If a train runs at $40 k m p h$ it reaches its destination late by $11 m i n u t e s$ . But if it runs at $50 k m p h$ it is late by $5 m i n u t e s$ only. Find the distance covered by the train.

Open in App

Solution

Let the correct time for the train to complete the journey be $x h o u r s$
Time taken by train if it runs at $40 k m p h = x h o u r s 11 m i n u t e s = (x + \frac{11}{60}) h o u r s$
So, distance covered by train if it runs at $40 k m p h$
$= 40 \times (x + \frac{11}{60}) k m$ [Distance = Speed $\times$ Time] ------- (1)
Time taken by train if it runs at $50 k m p h = x h o u r s 5 m i n u t e s = (x + \frac{5}{60}) h o u r s$
So, the distance covered by train if it runs ar $50 k m p h$
$= 50 \times (x + \frac{5}{60}) k m$ ------ (2)
Since both distances are equal.
$40 \times (x + \frac{11}{60}) = 50 \times (x + \frac{5}{60})$ [From (1) and (2)]
$\Rightarrow 40 \times (\frac{60 x + 11}{60}) = 50 \times (\frac{60 x + 5}{60})$
$\Rightarrow 40 (60 x + 11) = 50 (60 x + 5)$
$\Rightarrow 2400 x + 440 = 3000 x + 250$
$\Rightarrow 440 - 250 = 3000 x - 2400 x$
$\Rightarrow 600 x = 190$
$\Rightarrow x = \frac{190}{600}$
$\Rightarrow x = \frac{19}{60} h o u r s$
$\Rightarrow x = \frac{19}{60} \times 60 m i n u t e s$
$\Rightarrow x = 19 m i n u t e s$
From eq (1),
Distance = $40 \times (x + \frac{11}{60}) k m$
$= 40 \times (\frac{19}{60} + \frac{11}{60}) k m$ [ $∵ x = 19$ ]
$= 40 \times \frac{19 + 11}{60} k m$
$= 40 \times \frac{30}{60} k m$
$= 40 \times \frac{1}{2} k m$
$= 20 k m$
Therefore, the distance covered by train is $20 k m$ .

Suggest Corrections

Similar questions

40

11

50

5

Join BYJU'S Learning Program