Question

$A$ takes $3$ hour more than $B$ to walk $30km$. But if $A$ doubles his pace, he is ahead of $B$ by $\frac{3}{2}hours$. Find their speed of walking.

Answer 1

Step 1. Finding equation using first condition

Let speed of $A$ $=xkm/h$ & speed of B $=ykm/h$

Therefore ,Time taken by $A$ to cover $30km$$=\frac{30}{x}h$ [$Time=\frac{Dis\tan ce}{time}$]

and time taken by $B$ to cover $30km$ $=\frac{30}{y}h$

According to question,

$\frac{30}{x}=\frac{30}{y}+3$

$\frac{30}{x}-\frac{30}{y}=3$--------------------------$\left(i\right)$

Step 2. Finding equation using second condition

When $A$ doubles his speed, then the time taken by $A$$=\frac{30}{2x}$

According to question,

$\frac{30}{y}=\frac{30}{2x}+\frac{3}{2}$

$\Rightarrow$$-\frac{30}{2x}+\frac{30}{y}=\frac{3}{2}$-

$\Rightarrow$ $-\frac{15}{x}+\frac{30}{y}=\frac{3}{2}$-------------------------$\left(ii\right)$

Step 3. Solving equations

Let us assume that $\frac{1}{x}=a\&\frac{1}{y}=b$

So equation (i) and equation (ii) becomes
$30a-30b=3$----------------------$\left(iii\right)$
$-15a+30b=\frac{3}{2}$---------------------$\left(iv\right)$

On adding equation$\left(iii\right)$ and $\left(iv\right)$ we get,

$$\begin{gathered} 15a=\frac{9}{2} \\ \Rightarrow a=\frac{9}{2x15} \\ \Rightarrow a=\frac{3}{10} \end{gathered}$$

Substituting $a=\frac{3}{10}$ in equation$\left(iii\right)$ we get

$\Rightarrow$$30\left(\frac{3}{10}\right)–30b=3$

$\Rightarrow$ $9–30b=3$

$$\begin{gathered} \Rightarrow -30b=-6 \\ \Rightarrow b=\frac{1}{5} \end{gathered}$$

Now as we assumed earlier , $\frac{1}{y}=b$

So we have $\frac{1}{y}=\frac{1}{5}$

$\Rightarrow$ $y=5km/h$

And $\frac{1}{x}=a$

$\Rightarrow$ $\frac{1}{x}=\frac{3}{10}$

$\Rightarrow$ $x=\frac{10}{3}km/h$

Hence
Speed of $A=\frac{10}{3}km/h$
Speed of $B=5km/h$