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If a taxicab charges x cents for the first 1/9 mile and x/5 cents for

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If a taxicab charges x cents for the first 1/9 mile and x/5 cents for  [#permalink]

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New post 22 Feb 2018, 20:58
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Question Stats:

57% (02:52) correct 43% (03:23) wrong based on 62 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

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If a taxicab charges x cents for the first 1/9 mile and x/5 cents for  [#permalink]

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New post 22 Feb 2018, 21:13
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Bunuel wrote:
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


The taxicab charges x cents for the first 1/9 miles and x/5 for each additional 1/9 mile.
The simplest way is to assign values for x and y. Let x=10 and y=2 miles

To cover 1 mile, the taxicab would charge (x + 8x/5) = 10 + 16 = 26 cents
For the second mile, the taxicab would charge 9x/5 = 18 cents
Therefore, the taxicab charges 44 cents to cover 2 miles.

Evaluating answer options,
A. x + (xy - x)/45 = 10 + (20 - 10)/45 = 460/45
B. x - (xy - x)/45 = 10 - (20 - 10)/45 = 440/45
C. (2x + 9y)/5 = (20 + 18)/5 = 38/5
D. x + (9x - y)/5 = 10 + (90-2)/5 = 138/5

E. x + (9xy - x)/5 = 10 + (180 - 10)/5 = 10 + 34 = 44(Option E)
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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for  [#permalink]

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New post 23 Feb 2018, 02:17
Bunuel wrote:
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


x=10
y=1

1/9 10 cents

8/9 2 x8 = 16 cents

1mile = 26 cents

substitute x and y in options and we get x+9(yx-x)/5

(E) imo
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If a taxicab charges x cents for the first 1/9 mile and x/5 cents for  [#permalink]

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New post 07 Mar 2018, 11:03
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increment is 1/9 and total miles is y miles
to get rid of fraction, let us multiply by 9, so increment is 1 miles and total miles = 9y

total cost = x + (9y-1)x/5 => x+ (9xy - x)/5 => E
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If a taxicab charges x cents for the first 1/9 mile and x/5 cents for   [#permalink] 07 Mar 2018, 11:03
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