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Rajan Kumar Pasi 5 years, 7 months ago
Sol: Let the speed of the train = x kmph and that of the car = y kmph.
Total distance = 400 kms.
Time 1 = t1 = 4 hrs 30 mins = 4.5 hrs.
Time 2 = t2 = 4 hrs 30 mins + 15 mins = 4 hrs 45 mins = 4.75 hrs.
Physics formula used = Distance = speed x time.
Case 1:
If he covers half the distance by train and the rest of by the car then he travels 200 kms by each.
total time taken = distance/speed of train + distance/speed of car
{tex}\huge {=> \ 4.5= \frac {200}{x} + \frac {200}{y}}\\ \huge {=> \ \frac{4.5}{200}= \frac {1}{x} + \frac {1}{y}}\\ \text{Let u = }{\frac1x} \text{ and v = }{\frac1y}\\ \huge {=> \ \{\frac{4.5}{200}= u+v}\}.............eq.1\\{/tex}
Case 2:
if he travels 100km by train and rest by the car, then he travels 300 kms by car.
total time taken = distance/speed of train + distance/speed of car
{tex}\huge {=> \ 4.75= \frac {100}{x} + \frac {300}{y}}\\ \huge {=> \ \{\frac{4.75}{100}= u + 3v}\}.............eq.2\\{/tex}
Subtracting eq. 2 from eq. 1, we will get:
{tex}\huge {=> \ \frac{4.75}{100}-\frac{4.5}{200}= 2v}\\ \huge {=> \ \frac{1}{40}= 2v}\\ \huge {=> \ v=\frac{1}{80}}\\ \text{Therefore, putting value of v in eq. 1 we will get:}\\ \huge {=> \ u=\frac{1}{100}}\\{/tex}
{tex}\huge {=> \ u=\frac1x=\frac{1}{100}}\\ .\\ \text{Therefore, } \large \boxed{x=100} \ \ and,\\ \huge {=> \ v=\frac1x=\frac{1}{80}}\\ .\\ \text{Therefore, } \large \boxed{y=80}\\ \text{Therefore, } \large \text{ speed of the train = 100 kmph and speed of the car = 80 kmph.}\\ {/tex}
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