Question

hi there

need help with maths aryabhatta

The taxi charges in Hyderabad are fixed, along with the charge for the distance covered. For a distance of 10 km., the charge paid is D220. For a journey of 15km. the charge paid is D310.
i. What are the fixed charges and charge per km?
ii. How much does a person have to pay for travelling a distance of 25 km?

Answer 1

Hello!!

Solution:

Let fixed charge be p

And charge per km be q

So,  p + 10q = 220    ....    (1)

and  p + 15q = 310    ....    (2)

Solving the equations,

=> equation(2) - equation(1)

           p+15q  - p - 10q = 310 -220

                           5q =90

                             q=Rs.18

put value of q in equation(1)

p+10*18=220

      p=220-180

      p=Rs.40  

We have

q=Rs18, p=Rs 40

For a distance of 25 km,

A person have to pay (40+25*18)=Rs 490.

Hope it helps!

Answer 2

$<b><i>$i. Let the fixed charge be x.

Let the charges per km be y.

Total charges to be paid will be the sum of the fixed charges and the charges per km.

For 10kms, the amount paid is ₹220.

x + 10y = 220 …I

For 15kms, the amount paid is ₹ 310

x + 15y = 310 …II

Solving I and II we get

x + 10y = 220

x + 15y = 310

(-) (-) (-)

......................

-5y = -90

y = 18

Substituting the value of y in Eq. I we get

x + 10(18) = 220

x = 220 – 180

x = 40

The Fixed charge are ₹ 40 and the charges per km are ₹18

ii. For 25 km ,

total = 40 + 25(18) = 40 + 450 = 490

$<u>$hence, for 25 km, the taxi fare is 490