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This is a good problem I found, though good to share. If a

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Director
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This is a good problem I found, though good to share. If a [#permalink]  30 Jun 2008, 22:32
This is a good problem I found, though good to share.

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-y) / 5}

I will post OA and explanation later.
Manager
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Re: Taxi Charging More.... [#permalink]  30 Jun 2008, 22:46
x for the first 1/9
x/5 for the subsequent y-1/9 miles

x + ( x/5 ) (y - 1/9)
then,
x + {(9xy-x) / 45}

Edit:
Initially i calculated wrong.. now correcting
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Re: Taxi Charging More.... [#permalink]  01 Jul 2008, 01:06
code19 wrote:
x for the first 1/9
x/5 for the subsequent y-1/9 miles

x + ( x/5 ) (y - 1/9)
then,
x + {(9xy-x) / 45}

Edit:
Initially i calculated wrong.. now correcting

I think you miss a factor 9 in the second term: since the price is x/5 for 1/9 mile, the price for 1 mile is 9*x/5

Answer is therefore x + {(9xy-x) / 5}

abhijit_sen, are you sure answer (E) is well written ?
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Re: Taxi Charging More.... [#permalink]  01 Jul 2008, 01:09
abhijit_sent,

is there anhy type in answer choice E?

I got the answer as below.

x + (y-1/9)*9x/5

= x + (9xy-x)/5
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Re: Taxi Charging More.... [#permalink]  01 Jul 2008, 04:39
Quote:
I got the answer as below.

x + (y-1/9)*9x/5

= x + (9xy-x)/5

I am sorry but could you pls explain how you got this answer?
My one is the same as the previous one: x+(9xy-x)/45
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Re: Taxi Charging More.... [#permalink]  01 Jul 2008, 04:43
nlutsenko wrote:
I am sorry but could you pls explain how you got this answer?
My one is the same as the previous one: x+(9xy-x)/45

First 1/9 mile : $$x$$ cents
Following (y-1/9) miles : $$x/5$$ cents per $$1/9$$ mile i.e. $$(y-1/9)*\frac{x/5}{1/9} = (9xy-x)/5$$
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Re: Taxi Charging More.... [#permalink]  01 Jul 2008, 05:05
I was going to say this is a perfect problem to just choose numbers. I chose x = 100, y = 2. This makes the trip cost 440 cents. I noticed that all I had to do was make sure the answer came out evenly. So I knew that

(C) (2x+9y) / 5
(D) x + {(9x-y) / 5}
(E) x + {(9xy-y) / 5}

Could not be the answer because I would be dividing by 5 a number that I just subtracted 2 from... that would not be even. So I eliminated those off the bat. But as stated above, neither of the other two add up either. Hopefully on the test there is not a question that has no answer
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Re: Taxi Charging More.... [#permalink]  01 Jul 2008, 05:07
I think there is a typo in answer (E).
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Re: Taxi Charging More.... [#permalink]  01 Jul 2008, 06:52
OA is E.
Even I think answer is wrongly typed , but I got it as it is. So wanted to confirm my answer as well. Anyways "Oski" has explained it well.
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Re: Taxi Charging More.... [#permalink]  01 Jul 2008, 07:09
Oski wrote:
I think you miss a factor 9 in the second term: since the price is x/5 for 1/9 mile, the price for 1 mile is 9*x/5

Thanks..i too was missing this.
Re: Taxi Charging More....   [#permalink] 01 Jul 2008, 07:09
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