January 21, 2019 January 21, 2019 10:00 PM PST 11:00 PM PST Mark your calendars  All GMAT Club Tests are free and open January 21st for celebrate Martin Luther King Jr.'s Birthday. January 22, 2019 January 22, 2019 10:00 PM PST 11:00 PM PST In case you didn’t notice, we recently held the 1st ever GMAT game show and it was awesome! See who won a full GMAT course, and register to the next one.
Author 
Message 
TAGS:

Hide Tags

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]
Show Tags
Updated on: 20 Jul 2010, 14:16
Question Stats:
63% (02:20) correct 37% (02:11) wrong based on 419 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 + (xyx)/45 (B) x  (xyx)/45 (C) 2x+9y/5 (D) x + (9xy)/5 (E) x + (9xyx)/5
Originally posted by Safiya on 19 Jul 2010, 15:17.
Last edited by Safiya on 20 Jul 2010, 14:16, edited 2 times in total.




Math Expert
Joined: 02 Sep 2009
Posts: 52344

Re: Please help!
[#permalink]
Show Tags
20 Jul 2010, 01:13
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Manager
Joined: 20 Mar 2010
Posts: 78

Re: Please help!
[#permalink]
Show Tags
19 Jul 2010, 18:07
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
_________________
___________________________________ Please give me kudos if you like my post



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

Re: Please help!
[#permalink]
Show Tags
19 Jul 2010, 18:57
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[/quote] Hi Harsha, I double checked  the answer choices are correct



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

Re: Please help!
[#permalink]
Show Tags
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!



Math Expert
Joined: 02 Sep 2009
Posts: 52344

Re: If a taxicab charges x cents for the first 1/9 mile and x/5
[#permalink]
Show Tags
24 Jun 2013, 03:03



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8803
Location: Pune, India

Re: If a taxicab charges x cents for the first 1/9 mile and x/5
[#permalink]
Show Tags
18 Jul 2013, 21:28
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)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 21 Nov 2013
Posts: 39

Re: If a taxicab charges x cents for the first 1/9 mile and x/5
[#permalink]
Show Tags
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+ (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.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8803
Location: Pune, India

Re: If a taxicab charges x cents for the first 1/9 mile and x/5
[#permalink]
Show Tags
01 Dec 2013, 22:37
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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 17 Jul 2015
Posts: 5

Re: If a taxicab charges x cents for the first 1/9 mile and x/5
[#permalink]
Show Tags
03 Aug 2015, 15:29
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)



Intern
Joined: 27 May 2017
Posts: 1

Re: If a taxicab charges x cents for the first 1/9 mile and x/5
[#permalink]
Show Tags
02 Jan 2018, 09:01
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




Re: If a taxicab charges x cents for the first 1/9 mile and x/5 &nbs
[#permalink]
02 Jan 2018, 09:01






