SSC CGL 20161)X and Y are two stations which are 280 km apart. A train starts at a certain time from X and travels towards Y at 60 kmph. After 2hours, another train starts from Y and travels towards X at 20 kmph. After how many hours does the train leaving from X meet the train which left from Y?

Correct Option: D

4 hours

Distance between X and Y = 280 km

Distance covered by train starting from X in 2 hours = 2 x 60 = 120km. Remaining distance = 280 - 120 = 160km Relative speed = 60 + 20 = 80kmph Time taken to cover 160km 160/80 = 2hours Required time of meeting = 2+2 = 4hours

SSC CGL 20192)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 20193)Abhi rows upstream a distance of 28 km in 4 hours and rows downstream a distance of 50 km in 2 hours. To row a distance of 44.8 km in still water, he will take :

Correct Option: A

2.8 hours

Let the speed of boat be x and speed of stream be y.

Abhi rows upstream a distance of 28 km in 4 h.

So, speed in upstream = 28/4 = 7 km/hr;

x - y = 7 ---(1);

Abhi rows downstream a distance of 50 km in 2 h.

So, speed in downstream = 50/2 = 25 km/hr;

x + y = 25 ---(2);

On eq (1) + (2),

2x = 32;

x = 16;

Time taken to row a distance of 44.8 km in still water = 44.8/16 = 2.8 hr

SSC CGL 20194)A train travelling at the speed of x km/h crossed a 200 m long platform in 30 seconds and overtook a man walking in the same direction at the speed of 6 km/h in 20 seconds. What is the value of x ?

Correct Option: D

60

Length the length off the traiin be l m.

A train travelling at the speed of x km/h crossed a 200 m long platform in 30 seconds;

So, length = speed \times time;

(l + 200) = 30x;

l = 30x - 200 ---(1);

Speed of man = 6 km/hr = \(6 \times \frac{5}{18} \)= 5/3 m/sec;

Relative speed = x - 5/3;

\(\frac{l}{20} = \frac{3x - 5}{3};\)

put the value of l,

\(\frac{30x - 200}{20} = \frac{3x - 5}{3}; 90x - 600 = 60x - 100;\)

30x = 500;

x = 50/3 m/sec = \(\frac{50}{3} \times \frac{18}{5} \)= 60 km/hr

SSC CGL 20195)A and B started their journeys from X to Y and Y to X, respectively. After crossing each other, A and B completed the remaining parts of their journeys in \( 6\frac{1}{8}\) hours and 8 hours respectively. If the speed of B is 28 km/h, then the speed (in km/h) of A is:

Correct Option: C

32

By the formula,

Ratio of the speed =\( \sqrt{\ inverse\ \ ratio\ \ of\ \ the time}\);

\(\frac{S_a}{S_b} = \sqrt{\frac{t_b}{t_a}};\)

\(S_b = 28 km/hr;\)

\(t_a \)= \(6\frac{1}{8} = \frac{49}{8};\)

\(t_b \)= 8;

\(\Rightarrow \frac{S_a}{28} = \sqrt{\frac{8}{\frac{49}{8}}};\)

\(\Rightarrow \frac{S_a}{28} = \sqrt{\frac{64}{49}};\)

\(\Rightarrow \frac{S_a}{28} = \frac{8}{7};\)

\(S_a = 32 km/hr\)

SSC CGL 20196)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 20197)Pipes A and B are filling pipes while pipe C is an emptying pipe. A and B can fill a tank in 72 and 90 minutes respectively. When all the three pipes are opened together, the tank gets filled in 2 hours. A and B are opened together for 12 minutes, then closed and C is opened, The tank will be empty after:

Correct Option: B

18 minutes

SSC CGL 20198)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 20199)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

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

SSC CGL 201910)A man can row a distance of 900 metres against the stream in 12 minutes and returns to the starting point in 9 minutes. What is the speed (in km/h) of the man in still water?

Correct Option: C

\(5\frac{1}{4}\)

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