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Question: The taxi charges in a city consist of a fixed charge together with the charge for the distance cover...

The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 1010 km, the charge paid is Rs.105Rs.105 and for a journey of 1515 km, the charge paid is Rs.155Rs.155. What are the fixed charges and the charge per kilometer? How much does a person have to pay for traveling a distance of 2525 km?
A)A) Fixed charge is Rs.5Rs.5 and the charge per kilometer is Rs.10Rs.10 for 2525km person has to pay 255255.
B)B) Fixed charge is Rs.10Rs.10 and the charge per kilometer is Rs.5Rs.5 for 2525km person has to pay 135135.
C)C) Fixed charge is Rs.15Rs.15 and the charge per kilometer is Rs.5Rs.5 for 2525km person has to pay 140140.
D)D) Fixed charge is Rs.50Rs.50 and the charge per kilometer is Rs.20Rs.20 for 2525km person has to pay 5050.

Explanation

Solution

In this question we have been provided with the data that the taxi charges consist of a fixed charge and the cost for the distance covered. We will consider the fixed cost to be xx and the cost for the distance covered to be yy per kilometer. We will then make two equations based on the two cases given to us and then solve the equations as a set of simultaneous equations and get the values. We will then calculate the value for a journey of 2525km.

Complete step by step solution:
Let the fixed charge and the cost for the distance covered be xx and yy respectively.
Now For a distance of 1010 km, the charge paid is Rs.105Rs.105 therefore, we can write:
x+10y=105(1)\Rightarrow x+10y=105\to \left( 1 \right)
And for a journey of 1515 km, the charge paid is Rs.155Rs.155.
x+15y=155(2)\Rightarrow x+15y=155\to \left( 2 \right)
On subtracting equation (1)\left( 1 \right) from (2)\left( 2 \right), we get:
5y=50\Rightarrow 5y=50
On dividing both the sides by 55, we get:
y=10\Rightarrow y=10, which is the cost per kilometer.
On substituting y=10y=10 in equation (1)\left( 1 \right), we get:
x+10(10)=105\Rightarrow x+10\left( 10 \right)=105
On multiplying, we get:
x+100=105\Rightarrow x+100=105
On transferring 100100 to the right-hand side, we get:
x=105100\Rightarrow x=105-100
On simplifying, we get:
x=5\Rightarrow x=5, which is the fixed charge.
Now to find the cost for a distance of 2525km, we will multiply 2525 with the charge per kilometer and then add the fixed charge. Mathematically, we can write it as:
25×10+5\Rightarrow 25\times 10+5
On simplifying, we get:
255\Rightarrow 255, which is the required cost to travel 2525 kms therefore, the correct option is (A)\left( \text{A} \right).

Note: It is to be remembered that in any given equation multiplying or dividing the equation by a specific constant doesn’t change the value of the equation.
In the given question we had two variables which are xx and yy, therefore they can be solved by using elimination, where there are more than three variables, and the matrix is used to solve them.