# Calculate taxi fare by giving multiple inputs and single output

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Divyang Jain il 25 Mag 2019
Risposto: Karan il 25 Ott 2023
Write a function called taxi_fare that computes the fare of a taxi ride. It takes two inputs: the distance in kilometers (d) and the amount of wait time in minutes (t). The fare is calculated like this:
• the first km is \$5
• every additional km is \$2
• and every minute of waiting is \$0.25.
Once a km is started, it counts as a whole (Hint: consider the ceil built-in function). The same rule applies to wait times. You can assume that d >0 and t >= 0 but they are not necessarily integers. The function returns the fare in dollars. For example, a 3.5-km ride with 2.25 minutes of wait costs \$11.75. Note that loops and if-statements are neither necessary nor allowed.
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SALMA HACHIMY il 14 Feb 2022
what about random input what we gonna do??
Walter Roberson il 14 Feb 2022
You are going to write correct code so that your code passes automatic grading? That's what you are going to do about random input ?

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### Risposte (12)

Kalpesh Shah il 22 Set 2019
Its a Simple Question. Use 'ceil' as suggested in hint and it works fine
function fare = taxi_fare(d,t)
d = ceil(d)
t = ceil(t)
fare=5+(2*(d-1))+(t*0.25)
##### 9 CommentiMostra 7 commenti meno recentiNascondi 7 commenti meno recenti
Quyen Le il 30 Mag 2021
@Thato Mokhothu idk why but you have to arrange like that way : func (1st row), ceil for 2nd and 3rd rows, then the equation in the forth row, then it works :)) i don't really get why it can not run when I let the equation on the 2nd row.
LIke this
function fare = taxi_fare (d,t)
d = ceil(d);
t = ceil(t);
fare = 5 + 2*(d-1) + 0.25*t;
end
function fare = taxi_fare (d,t)
fare = 5 + 2*(d-1) + 0.25*t;
d = ceil(d);
t = ceil(t);
end
Walter Roberson il 30 Mag 2021
Suppose you do
A = 1
A = 1
B = A * 5
B = 5
A = 2
A = 2
After that series of statements, what is B ?
B
B = 5
B did not change. When you wrote B = A * 5, you were not creating a formula for B: instead MATLAB copies the current value for A as of the time of the assignment and uses that to calculate B, and after that B has no knowledge of how it got to have the value it has.
If you were writing formulas instead of expressions, then consider the formula
B = B + 1
B = 6
As a formula, it would have to be interpreted as forcing B to be some value such that B and (B+1) were the same value. There are only three values such that B and (B+1) are the same value:
-inf == -inf + 1
ans = logical
1
inf == inf + 1
ans = logical
1
[nan, nan+1]
ans = 1×2
NaN NaN
I did not try to compare nan and nan+1 because comparisons with nan are always false.
If we understand
B = B + 1
as copying the current value of B, adding 1, and making the result the new value of B, then we get something that is very useful in programming. But if we think we are writing formulas, that
B = B + 1
is defining a formula saying that B and B + 1 must be the same value, then that is something that has very few uses.
Why does it matter? Well, if you define
fare = 5 + 2*(d-1) + 0.25*t;
and then afterwards change d and t, then if you are defining fare as a formula, it might make sense to change d and t afterwards -- but as the B = B + 1 case shows, expecting those to be formulas is often not very useful. If you instead understand it as copying the current values of d and t and using those to compute fare and then forgetting all information about how that exact value came to be, then when you change d and t afterwards, then those changes to d and t do not matter.
Furthermore, if we were interpreting
fare = 5 + 2*(d-1) + 0.25*t;
as being a formula relating fare to whatever value of d and t existed at the time we asked about the value of fare, then we would also have to understand
d = ceil(d);
as being a formula -- and that is a formula that could only be true if d is already an integer. If the input d were, for example, 3.8, then as a formula
3.8 == ceil(3.8)
would be false, and instead of doing something useful, you could only be telling MATLAB that the function as a whole only applies if d was already an integer... and then what would you expect MATLAB to do if the user wanted to calculate the fare for 3.8 miles ?

