=3^2\). Possible pairs of (p,q):-(5,4), (5,-2),(-3,4), (-3,-2)---(4 pairs)
3)\((p-1)^2+(q-1)^2=5^2=3^2+4^2\);so \((p-1)^2=3^2 and (q-1)^2=4^2\). Possible pairs of (p,q):-(4,5), (-2,5),(4,-3), (-2,-3)---(4 pairs)
N,B,:- If \((x-a)^2=k\) , then \(\sqrt{(x-a)^2}=|x-a|=+k\) or \(-k\)
Total number of (p,q) pairs=4+4+4=12
Ans. (D)
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Re: In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5  [#permalink]

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New post 13 Aug 2018, 06:29
=>

(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
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Re: In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5 &nbs [#permalink] 13 Aug 2018, 06:29
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