Question

A man misses a bus by 40 minutes if he travels at 30 kmph. If he travels at 40 kmph, then also he misses the bus by 10 minutes. What is the minimum speed required to catch the bus on time?

Answer 1

Define x:

Let the total distance be x.

Find the time needed in term of x for speed at 30 km/h:

$\text {Time } = \dfrac{\text{distance}}{\text{Speed}}$

$\text{Time } =\dfrac{x}{30} \text { hr}$

Find the time needed in term of x for 40 km./h:

$\text {Time } = \dfrac{\text{distance}}{\text{Speed}}$

$\text{Time } =\dfrac{x}{40} \text { hr}$

Find the difference in time:

$\text{Difference in time }= 40 \text { mins} - 10 \text { mins}$

$\text{Difference in time }= 30 \text { mins}$

$\text{Difference in time }= \dfrac{30}{60} \text { hr}$

$\text{Difference in time }= \dfrac{1}{2} \text { hr}$

Solve x:

$\dfrac{x}{30} - \dfrac{x}{40} = \dfrac{1}{2}$

$\dfrac{40x - 30x}{1200} = \dfrac{1}{2}$

$\dfrac{10x}{1200} = \dfrac{1}{2}$

$\dfrac{x}{120} = \dfrac{1}{2}$

$2x=120$

$x = 60 \text{ km}$

Find the total time needed to for the whole journey:

$\text{Distance } = 60 \text { km}$

$\text{Speed } = 30 \text { km/h}$

$\text {Time } = \dfrac{\text{60}}{\text{30}}$

$\text {Time } = 2 \text { hours}$

Given that he was 40 mins late:

$\text{Time needed} = 2 \text{ hrs} - 40 \text { mins}$

$\text{Time needed} = 1 \text{ hrs } 20 \text { mins}$

$\text{Time needed} = \dfrac{4}{3} \text{ hrs }$

Find the minimum speed needed:

$\text {Speed} = \text{Distance } \div \text{time }$

$\text {Speed} =60 \div \dfrac{4}{3}$

$\text {Speed} =60 \times \dfrac{3}{4}$

$\text {Speed} = 45 \text{ km/h}$

Answer: The minimum speed is 45 km/h