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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
11 Jan 2024, 16:24
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Question Stats:
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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
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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