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

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

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19 Jul 2010, 15:17
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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

Last edited by Safiya on 20 Jul 2010, 14:16, edited 2 times in total.

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19 Jul 2010, 18:07
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Safiya wrote:
Hi All,

I hope someone can explain me the solution of the below problem;

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

Total cost for y miles = cost for first 1/9 mile + cost for the additional (y-1/9) miles
= x + (y-1/9)* x (Since each additional 1/9 th mile costs x/9 cents , each additional mile costs x cents and (y-1/9) additional miles cost (y-1/9)*x cents)
= x + xy -x/9 which i don't see in the answers
Are you sure the answer choices are accurate?

Thanks
Harsha
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19 Jul 2010, 18:57
Total cost for y miles = cost for first 1/9 mile + cost for the additional (y-1/9) miles
= x + (y-1/9)* x (Since each additional 1/9 th mile costs x/9 cents , each additional mile costs x cents and (y-1/9) additional miles cost (y-1/9)*x cents)
= x + xy -x/9 which i don't see in the answers
Are you sure the answer choices are accurate?

Thanks
Harsha[/quote]

Hi Harsha,

I double checked - the answer choices are correct

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20 Jul 2010, 01:13
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Safiya wrote:
Hi All,

I hope someone can explain me the solution of the below problem;

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

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;

$$x+(9y-1)*\frac{x}{9}=x+\frac{9xy-x}{9}$$.

If you say that OA is E, then charge for each additional 1/9 mile should be x/5 cents instead of x/9.

Hope it helps.
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20 Jul 2010, 14:18
Crack 700 and Bunuel ;

I am so sorry it was written x/9 instead of x/5 , I've just edited the question. Many thanks for the answer!

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

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24 Jun 2013, 03:03
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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18 Jul 2013, 21:28
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Safiya 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

Responding to a pm:

You can assume values for x and y to get the answer.

Charges for first 1/9 mile = x = 5 cents (assume)
Charges for each subsequent 1/9 mile = x/5 = 1 cent (if x = 5 cents)

So say, we need to cover a total distance of y = 1 mile. How much would be the charge?
5 cents for the first 1/9 mile. Now we are left with 8/9 mile distance. 1 cent for every subsequent 1/9 mile i.e. 8 cents.
So total charge = 5 + 8 = 13 cents

Now put x = 5 and y = 1 in the options. The moment you put y = 1, you see that options (A), (B) and (C) are out of the picture since they give very small values. In option (D), 9x - y will be 1 less than a multiple of 5 so it will not be divisible by 5. Only (E) gives you 13.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 18477 [4], given: 237 Intern Joined: 21 Nov 2013 Posts: 39 Kudos [?]: 144 [0], given: 3 Re: If a taxicab charges x cents for the first 1/9 mile and x/5 [#permalink] ### Show Tags 30 Nov 2013, 02:18 VeritasPrepKarishma wrote: Safiya 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 Responding to a pm: You can assume values for x and y to get the answer. Charges for first 1/9 mile = x = 5 cents (assume) Charges for each subsequent 1/9 mile = x/5 = 1 cent (if x = 5 cents) So say, we need to cover a total distance of y = 1 mile. How much would be the charge? 5 cents for the first 1/9 mile. Now we are left with 8/9 mile distance. 1 cent for every subsequent 1/9 mile i.e. 8 cents. So total charge = 5 + 8 = 13 cents Now put x = 5 and y = 1 in the options. The moment you put y = 1, you see that options (A), (B) and (C) are out of the picture since they give very small values. In option (D), 9x - y will be 1 less than a multiple of 5 so it will not be divisible by 5. Only (E) gives you 13. Answer (E) I tried solving it this way, however failed. What am I doing wrong? Many thanks. I did the following: x=5 y=9 Charges for the first 1/9 mile: 1/9 * 9 * 5 = 5 Charges for the remaining 1/9 miles: 8/9 * 9 * 5/5 = 8 The sum of both is indeed: 5+8 = 13 Putting this into answer E: x+(9xy-x) / 5 yields the following: 5 + (9*5*9 - 5)/5 = 5 + (405 - 5)/5 = 5 + 80 = 85 ...thus, not 13. Kudos [?]: 144 [0], given: 3 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7866 Kudos [?]: 18477 [1], given: 237 Location: Pune, India Re: If a taxicab charges x cents for the first 1/9 mile and x/5 [#permalink] ### Show Tags 01 Dec 2013, 22:37 1 This post received KUDOS Expert's post BabySmurf wrote: VeritasPrepKarishma wrote: Safiya 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 Responding to a pm: You can assume values for x and y to get the answer. Charges for first 1/9 mile = x = 5 cents (assume) Charges for each subsequent 1/9 mile = x/5 = 1 cent (if x = 5 cents) So say, we need to cover a total distance of y = 1 mile. How much would be the charge? 5 cents for the first 1/9 mile. Now we are left with 8/9 mile distance. 1 cent for every subsequent 1/9 mile i.e. 8 cents. So total charge = 5 + 8 = 13 cents Now put x = 5 and y = 1 in the options. The moment you put y = 1, you see that options (A), (B) and (C) are out of the picture since they give very small values. In option (D), 9x - y will be 1 less than a multiple of 5 so it will not be divisible by 5. Only (E) gives you 13. Answer (E) I tried solving it this way, however failed. What am I doing wrong? Many thanks. I did the following: x=5 y=9 Charges for the first 1/9 mile: 1/9 * 9 * 5 = 5 Charges for the remaining 1/9 miles: 8/9 * 9 * 5/5 = 8 The sum of both is indeed: 5+8 = 13 Putting this into answer E: x+(9xy-x) / 5 yields the following: 5 + (9*5*9 - 5)/5 = 5 + (405 - 5)/5 = 5 + 80 = 85 ...thus, not 13. The first 1/9 mile is charged at 5 cents (as per the numbers assumed) Now every remaining 1/9th of a mile will be charged at 1 cent. If you have to cover a distance of total 9 miles, the first 1/9th of a mile will be charged ta 5 cents. Now remaining distance is 8 miles and 8/9th of a mile. For every 1 mile now the fare will be 9 cents (1 cent per 1/9th of a mile) and for the 8/9th of a mile, the fare will be 8 cents. So total fare of the remaining distance will be 9*8 + 8 = 80 Total fare of first 1/9th of a mile + additional 8/9th of a mile and 8 more miles = 5 + 80 = 85 Note that I had assumed y = 1. You assumed y = 9. Hence total fare for both cannot be 13 cents. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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03 Aug 2015, 15:29

First 1/9 mile:
x * 1/9

additional miles past the first 1/9:
x/5 * (y - 1/9)

x/9 + x/5(y - 1/9) = x/9 + xy/5 - x/45

Now remember that E is:
x + (9xy-x)/5

so if we multiple our equation by 9, we get:
x + 9/5 * xy - x/5

factor out the 1/5 in second part of the equation:
x + 1/5 ( 9xy - x)

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

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02 Jan 2018, 09:01
Here is how, I solved it.
Fare is charged in fractions of 1/9 miles. For y miles you have 9y such fractions.

For first 1/9mile, the fare is x.
Remaining number of fraction miles 9y-1. Fare charged = (9y-1)x/5 = (9xy-x)/5

Total = x + (9xy-x)/5. Ans. E

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Re: If a taxicab charges x cents for the first 1/9 mile and x/5   [#permalink] 02 Jan 2018, 09:01
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# If a taxicab charges x cents for the first 1/9 mile and x/5

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