Question

The taxi charge in a city consists of fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge 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.? ​

Answer 1

Given :-

• Charged paid for travelling a distance of 10km is = Rs 105

• Charged paid for travelling a distance of 15 km is = Rs 155

To Find :-

• Charge paid per km

• The fixed charge

• Total charge to be paid for travelling a distance of 25 km.

Solution :-

Let the fixed charge be ₹ x.

Charge per km be ₹ y.

As per question :-

Given that,

Charged paid for travelling a distance of 10km is = Rs 105.

Therefore,

x + 10y = 105......... eq(1)

Again, given that

Charged paid for travelling a distance of 15 km is = Rs 155

Hence,

x + 15 y = 155.......... eq(2)

Find the value of x from eq(1)

x + 10y = 105

⟼ x = 105 - 10y...........eq (3)

Now, put the value of x in eq (2)

x + 15 y = 155

⟼ 105 - 10y + 15 y = 155

⟼ 105 + 5y = 155

⟼ 5y = 50

⟼ y = 10

After putting the value of y in eq(3), we get

x = 105 - 10y

⟼ x = 105 - 100

$⟼x = 5$

Charge for 25 km is = x + 25 y

Substituting the value of x and y, we get

Charge for 25 km is = 5 + 25 × 10

= ₹ 255

Hence,

fixed charge is = ₹ 5

Per km charge = ₹ 10

Charge for 25 km = ₹ 255

Answer 2

Answer:

Given :-

• Charged paid for travelling a distance of 10km is = Rs 105

• Charged paid for travelling a distance of 15 km is = Rs 155

To Find :-

• Charge paid per km

• The fixed charge

• Total charge to be paid for travelling a distance of 25 km.

Solution :-

Let the fixed charge be ₹ x.

Charge per km be ₹ y.

As per question :-

Given that,

Charged paid for travelling a distance of 10km is = Rs 105.

Therefore,

x + 10y = 105......... eq(1)

Again, given that

Charged paid for travelling a distance of 15 km is = Rs 155

Hence,

x + 15 y = 155.......... eq(2)

Find the value of x from eq(1)

x + 10y = 105

⟼ x = 105 - 10y...........eq (3)

Now, put the value of x in eq (2)

x + 15 y = 155

⟼ 105 - 10y + 15 y = 155

⟼ 105 + 5y = 155

⟼ 5y = 50

⟼ y = 10

After putting the value of y in eq(3), we get

x = 105 - 10y

⟼ x = 105 - 100

⟼x = 5⟼x=5

Charge for 25 km is = x + 25 y

Substituting the value of x and y, we get

Charge for 25 km is = 5 + 25 × 10

= ₹ 255

Hence,

fixed charge is = ₹ 5

Per km charge = ₹ 10

Charge for 25 km = ₹ 255