A train increases its usual speed by 16% and reaches its destination 30 minutes early. What is the time taken (in hours) normally by the train in the journey?

Correct option is C

Given:

Increase in speed of train = 16%

Train reaches early by 30 minutes

Formula Used:

Distance = Speed $$\times$$​ Time

Solution:

Let the speed of train and time taken to reach be S and T respectively

New speed = $$S \left(1+\frac{16}{100}\right)$$​​

New time = $$T - \frac{30}{60}$$​​

Then according to question:

​$$S\times T = S\left (1+\frac{16}{100}\right)\times \left(T - \frac{30}{60}\right)$$​  (Distance is constant)

​$$T =\left(\frac{29}{25}\right)\times\left(T - \frac{1}{2}\right)$$​​

​$$50T = 29(2T-1)$$​​

​$$50T = 58T - 29$$​​

​$$29 = 8T$$​​

​$$T = \frac{29}{8} = \bf 3\frac{5}{8}\ hours$$  

Thus, Usual time taken by train is $$\bf 3\frac{5}{8}\ hours$$.​