1. Ram can cover a certain distance in 1 hr 24 min. By covering two third of the distance at 4 kmph and the rest at 5 kmph. Find the total distance covered ?
Solution:
time=distance/speed
let total distance travelled be x in 84/60 hrs
2/3 rd of x travelled in 4 km/hr
1/3 rd of distance travelled in 5 km/hr
2x/(3*4)+x/(3*5)=84/60
x=6 km
2. Walking at 4 / 5 of its normal speed, a school bus is 10 minutes late. Find its usual time to cover the journey
Solution:
New Speed = 4 / 5 of original speed,
since the speed and time has inverse relation so,
New Time taken by bus = 5 / 4 of the normal time
( 5 / 4 of usual time ) - ( usual time ) = 10 min.
1 / 4 of the normal time = 10 min
normal time = 40 min
3. The distance between two stations Delhi and Lucknow is 500 km. A train starts at 5 pm from delhi and moves towards Lucknow at an average speed of 50 km / hr, Another train starts at 4.20 pm and moves towards delhi at an average speed of 70 km/hr. How far from delhi the two trains meet and at what time ?
Solution :
Let the two trains meets at a distance of x km from Delhi.
Now ,
Time taken by train from Lucknow to cover (500-x)km - Time taken by train from Delhi to cover x km = 40 / 60
so,
((500 - x)/70) - x/50 = 40/60
solving this we get,
x= 188.9 kms
so the distance is 188.9 kms
4. Bullcart A cover a certain distance at the speed of 15 km/hr, another bullcart B covers the same distance at the speed of 16 km/hr. If Bull cart A takes 16 minutes longer than B to cover the same distance find the distance?
Solution:
Let the required distance be x km.
or 16x – 15x = 64 or x = 64
Hence, the required distance = 64 km
5. A train travells at average speed of 100 km / hr, it stops for 3 mins after travelling 75 kms of diatance. How long it takes to reach 600 kms from the starting point.
Solution :
Time taken to travel 600 kms= 600 / 100 = 6 hrs
But it stop after travelling 75 kms , so number of stoping point in 600 kms will be
= 600 / 75 = 8,
but the last stoping point is actual end stoping so
Number of stoping point will be 7, and time taken
= 7*3= 21 minutes
So total time = 6 hrs 21 mins
6. A is faster than B . A and B each walk 24 km. The sum of their speeds is 7 km / hr and the sum of their time taken is 14 hrs. Then A's speed is equal to :
Solution:
Let A's speed = x km / hr.
Then, B's speed = (7 - x) km / hr.
So,
(24 / x) + 24 / (7−x)= 14
=> 24 (7 - x) + 24x = 14x (7 - x)
=> 14x2 - 98x + 168 = 0
=> x2 - 7x + 12 = 0
=> (x - 3) (x - 4) = 0
=> x = 3 or x = 4.
Since, A is faster than B, so A's speed = 4 km / hr and B's speed = 3 km / hr.
7. A man on tour travels first 160 km at 60 km / hr and next 180 km at speed of 80 km / hr . The average speed of first 360 km of the tour is:
Solution:
| Total time taken = |  | 160 | + | 160 |  hrs. | = | 9 | hrs. |
| 64 | 80 | 2 |
 Average speed = | | 320 x | 2 | km/hr | = 71.11 km/hr |
8. A train running at 3 / 7 of its own speed reached the destination in 14 hours, how much time could be saved if the train would have run at its own speed ?
Solution :
New speed = 3 / 7 of normal speed
So, New Time will be = 7 / 3 of normal time. (Invers relation )
As 7 / 3 of normal time is = 14 hours
So, normal time = (14 * 3 / 7 ) = 12 hrs
So time saved = 14 - 12 = 2 hours
9. Speed ratio of two school buses A and B in covering a certain distance is 4 : 5, If A takes 30 minutes more than B covering the distance, then time taken by B to reach the destination is
Solution:
Since speed ratio is 4 : 5
Time ratio will be 5 : 4 ,
let A takes 5x hrs and B takes 4x hrs to reach the destination then ,
5x - 4x= 30/60 =1/2
x =1/2
Time taken by B =4* (1/2)
=2hrs
10. Walking 6/7th of his usual speed, a man is 12 minutes too late. The usual time taken by him to cover that distance is:
a) 1 hour
b) 1 hr 12min
c) 1 hr 15 min
d) 1 hr 20 min
Solution:
6/7 of speed
so 7/6 of time we get.
