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