TIPS AND FORMULAS ON TIME AND DISTANCE:
1. Speed = Distance/Time
2. Time = Distance /Speed
3. Distance = Speed x Time
4. km/hr to m/s conversion:
1 kmph = 5/18 m/s
5. m/s to km/hr conversion:
1 kmph = 5/18 m/s
5. m/s to km/hr conversion:
6. If the ratio of speeds of train A and B is a : b, then the ratio of time taken by them to cover the same distance = b : a.
7. If a man covers a certain distance at x km/hr and an equal distance at y km/hr. Then,the average speed during the whole journey is
2xy/(x + y) km/h
8. The time taken by a train in passing a pole or standing man is the same as the time taken by the train to cover a distance equal to its own length.
9. The time taken by a train of length 'L' metres in passing a stationary object of length 'B' metres is equal to the time taken by the train to cover a distance equal to (L + B) m.
10. If two trains are moving in the same directions at u m/s and v m/s, where u > v, then their relative speed will be equal to the difference of their speeds i.e. (u - v) m/s.
11. If two trains are moving in the opposite directions at u m/s and v m/s, then their relative speed will be equal to the sum of their speeds i.e. (u + v) m/s.
12.If two trains of length 'a' metres and 'b' metres are moving in the same directions at u m/s and v m/s respectively, then:
The time taken by the faster train to cross the slower train is
(a + b)/(u - v) sec.
13.If two trains of length 'a' metres and 'b' metres are moving in the opposite directions at u m/s and v m/s respectively, then:
The time taken by the faster train to cross the slower train is
(a + b)/(u + v) sec.
14.If two trains start at the same time from points A and B towards each other and after crossing they take 'a' and 'b' hour in reaching B and A respectively, then:
(A's speed) : (B's speed) = Öb : Öa
Questions based on above formula:
1.A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is
Questions based on above formula:
1.A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is
A. 11 hrs
B. 8 hrs 45 min
C. 7 hrs 45 min
D. 9 hrs 20 min
2. Two men P and Q start a journey from same place at a speed of 3 1/2 km/hr and 3 km/hr respectively. If they move in the same direction then what is the distance between them after 4 hours?
A. 3 km
B. 2 1/2 km
C. 2 km
D. 3 1/2 km
3.A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:
A. 100 kmph
B. 110 kmph
C. 120 kmph
D. 130 kmph
4. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
A. 12
B. 11
C. 10
D. 9
5. A student walks from his house at 4 km/hr and reaches his school 5 minutes too late. If his speed had been 5 km/hr, he would have reached 10 minutes too early. The distance of the school from his house is:
A. 5/3 km
B. 5/27 km
C. 5 km
D. 5/9 km
6. In a journey of 160 km, a train covers the first 120 km at a speed of 80 km/h and the remaining distance at 40 km/h. The average speed of the train for the whole journey is:
A. 60 km/h
B. 64 km/h
C. 68 km/h
D. 72 km/h
7. Two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction. What time will they take to be 8.5 km apart?
A. 17 hr
B. 14 hr
C. 12 hr
D. 19 hr
8. A tractor is moving with a speed of 20 km/h, x km ahead of a truck moving with a speed of 35 km/h. If it takes 20 minutes for the truck to overtake the tractor, then x is equal to:
A. 5 km
B. 10 km
C. 15 km
D. 20 km
9. By walking at 4/5 of his normal speed, a man reaches his office 10 minutes late. How much time he normally takes to reach his office?
A. 40 minutes
B. 45 minutes
C. 50 minutes
D. 60 minutes
10. Walking 6/7th of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance?
A. 1 hr 42 min
B. 1 hr
C. 2 hr
D. 1 hr 12 min
ANSWERS AND SOLUTION:
1(C)Explanation :
Given that time taken for riding both ways will be 2 hours lesser than
the time needed for waking one way and riding back
From this, we can understand that
time needed for riding one way = time needed for waking one way - 2 hours
Given that time taken in walking one way and riding back = 5 hours 45 min
Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min
2(C)Explanation:
If two trains are moving in the same directions at u m/s and v m/s, where u>v, then their relative speed will be equal to the difference of their speeds i.e.(u - v) m/s
Let u = 3 1/2 km/hr and v = 3 km/hr
Thus, Relative speed = 3 1/2 - 3
= 1/2 km/hr
Therefore, Required distance = Speed x Time = 1/2 x 4
= 2 km
3(C)Explanation:
Let speed of the car be x kmph.
Then, speed of the train = 150/100 x = 3/2 x kmph.
So 75/x - 75/(3/2)x = 125/(10 x 60)
=> 75/x - 50/x = 5/24
=> x = (25 x 24)/5 = 120 kmph.
