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

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carcass
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Marion rented a car for $18.00 plus $0.10 per mile driven. Craig rente [#permalink] New post   21 May 2017, 03:11
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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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D
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Marion rented a car for $18.00 plus $0.10 per mile driven. Craig rente [#permalink] New post   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:

\(18 + 0.1d = 25 + 0.05d\);

\(0.05d=7\);

\(\frac{5}{100}*d=7\);

\(d = 140\).

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

At a certain distance (d miles) for the rentals of both Marion and Craig to become equal, the initial difference of $7 (700 cents) Must be compensated by the 5 cents per mile difference

So d = 700/5 = 140 miles
Thus it takes 140 miles for the initial fixed cost difference (700 cents) to be compensated by variable per mile cost (5 cents)

Hence answer is D
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Re: Marion rented a car for $18.00 plus $0.10 per mile driven. Craig rente [#permalink] New post   25 May 2017, 15:53
Expert Reply
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


We can create the following equation:

18 + 0.1d = 25 + 0.05d

0.05d = 7

d = 7/0.05 = 140

Answer: D
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Re: Marion rented a car for $18.00 plus $0.10 per mile driven. Craig rente [#permalink] New post   27 Jan 2020, 19:42
Can we check the answer choices?
Like there is a difference of cost is 25 - 18 = 7 and 140 is the only multiple of 7..Didn't consider decimal as the difference is too small. Is my approach ok?
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Re: Marion rented a car for $18.00 plus $0.10 per mile driven. Craig rente [#permalink] New post   02 Jun 2020, 00:24
Farina wrote:
Can we check the answer choices?
Like there is a difference of cost is 25 - 18 = 7 and 140 is the only multiple of 7..Didn't consider decimal as the difference is too small. Is my approach ok?

Farina
You should not do this - the highlighted part.
Had the second car rental been 0.03$ per mile then you would have been in trouble. The answer would have been 100(A).
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Re: Marion rented a car for $18.00 plus $0.10 per mile driven. Craig rente [#permalink] New post   25 Oct 2022, 03:05
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Marion rented a car for $18.00 plus $0.10 per mile driven. Craig rente [#permalink]
25 Oct 2022, 03:05
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