Question

A cyclist cycles non-stop from $A$ to $B$ a distance of $14$ km at a certain average speed. If his average speed reduces by $1$ km/hr, then he takes $\frac{1}{3}$ hour more to cover the same distance. What was the original average speed of the cyclist?

• A. $3$ km/hr
• B. $5$ km/hr
• C. $7$ km/hr
• D. $11$ km/hr

Answer 1

Answer: (C)

$7$ km/hr

Let the original average speed of the cyclist be $x$ km/hr

$Time=\frac{Distance}{Speed}$

Then as per given condition $\frac{14}{(x-1)}=\frac{14}{x}+\frac{1}{3}$

$\Rightarrow \frac{14x-14(x-1)}{x(x-1)}=\frac{1}{3}$

$\Rightarrow \frac{14}{x^2-x}=\frac{1}{3}$

$\Rightarrow x^2-x-42=0$

$\Rightarrow x^2-7x+6x-42=0$

$\Rightarrow x(x-7)+6(x-7)=0$

$\Rightarrow (x-7)(x+6)=0$

$\Rightarrow x=7$ or $-6$

Since speed cannot be negative, so $x=7$ km/hr.