Question

A train covers a distance of 90 km at a uniform speed. Had the speed been $15\dfrac{km}{hr}$ more, it would have been taken 30 minutes less for the journey. Find the original speed of the train.

Answer 1

Hint: Suppose the original speed of the train as a variable. Use the relation among speed, distance and time which is given as $\text{Speed}\ \text{=}\ \dfrac{\text{Distance}}{\text{Time}}$ to get the mathematical equations in terms of time and the original speed using the given information in the problem. Now, solve the equations further to get the original speed of the train.

Complete step-by-step answer:
Let the original speed of the train is v $\dfrac{km}{hr}$.
So, from the given problem, it is given that the train is covering a distance of 90 km with a uniform speed and if the speed will increase by 15 $\dfrac{km}{hr}$, it is taking 30 minutes less for the same distance.
We know, there are 60 minutes in 1 hour
 So, we get
$60\ \text{minute =}\ \text{1 hour}$
$1\ \text{minute =}\ \dfrac{\text{1}}{60}\text{ hour}$
$30\ \text{minute =}\ \dfrac{\text{1}}{60}\times \text{30 = }\dfrac{1}{2}\ \text{hour}$
Hence, in other words, we can say that train is taking $\dfrac{1}{2}\ \text{hour}$less to cover the distance 90 km if the original speed will increase by 15 $\dfrac{km}{hr}$.
So, let us suppose the time taken by the train for covering the 90 km with uniform speed ‘v’ be t hour. Hence, time taken to cover the distance 90 km would be $\left( t-\dfrac{1}{2} \right)\ \text{hours}$if speed will increase by 15 $\dfrac{km}{hr}$.
As we know the relation among speed, distance and time is given as
$\text{Speed}\ \text{=}\ \dfrac{\text{Distance}}{\text{Time}}$ …………………………………………………………………………………….(i)
So, we can apply the above formula for both of the conditions given in the problem or discussed above.
For the first case, we have speed ‘v’ and time ‘t’ and the distance covered is 90 km. so, we can put these values in equation (i) and hence get
$v\ =\ \dfrac{90}{t}$ …………………………………………………(ii)
And now for the second case we are increasing the speed by 15 $\dfrac{km}{hr}$ and time is decreasing to $\left( t-\dfrac{1}{2} \right)\ \text{hours}$ and distance remains the same i.e. 90 km. Hence, we have
New velocity $=\ v+15$
New time $=\ \left( t-\dfrac{1}{2} \right)$
Distance $=\ 90\ \text{km}$(same)
So, we can put these values to equation (i) and get
$v+15=\ \dfrac{90}{\left( t-\dfrac{1}{2} \right)}$ ……………………………………………………..(iii)
Now, we get value of ‘t’ from equation (ii)
$t=\ \dfrac{90}{v}$ …………………………………………….(iv)
Now, put the value of ‘t’ for equation (iv) to equation (iii). Hence, we get
$v+15=\ \dfrac{90}{\left( \dfrac{90}{v}-\dfrac{1}{2} \right)}$
Now, on cross multiplying the above equation, we get
$\left( v+15 \right)\left( \dfrac{90}{v}-\dfrac{1}{2} \right)=\ 90$
Now, take the L.C.M. of denominator in the second bracket; Hence, we get
$\left( v+15 \right)\left( \dfrac{180-v}{2v} \right)=\ 90$
$\Rightarrow 180v-{{v}^{2}}+15\times 180-15v=180v$
Cancelling the terms 180v from both the sides, we get
$-{{v}^{2}}-15v+15\times 180=0$
$\Rightarrow {{v}^{2}}+15v-2700=0$ ………………………………………………………(v)
Now, we can get roots of above quadratic by the quadratic formula; so,
We, know that roots of quadratic equation $\text{A}{{x}^{2}}\text{+B}x\text{+C}=0$ can be given by relation
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So, we can get values of ‘v’ from equation (v) as
$v=\dfrac{-15\pm \sqrt{{{\left( 15 \right)}^{2}}-4\times \left( 1 \right)\left( 2700 \right)}}{2\times 1}$
$v=\dfrac{-15\pm \sqrt{225+10800}}{2}$
$v=\dfrac{-15\pm \sqrt{11025}}{2}$
Now, we can find $\sqrt{11025}$ by prime factorization method as
$\begin{align}
  & 3\left| 11025 \right. \\
 & 3\left| 3675 \right. \\
 & 3\left| 1225 \right. \\
 & 5\left| 245 \right. \\
 & 7\left| 49 \right. \\
 & 7\left| 7 \right. \\
 & \\
\end{align}$
$11025=\overline{3\times 3}\times \overline{5\times 5}\times \overline{7\times 7}$
So, $\sqrt{11025}=3\times 5\times 7=105$
So, value of v can be given as
$v=\dfrac{-15\pm \sqrt{11025}}{2}$
$v=\dfrac{-15\pm 105}{2}$
Hence, we can get two values of v as
$v=\dfrac{-15-105}{2}$ $\Rightarrow v=\dfrac{-15+105}{2}$
$v=\dfrac{-120}{2}$ $\Rightarrow v=\dfrac{90}{2}$
$v=-60$ or $v=\ 45$
Now, we can ignore $v=-60$ as speed will never be negative. Hence, the value of speed is 45 $\dfrac{km}{hr}$.
So, the original speed is 45 $\dfrac{km}{hr}$.


Note: Don’t put time taken by the train as $\left( t-30 \right)\ \text{hours}$ in the second case where speed will increase by 15 $\dfrac{km}{hr}$. As, t is in hours and it is given that the train is taking 30 minutes less than ‘t’ for the second case. So, first we need to convert ‘t’ to minutes or 30 minutes to $\dfrac{1}{2}$hours. So, be careful with nits in the question as well.
One may go wrong with the relation $\text{Speed}\ \text{=}\ \dfrac{\text{Distance}}{\text{Time}}$. He / She may get confused with the formula and apply $\text{Time}\ \text{=}\ \text{Speed}\times \text{Distance}$ or $\text{Speed}\ \text{=}\ \text{Time}\times \text{Distance}$which are wrong. So, be clear with relation of speed, time and distance.
One may factorize ${{v}^{2}}+15v-2700=0$ as well for finding the roots or the values of v. We have to split the middle term as
${{v}^{2}}+60v-45v-2700=0$
$\left( v+60 \right)\left( v-45 \right)=0$