QUESTION

# A started on my bicycle at 7 a.m. to reach a certain place. After going a certain distance his bicycle went out of order. Consequently, A rested for 35 minutes and came back to his house walking all the way A reached my house at 1 p.m. if his cycling speed is 10 kmph and his walking speed is 1 kmph, then on his bicycle A covered a distance of:${\text{A}}{\text{. 4}}\dfrac{{61}}{{66}}km$${\text{B}}{\text{. 13}}\dfrac{4}{9}km$${\text{C}}{\text{. 14}}\dfrac{3}{8}km$${\text{D}}{\text{. 15}}\dfrac{{10}}{{21}}km$

Hint: Let the distance reached when the bicycle went out of order be $10 \times t$, where t is the time taken to reach there. Also, time taken by walking + time taken for resting $= 6 - \dfrac{{35}}{{60}} - t$. So, use this concept to solve.

Given in the question:
Walking speed is 1kmph.
Cycling speed is 10 kmph.
He rested for 35 minutes.
Lets us assume the distance reached when bicycle went out of order be-
$10 \times t$, where t is the time to reach there and 10kmph is the cycling speed.
Now, if we see that the time, he rested$= 35\min = \dfrac{{35}}{{60}}hr$
Now, Time taken by walking + Time for resting $= 6 - \dfrac{{35}}{{60}} - t$.
So, the distance covered on walking will be, $= 1 \times \left( {6 - \dfrac{{35}}{{60}} - t} \right)$, as the walking speed is 1kmph.
Now, if we compare the distance covered, we get-
$10 \times t$$= 1 \times \left( {6 - \dfrac{{35}}{{60}} - t} \right)$
Solving to find the value of t.
$10t = \left( {6 - \dfrac{{35}}{{60}} - t} \right) \\ \Rightarrow 10t = \left( {\dfrac{{325}}{{60}} - t} \right) \\ \Rightarrow t = \left( {\dfrac{{325}}{{660}}} \right) \\$
Therefore, the distance he covered by bicycle is $10 \times t$, put the value of t, we get-
$10t = 10 \times \dfrac{{325}}{{660}} = 4\dfrac{{61}}{{66}}$

Therefore, the distance covered by A on his bicycle is $4\dfrac{{61}}{{66}}$km.
Hence, the correct option is option (A).

Note: Whenever such types of questions appear, note down the things that are given in the question. Form the equation by the data given in the question and by using the standard formula of distance = speed $\times$ time, solve the question.