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

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 :

Correct Option: C

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:

Correct Option: A

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:

Correct Option: D

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?

Correct Option: A

\(33{1\over3} \)%

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?

Correct Option: A

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

showing 1 - 10 results of 60 results