Question

It took a messenger $30$ minutes to deliver a message from point $\mathrm{A}$ to point $\mathrm{B}$. On the way back he reduced his speed by $1\mathrm{mph}$ and it took him $36$ minutes to get back. Find the speed with which the messenger was going from point $\mathrm{A}$ to point $\mathrm{B}$.

Answer 1

Step-1: Calculating the distance between point $\mathrm{A}\&\mathrm{B}$ in terms of speed $s$:

Let the distance be $d$
$\Rightarrow d=s\times t$, here $t$ is time
At first it took a messenger $30\mathrm{minutes}$to deliver a message from point $\mathrm{A}$ to point $\mathrm{B}$.
$$\begin{gathered} \Rightarrow d=s\times \frac{30}{60} \\ =\frac{1}{2}s...\left(i\right) \end{gathered}$$
Then On the way back he reduced his speed by $1\mathrm{mph}$ and it took him $36\mathrm{minutes}$ to get back but the distance will remain same.
$$\begin{gathered} \Rightarrow d=(s-1)\times \frac{36}{60} \\ =\frac{3}{5}(s-1)...\left(ii\right) \end{gathered}$$

Step-2 : Calculating the speed by equating both equation:

$$\begin{gathered} \Rightarrow \frac{1}{2}s=\frac{3}{5}(s-1) \\ \Rightarrow 5s=6s-6 \\ \Rightarrow 6s-5s=6 \\ \Rightarrow s=6\mathrm{mph} \end{gathered}$$

Hence, the speed of the messenger is $6\mathrm{mph}$.