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Vijay G il 17 Ago 2020
Modificato: Vijay G il 17 Ago 2020
function fare = taxi_fare(x,y)
d = 5+(((ceil (x))-1)*2);
t = ((ceil(y))*.25);
fare = d+t;
end
letme know whether this helped you. Thanks
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Raj il 27 Mag 2019
Seems quite straightforward. Where exactly are you having problem? Your formulation should look like this:
Fare=5+(2*(d-1))+(t*0.25) % Minimum fare of \$5 for first KM plus \$2 for every additional KM plus waiting time charge
Just pass the inputs d and t to the function named taxi_fare and use ceil on both before putting them in the above mentioned equation. You will get your fare which you can pass as output of the function. Good luck!!
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Sejal Syed il 21 Set 2019
Doesnt work tried it
Walter Roberson il 22 Set 2019
Sejal Syed please explain what it was you tried, and what error you encountered.

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amjad khan il 5 Apr 2020
Modificato: DGM il 4 Mar 2023
function fare=taxi_fare(distance,time)
distance=ceil(distance-1)
time=ceil(time)
fare=5*(2+(distance-1)+0.25*time
end
% "d"can be used instead of distance and "t" instead of time,
##### 3 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
Jessika Lisboa il 12 Mag 2020
Thank you
Walter Roberson il 13 Mag 2020
the first km is in the base price. You need to calculate "additional" kilometres, after the first, so you subtract the 1 initial kilometre

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Arooba Ijaz il 1 Mag 2020
function [fare] =taxi_fare (d,t)
a=ceil(d)
b=ceil(t)
fare=(b*0.25)+((a-1)*2)+5
end
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AYUSH MISHRA il 25 Mag 2020
function fare= taxi_fare(d,t)
d=ceil(d); %d=total distance cover by taxi
t=ceil(t); %t=total time taken to complete distance
RD=ceil(d-1); %RD= remaning distace after 1 km distance
fare=5+2*RD+0.25*t;%fare=1st km charge + Remaning km Distance charge + waiting time charge
end
solution of question
fare=taxi_fare(3.5,2.25)
fare =
11.7500
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DGM il 21 Feb 2023
After the first line, d is an integer, so d-1 is also an integer. Using ceil() on d-1 does nothing.
I'll give you credit for using comments though. That's good.

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Dante Gallo Torres il 14 Giu 2023
You can use matrix for this problem:
function fare = taxi_fare(d, t)
fare = [1 ceil(d-1) ceil(t)] * [5; 2; 0.25]
end
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Rik il 14 Giu 2023
Neat solution to 'hide' the addition in the matrix product. The only things missing in your answer are documentation of this function and a semicolon.

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Roweida Bawab il 5 Mag 2020
function fare = taxi_fare(d, t)
d = ceil(d);
t = ceil(t);
fare = 5 + (d-1)*2 + t*0.25
end
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muyiwabowen il 8 Mag 2020
function rare = taxi_fare(d,t)
rare = 5 + 2*ceil((d-1)) + 0.25*ceil(t);
end
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Muhammad Talha Malik il 14 Giu 2020
Modificato: Walter Roberson il 14 Giu 2020
can anyone tell me what is wrong with this code?
function fare = taxi_fare(d,t)
fare = (5+2*(d-1))+0.25*d;
end
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RAHWA ZESLUS il 9 Feb 2021
how can i solve rondom input?
Walter Roberson il 9 Feb 2021
@RAHWA ZESLUS by writing correct code?
What happens when you test your code with random input? Is it giving the same answer you get when you manually calculate for the same input?

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Stavan il 25 Lug 2023
Modificato: Stavan il 25 Lug 2023
I wrote this code for the test, but somehow it isnt working with the test, i am getting the right answers but it isnt letting me pass, can anyone show what could be the problem
function total_taxi=taxi_fare(d,t)
total_taxi= 5+2*ceil(d-1)+0.25*ceil(t)
end
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Stavan il 26 Lug 2023
okay so what should be written instead of -1 according to you
Also the grader just says the answer is incorrect, even though the code works and gives out correct answers
Walter Roberson il 26 Lug 2023

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Karan il 25 Ott 2023
I would say this code is more general and can be use for any km if not counted as a whole number, like one can use it for 0.5km.
function fare = taxi_fare(d, t)
d = ceil(d);
t = ceil(t);
fare = 5*(d - (d-1)) + (d-1)*2 + t*0.25
end
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