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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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Safiya
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If a taxicab charges x cents for the first 1/9 mile and x/5 cents for [#permalink] New post   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 + (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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E

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 [#permalink] New post   20 Jul 2010, 02:13
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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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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for [#permalink] New post   18 Jul 2013, 22:28
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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.
Answer (E)
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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for [#permalink] New post   19 Jul 2010, 19:07
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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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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for [#permalink] New post   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 [#permalink] New post   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+ (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. :?:
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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for [#permalink] New post   01 Dec 2013, 23:37
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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.
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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for [#permalink] New post   03 Aug 2015, 16:29
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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 + (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 cents for [#permalink] New post   02 Jan 2018, 10:01
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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 cents for [#permalink] 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] 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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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for [#permalink] New post   16 Aug 2019, 17:35
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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)

\(y=x+\frac{9x}{5}(y-\frac{1}{9})\)
\(y=x+\frac{9x}{5}(\frac{9y}{9}-\frac{1}{9})\)
\(y=x+\frac{9x}{5}·\frac{(9y-1)}{9}\)
\(y=x+\frac{x}{5}·\frac{(9y-1)}{1}\)
\(y=x+\frac{x(9y-1)}{5}\)
\(y=x+\frac{(9xy-x)}{5}\)

E
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If a taxicab charges x cents for the first 1/9 mile and x/5 cents for [#permalink] New post   18 Dec 2019, 11:46
VeritasKarishma

hellosanthosh2k2

I formulated the right equation, but struggled to identify that 9 can be factored out. How can we be sure that 9 can be factored out from both the terms?

Is it only because y is given as a whole number or could this have been done in any case?
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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for [#permalink] New post   18 Dec 2019, 12:05
This is ambiguous,

x cents per mile for the first 1/9 mile

NOT a flat fee of x cents for the first 1/9 mile

allenh98 wrote:
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 + (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 cents for [#permalink] New post   20 Feb 2022, 00:48
Asked: 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?

Taxicab charges for y miles = x + (y-1/9)*9x/5 = x + (9xy - x)/5

IMO E
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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for [#permalink] New post   24 Jun 2022, 22:18
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



What if y=0?

Does option E still satisfy?
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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for [#permalink] New post   24 Jun 2022, 22:57
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udaypratapsingh99 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



What if y=0?

Does option E still satisfy?


If y=0, get the heck out of my taxicab, you clown!!!
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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for [#permalink] New post   25 Jun 2022, 00:39
ThatDudeKnows wrote:
udaypratapsingh99 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



What if y=0?

Does option E still satisfy?


If y=0, get the heck out of my taxicab, you clown!!!


Haha!
I wish, the equation had itself said that.
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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 cents for [#permalink] New post   25 Aug 2023, 20:02
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