Question

A cyclist, after riding a certain distance, stopped for half an hour to repair his bicycle, after which he completed the whole journey of 30 km at half speed in 5 hours. If the breakdown had occurred 10 km farther off, he would have done the whole journey in 4 hours. Find where the breakdown occurred and his original speed. Ignore the time he took off for the repair of the cycle.

Answer 1

Hint: If we read the question carefully, we would see that two linear equations in two variables are hidden here. And some more pieces of information are given which we will use all together to form two equations and then solve them both to get the desired result. For example, we are given that the whole journey is 30km which we can use. Moreover, we are given an alternate situation of the breakdown occurring which would help us create another equation. So, using the formula of speed distance and time, we would solve this question.

Complete step-by-step solution:
We should be aware of the basic formula relating speed, distance and time.
$\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}$
Using this we can say:
$\text{Time}=\dfrac{\text{Distance}}{\text{Speed}}$
We now start the making of the first equation:
Let $x$ be the distance in km where the breakdown occurred. Let $y$ be the original speed of the cycle. Then, according to question $x$ km has been covered with $y$ speed and the remaining distance $30-x$ has been covered with half the speed i.e. $\dfrac{y}{2}$ and this whole incident took place in 5 hours, so the first equation becomes:
$\dfrac{x}{y}+\dfrac{30-x}{\dfrac{y}{2}}=5$
$\Rightarrow \dfrac{x}{y}+\dfrac{60-2x}{y}=5........\left( 1 \right)$
Now, we create the second equation:
If the breakdown had occurred at $x+10$ km and the remaining distance $30-\left(x+10\right)$ would have been covered with half the original speed then it would have taken 4 hours. So, the second equation becomes:
$\dfrac{x+10}{y}+\dfrac{30-\left( x+10 \right)}{\dfrac{y}{2}}=4........\left( 2 \right)$
Simplifying the first equation:
$x+5y=60........\left( 3 \right)$
And simplifying the second equation:
$x+4y=50........\left( 4 \right)$
Subtracting $\left(4\right)$ from $\left(3\right)$, we get:
$y=10$
Substituting this in any of the equations we get $x=10$
So, the breakdown had occurred at 10kms from the start and the original speed was 10km/hr.


Note: While putting the formula of speed distance and time relation, it is common to put the wrong fraction. Moreover, while creating the equation make sure that you divide the speed by 2 because it has been told in the question that the speed has been reduced to half. Also, while creating the second equation, you need to subtract the distance $x+10$ and not only $x$.