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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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Updated on: 17 Aug 2019, 03:23
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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 + (xyx)/45 (B) x  (xyx)/45 (C) 2x+9y/5 (D) x + (9xy)/5 (E) x + (9xyx)/5
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Originally posted by Safiya on 19 Jul 2010, 16:17.
Last edited by Bunuel on 17 Aug 2019, 03:23, edited 3 times in total.
Edited the question.




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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for
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20 Jul 2010, 02:13
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+ xyx /45 (B) x xyx /45 (C) 2x+9y / 5 (D) x+ 9xy / 5 (E) x+ 9xyx / 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) = \(9y1\) parts of \(\frac{1}{9}\) miles to be charged additionally. \(\frac{x}{9}\) cents per part = \((9y1)*\frac{x}{9}\) cents; \(x+(9y1)*\frac{x}{9}=x+\frac{9xyx}{9}\). 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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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for
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19 Jul 2010, 19:07
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+ xyx /45 (B) x xyx /45 (C) 2x+9y / 5 (D) x+ 9xy / 5 (E) x+ 9xyx / 5 Total cost for y miles = cost for first 1/9 mile + cost for the additional (y1/9) miles = x + (y1/9)* x (Since each additional 1/9 th mile costs x/9 cents , each additional mile costs x cents and (y1/9) additional miles cost (y1/9)*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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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for
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20 Jul 2010, 15:18
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 cents for
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18 Jul 2013, 22:28
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+ (xyx) /45 (B) x (xyx) /45 (C) 2x+9y / 5 (D) x+ (9xy) / 5 (E) x+ (9xyx) / 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)
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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for
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30 Nov 2013, 03:18
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+ (xyx) /45 (B) x (xyx) /45 (C) 2x+9y / 5 (D) x+ (9xy) / 5 (E) x+ (9xyx) / 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+(9xyx) / 5 yields the following: 5 + (9*5*9  5)/5 = 5 + (405  5)/5 = 5 + 80 = 85 ...thus, not 13.



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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for
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01 Dec 2013, 23:37
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+ (xyx) /45 (B) x (xyx) /45 (C) 2x+9y / 5 (D) x+ (9xy) / 5 (E) x+ (9xyx) / 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+(9xyx) / 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.
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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for
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03 Aug 2015, 16:29
Answer is E
First 1/9 mile: x * 1/9
additional miles past the first 1/9: x/5 * (y  1/9)
added together: x/9 + x/5(y  1/9) = x/9 + xy/5  x/45
Now remember that E is: x + (9xyx)/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 cents for
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02 Jan 2018, 10:01
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 9y1. Fare charged = (9y1)x/5 = (9xyx)/5
Total = x + (9xyx)/5. Ans. E



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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for
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22 Feb 2018, 21:13
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 + (902)/5 = 138/5E. 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
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07 Mar 2018, 11:03
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 + (9y1)x/5 => x+ (9xy  x)/5 => E



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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for
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16 Aug 2019, 17:35
x is cost for first \(\frac{1}{9}\) miles y is total miles y\(\frac{1}{9}\) is number of miles at the new rate \(\frac{9x}{5}\) is new rate per mile (not per \(\frac{1}{9}\) of a mile)
\(y=x+\frac{9x}{5}(y\frac{1}{9})\) \(y=x+\frac{9x}{5}(\frac{9y}{9}\frac{1}{9})\) \(y=x+\frac{9x}{5}·\frac{(9y1)}{9}\) \(y=x+\frac{x}{5}·\frac{(9y1)}{1}\) \(y=x+\frac{x(9y1)}{5}\) \(y=x+\frac{(9xyx)}{5}\)
E




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