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SSC CHSL Model Paper 8

Question : 56 of 100

Marks: +1, -0

\frac{3}{4}

Solution:

Let the distance between A and B be D km and the real speed of the car be S km/hr.

First time the car takes 90 minutes more and second time the car takes 75 minutes more than the scheduled time.

So,

T_1 - T_2 = \frac{15}{60}

Difference in time taken during the two journeys:

\frac{15}{60} = \left( \frac{100}{s} + \frac{D-100}{\frac{3s}{4}} \right)

-\left( \frac{160}{s} + \frac{D-160}{\frac{3s}{4}} \right)

⇒

\frac{15}{60} = \frac{100}{s} + \frac{4D}{3s} - \frac{400}{3s} - \frac{160}{s}

-\frac{4D}{3s} + \frac{640}{3s}

⇒

\frac{15}{60} == \frac{240}{3s} - \frac{60}{s}

⇒

\frac{15}{60} = \frac{20}{s}

⇒ s = 80 km/hr

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