Question

A car covers a distance of 240 km with some speed. If the speed is increased by 20 km/hr, it will cover the same distance in 2 hours less. Find the speed of the car.
$
  (a){\text{ 40}} \\
  (b){\text{ 45}} \\
  (c){\text{ 50}} \\
  (d){\text{ 60}} \\
 $

Answer 1

Hint: In this question let the speed of the car be v km/hr. and the time taken to cover the distance be t hrs. Now the distance of 240km is covered with this speed and then increased to (v+20) km/hr. use the constraint of the question to formulate an equation involving the variable.

Complete step-by-step answer:

Let the speed of the car be V km/hr.
And the time it takes to cover a distance be (t) hr.
Now it is given that the car covers a distance (d) = 240 km with some speed.
Now it is given that the speed is increased by 20 km/hr. it will cover the same distance in 2 hours less.
Therefore new speed = (V + 20) km/hr.
And the new time = (t – 2) hr.
Now the distance is the same in both the cases.
And we know that the distance, speed and time is related as
Distance = speed $ \times $ time.
Therefore
$ \Rightarrow 240 = V.t$ Km............................. (1)
And
\[ \Rightarrow 240 = \left( {V + 20} \right)\left( {t - 2} \right)\] Km.............................. (2)
So equate these two equation we have,
\[ \Rightarrow \left( {V + 20} \right)\left( {t - 2} \right) = V.t\]
Now simplify the above equation we have,
\[ \Rightarrow V.t - 2V + 20t - 40 = V.t\]
\[ \Rightarrow - 2V + 20t - 40 = 0\]
Now from equation (1) \[t = \dfrac{{240}}{V}\] so substitute this value in above equation we have,
\[ \Rightarrow - 2V + 20 \times \dfrac{{240}}{V} - 40 = 0\]
Now simplify the above equation we have,
\[ \Rightarrow - 2{V^2} + 4800 - 40V = 0\]
Divide by -2 throughout we have,
\[ \Rightarrow {V^2} + 20V - 2400 = 0\]
Now factorize this equation we have,
\[ \Rightarrow {V^2} + 60V - 40V - 2400 = 0\]
$ \Rightarrow V\left( {V + 60} \right) - 40\left( {V + 60} \right) = 0$
$ \Rightarrow \left( {V + 60} \right)\left( {V - 40} \right) = 0$
$ \Rightarrow V = 40, - 60$
Negative speed is not possible.
So the speed of the car is 40 km/hr.
Hence option (A) is correct.

Note: It is advised to remember the relation between distances, time and speed that is Distance = speed $ \times $ time. The important step here was the factorization of the quadratic equation formed, use the middle term splitting method or even the direct method of Dharacharya formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ to solve the quadratic.