SSC CGL 20191)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?
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 20192)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:
\(9\frac{3}{8}\)
SSC CGL 20193)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:
\(2\frac{1}{2}\)
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 20194)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 :
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