Question

Ankita travels 14km to her home partly by Rickshaw and partly by bus. She takes half an hour if she travels 2km by Rickshaw, and the remaining distance by Bus. On the other hand, if she travels 4km by rickshaw and the remaining distance by bus, she takes 9 minutes longer. Find the speed of Rickshaw in km per hour
$\begin{align}
  & \text{a) 10} \\
 & \text{b) 14} \\
 & \text{c) 13} \\
 & \text{d) 7 } \\
\end{align}$

Answer 1

Hint: Now let us assume the speed of the rickshaw is x and the speed of the bus is y. Now we are given that Ankita takes half an hour if she travels 2km by Rickshaw and the remaining distance by Bus. Using this condition and the formula $\dfrac{\text{distance}}{\text{speed}}\text{ = time}$ we will form our first equation. Now again we have if she travels 4km by rickshaw and the remaining distance by bus, she takes 9 minutes longer. Hence using this we will form our second equation. Now we solve the two equations simultaneously.

Complete step-by-step solution:
Let us say the speed of the Rickshaw is x km per hour and the speed of the bus is y km per hour.
Now we are given that Ankita travels 14km in total.
Now she travels 2km by rickshaw and hence she travels 14 – 2 = 12km by Bus
Now we know that$\dfrac{\text{distance}}{\text{speed}}\text{ = time}$ .
Hence we get,
$\dfrac{2}{x}+\dfrac{12}{y}=\dfrac{1}{2}..........\left( 1 \right)$
Now if she travels 4km by rickshaw and the remaining distance by bus, she takes 9 minutes longer.
We know that 9 minutes is $\dfrac{9}{60}$ hours.
Now again if she travels 4km by rickshaw then she travels 14 – 4 = 10 km by Bus.
Hence we have,
$\dfrac{4}{x}+\dfrac{10}{y}=\dfrac{1}{2}+\dfrac{9}{60}.........\left( 2 \right)$
Now multiplying 2 to equation (1) and then subtracting it from equation (2) we get.
\[\begin{align}
  & \dfrac{4}{x}+\dfrac{10}{y}-\dfrac{4}{x}-\dfrac{24}{y}=\dfrac{1}{2}+\dfrac{9}{60}-1 \\
 & \Rightarrow \dfrac{-14}{y}=\dfrac{9}{60}-\dfrac{1}{2} \\
\end{align}\]
\[\Rightarrow \dfrac{-14}{y}=\dfrac{9}{60}-\dfrac{30}{60}\]
\[\Rightarrow \dfrac{-14}{y}=\dfrac{9-30}{60}\]
\[\begin{align}
  & \Rightarrow \dfrac{-14}{y}=\dfrac{-21}{60} \\
 & \Rightarrow y=\dfrac{-14\times 60}{-21} \\
 & \Rightarrow y=\dfrac{2\times 20}{1} \\
\end{align}\]
Hence we get y = 40.
Now substituting y = 40 in equation (1) we get,
$\begin{align}
  & \dfrac{2}{x}+\dfrac{12}{40}=\dfrac{1}{2} \\
 & \Rightarrow \dfrac{2}{x}=\dfrac{1}{2}-\dfrac{12}{40} \\
 & \Rightarrow \dfrac{2}{x}=\dfrac{20}{40}-\dfrac{12}{40} \\
 & \Rightarrow \dfrac{2}{x}=\dfrac{8}{40} \\
 & \Rightarrow 2\times 40=8x \\
 & \Rightarrow x=10 \\
\end{align}$
Hence the speed of a rickshaw is 10 km per hour. Option a is the correct option.

Note: Now first note that when we get equation (1) and (2) we do not cross multiply the fractions and simplify as it will complicate the solutions. Also here when adding and subtracting fractions remember that the denominators must be the same before carrying out the operation.