A taxi charge \[{\text{Rs}}.20\] for the first \[km\]& @ \[{\text{Rs}}{\text{.1}}2\] per \[km\] for subsequent distance covered. Taking the distance as \[x{\text{ }}km\] and a total fare \[{\text{Rs}}{\text{. }}y\], write a linear equation depicting the relation in \[x{\text{ and }}y\]. From your graph find the taxi charges for covering \[{\text{16 }}km\].
Answer
556.9k+ views
Hint: In this question, the linear equation is formed by equating the taxi charges and the distance covered multiplied with their respective charges. Then plot the graph for the obtained linear equation in two variables and then find the point at which the distance covered is \[{\text{16 }}km\] to find the required taxi charges.
Complete step-by-step answer:
Given taxi charges for the first \[km\] = \[{\text{Rs}}.20\]
Taxi charges for next subsequent distance in \[km\] = \[{\text{Rs}}{\text{.1}}2\]
The total distance covered = \[x{\text{ }}km\]
The total fare = \[{\text{Rs}}{\text{. }}y\]
So, the total fare is given by
\[ \Rightarrow y = 20\left( {{\text{for first }}km} \right) + 12\left[ {\left( {{\text{total distance}} - 1} \right){\text{ in }}km} \right]\]
By using the above data, we have
\[
\Rightarrow y = 20 + 12\left( {x - 1} \right) \\
\Rightarrow y = 20 + 12x - 12 \\
\therefore y = 12x + 8 \\
\]
Thus, the linear equation depicting the relation in \[x{\text{ and }}y\] is \[y = 12x + 8\].
Drawing graph to the linear equation, by using the given below table, we have
So, the graph is shown below:
Now we have to plot the point at which the distance covered is \[16{\text{ }}km\]i.e., \[x = 16\]
So, clearly from the graph \[y = 200\] when \[x = 16\].
Therefore, the taxi charges are Rs. 200.
Note: The graph is taken on the bases of the scale 1 unit = \[5{\text{ }}km\]on x-axis at which distance is covered and 1 unit = \[{\text{Rs}}.50\] on y-axis at which taxi charges are mentioned. The taxi charges cannot be negative as the distance covered is positive.
Complete step-by-step answer:
Given taxi charges for the first \[km\] = \[{\text{Rs}}.20\]
Taxi charges for next subsequent distance in \[km\] = \[{\text{Rs}}{\text{.1}}2\]
The total distance covered = \[x{\text{ }}km\]
The total fare = \[{\text{Rs}}{\text{. }}y\]
So, the total fare is given by
\[ \Rightarrow y = 20\left( {{\text{for first }}km} \right) + 12\left[ {\left( {{\text{total distance}} - 1} \right){\text{ in }}km} \right]\]
By using the above data, we have
\[
\Rightarrow y = 20 + 12\left( {x - 1} \right) \\
\Rightarrow y = 20 + 12x - 12 \\
\therefore y = 12x + 8 \\
\]
Thus, the linear equation depicting the relation in \[x{\text{ and }}y\] is \[y = 12x + 8\].
Drawing graph to the linear equation, by using the given below table, we have
| For equation \[y = 12x + 8\] | |||||
| \[x\] | - 2 | - 1 | 0 | 1 | 2 |
| \[y\] | - 16 | - 4 | 8 | 20 | 32 |
So, the graph is shown below:
Now we have to plot the point at which the distance covered is \[16{\text{ }}km\]i.e., \[x = 16\]
So, clearly from the graph \[y = 200\] when \[x = 16\].
Therefore, the taxi charges are Rs. 200.
Note: The graph is taken on the bases of the scale 1 unit = \[5{\text{ }}km\]on x-axis at which distance is covered and 1 unit = \[{\text{Rs}}.50\] on y-axis at which taxi charges are mentioned. The taxi charges cannot be negative as the distance covered is positive.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Two of the body parts which do not appear in MRI are class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

