1. A person travels 285 km in 6 hr.
In the first part of the journey, he travels at 40 km/hr by bus. In the second
part, he travels at 55 km/hr by train. The distance traveled by train is
(a) 156 km
(b) 165 km
(c) 615 km
(d) 561 km
2. A car covers four
successive 6 km stretches at speeds of 25 kmph, 50 kmph, 75 kmph and 150 kmph
respectively. Its average speed over this distance is
(a) 55 kmph
(b) 50 kmph
(c) 75 kmph
(d) 150 kmph
3. The speeds of three cars are in
the ratio 3 : 4 : 5. The time taken by each of the them to travel the same
distance is in the ratio;
(a) 12 : 15 : 20
(b) 3 : 4 : 5
(c) 20 : 15 : 12
(d) 5 : 4 : 3
4. A certain distance is covered by
a cyclist at a certain speed. If a jogger covers half the distance in double
the time, the ratio of the speed of the jogger to that of the cyclist
(a) 1 : 2
(b) 2 : 1
(c) 1: 4
(d) 4 : 1
5. Walking at 5 km/hr a student
reaches his school from his house 15 minutes early and walking at 3 km/hr is
late by 9 minutes. What is the distance between his school and his house?
(a) 5 km
(b) 8 km
(c) 3 km
(d) 2 km
6. A man has to be at a certain
place at a certain time. He finds that he shall be 20 minutes late if he walks
at 3 km/h speed and 10 minutes earlier if he walks at a speed of 4 km/h. The
distance he has to walk is
(a) 24 km
(b) 12. 5 km
(c) 10 km
(d) 6 km
7. Shri X goes to his office by
scooter at a speed of 30 km/h and reached 6 minutes earlier. If he goes at a
speed of 24 km/h, he reaches 5 minutes late. The distance to his office
is
(a) 20 km
(b) 21 km
(c) 22 km
(d) 24 km
8. If a man walks 3 km/hour, he is
late to his office by 20 minutes. If he increases his speed to 6 km/hour, he
reaches the office 30 minutes early. The distance of his office from the
starting place is
(a) 6 km
(b) 5 km
(c) 5.5 km
(d) 4 km
9. A and B walk on a circular path
whose circumference is 35 km. They start walking from the same place in the
same direction at the same time. Speed of A is 4 km/hr and speed of B is 5
km/hr. They will meet earliest again after.
(a) 15 hours
(b) 21 hours
(c) 35 hours
(d) 42 hours
10. Four runners started running
simultaneously from a point on a circular track. They took 200 sec, 300 sec,
360 sec and 450 sec to complete one round. After how much time to they meet at
the starting point for the first time?
(a) 1800 Seconds
(b) 3600 Seconds
(c) 2400 Seconds
(d) 4800 Seconds
Answers and solutions:
1. (b)
Let distance traveled by train be x km
so (285-x)/40 + x/55 = 6
=> x = 165 km
2. (b)
Total distance = 4*6 = 24 km
Total time = 6/25 + 6/50 + 6/75 + 6/150 = 72/150 hr.
so The average speed = total distance /total time = 24/(72/150) = (24*150)/72 = 50 km/hr.
3. (c)
4. (d)
If the time taken by cyclist to cirtain distance is t
then jogger covers half distance in 2t time
then jogger covers total distance in 4t time
then the ratio between their time = t:4t = 1:4
so The ratio between their speeds = 4:1
5. (c)
Distance = Product of speed/Difference of speeds * time difference
= (5*3)/2 * 24/60 = 3 km
6. (d)
7. (c)
The distance= Product of speed/Difference of speeds * time difference
= (30*24)/6 * 11/60 = 22 km
8. (b)
9. (c)
The time taken by A in 1 round = 35/4 hrs.
The time taken by B in 1 round = 35/5 hrs.
so L.C.M. of 35/4 and 35/5 = 35
they will meet earlist again after 35 hours.
10.(a)
Let distance traveled by train be x km
so (285-x)/40 + x/55 = 6
=> x = 165 km
2. (b)
Total distance = 4*6 = 24 km
Total time = 6/25 + 6/50 + 6/75 + 6/150 = 72/150 hr.
so The average speed = total distance /total time = 24/(72/150) = (24*150)/72 = 50 km/hr.
3. (c)
4. (d)
If the time taken by cyclist to cirtain distance is t
then jogger covers half distance in 2t time
then jogger covers total distance in 4t time
then the ratio between their time = t:4t = 1:4
so The ratio between their speeds = 4:1
5. (c)
Distance = Product of speed/Difference of speeds * time difference
= (5*3)/2 * 24/60 = 3 km
6. (d)
7. (c)
The distance= Product of speed/Difference of speeds * time difference
= (30*24)/6 * 11/60 = 22 km
8. (b)
9. (c)
The time taken by A in 1 round = 35/4 hrs.
The time taken by B in 1 round = 35/5 hrs.
so L.C.M. of 35/4 and 35/5 = 35
they will meet earlist again after 35 hours.
10.(a)
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