**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 m/s = 18/5 m/s****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**

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