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?
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?
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:
\(9\frac{3}{8}\)
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:
\(2\frac{1}{2}\)
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 :
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 20206)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 20207)A train crosses a platform 180 m long in 60 sec at a speed of 72 km/h. The time taken by the train to cross an electric pole is:
51 seconds
Length of platform = 180 m;
Time = 60 sec;
Speed = 72 km/hr = \(72\times 5\over18\) = 20 metre/second;
Total length = length of platform + length of train = 180 + length of train;
Speed = length/time;
180 + length of train = \(20 \times60\);
Length of train =1200 - 180 = 1020 m;
The time taken by the train to cross an electric pole = 1020/20 = 51 seconds
SSC CGL 20208)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 20209)A student takes 1.25 hours to travel from home to school at a speed of 4 km/h. By what percentage should he increase his speed to reduce the time by 25% to cover the same distance from school to home?
\(33{1\over3} \)%
Decrease in time = 25% = x; \(\therefore \) Percentage increase in speed = \({x\over 100-x}\times100={25\times100\over75}=33{1\over3}\)%
SSC CGL 202010)The distance between two stations, A and B, is 428 km.A train starts from station ‘A’ at 6:00 a.m. and moves towards station ‘B’ at an average speed of 48 km/h. Another train starts from station ‘B’ at 6:20 a.m. and moves towards station ‘A’ at an average speed of 55 km/h. At what time will the trains meet?
10 : 20 a.m.
Distance travelled by train in 20 min = \(48\times{20\over60}\) = 16 km. ; Now when both train starts moving the distance between them is = 428 - 16 = 412 km. ;
Relative speed = 55 + 48 = 103 km/ hr ; Time = \(412\over103\) = 4 hr. ; Hence both train will meet at 6:20 + 4 = 10:20 am