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.

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Solution

## Step 1. Finding equation using first conditionLet speed of $A$ $=xkm/h$ & speed of B $=ykm/h$Therefore ,Time taken by $A$ to cover $30km$$=\frac{30}{x}h$ [$Time=\frac{Dis\mathrm{tan}ce}{time}$] and time taken by $B$ to cover $30km$ $=\frac{30}{y}h$According to question, $\frac{30}{x}=\frac{30}{y}+3$ $\frac{30}{x}-\frac{30}{y}=3$--------------------------$\left(i\right)$Step 2. Finding equation using second conditionWhen $A$ doubles his speed, then the time taken by $A$$=\frac{30}{2x}$According to question, $\frac{30}{y}=\frac{30}{2x}+\frac{3}{2}$$⇒$$-\frac{30}{2x}+\frac{30}{y}=\frac{3}{2}$-$⇒$ $-\frac{15}{x}+\frac{30}{y}=\frac{3}{2}$-------------------------$\left(ii\right)$Step 3. Solving equationsLet 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,$15a=\frac{9}{2}\phantom{\rule{0ex}{0ex}}⇒a=\frac{9}{2x15}\phantom{\rule{0ex}{0ex}}⇒a=\frac{3}{10}$Substituting $a=\frac{3}{10}$ in equation$\left(iii\right)$ we get$⇒$$30\left(\frac{3}{10}\right)–30b=3$$⇒$ $9–30b=3$$⇒-30b=-6\phantom{\rule{0ex}{0ex}}⇒b=\frac{1}{5}$ Now as we assumed earlier , $\frac{1}{y}=b$So we have $\frac{1}{y}=\frac{1}{5}$$⇒$ $y=5km/h$And $\frac{1}{x}=a$$⇒$ $\frac{1}{x}=\frac{3}{10}$$⇒$ $x=\frac{10}{3}km/h$HenceSpeed of $A=\frac{10}{3}km/h$ Speed of $B=5km/h$

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