objective Ques (34 results)
SSC CGL 201911)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?
\(11\frac{1}{4}\)
SSC CGL 201912)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:
150
SSC CGL 201913)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?
41
SSC CGL 201914)A field roller, in the shape of a cylinder, has a diameter of 1 m and length of \(1\frac{1}{4}\) m. If the speed at which the roller rolls is 14 revolutions per minute, then the maximum area (in square m ) that it can roll in 1 hour is: (Take \(\pi={22\over7}\) )
3300
Area covered in single revolution = \(2 \pi rh\) ;
Area covered in 1 hour = \({2\times {22\over 7} \times {1\over 2} \times {5\over 4} \times 14 \times 60} = 3300 cm^2\)
SSC CGL 201915)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 :
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 202016)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 202017)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 :
10 km.
SSC CGL 202018)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 202019)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 202020)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
