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If no car in 1993 was involved in more than four accidents, what is th

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If no car in 1993 was involved in more than four accidents, what is th [#permalink]

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28 May 2017, 07:07
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If no car in 1993 was involved in more than four accidents, what is the minimum number of cars that could have been in accidents in 1993?

(A) 50 thousand
(B) 60 thousand
(C) 70 thousand
(D) 80 thousand
(E) 90 thousand
[Reveal] Spoiler: OA

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If no car in 1993 was involved in more than four accidents, what is th [#permalink]

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08 Aug 2017, 02:56

To find the minimum number of cars that would be in accidents,
we need to assume maximum cars are involved in exactly four accidents.

Since there are a total of 36*10000 accidents that take place in 1993,
the number of accidents taking place in 1993 is $$\frac{36*10000}{4} = 90$$ thousand(Option E)
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08 Aug 2017, 18:00
Bunuel wrote:

If no car in 1993 was involved in more than four accidents, what is the minimum number of cars that could have been in accidents in 1993?

(A) 50 thousand
(B) 60 thousand
(C) 70 thousand
(D) 80 thousand
(E) 90 thousand

This prompt's lesson: look carefully at the legends.

The uselessness of the second graph should be apparent immediately from its legend "in millions."

Accidents are reported in "teen thousands." Division is necessary, and there are no fractional accidents.

1993 had 360,000 accidents. At four per car, minimum number of cars involved is 90,000.

Kudos [?]: 335 [0], given: 590

If no car in 1993 was involved in more than four accidents, what is th   [#permalink] 08 Aug 2017, 18:00
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