Question

Arjun travels 240miles at a certain speed. If he were to increase his speed by 20mph, it would have taken him 2 hours less to travel the same distance. What was his speed in mph?
A) 30mph
B) 40mph
C) 50mph
D) 60mph

Answer 1

Hint: Here there will be two equations one equation would be for the initial speed and initial time and another one would be for the final time and final speed. Here, in the two equations the distance would be similar. Use the same distance for both of the equations and find the initial speed.

Formula Used:
$D = V \times T$
Where:
D= Distance/ Displacement; (240 miles)
V= Velocity/Speed.
T= Time

Complete step by step answer:
Step1: Write the initial equation and the final equation and find a relation between them.
$D = V \times T$;
Put the value of distance in the above equation
$240 = V \times T$;
Take V to LHS
\[\dfrac{{240}}{V} = T\];
Now, the second equation
Given:
\[V' = V + 20\] ;
$T' = T - 2$ ;
Step2:
Now, Put the above value in the below given equation
\[D = V' \times T'\]
Solve,
\[D = (V + 20) \times (T - 2)\] ;
Multiply the brackets,
$240 = V \times T + 20T - 2V - 40$ ;
Here; $V \times T = 240$; put this value in above equation
$240 = 240 + 20T - 2V - 40$;
$20T - 2V - 40 = 0$;
Now\[T = \dfrac{{240}}{V}\];
$20 \times \dfrac{{240}}{V} - 2V - 40 = 0$;
Divide the whole equation by V;
$20 \times 240 - 2{V^2} - 40V = 0$;
We have, after dividing the whole equation by 20,
${V^2} + 20V - 2400 = 0$;
Factories the above formed equation:
\[{V^2} - 40V + 60V - 2400 = 0\]
Solve the above equation,
\[(V - 40)V + 60(V - 40) = 0\];
$(V - 40)(V + 60) = 0$ ;
Simplify even more,
$V = 40;V = - 60$ ;
So, here the spee can’t be negative so,
$V = 40mph$;

Final Answer: The initial velocity is 40mph, Option B

Note: It is important to note that, here we are only dealing with speed and distance and not with velocity and displacement and hence while solving quadratic equations, always choose the positive value for speed and distance.