objective Ques (32 results)
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?
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?
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 :
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 ?
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:
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?
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:
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:
\(9\frac{3}{8}\)
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:
\(2\frac{1}{2}\)
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?
\(5\frac{1}{4}\)
