One can use three different transports which move at 10, 20, and 30 kmph, respectively to reach from A to B. Amal took each mode of transport for $$\frac{1}{3}^{rd}$$ of his total journey time, while Bimal took each mode of transport for $$\frac{1}{3}^{rd}$$ of the total distance. The percentage by which Bimal’s travel time exceeds Amal’s travel time is nearest to

Assume the total distance between A and B as d and time taken by Amal = t

Since Amal travelled $$\frac{1}{3}^{rd}$$ of his total journey time in different speeds
d = $$\ \frac{\ t}{3}\times\ 10+\ \frac{\ t}{3}\times\ 20+\frac{\ t}{3}\times\ 30\ \ =\ 20t$$

$$\text{Total time taken by Bimal} = \ \frac{d_1}{s_1}+\frac{d_2}{s_2}+\frac{d_3}{s_3}$$

$$=\ \frac{20t}{3}\times\ \frac{1}{10}+\frac{20t}{3}\times\ \frac{1}{20}+\frac{20t}{3}\times\ \frac{1}{30}\ \ =\frac{20t\left(6+3+2\right)}{3\ \times30}\ =\frac{11}{9}t$$

Hence, the ratio of time taken by Bimal to time taken by Amal = $$\frac{\frac{11t}{9}}{t}=\frac{11}{9}$$
Therefore, Bimal will exceed Amal's time by $$\ \ \ \frac{\ \ \frac{\ 11t}{9}-t}{t}\times\ 100 = 22.22%$$