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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] New post 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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Re: Please help! [#permalink] New post 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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Re: Please help! [#permalink] New post 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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Re: Please help! [#permalink] New post 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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Re: Please help! [#permalink] New post 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] New post 24 Jun 2013, 03:03
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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 [#permalink] New post 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.
Answer (E)
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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 [#permalink] New post 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. :?:
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Re: If a taxicab charges x cents for the first 1/9 mile and x/5 [#permalink] New post 01 Dec 2013, 22:37
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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   [#permalink] 01 Dec 2013, 22:37
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