A man travels 600km partly by …

Let speed of the train be x km/hr and speed of the car be y km/hr

Total distance = 600 Km

In first case, he travelled 320 km by train and rest i.e. 600 -320 =280 Km by Car.

In 2nd case, he travelled 200 Km by train and rest i.e. 400 Km by Car.
Using formula ; {tex}time=\frac{distance}{speed}{/tex} & according to question,

{tex}\frac{320}{x}+\frac{280}{y}=\frac{26}{3}{/tex} ..(i) (first case, here time = 8hr 40 minutes= 26/3 hrs)
{tex}\frac{200}{x}+\frac{400}{y}=\frac{55}{6}{/tex} ..(ii) (2nd case, here time = 30 minutes i.e. 1/2 hrs greater than that of 1st case)

multiplying eq. (i) by 5 and eq. (ii) by 8, we get
{tex}\frac{1600}{x}+\frac{1400}{y}=\frac{130}{3}{/tex} ..(iii)
{tex}\frac{1600}{x}+\frac{3200}{y}=\frac{220}{3}{/tex} ..(iv)
subtracting (iii) by (iv), we get
{tex}\frac{-1800}{y}=\frac{-90}{3}{/tex}
y = {tex}\frac{1800 \times 3}{90}{/tex}
y = 60 km/hr
Putting y = 60 in eq. (i), we get
{tex}\frac{320}{x}+\frac{280}{60}=\frac{26}{3}{/tex}
{tex}\frac{320}{x}=\frac{26}{3}-\frac{14}{3}{/tex}
{tex}\frac{320}{x}=\frac{12}{3}{/tex}
x = {tex}\frac{320 \times 3}{12}{/tex}
x = 80 km/hr

Hence, speed of the train = 80 Km/hr
speed of the car = 60 Km/hr