4(C)Explanation :
speed of the bus excluding stoppages = 54 kmph
speed of the bus including stoppages = 45 kmph
Loss in speed when including stoppages = 54 - 45 = 9kmph
=> In 1 hour, bus covers 9 km less due to stoppages
Hence, time that the bus stop per hour = time taken to cover 9 km
=distance/speed = 9/54 hour =1/6 hour = 60/6 min=10 min
5(C)Explanation:
Let the distance of the school from his house be x km.
Time = Distance/Speed
So x/4 - x/5 = 15/60
or, (5x - 4x)/20 - 1/4
or, x = 5 km
6(B)Explanation:
Time = Distance/Speed
Total time taken for the journey = 120/80 + 40/40 = 5/2 hours
So, Average speed = 160 x 2/5
= 64 km/hr
7(A)Explanation :
Relative speed = 5.5 - 5 = .5 kmph (because they walk in the same direction)
distance = 8.5 km
time = distance/speed=8.5/.5=17 hr
8(A)Explanation:
Distance = Speed x Time
Distance covered by the truck in 20 minutes = 35 x 20/60 = 35/3 km
Distance covered by the tractor in 20 minutes = 20 x 20/60 = 20/3 km
So 20/3 + x = 35/5
or, x = 5 km
9(A)Explanation:
Walking at 4/5 of the normal speed means that the time taken would be 5/4 of the normal time.
Let the normal time taken to reach the office be "t" minutes,
So 5/4t - t 10 minutes
or , t/4 = 10
or, t = 40 minutes
10(D)Explanation :
New speed = 6/7 of usual speed
Speed and time are inversely proportional.
Hence new time = 7/6 of usual time
Hence, 7/6 of usual time - usual time = 12 minutes
=> 1/6 of usual time = 12 minutes
=> usual time = 12 x 6 = 72 minutes = 1 hour 12 minutes
ANSWERS AND SOLUTION:
1(C)Explanation :
Given that time taken for riding both ways will be 2 hours lesser than
the time needed for waking one way and riding back
From this, we can understand that
time needed for riding one way = time needed for waking one way - 2 hours
Given that time taken in walking one way and riding back = 5 hours 45 min
Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min
2(C)Explanation:
If two trains are moving in the same directions at u m/s and v m/s, where u>v, then their relative speed will be equal to the difference of their speeds i.e.(u - v) m/s
Let u = 3 1/2 km/hr and v = 3 km/hr
Thus, Relative speed = 3 1/2 - 3
= 1/2 km/hr
Therefore, Required distance = Speed x Time = 1/2 x 4
= 2 km
3(C)Explanation:
Let speed of the car be x kmph.
Then, speed of the train = 150/100 x = 3/2 x kmph.
So 75/x - 75/(3/2)x = 125/(10 x 60)
=> 75/x - 50/x = 5/24
=> x = (25 x 24)/5 = 120 kmph.
4(C)Explanation :
speed of the bus excluding stoppages = 54 kmph
speed of the bus including stoppages = 45 kmph
Loss in speed when including stoppages = 54 - 45 = 9kmph
=> In 1 hour, bus covers 9 km less due to stoppages
Hence, time that the bus stop per hour = time taken to cover 9 km
=distance/speed = 9/54 hour =1/6 hour = 60/6 min=10 min
5(C)Explanation:
Let the distance of the school from his house be x km.
Time = Distance/Speed
So x/4 - x/5 = 15/60
or, (5x - 4x)/20 - 1/4
or, x = 5 km
6(B)Explanation:
Time = Distance/Speed
Total time taken for the journey = 120/80 + 40/40 = 5/2 hours
So, Average speed = 160 x 2/5
= 64 km/hr
7(A)Explanation :
Relative speed = 5.5 - 5 = .5 kmph (because they walk in the same direction)
distance = 8.5 km
time = distance/speed=8.5/.5=17 hr
8(A)Explanation:
Distance = Speed x Time
Distance covered by the truck in 20 minutes = 35 x 20/60 = 35/3 km
Distance covered by the tractor in 20 minutes = 20 x 20/60 = 20/3 km
So 20/3 + x = 35/5
or, x = 5 km
9(A)Explanation:
Walking at 4/5 of the normal speed means that the time taken would be 5/4 of the normal time.
Let the normal time taken to reach the office be "t" minutes,
So 5/4t - t 10 minutes
or , t/4 = 10
or, t = 40 minutes
10(D)Explanation :
New speed = 6/7 of usual speed
Speed and time are inversely proportional.
Hence new time = 7/6 of usual time
Hence, 7/6 of usual time - usual time = 12 minutes
=> 1/6 of usual time = 12 minutes
=> usual time = 12 x 6 = 72 minutes = 1 hour 12 minutes
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