SSC CGL 20201)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 :
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 20202)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 ?
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
SSC CGL 20203)In a stream running at 3 km/h, a motorboat goes 12 km upstream and back to the starting point in 60 minutes. Find the speed of the motorboat in still water. (in km/h)
\(3(4+\sqrt{17})\)
Speed of stream = 3 km/hr;
Let the speed of motor boat be v.
Relative speed in upstream = v - 3 km/hr;
Relative speed in upstream = v + 3 km/hr;
Distance = 12 km;
Time = 60 min = 1 hr;
Time = Distance/speed;
\({12\over v-3}+{12\over v+3} =1\);
12(v + 3) + 12(v - 3) = (v + 3)(v - 3);
24v = \(v^2-3^2\);
\(v^2 − 24v − 9 = 0\);
From Dharacharya Formula, \(x= {-b\pm\sqrt{b^2-4ac}\over2a}\); \(x= {-(-24)\pm\sqrt{(-24)^2-4\times1(-9)}\over2}\); \(x = {24\pm6\sqrt{17}\over2}\); As x cannot be negative \(x ={ 6(4+\sqrt{17})\over2} = 3(4+\sqrt{17})\)kmph.
SSC CGL 20204)The two trains leave Varanasi for Lucknow at 11:00 a.m. and at 11:30 a.m., respectively and travel at speeds of 110 km/h and 140 km/h, respectively. How many kilometers from Varanasi will both trains meet?
\(256{2\over3}\)km
Let the train which leaves Varanasi at 11:00 be A and another train be B.
Speed of train A = 110 km/hr;
Speed of train B = 140 km/hr;
Distance covered by train A in 30 min(30/60 = 1/2 hr) = speed x time = 110 x 1/2 = 55 km;
At 11:30, distance between both trains will be 55 km.
Relative speed of train B = 140 - 110 = 30 km/hr;
Time taken by train B to meet = 55/30 = 11/6 hr;
Distance covered by train B in 11/6 hr = 140 x 11/6 = \(256{2\over3}\)km
SSC CGL 20205)Two cars A and B leave Delhi at 8:30 a.m. and at 9 a.m. for Shimla, respectively. They travel at the speeds of 40km/h and 50 km/h respectively. How many kilometres away from Delhi will the two cars be together?
100 km
Cars A and B leave Delhi at 8:30 a.m. and at 9 a.m. for Shimla, respectively so, Distance traveled by car A in 30 min = \(speed \times times = 40 \times 1/2 \) = 20 km;
(30 min = 30/60 hr = 1/2 hr)
Relative speed of car B = 50 - 40 = 10 km/hr;
Time taken by car B to cover 20 km = 20/10 = 2 hr;
(Time = distance/speed)
Distance covered in 2 hr = 50 \times 2 = 100 km
SSC CGL 20206)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
SSC CGL 20207)A boat can go 3.6 km upstream and 5.4 km downstream in 54 minutes, while it can go 5.4 km upstream and 3.6 km downstream in 58.5 minutes. The time (in minutes) taken by the boat in going 10 km downstream is :
50
Let speed of boat in still water be x kmph and speed of stream be y kmph.
\({3.6\over x-y}+{5.4\over x+y}={54\over60}\) ; Again, \({5.4\over x-y}+{3.6\over x+y}={58.5\over60}\) ; ⇒ x + y = 12 kmph ;
Required time = \({10\over12}\times 60 = 50 \space minutes\)