Question

the taxi charges in a city of a consist charge together with the charge for the distance covered. for a distance of 10 km , for a charged paid is 315 and fora journey of 15 km the charge is paid 465..... what are the fixed charges and the charge per kilometre? how much does a have to pay for a trip of 32km​

Answer 1

ANSWER

Let fixed charge be Rs x and charge per km be Rs y.

 ⇒x+10y=105...(1)

 ⇒x+15y=155..(2).

Now From eq 2

⇒15y=155−x

⇒y=15155−x...(3)

Substituting y from eq3 in eq1

⇒x+1510(155−x)=105

⇒15x+1550−10x=1575

⇒5x=1575−1550

⇒5x=25

⇒x=5

Substituting x in eq2

⇒5+15y=155

⇒15y=155−5

⇒15y=150

⇒y=10

Hence x=5    and  y=10

Fixed charge is Rs.55 and the charge per kilometer is Rs.1010 

For 2525 km person have to pay = 5+10×25=255Rs.

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Answer 2

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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 kilometer? How much does a person have to pay for traveling a distance of 25 km?

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Let fixed charge be Rs x and charge per km be Rs y.

$⇒x+10y=105...(1)$

$⇒x+15y=155..(2)$

Now From eq (2)

$⇒15y = 155 - x$

$⇒y = \frac{155 - x}{15} .....(3)$

Substituting y from eq(3) in eq(1)

$⇒x + \frac{10(155 - x)}{15} = 105$

$⇒15x + 1550 - 10x = 1575$

$⇒5x = 1575 - 1550$

$⇒5x = 25$

$⇒x = 5$

Substituting $x$in eq (2)

$⇒5 + 15y = 155$

$⇒15y = 10$

$⇒y = 10$

Hence $x = 5$and $y = 10$

Fixed charge is Rs.55 and the charge per kilometer is Rs.1010

For 2525 km person have to pay$⇒5 + 10 \times 25 \\ ⇒225Rs$

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