Question

A man covers a distance at some speed. If he had moved 3 KMPH faster, he would have taken 40 minutes less. If he had moved 2 KMPH slower, he would have taken 40 minutes more. The distance in KM is
A. 35
B. \[36\dfrac{2}{3}\]
C. 37
D. 40

Answer 1

Hint: Use the set of hints to form 3 equations with 3 variables then eliminate one of the equations to get the rest two with 2 variables left Now use the substitution method to get the whole equation in one variable and then substitute again to get the value of distance.

Complete step by step answer:
Let the speed of the car be y KMPH, t be the time taken and x be the total distance travelled.
Now we know that
\[y = \dfrac{x}{t}.....................................................(i)\]
And by the second hint in the question we are getting a equation
\[\dfrac{x}{{t - \dfrac{2}{3}}} = y + 3.............................................(ii)\]
And now by using the third hint i am getting
\[\dfrac{x}{{t + \dfrac{2}{3}}} = y - 2.............................................(iii)\]
Substituting the value of (i) in (ii) and (iii) i am getting \[\dfrac{x}{{t - \dfrac{2}{3}}} = \dfrac{x}{t} + 3\] and \[\dfrac{x}{{t + \dfrac{2}{3}}} = \dfrac{x}{t} - 2\] respectively So by solving a bit more getting equation (ii) as
\[\begin{array}{l}
\therefore \dfrac{x}{{t - \dfrac{2}{3}}} = \dfrac{x}{t} + 3\\
 \Rightarrow \dfrac{x}{{\dfrac{{3t - 2}}{3}}} - \dfrac{x}{t} = 3\\
 \Rightarrow \dfrac{{3x}}{{3t - 2}} - \dfrac{x}{t} = 3\\
 \Rightarrow \dfrac{{3xt - 3xt + 2x}}{{t(3t - 2)}} = 3\\
 \Rightarrow \dfrac{{2x}}{{t(3t - 2)}} = 3\\
 \Rightarrow 2x = 3t(3t - 2)
\end{array}\]
Now the equation (iii) will become
\[\begin{array}{l}
\therefore \dfrac{x}{{t + \dfrac{2}{3}}} = \dfrac{x}{t} - 2\\
 \Rightarrow \dfrac{x}{{\dfrac{{3t + 2}}{3}}} - \dfrac{x}{t} = - 2\\
 \Rightarrow \dfrac{{3x}}{{3t + 2}} - \dfrac{x}{t} = - 2\\
 \Rightarrow \dfrac{{3xt - 3xt - 2x}}{{t(3t + 2)}} = - 2\\
 \Rightarrow \dfrac{{ - 2x}}{{t(3t + 2)}} = - 2\\
 \Rightarrow x = t(3t + 2)
\end{array}\]
So let us substitute the value of x in the modified version of equation (ii)
We will get it as
\[\begin{array}{l}
\therefore 2t(3t + 2) = 3t(3t - 2)\\
 \Rightarrow 2(3t + 2) = 3(3t - 2)\\
 \Rightarrow 6t + 4 = 9t - 6\\
 \Rightarrow 10 = 3t\\
 \Rightarrow t = \dfrac{{10}}{3}
\end{array}\]
So now as we have the value of t we will put it in \[x = t(3t + 2)\]
So we are getting
\[\begin{array}{l}
\therefore x = t(3t + 2)\\
 \Rightarrow x = \dfrac{{10}}{3}\left( {3 \times \dfrac{{10}}{3} + 2} \right)\\
 \Rightarrow x = \dfrac{{10}}{3}\left( {10 + 2} \right)\\
 \Rightarrow x = \dfrac{{10}}{3} \times 12\\
 \Rightarrow x = 10 \times 4\\
 \Rightarrow x = 40
\end{array}\]

So, the correct answer is “Option D”.

Note: Here in the question it was given 40 minutes but as speed was in KMPH so i have to convert minutes into hours, as we know that there is a total of 60 minutes in 1 hour then 40 mins will be equal to \[\dfrac{{40}}{{60}} = \dfrac{2}{3}\] which is directly used in the answer.