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

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49% (03:17) correct
49% (01:53) wrong based on 68 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.

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
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___________________________________ Please give me kudos if you like my post

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?

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.

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

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Expert's post

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)
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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]
01 Dec 2013, 22:37

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.
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