Question

A person travelled $25{\text{km}}$ by steamer, ${\text{40km}}$ by train, ${\text{30km}}$ by horse. It took $7{\text{ hours}}$. If the rate of the train is $4$ times that of the horse and $2$ times that of the steamer. Find the rate of horses.
A) $15{\text{km/h}}$
B) $7\dfrac{1}{2}{\text{km/h}}$
C) $30{\text{ km/h}}$
D) ${\text{16 km/h}}$

Answer 1

Hint: let the rate of train be $x{\text{ km/h}}$
Then the rate of horse will be $\dfrac{x}{4}{\text{ km/h}}$
Also the rate of steamer will be $\dfrac{x}{2}{\text{ km/h}}$
We know that the distance travelled by steamer, train and the horse. Hence we can time of each by the formula ${\text{distance}} = {\text{speed}} \times {\text{time}}$

Complete step by step solution:
In the question, we are given that a person travelled $25{\text{km}}$ by steamer, ${\text{40km}}$ by train, ${\text{30km}}$ by horse and the total time given is $7{\text{ hours}}$. We are also told that the rate of the train is $4$ times that of the horse and $2$ times that of the steamer.
So if we assume that the rate of train be $x{\text{ km/h}}$
Then the rate of horse will be $\dfrac{x}{4}{\text{ km/h}}$
Also the rate of steamer will be $\dfrac{x}{2}{\text{ km/h}}$
Now let ${t_1}$ be the time taken by travelling through a steamer, ${t_2}$ is the time through the train and ${t_3}$ be the time through horse. As we are given that the total time taken to travel is $7{\text{ hours}}$
So ${t_1} + {t_2} + {t_3} = 7 - - - - - - - (1)$
Now as we know that ${t_1}$ is the time taken to travel $25{\text{ km}}$ by the steamer. So as we know that
${\text{distance}} = {\text{speed}} \times {\text{time}}$
Speed is equivalent to rate. Rate we assumed was $\dfrac{x}{2}{\text{ km/h}}$
$25 = \dfrac{x}{2}({t_1})$
So we get ${t_1} = \dfrac{{50}}{x}{\text{ hour}} - - - - - (2)$

Now we also know that it takes ${t_2}$ time to travel ${\text{40 km}}$ by the train at the rate of \[x{\text{ km/h}}\]. So as we know that
${\text{distance}} = {\text{speed}} \times {\text{time}}$
$40 = x({t_2})$
${t_2} = \dfrac{{40}}{x}{\text{ hour}} - - - - - (3)$
We also know that it takes ${t_3}$ time to travel ${\text{30 km}}$ by the horse at the rate of \[\dfrac{x}{4}{\text{ km/h}}\]. So as we know that
${\text{distance}} = {\text{speed}} \times {\text{time}}$
$30 = \dfrac{x}{4}({t_3})$
${t_3} = \dfrac{{120}}{x}{\text{ hour}} - - - - - (4)$
Now putting the values of \[{t_1},{t_2},{t_3}\] in the equation (1)
${t_1} + {t_2} + {t_3} = 7$
$\dfrac{{50}}{x} + \dfrac{{40}}{x} + \dfrac{{120}}{x} = 7$
Taking the LCM, we get that
$\dfrac{{50 + 40 + 120}}{x} = 7$
$7x = 210$
$x = 30$
So $x$ is the rate of train which is $30{\text{ km/h}}$
Now we know that the rate of the horse is \[\dfrac{x}{4}{\text{ km/h}}\]

So the rate of horse$ = \dfrac{{30}}{4}{\text{ km/h}} = 7\dfrac{1}{2}{\text{ km/h}}$

Note: ${\text{(distance}} = {\text{speed}} \times {\text{time)}}$ this is valid only when people or particles are moving with the constant speed or with the zero acceleration. Acceleration is the rate of the change of velocity. So if the acceleration is non-zero then speed cannot be constant.