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

Can some experts show the best way to solve this problem?

OA D

Hi lheiannie07,
Let's take a look at your question.

Marion rented a car for $18.00 plus $0.10 per mile driven. If 'c' represents the amount charges and 'd' represents the miles driven, we can write:
$$c=18+0.10d$$

Craig rented a car for $25.00 plus $0.05 per mile driven. We can write it as,
$$c=25+0.05d$$

If each was charged exactly the same amount for the rental, then,
$$18+0.10d=25+0.05d$$
$$0.10d-0.05d=25-18$$
$$0.05d=7$$
$$d=\frac{7}{0.05}=140$$

Therefore, Option D is correct.

Hope it helps.
I am available if you'd like any follow up.

lheiannie07 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

Can some experts show the best way to solve this problem?

OA D

we can easily solve this algebrically. but if give a little thinking we can do it orally as well.
Craig is initially taking $7 more than Marion . For both to get the same amount Marion should compensate from the rate
per kilometer which he charges more than Craig, As Marion is charging $ .05 per kilometer more than Craig he should drive
7/.05=140 kilometer for both to get the same amount .
Hence option D