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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
SSC CGL 20205)Amit travelled a distance of 50 km in 9 hours. He travelled partly on foot at 5 km/h and partly by bicycle at 10 km/h. The distance travelled on the bicycle is :
10 km.
SSC CGL 20206)Two cyclists X and Y start at the same time from place A and go towards place B at a speed of 6 km/h and 8 km/h,respectively. Despite stopping for 15 minutes during the journey, Y reaches 10 minutes earlier than X. The distance between the places A and B is:
10 km
Let the distance between A and B be x km. According to the question, \({x\over6}-{x\over8}= {15+10\over60}\); x = 10 km.
SSC CGL 20207)A train takes \(2{1\over2}\) hours less for a journey of 300 km, if its speed is increased by 20 km/h from its usual speed. How much time will it take to cover a distance of 192 km at its usual speed?
4.8 hours
Let the usual peed of train be x km/hr.
Distance = 300 km;
Time = \(2\frac{1}{2} hours = \frac{5}{2}\) = 2.5 hr;
Time = distance/speed;
According to quetion,
\(\frac{300}{x} - \frac{300}{x + 20} = 2.5\);
\((x + 20) \times 120 - 120x = x(x + 20)\);
\(x^2 + 20x - 2400 = 0\);
\(x^2 + 60x - 40x - 2400 = 0\);
x(x + 60) - 40(x + 60) = 0;
(x + 60)(x- 40) = 0;
x = 40;
Distance = 192 km;
Time taken to cover distance by usually speed = 192/40 = 4.8 hours
SSC CHSL 20218)A car covers a distance of 48 km at a speed of 40 km/h and another 52 km with a speed of 65 km/h. What is the average speed of the car (in km/h) for the total distance covered?
50
SSC CHSL 20219)A truck covers a distance of 480 km at a certain speed. If the speed is decreased by 8 km/h, it would have taken 2 hours more to cover the same distance. 50% of its original speed (in km/h) is:
24
SSC CHSL 202110)Ashok completes a journey in 7 hours. He travels the first half of the journey at the speed of 24 km/h and the second half of it at the speed of 32 km/h. The total distance of the journey is:
192 km