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Intern  Joined: 06 Aug 2007
Posts: 29
Location: Montreal
If a taxicab charges x cents for the first 1/9 mile and x/5  [#permalink]

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Question Stats: 64% (02:20) correct 36% (02:12) wrong based on 526 sessions

### HideShow timer Statistics 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

Originally posted by Safiya on 19 Jul 2010, 16:17.
Last edited by Safiya on 20 Jul 2010, 15:16, edited 2 times in total.
Math Expert V
Joined: 02 Sep 2009
Posts: 56369

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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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Manager  Joined: 20 Mar 2010
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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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Intern  Joined: 06 Aug 2007
Posts: 29
Location: Montreal

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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  Intern  Joined: 06 Aug 2007
Posts: 29
Location: Montreal

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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!
Math Expert V
Joined: 02 Sep 2009
Posts: 56369
Re: If a taxicab charges x cents for the first 1/9 mile and x/5  [#permalink]

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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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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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Karishma
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Intern  Joined: 22 Nov 2013
Posts: 39
Re: If a taxicab charges x cents for the first 1/9 mile and x/5  [#permalink]

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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.

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. Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9458
Location: Pune, India
Re: If a taxicab charges x cents for the first 1/9 mile and x/5  [#permalink]

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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.

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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Karishma
Veritas Prep GMAT Instructor

Intern  Joined: 17 Jul 2015
Posts: 5
Re: If a taxicab charges x cents for the first 1/9 mile and x/5  [#permalink]

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1

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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Joined: 27 May 2017
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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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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]

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_________________ Re: If a taxicab charges x cents for the first 1/9 mile and x/5   [#permalink] 04 Mar 2019, 05:13
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