then new time=7/6 t
usual time =t
then 7/6-t=12/60
(1/6)t=1/5
t=6/5hours
i.e 6/5*60=72 mins=1 hour 12 mins
11. A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively ?
a) 2 : 1
b) 3 : 2
c) 8 : 3
d) Cannot be determined Option
Solution:
Let the man's rate upstream be x kmph and that downstream be y kmph.
a) 37 seconds
b) 48 seconds
c) 72 seconds
d) 68 seconds
Solution:
Relative speed=60-35=25 Km/hr
For car to be ahead of bus by 250 meter,distance to be covered
=250+250=500m=0.5km
So time taken=0.5/25=1/50 Hr (or)
3600/50=72 seconds
13. A man covers a distance of 1200 km in 70 days resting 9 hours a day, if he rests 10 hours a day and walks with speed 1½ times of the previous in how many days will he cover 750 km?
a) 30
b) 31.25
c) 31
d) 33
Solution:
Case 1:
Total Distance = 1200km
Total time=70*(24-9)=1050
Speed=D/T=1200/1050
Case 2:
Speed = (old speed*1.5)= (1200/1050)*1.5
Distance = 750
Time = Dis/Speed
= 750/((1200/1050)*1.5)=437.5
Now resting 10 hrs a day. working 14 hrs a day
437.5/14=31.25
14. A ship is moving at a speed of 30 kmph. To know the depth of the ocean beneath it, it sends a radiowave which travels at a speed 200 m/s. The ship receives back the signal after it has moved 500 m. What is the depth of the ocean?
a) 4 km
b) 8 km
c) 6 km
d) 12 km
Solution:
speed of ship = 30km/hr so 150/18 m/s
after 500 m took time = 500/150*18 = 60s
since radiowave travels at 200 m/s so in 60 s it travel 200m/s*60 s
= 12000m = 12km (for both direction)
hence depth of ocean is = 12km/2 = 6 km.
15. A man rows a boat at a speed of 5 km/hr in still water. Find the speed of a river if it takes him 1 hr to row a boat to a place 2.4 km away and return back.
a) 1 km/hr
b) 6 km/hr
c) 3 km/hr
d) 4 km/hr
Solution:
speed of river be y km/h
up=5-y
down=5+y
(2.4/(5-y))+(2.4/(5+y))=1
y^2 - 1=0
y=1
16. A boat covers 40 km upstream and 90 km downstream in 5 hr. It can also cover 60 km upstream and 60 km downstream in 5 hr. The speed of the water current is
a) 4 km/hr
b) 5 km/hr
c) 20 km/hr
d) 25 km/hr
Solution:
If Speed(in Km/Hr.) of the Stream or water current = x & boat = y, then
Total upstream speed=y-x & downstram speed=y+x
1st Condition:
[40/(y-x)] +[90/(y+x)]= 5 or 130y-50x= 5(y-x)(y+x) ----(i) &
2nd Condition:
[60/(y-x)] +[60/(y+x)] =5 or 120y=5(y-x)(y+x) ----(ii)
Comparing (i) & (ii), y=5x
Substituting value of y=5x in (i), x=5 , y=25
17. Two champion swimmers start a two-length swimming race at the same time, but from opposite ends of the pool. They swim at constant but different speeds. They first pass at a point 18.5 m from the deep end. Having completed one length, each swimmer take
a) 90 m
b) 45 m
c) 26.5m
d) Data insufficient
Solution:
45 m
18. A locomotive engine, without any wagons attached to it, can go at a speed of 40 km/hr. Its speed is diminished by a quantity that varies proportionally as the square root of the number of wagons attached. With 16 wagons, its speed is 28 km/hr. The
a) 99
b) 100
c) 101
d) 120
Solution:
100m
Solution :
New speed = 3 / 7 of normal speed
So, New Time will be = 7 / 3 of normal time. (Invers relation )
As 7 / 3 of normal time is = 14 hours
So, normal time = (14 * 3 / 7 ) = 12 hrs
So time saved = 14 - 12 = 2 hours
9. Speed ratio of two school buses A and B in covering a certain distance is 4 : 5, If A takes 30 minutes more than B covering the distance, then time taken by B to reach the destination is
Solution:
Since speed ratio is 4 : 5
Time ratio will be 5 : 4 ,
let A takes 5x hrs and B takes 4x hrs to reach the destination then ,
5x - 4x= 30/60 =1/2
x =1/2
Time taken by B =4* (1/2)
=2hrs
10. Walking 6/7th of his usual speed, a man is 12 minutes too late. The usual time taken by him to cover that distance is:
a) 1 hour
b) 1 hr 12min
c) 1 hr 15 min
d) 1 hr 20 min
Solution:
6/7 of speed
so 7/6 of time we get.
then new time=7/6 t
usual time =t
then 7/6-t=12/60
(1/6)t=1/5
t=6/5hours
i.e 6/5*60=72 mins=1 hour 12 mins
11. A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively ?
a) 2 : 1
b) 3 : 2
c) 8 : 3
d) Cannot be determined Option
Solution:
Let the man's rate upstream be x kmph and that downstream be y kmph.
