, Avg: 02:58 , 75%✓ SSC CGL 20201)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.
, Avg: 01:36 , 36%✓ SSC CGL 20202)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.
, Avg: 03:16 , 75%✓ SSC CGL 20203)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 = \({x\over 100-x}\times100={25\times100\over75}=33{1\over3}\)%
, Avg: 01:56 , 53%✓ SSC CGL 20204)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