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Mike rented a car for $18 plus $x per mile driven. Tom rented a car fo
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11 Jan 2024, 16:24

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Mike rented a car for $18 plus $x per mile driven. Tom rented a car for $25 plus $y per mile driven. If each drove d miles and each was charged exactly the same amount for the rental then what was the charge, in dollars, each had to pay?

Indicate all possible answers.

A. \(25+\frac{7x}{x-y}\)

B. \(\frac{18x}{x-y}\)

C. \(18+\frac{7x}{x-y}\)

D. \(\frac{32x-25y}{x-y}\)

E. \(\frac{25x-18y}{x-y}\)

Indicate all possible answers.

A. \(25+\frac{7x}{x-y}\)

B. \(\frac{18x}{x-y}\)

C. \(18+\frac{7x}{x-y}\)

D. \(\frac{32x-25y}{x-y}\)

E. \(\frac{25x-18y}{x-y}\)

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Re: Mike rented a car for $18 plus $x per mile driven. Tom rented a car fo
[#permalink]
18 Jan 2024, 23:13

1

The charge that Mike paid = 18 + xd

The charge that Tom paid = 25 + yd

Given that the charges were equal,

\(18 + xd = 25 + yd \)

\(xd = 7 + yd \)

\(d(x - y) = 7\)

\(d = \frac{7}{x-y}\)

Plug d back into the initial equations for charges:

\(18 + xd = 18 + x(\frac{7}{x-y}) = 18 + \frac{7x}{x-y}\)

This is option C.

Multiply 18 by the denominator, x-y, to get:

\(18(\frac{x-y}{x-y}) + \frac{7x}{x-y} = \frac{18x-18y+7x}{x-y} = \frac{25x-18y}{x-y} \)

This is option E.

The charge that Tom paid = 25 + yd

Given that the charges were equal,

\(18 + xd = 25 + yd \)

\(xd = 7 + yd \)

\(d(x - y) = 7\)

\(d = \frac{7}{x-y}\)

Plug d back into the initial equations for charges:

\(18 + xd = 18 + x(\frac{7}{x-y}) = 18 + \frac{7x}{x-y}\)

This is option C.

Multiply 18 by the denominator, x-y, to get:

\(18(\frac{x-y}{x-y}) + \frac{7x}{x-y} = \frac{18x-18y+7x}{x-y} = \frac{25x-18y}{x-y} \)

This is option E.

gmatclubot

Re: Mike rented a car for $18 plus $x per mile driven. Tom rented a car fo [#permalink]

18 Jan 2024, 23:13
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