Then, distance covered upstream in 8 hrs 48 min = Distance covered downstream in 4 hrs.
 | | x x 8 | 4 | | = (y x 4) |
| 5 |
| 44 | x =4y |
| 5 |
| y = | 11 | x. |
| 5 |
| Required ratio = | | y + x | | : | | y - x | |
| 2 | 2 |
| = | | 16x | x | 1 | | : | | 6x | x | 1 | |
| 5 | 2 | 5 | 2 |
| = | 8 | : | 3 |
| 5 | 5 |
= 8 : 3
12. A car is 250 metres behind the bus. The car and bus are moving with speed 60 km/hr and 35 km/hr respectively. The car will be ahead of bus by 250 metres in: a) 37 seconds
b) 48 seconds
c) 72 seconds
d) 68 seconds
Solution:
Relative speed=60-35=25 Km/hr
For car to be ahead of bus by 250 meter,distance to be covered
=250+250=500m=0.5km
So time taken=0.5/25=1/50 Hr (or)
3600/50=72 seconds
13. A man covers a distance of 1200 km in 70 days resting 9 hours a day, if he rests 10 hours a day and walks with speed 1½ times of the previous in how many days will he cover 750 km?
a) 30
b) 31.25
c) 31
d) 33
Solution:
Case 1:
Total Distance = 1200km
Total time=70*(24-9)=1050
Speed=D/T=1200/1050
Case 2:
Speed = (old speed*1.5)= (1200/1050)*1.5
Distance = 750
Time = Dis/Speed
= 750/((1200/1050)*1.5)=437.5
Now resting 10 hrs a day. working 14 hrs a day
437.5/14=31.25
14. A ship is moving at a speed of 30 kmph. To know the depth of the ocean beneath it, it sends a radiowave which travels at a speed 200 m/s. The ship receives back the signal after it has moved 500 m. What is the depth of the ocean?
a) 4 km
b) 8 km
c) 6 km
d) 12 km
Solution:
speed of ship = 30km/hr so 150/18 m/s
after 500 m took time = 500/150*18 = 60s
since radiowave travels at 200 m/s so in 60 s it travel 200m/s*60 s
= 12000m = 12km (for both direction)
hence depth of ocean is = 12km/2 = 6 km.
15. A man rows a boat at a speed of 5 km/hr in still water. Find the speed of a river if it takes him 1 hr to row a boat to a place 2.4 km away and return back.
a) 1 km/hr
b) 6 km/hr
c) 3 km/hr
d) 4 km/hr
Solution:
speed of river be y km/h
up=5-y
down=5+y
(2.4/(5-y))+(2.4/(5+y))=1
y^2 - 1=0
y=1
16. A boat covers 40 km upstream and 90 km downstream in 5 hr. It can also cover 60 km upstream and 60 km downstream in 5 hr. The speed of the water current is
a) 4 km/hr
b) 5 km/hr
c) 20 km/hr
d) 25 km/hr
Solution:
If Speed(in Km/Hr.) of the Stream or water current = x & boat = y, then
Total upstream speed=y-x & downstram speed=y+x
1st Condition:
[40/(y-x)] +[90/(y+x)]= 5 or 130y-50x= 5(y-x)(y+x) ----(i) &
2nd Condition:
[60/(y-x)] +[60/(y+x)] =5 or 120y=5(y-x)(y+x) ----(ii)
Comparing (i) & (ii), y=5x
Substituting value of y=5x in (i), x=5 , y=25
17. Two champion swimmers start a two-length swimming race at the same time, but from opposite ends of the pool. They swim at constant but different speeds. They first pass at a point 18.5 m from the deep end. Having completed one length, each swimmer take
a) 90 m
b) 45 m
c) 26.5m
d) Data insufficient
Solution:
45 m
18. A locomotive engine, without any wagons attached to it, can go at a speed of 40 km/hr. Its speed is diminished by a quantity that varies proportionally as the square root of the number of wagons attached. With 16 wagons, its speed is 28 km/hr. The
a) 99
b) 100
c) 101
d) 120
Solution:
100m
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