Question

A cyclist cycles non-stop from $A$ and $B$ a distance of $14\,km$ at a certain average speed. If his average speed reduces by $1\,kmh{r^{ - 1}}$, then he takes $\dfrac{1}{3}\,hr$ more to cover the same distance. What was the original average speed of the cyclist?
(A) $3\,kmh{r^{ - 1}}$
(B) $5\,kmh{r^{ - 1}}$
(C) $7\,kmh{r^{ - 1}}$
(D) $11\,kmh{r^{ - 1}}$

Answer 1

Hint: The speed of the cycle can be determined by using the distance and time relation. In the question, the speed relation is given and then the time relation is given. By using the relations which are given in the question, the original average speed of the cyclist can be determined.

Formulae Used:
The time, distance and speed relation are given by,
$t = \dfrac{d}{s}$
Where, $t$ is the time, $d$ is the distance and $s$ is the speed.

Complete step-by-step solution:
Given that,
Distance between two points, $d = 14\,km$,
Let assume the original average speed is $x$,
Given that, the average speed is reduced by $1\,kmh{r^{ - 1}}$, $\left( {x - 1} \right)$
The time, distance and speed relation are given by,
$t = \dfrac{d}{s}$
Time of the cycle at initial, $ \Rightarrow \dfrac{{14}}{x}$
The average speed is reduced by $1\,kmh{r^{ - 1}}$, then the time is $ \Rightarrow \dfrac{{14}}{{\left( {x - 1} \right)}}$
It is given that, the average speed is reduced by $1\,kmh{r^{ - 1}}$, then the time takes by $\dfrac{1}{3}\,hr$ with the initial time of the cyclist. So, the time when the speed is reduced is equal to the sum of the original time and the extra time.
By this information, the equation is made,
$\dfrac{{14}}{{\left( {x - 1} \right)}} = \dfrac{{14}}{x} + \dfrac{1}{3}\,...................\left( 1 \right)$
By taking the $x$ terms in one side and other terms in other side, then
$\dfrac{{14}}{{\left( {x - 1} \right)}} - \dfrac{{14}}{x} = \dfrac{1}{3}$
By cross multiplying the terms in LHS, then the above equation is written as,
$\dfrac{{14x - 14\left( {x - 1} \right)}}{{x\left( {x - 1} \right)}} = \dfrac{1}{3}$
On multiplying the terms in the above equation, then
$\dfrac{{14x - 14x + 14}}{{{x^2} - x}} = \dfrac{1}{3}$
By cancelling the same terms which are having different signs, then
$\dfrac{{14}}{{{x^2} - x}} = \dfrac{1}{3}$
On cross multiplying the terms in LHS and RHS, then
$14 \times 3 = {x^2} - x$
By rearranging the terms, then the above equation is written as,
${x^2} - x - 42 = 0\,...............\left( 2 \right)$
On factorising the above equation, then
$\left( {x - 7} \right)\left( {x + 6} \right) = 0$
Then, the values of $x$ are $x = 7$ and $x = - 6$
The speed value cannot be negative. So, the original actual speed is $x = 7\,kmh{r^{ - 1}}$
Hence, the option (C) is correct.

Note:- In equation (1), the term $\dfrac{1}{3}\,hr$ is added because, if the speed is decreased then the time is increased. So, the time is added in that equation. In the equation (2), the speed is determined by factoring the equation. This equation can be solved by another form also, by using the quadratic formula the speed can also be determined.