Question

A bus left point x for point y. two hours later a car left point x for y and arrived at y at the same time as the bus. if the car and the bus left simultaneously from the opposite ends x and y towards each other, they would meet 1.33 hours after the start. how much time did it take the bus to travel from x to y?

Answer 1

Time  taken by bus to travel from x to y is 4 hours

Step-by-step explanation:

Let the speed of bus is S kmph and speed of car is s kmph.

Let the total distance between x and y point is d  km.

And let the time taken by bus to cover the distance d is t hours.

So, the  time taken by car to cover the distance d is (t - 2) hours.

We know that   $\textbf{\large Speed = } \frac{Distance}{Time}$

by above given details in question

$S = \frac{d}{t}\\\\s = \frac{d}{t - 2}$                                                                       (equation 1)

When both vehicles travel from opposite sides, they meet at a point

after 1.33 hours = $\frac{4}{3} hours$

So the total distance covered by both vehicles = d.

and time taken =  $\frac{4}{3} hours$

and total speed will be ( S + s ) =   $\frac{d}{\frac{4}{3}}$ = $\frac{3d}{4}$                 (equation 2)

By above values of equation 1 and equation 2

$\frac{3d}{4} = \frac{d}{t} + \frac{d}{t-2}$

cancelling d from LHS and RHS

$\frac{3}{4} = \frac{1}{t} + \frac{1}{t-2}$

Solving this equation

$\frac{3}{4} = \frac{t - 2 + t}{t \times (t-2)} =\frac{2t - 2 }{t \times (t-2)}$

on solving

$3t^2 - 6t = 8t - 8\\\\3t^2 - 14t + 8 = 0$

$(3t - 2)\times (t - 4) = 0\\\\So,\\\\\textbf{\large t = 4 or t = }\frac{\textbf{\large 2}}{\textbf{\large 3}}$

t ≠ $\frac{2}{3}$                              

  (It is saying that bus alone can move faster than the case when both bus and car move through opposite ends at same time,  which is not possible)

$\textbf{\Large So time taken by bus to travel from x to y}\\\\\textbf{\Large t = 4 hours}$