**Type 1:**

When you are given two different speeds (s1 and s2) for travelling through a certain distance, and total time (t) for these two journeys:

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**Formula:**

###
** Distance = (S1*S2)/[(S1+S2)t]**

**Ex 1:** A boy/train goes from A to B at 3 km/hr, back from B to A at 2 km/hr. Total time for these two journeys is 5 hours, then distance from A to B is given by:

**Sol:**

Distance = Product of speeds/Addition of speeds * Time

D = 3*2/(3+2) * 5 = 6 km

**D = 6km**
**Ex 2:** A boy/train travels from A to B. He covers half distance at 3 km/hr, and other half at 2 km/hr. Total time taken is 5 hours, then distance from A to B is given by:

**Sol:**

So this time total distance from A to B = (2* Product of speeds)/(Addition of speeds * Time)

D = 2 * [3*2/(3+2) * 5 ]= 12 km

**D = 12km**
**Ex 3:** A boy/train travels from A to B. He covers 2/3rd distance at 2 km/hr, and rest at 3 km/hr. Total time taken is 24 hours, then distance from A to B is given by:

**Sol:**

S1 = 2 km/hr, S2 = 3 km/hr, Total time, T = 24 hrs

Distance left = 1 – 2/3 = 1/3

Reciprocate both distances, i.e. D1 = 3/2 and other D2 = 3/1

Distance from A to B = D1*S1 * D2*S2/ (D1*S1 + D2*S2) * T

i.e. 3/2*2 * 3/1*3/ (3/2*2 + 3/1*2) * 24 = 54 km

**D = 54km**
**Type 2:**

When you are given two different speeds (s1 and s2) for travelling through a certain distance, and total difference in time (t) is given for these two journeys:

**Formula:**

###
** Distance = (S1*S2)/[(S1-S2)t]**

**Ex 1:** A boy/train goes from A to B. If speed is 30 km/hr, he/it is late by 10 minutes. If speed is 40 km/hr, he/it reaches 5 minutes earlier, then distance from A to B is given by:

**Sol:**

Difference in time = 10 – (-5) = 15 minutes or 1/4 hr……..earlier(-5)

Distance = Product of speeds/Difference of speeds * Difference in time

= 30*40/(40-30) * 1/4 = 30 km

**D = 30km**
**Ex 2:** A train leaves Delhi at 9 AM at 25 km/hr, another train at 35 km/hr leaves Delhi at 2 PM in same direction. How many kilometres from Delhi they will be together?

**Sol:**

Difference in times = 2 PM – 9 AM = 5 hrs

Distance from Delhi = 25*35/(35-25) * 5

**D = 17.5km**
**Type 3:**

When different speed and time are given to find the original time.**Formula:**

Original time = (Reciprocal of given speed*t) – t(late)

Original time = (Reciprocal of given speed*t) + t(early)
**Ex 1:** Walking at 3/4th of usual speed, person is late by 10 minutes. Find the usual time.

**Sol:**

Let t be usual time.

Reciprocate speed for time, i.e. 4/3 * t

Person is late by 10 minutes, so (4/3 * t) – t = 10

**t = 30 min**
**Type 4:****To find the time of halts when the speed with stoppages (s1)and speed without stoppages (s2) is given.**

Formula:

Halt time = (s2-s1)/s2
Ex 1: A man/train leaves from A to B. Speed with stoppages = 60 km/hr, speed without stoppages = 80 km/hr. Find the time man/train stop?

**Sol:**

Minutes per hour the man/ train stops = (80-60)/80 = 1/4 hr = 15 min

**Time = 15 min**
**Type 5 :**

A & B start at same time towards each other. After meeting each other, they reach their destinations after ‘a’ and ‘b’ hrs respectively.

**Formula:**

###
** A’s Speed : B’s Speed = √b : √a**

**Ex 1:** Two, trains, one from Delhi to Mumbai and the other from Mumbai to Delhi, start simultaneously. After they meet, the trains reach their destinations after 4 hours and 9 hours respectively. The ratio of their speeds is:

**Sol:**

A:B = √b : √a = 3 : 2

**A:B = 3:2**
**Type 6 :**

A man takes ‘a’ hrs to walk to a certain distance and ride back.
- If riding both ways takes him b hrs, then Walking both ways takes him

** [a + (a-b)] hrs**
- If walking both ways takes him b hrs, then Riding both ways takes him

** [a – (b-a)] hrs**

**Ex 1:** A man takes 6 hrs to walk to a certain distance and ride back. If riding both ways takes him 4 hrs, then Walking both ways takes him how much time?

**Sol:**

T = [a + (a-b)] hrs

T = 6 +2

**T = 8hrs**
**Ex 2:** A man takes 5 hrs to walk to a certain distance and ride back. If walking both ways takes him 7 hrs, then Riding both ways takes him how much time?

**Sol:**

T = [a – (b-a)] hrs

T = 5 – 2

**T = 3hrs**
**Type 7 :**

A man/train leaves from A at x1 AM and reaches Q at y1 AM. Another man/train leaves from Q at x2 AM and reaches P at y2 PM. When they will meet?

