The taxi charges in Hyderabad are fixed, along with the charge for the distance covered. For a distance of $10 k m$ , the charge paid is Rs. $220$ . For a journey of $15 k m$ , the charge paid is Rs. $310$
(i) What are the fixed charges and charge per km?
(ii) How much does a person have to pay for travelling a distance of $25 k m$ ?

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Solution

(i) Let the fixed be $R s . X$ and charge per $k m = y$

$x + 10 y = 220$ ......(1)

$x + 15 y = 310$ .......(2)
--___________
$+ 5 y = 90$

$y = 18$

(ii) for $25 k m$

$= x + 25 y$

$\Rightarrow 40 + 25 (18)$ $= R s .490$

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Q. Question 3 (iv)
Form the pair of linear equations for the following problems and find their solution by substitution method:
The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

The taxi charges in a city consists of a fixed charge along with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹105 and for a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charge per km? How much does a person have to pay for traveling a distance of 25 km?

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(i) Let the fixed be Rs.X and charge per km=yx+10y=220 ......(1)x+15y=310 .......(2)__-____-_____________ +5y=90 y=18(ii) for 25km=x+25y⇒40+25(18) =Rs.490

(i) Let the fixed be $R s . X$ and charge per $k m = y$

$x + 10 y = 220$ ......(1)

$x + 15 y = 310$ .......(2)
--___________
$+ 5 y = 90$

$y = 18$

(ii) for $25 k m$

$= x + 25 y$

$\Rightarrow 40 + 25 (18)$ $= R s .490$