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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7609
GMAT 1: 760 Q51 V42 GPA: 3.82
In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 45% (01:57) correct 55% (01:56) wrong based on 38 sessions

### HideShow timer Statistics [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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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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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7609
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5  [#permalink]

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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).

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