**Formula:**

They will meet at** = (x1+y2)/2**
**Ex 1:** A man/train leaves from P at 6 AM and reaches Q at 10 AM. Another man/train leaves from Q at 8 AM and reaches P at 12 PM. When they will meet

**Sol:**

Since both takes 4 hrs to reach their destination, so they will exactly at the middle of 6 AM and 12 PM, i.e. at **9 AM.**

**Formula:**

###
** Distance = (S1*S2)/[(S1+S2)t]**

**Ex 1:**A boy/train goes from A to B at 3 km/hr, back from B to A at 2 km/hr. Total time for these two journeys is 5 hours, then distance from A to B is given by:

**Sol:**

Distance = Product of speeds/Addition of speeds * Time

D = 3*2/(3+2) * 5 = 6 km

**D = 6km**

**Ex 2:**A boy/train travels from A to B. He covers half distance at 3 km/hr, and other half at 2 km/hr. Total time taken is 5 hours, then distance from A to B is given by:

**Sol:**

So this time total distance from A to B = (2* Product of speeds)/(Addition of speeds * Time)

D = 2 * [3*2/(3+2) * 5 ]= 12 km

**D = 12km**

**Ex 3:**A boy/train travels from A to B. He covers 2/3rd distance at 2 km/hr, and rest at 3 km/hr. Total time taken is 24 hours, then distance from A to B is given by:

**Sol:**

S1 = 2 km/hr, S2 = 3 km/hr, Total time, T = 24 hrs

Distance left = 1 – 2/3 = 1/3

Reciprocate both distances, i.e. D1 = 3/2 and other D2 = 3/1

Distance from A to B = D1*S1 * D2*S2/ (D1*S1 + D2*S2) * T

i.e. 3/2*2 * 3/1*3/ (3/2*2 + 3/1*2) * 24 = 54 km

**D = 54km**

**Type 2:**

When you are given two different speeds (s1 and s2) for travelling through a certain distance, and total difference in time (t) is given for these two journeys:

When you are given two different speeds (s1 and s2) for travelling through a certain distance, and total difference in time (t) is given for these two journeys:

**Formula:**

###
** Distance = (S1*S2)/[(S1-S2)t]**

**Ex 1:**A boy/train goes from A to B. If speed is 30 km/hr, he/it is late by 10 minutes. If speed is 40 km/hr, he/it reaches 5 minutes earlier, then distance from A to B is given by:

**Sol:**

Difference in time = 10 – (-5) = 15 minutes or 1/4 hr……..earlier(-5)

Distance = Product of speeds/Difference of speeds * Difference in time

= 30*40/(40-30) * 1/4 = 30 km

**D = 30km**

**Ex 2:**A train leaves Delhi at 9 AM at 25 km/hr, another train at 35 km/hr leaves Delhi at 2 PM in same direction. How many kilometres from Delhi they will be together?

**Sol:**

Difference in times = 2 PM – 9 AM = 5 hrs

Distance from Delhi = 25*35/(35-25) * 5

**D = 17.5km**

**Type 3:**

When different speed and time are given to find the original time.

When different speed and time are given to find the original time.

**Formula:**

Original time = (Reciprocal of given speed*t) – t(late)

Original time = (Reciprocal of given speed*t) + t(early)

Original time = (Reciprocal of given speed*t) – t(late)

Original time = (Reciprocal of given speed*t) + t(early)

**Ex 1:**Walking at 3/4th of usual speed, person is late by 10 minutes. Find the usual time.

**Sol:**

Let t be usual time.

Reciprocate speed for time, i.e. 4/3 * t

Person is late by 10 minutes, so (4/3 * t) – t = 10

**t = 30 min**

**Type 4:**

**To find the time of halts when the speed with stoppages (s1)and speed without stoppages (s2) is given.**

Formula:

Halt time = (s2-s1)/s2

Formula:

Halt time = (s2-s1)/s2

**Sol:**

Minutes per hour the man/ train stops = (80-60)/80 = 1/4 hr = 15 min

**Time = 15 min**

**Type 5 :**

A & B start at same time towards each other. After meeting each other, they reach their destinations after ‘a’ and ‘b’ hrs respectively.

A & B start at same time towards each other. After meeting each other, they reach their destinations after ‘a’ and ‘b’ hrs respectively.

**Formula:**

###
** A’s Speed : B’s Speed = √b : √a**

**Ex 1:**Two, trains, one from Delhi to Mumbai and the other from Mumbai to Delhi, start simultaneously. After they meet, the trains reach their destinations after 4 hours and 9 hours respectively. The ratio of their speeds is:

**Sol:**

A:B = √b : √a = 3 : 2

**A:B = 3:2**

**Type 6 :**

A man takes ‘a’ hrs to walk to a certain distance and ride back.

A man takes ‘a’ hrs to walk to a certain distance and ride back.

**[a + (a-b)] hrs****[a – (b-a)] hrs****Ex 1:**A man takes 6 hrs to walk to a certain distance and ride back. If riding both ways takes him 4 hrs, then Walking both ways takes him how much time?

**Sol:**

T = [a + (a-b)] hrs

T = 6 +2

**T = 8hrs**

**Ex 2:**A man takes 5 hrs to walk to a certain distance and ride back. If walking both ways takes him 7 hrs, then Riding both ways takes him how much time?

**Sol:**

T = [a – (b-a)] hrs

T = 5 – 2

**T = 3hrs**

**Type 7 :**

A man/train leaves from A at x1 AM and reaches Q at y1 AM. Another man/train leaves from Q at x2 AM and reaches P at y2 PM. When they will meet?

A man/train leaves from A at x1 AM and reaches Q at y1 AM. Another man/train leaves from Q at x2 AM and reaches P at y2 PM. When they will meet?

**Formula:**

They will meet at

**= (x1+y2)/2**

**Ex 1:**A man/train leaves from P at 6 AM and reaches Q at 10 AM. Another man/train leaves from Q at 8 AM and reaches P at 12 PM. When they will meet

**Sol:**

Since both takes 4 hrs to reach their destination, so they will exactly at the middle of 6 AM and 12 PM, i.e. at

**9 AM.**

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**Call :-**

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