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# Marion rented a car for $18.00 plus$0.10 per mile driven. Craig rente

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Joined: 01 Sep 2010
Posts: 3305
Marion rented a car for $18.00 plus$0.10 per mile driven. Craig rente  [#permalink]

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21 May 2017, 03:11
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Difficulty:

5% (low)

Question Stats:

94% (01:20) correct 6% (01:13) wrong based on 47 sessions

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Marion rented a car for $18.00 plus$0.10 per mile driven. Craig rented a car for $25.00 plus$0.05 per mile driven. If each drove d miles and each was charged exactly the same amount for the rental, then d equals

(A) 100

(B) 120

(C) 135

(D) 140

(E) 150

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Marion rented a car for $18.00 plus$0.10 per mile driven. Craig rente  [#permalink]

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21 May 2017, 03:39
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carcass wrote:
Marion rented a car for $18.00 plus$0.10 per mile driven. Craig rented a car for $25.00 plus$0.05 per mile driven. If each drove d miles and each was charged exactly the same amount for the rental, then d equals

(A) 100

(B) 120

(C) 135

(D) 140

(E) 150

For d miles Marion's cost would be: 18 + 0.1d and Craig's cost would be 25 + 0.05d.

Since we are told that each was charged exactly the same amount then:

$$18 + 0.1d = 25 + 0.05d$$;

$$0.05d=7$$;

$$\frac{5}{100}*d=7$$;

$$d = 140$$.

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Re: Marion rented a car for $18.00 plus$0.10 per mile driven. Craig rente  [#permalink]

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22 May 2017, 23:52
Another way to look at this problem:

Marion = $18 + 10 cents per mile Craig =$25 + 5 cents per mile

Initial fixed cost for Craig is $(25-18) =$7 more than that of Marion, but variable cost (per mile) for Craig is 5 cents less than that of Marion.