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If a taxicab charges x cents
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18 May 2020, 11:12
18 May 2020, 11:12
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Question Stats:
72% (02:10) correct
27% (01:23) wrong based on 22 sessions
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If a taxicab charges x cents for the first 1/9 mile and x/5 cents for each additional 1/9 mile or fraction thereof, what is the charge, in cents, for a ride of y miles, where y is a whole number?
(A) x + (xy-x)/45
(B) x - (xy-x)/45
(C) 2x+9y/5
(D) x + (9x-y)/5
(E) x + (9xy-x)/5
I would like to ask why the choosing number strategy here doesn't work?
I chose x= 10cents
Y= 1 mile
(A) x + (xy-x)/45
(B) x - (xy-x)/45
(C) 2x+9y/5
(D) x + (9x-y)/5
(E) x + (9xy-x)/5
I would like to ask why the choosing number strategy here doesn't work?
I chose x= 10cents
Y= 1 mile
Re: If a taxicab charges x cents
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18 May 2020, 12:10
18 May 2020, 12:10
Expert Reply
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Charge will be the sum of the following:
\(x\) cents for for the first \(\frac{1}{9}\) mile;
In 1 mile there are 9 parts of \(\frac{1}{9}\), hence in \(y\) miles (where \(y\) is a whole number) there are \(9y\) parts of \(\frac{1}{9}\) miles minus one part (first \(\frac{1}{9}\) mile) = \(9y-1\) parts of \(\frac{1}{9}\) miles to be charged additionally. \(\frac{x}{9}\) cents per part = \((9y-1)*\frac{x}{9}\) cents;
\(x+(9y-1)*\frac{x}{9}=x+\frac{9xy-x}{9}\).
PS: this is NOT a question contained in the GRE big book. I have the book and is NOT there
\(x\) cents for for the first \(\frac{1}{9}\) mile;
In 1 mile there are 9 parts of \(\frac{1}{9}\), hence in \(y\) miles (where \(y\) is a whole number) there are \(9y\) parts of \(\frac{1}{9}\) miles minus one part (first \(\frac{1}{9}\) mile) = \(9y-1\) parts of \(\frac{1}{9}\) miles to be charged additionally. \(\frac{x}{9}\) cents per part = \((9y-1)*\frac{x}{9}\) cents;
\(x+(9y-1)*\frac{x}{9}=x+\frac{9xy-x}{9}\).
PS: this is NOT a question contained in the GRE big book. I have the book and is NOT there
