A train takes 1$$\frac{1}{3}$$​ hours less for a journey of 528 km, if its speed is increased by 5$$\frac{1}{2}$$​ km/hour from its usual speed. The usual speed of the train, in km/hour, is:

Correct option is A

Given:

Distance = 528 km

Time saved = $$1 \frac{1}{3} = \frac{4}{3}$$​ hours

Increase in speed = $$5 \frac{1}{2} = \frac{11}{2}$$​ km/h

Formula Used:

Use time formula:

Time = $$\frac{\text{Distance}}{\text{Speed}}$$​​

Solution:

Let the usual speed be x km/h.
Then, increased speed = $$x + \frac{11}{2}$$​​

Time difference;

​$$\frac{528}{x} - \frac{528}{x + \frac{11}{2}} = \frac{4}{3} \\ \ \\528\left(\frac{1}{x} - \frac{2}{2x+11} \right) = \frac{4}{3} \\ \ \\\frac{2x +11 – 2x}{x(2x+11} = \frac{4}{3\times528} \\ \ \\\implies 2x^2 + 11x - 4356 = 0 \\ \ \\2x^2 - 88x + 99x - 4356 = 0 \\ \ \\2x(x - 44) + 99(x - 44) = 0 \\ \ \\(2x + 99)(x-44) =0$$​​

x = 44 or x = -44.5 (speed can’t be negative)

Thus, Speed = 44 km/h