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# GRE Math Challenge #101-In a series of races 10 toy cars

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GRE Math Challenge #101-In a series of races 10 toy cars [#permalink]  09 May 2015, 12:04
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In a series of races, 10 toy cars are raced, 2 cars at a time. If each car must race each of the other cars exactly twice, how many races must be held?

(A) 40
(B) 90
(C) 100
(D) 180
(E) 200
[Reveal] Spoiler: OA

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Re: GRE Math Challenge #101-In a series of races 10 toy cars [#permalink]  18 Dec 2017, 22:32
sandy wrote:
In a series of races, 10 toy cars are raced, 2 cars at a time. If each car must race each of the other cars exactly twice, how many races must be held?

(A) 40
(B) 90
(C) 100
(D) 180
(E) 200

Here total of 10 cars with 2 cars race at a time

so the total possibility will be 10C2 = 40

Now each car must race twice, then the total race = 45 * 2 =90 races.

Moreoevr another approach is -

Since there are 10 cars and each of them race against 9 cars(since the tenth car cannot race against itself) for twice = 10 * 9 * 2 =180.

But there are two cars common so total races will be = $$\frac{180}{2}$$ =90 races
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Re: GRE Math Challenge #101-In a series of races 10 toy cars   [#permalink] 18 Dec 2017, 22:32
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