Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If a taxicab charges x cents for the first 1/9 mile and x/5 [#permalink]
19 Jul 2010, 15:17

1

This post received KUDOS

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

60% (03:27) correct
40% (02:03) wrong based on 167 sessions

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.

Re: If a taxicab charges x cents for the first 1/9 mile and x/5 [#permalink]
18 Jul 2013, 21:28

4

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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?

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 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?

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.

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?

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 _________________

___________________________________ Please give me kudos if you like my post

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

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. _________________

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?

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

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.

Re: If a taxicab charges x cents for the first 1/9 mile and x/5 [#permalink]
07 Apr 2015, 19:36

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

McCombs Acceptance Rate Analysis McCombs School of Business is a top MBA program and part of University of Texas Austin. The full-time program is small; the class of 2017...