SSC CGL 20191)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 20192)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 20193)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 20194)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:
\(2\frac{1}{2}\)
Let the speed of the man be x and the usual time to reach the destination be t hrs
\(xt = x \times {60 \over 100}{(t + {100 \over 60})}\)
\(xt = x \times {60 \over 100}{(t + {100 \over 60})}\)
solving above equation we get t=2.5 hrs
SSC CGL 20195)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 20206)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 20207)Two cyclists X and Y start at the same time from place A and go towards place B at a speed of 6 km/h and 8 km/h,respectively. Despite stopping for 15 minutes during the journey, Y reaches 10 minutes earlier than X. The distance between the places A and B is:
10 km
Let the distance between A and B be x km. According to the question, \({x\over6}-{x\over8}= {15+10\over60}\); x = 10 km.
SSC CGL 20208)A student takes 1.25 hours to travel from home to school at a speed of 4 km/h. By what percentage should he increase his speed to reduce the time by 25% to cover the same distance from school to home?
\(33{1\over3} \)%
Decrease in time = 25% = x; \(\therefore \) Percentage increase in speed = \({x\over 100-x}\times100={25\times100\over75}=33{1\over3}\)%
SSC CGL 20209)A train takes \(2{1\over2}\) hours less for a journey of 300 km, if its speed is increased by 20 km/h from its usual speed. How much time will it take to cover a distance of 192 km at its usual speed?
4.8 hours
Let the usual peed of train be x km/hr.
Distance = 300 km;
Time = \(2\frac{1}{2} hours = \frac{5}{2}\) = 2.5 hr;
Time = distance/speed;
According to quetion,
\(\frac{300}{x} - \frac{300}{x + 20} = 2.5\);
\((x + 20) \times 120 - 120x = x(x + 20)\);
\(x^2 + 20x - 2400 = 0\);
\(x^2 + 60x - 40x - 2400 = 0\);
x(x + 60) - 40(x + 60) = 0;
(x + 60)(x- 40) = 0;
x = 40;
Distance = 192 km;
Time taken to cover distance by usually speed = 192/40 = 4.8 hours
SSC CHSL 202110)A car covers a distance of 48 km at a speed of 40 km/h and another 52 km with a speed of 65 km/h. What is the average speed of the car (in km/h) for the total distance covered?
50