Question

A car covers 300km at a constant speed. If its speed was 10kmph more, it would have taken 1 h less to travel the same distance. Find the speed of the car.
(a) 60 kmph
(b) 50 kmph
(c) 40 kmph
(d) 75 kmph

Answer 1

Hint: We will assume the initial speed of the car as x kmph. Time taken to cover distance $d=300km$ as t hours. Then using the formula $\text{speed=}\dfrac{\text{distance}}{\text{time}}$ , we will get one equation. Then for another statement given, we will get data as $s=\left( x+10 \right)kmph$ , $t=\left( t-1 \right)hour$ . So, again using the same formula we will get another equation. Now in both equations, we will make distance as subject variable and will equate it. We will make an equation in terms of x variable. Thus, on solving the quadratic equation, we will get the speed of the car.

Complete step-by-step answer:
Here, we are given the data i.e. distance $d=300km$ , initial speed was let say x which is represented as ${{s}_{1}}=x$ , time taken is let say t hours i.e. ${{t}_{1}}=t$ .
So, using the formula $\text{speed=}\dfrac{\text{distance}}{\text{time}}$ and substituting the values, we will get as
$x=\dfrac{300}{t}$ ……………………………….(1)
Now, it is said that if speed was 10kmph more then, time would be 1 hour less. So, we can write this as speed $s=\left( x+10 \right)kmph$ and time $t=\left( t-1 \right)hour$ . Again, we will use the formula $\text{speed=}\dfrac{\text{distance}}{\text{time}}$ we will get as
 $\left( x+10 \right)=\dfrac{300}{\left( t-1 \right)}$ …………………………(2)
In both the cases the distance is the same i.e. 300km. So, making distance as subject variable in equation (1) and (2), we get equation as
$300=xt$ ……………………(4)
$300=\left( x+10 \right)\left( t-1 \right)$ ……………………(5)
On equating this tow above equations, we will get equation as
$xt=\left( x+10 \right)\left( t-1 \right)$
On multiplying the brackets, we can write as
$xt=xt+10t-x-10$
On further simplification we will get as
$x=10t-10$
Taking 10 common from RHS, we will get
$x=10\left( t-1 \right)$
Now, from equation (2), we can write $\left( t-1 \right)$ as $\left( t-1 \right)=\dfrac{300}{\left( x+10 \right)}$ . So, we will substitute this value in above equation, and we will get as
$x=10\cdot \dfrac{300}{\left( x+10 \right)}$
$x=\dfrac{3000}{x+10}$
On further solving, we will get
${{x}^{2}}+10x-3000=0$
Here, we will solve the quadratic equation by splitting the middle term such that in addition it gives 10x and on multiplying we get $-3000$ . So, we can write it as
${{x}^{2}}+60x-50x-3000=0$
Taking common from first 2 terms and last2 terms. We will get
$x\left( x+60 \right)-50\left( x+60 \right)=0$
$\left( x-50 \right)\left( x+60 \right)=0$
Thus, we have two values of x i.e. $x=50,x=-60$ . Here, we will take x as 50kmph because the car is not moving in the opposite direction or there is no other car mentioned in the question.
Hence, option (b) is the correct answer.

Note: Another way of solving this problem is by finding time taken by car to cover 300km and then finding speed of car. After the equation $x=10t-10$ , we will put the value of x from equation (1) in order to make equation in terms of variable t. So, on putting the values and solving them we will get a quadratic equation as $10{{t}^{2}}-10t-300=0$ . On solving this we will get 2 values of t i.e. $t=6,t=-5$ . We know that time cannot be negative, so taking t as 5 and putting equation (1), we will get x as 50kmph.