SSC CGL 20191)Travelling at 60 km/h, a person reaches his destination in a certain time. He covers 60% of his journey in \({2 \over 5}\)th of the time . At what speed (in km/h) should he travel to cover the remaining journey so that he reaches the destination right on time?

Correct Option: A

40

Let the distance be 60 km and time taken to cover the distance be 1 hr.

Remaining distance = 40% of the total distance = \(60 \times \frac{40}{100}\) = 24 km;

Remaining time = 1 - 2/5 = 3/5 hr;

Speed = distance/time = \(\frac{24}{3/5}\) = 40 km/hr

SSC CGL 20192)To cover a distance of 416 km, a train A takes \(2\frac{2}{3}\) hours more than train B. If the speed of A is doubled, it would take \(1\frac{1}{3}\) hours less than B, What is the speed (in km/h) of train A?

Correct Option: C

52

Let the speed of Train A be x Km/hr and that of be B Y km/hr. Make relation b/w them in time;

. \( {416 \over x} - {416 \over y} = {8 \over 3}\) ---(1) ;

\( {416 \over y} - {416 \over 2x} = {4 \over 3}\) ----(2) ;

adding eqn (1) and (2) ;

\( {416 \over x} - {416 \over 2x} = {8 \over 3} + {4 \over 3}\) ;

x = 52 kmph

SSC CGL 20193)A person covers 40% of the distance from A to B at 8 km/h, 40% of the remaining distance at 9 km/h and the rest at 12 km/h. His average speed (in km/h) for the journey is:

Correct Option: C

\(9\frac{3}{8}\)

SSC CGL 20194)Walking at 60% of his usual speed, a man reaches his destination 1 hour 40 minutes late, His usual time (in hours) to reach the destination is:

Correct Option: A

\(2\frac{1}{2}\)

Let the speed of the man be x and the usual time to reach the destination be t hrs

\(xt = x \times {60 \over 100}{(t + {100 \over 60})}\)

\(xt = x \times {60 \over 100}{(t + {100 \over 60})}\)

solving above equation we get t=2.5 hrs

SSC CGL 20195)A man starts from his house and travelling at 30 km/h, he reaches his office late by 10 minutes, and travelling at 24 km/h, he reaches his office late by 18 minutes. The distance (in km) from his house to his office is :

Correct Option: B

16

Let the actual time required to reach office be t hrs, then equating distances;

\({30 (t+ {10\over 60})} = {24 (t+ {18\over 60})}\)

\(t = {11 \over 30} hrs\)

put value of t in above equation, distance will be 16 Km

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