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)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 20193)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 20194)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 20195)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

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

SSC CGL 20196)The speed of a boat in still water is 18 km/h and the speed of the current is 6 km/h. In how much time (in hours) will the boat travel a distance of 90 km upstream and the same distance downstream?

SSC CGL 20197)Renu was sitting inside train A, which was travelling at 50 km/h. Another train, B, whose length was three times the length of A crossed her in the opposite direction in 15 seconds. If the speed of train B was 58 km/h, then the length of train A (in m) is:

SSC CGL 20198)Places A and B are 396 km apart. Train X leaves from A for B and train Y leaves from B for A at the same time on the same day on parallel tracks. Both trains meet after \(5\frac{1}{2}\) hours. The speed of train Y is 10 km/h more than that of train X. What is the speed (in km/h) of train Y?

SSC CGL 20209)A train X travelling at 60 km/h overtakes another train Y, 225 m long, and completely passes it in 72 seconds. If the trains had been going in opposite directions, they would have passed each other in 18 seconds. The length (in m) of X and the speed (in kmph) of Y are, respectively :

Correct Option: C

255 and 36

Time taken to cross in same direction = 72 second;

Time taken to cross in opposite direction = 18 second;

Speed of train X = 60 km/hr =\( 60 \frac{5}{18}\) = 16.66 m/sec;

Length of train Y = 225 m;

Let the length of train X be l m and speed of train Y be x m/sec.

Total length = (225 + l) m;

Relative speed when trains run opposite direction = (16.66 + x) m/sec;

Length = speed x time;

225 + l = (16.66 + x) x 18;

225 + l = 300 + 18x;

l = 75 + 18x ---(1);

Relative speed when trains run opposite direction = (16.66 - x) m/sec;

225 + l = (16.66 - x) 72;

225 + l = 1200 - 72x;

l = 975 - 72x ---(2);

By eq(1) and (2),

75 + 18x = 975 - 72x;

90x = 900;

x = 10 m/sec;

Speed (in km/h) of Y = \(10 \frac{18}{5}\) = 36 km/hr;

Put the value of x in eq(1);

l = 75 + 18 10 = 255 m.

SSC CGL 202010)The distance between two stations A and B is 575 km. A train starts from station ‘A’ at 3:00 p.m. and moves towards station ‘B’ at an average speed of 50 km/h. Another train starts from station ‘B’ at 3:30 p.m. and moves towards station ‘A’ at an average speed of 60 km/h. How far from station ‘A’ will the trains meet ?

Correct Option: A

275 km

Distance between stations = 575 km;

Speed of train A = 50 km/hr;

Distance covered by train A in (30 min = 1/2 hr) =\( time \times speed = 50 \times 1/2 = 25 km\);

Trains are running in opposite direction so,

Relative speed = 50 + 60 = 110 km/hr;

Distance covered by both train = 575 - 25 = 550 km;

Time taken by both trains to meet = 550/110 = 5 hr;

Distance covered by train A in 5 hr =\( 5 \times 50 \)= 250 km;

Distance covered by train A from station A = distance covered by train A in 30 min + distance covered by train A in 5 = 250 + 25 = 275km

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