**Solve Time and Distance Problems using Time and Distance Formulas for IBPS PO, SSC CGL and SBI PO Exams**

Time and Distance problems are an integral part of the arithmetic
ability section of SSC CGL, IBPS PO and SBI PO Exams. You can solve Time and Distance
Problems within 35 seconds, if you have all the Time and Distance formulas up your
sleeves. This is a simple concept and you can solve all the time and distance problems
by just substituting the values in the formulas.

In this post we will discuss Time and Distance Problems
and Time and Distance Formulas which will help you solve these problems in
minimum time.

**List of Time and Distance Formulas to Solve Time and Distance Problems**

Relationship between Speed, Distance and Time.

**Variables for measurement and Conversion of these Measurement**

Generally speed is expressed as km/hr or m/sec, distance
is expressed in meters or km and time is expressed in seconds(sec) or hours(hr)

**Solution:**

**Step 1:**

To convert 72 km/hr into m/sec, we need to multiply,

72 x 5/18 = 20 m/sec

**Example 2:**Convert 15 m/s into km/hr

**Solution:**

**Step 1:**

To convert 15 m/sec into km/hr, we need to multiply,

15 x 18/5 = 54 km/hr

**Module 1: Basic Time and Distance Problems solve using Time and Distance Formulas.**

**Question:**A bus covers 216 km in 4 hours, convert the speed of the bus in m/s.

**Solution:**

**Step 1:**

To find the speed of the bus we use the time and distance
formula,

Speed = Distance / Time

By substituting the values in the time and distance formula,

Speed = 216/4

Speed= 54 km/hr

**Step 3:**

As we need to convert the speed from km/hr, we need use
the formula,

1 km/hr =5/18 m/sec

By substituting the values in the formula,

54 x 5/18

= 15 m/sec

Therefore, the speed of the bus is 15 m/sec.

**Example 2:**

**Question:**A man walks at the speed of 54 km/hr and runs at the speed of 10 km/hr. How much time will the man require to cover the distance of 28 km, if he covers half (first = 14 km/hr) his journey walking and half of his journey running?

**Solution:**

**Step 1:**

As you know, Time = Distance/Speed

We need to find the total time taken by the man to finish
his journey we need to add the time taken by him by walking and running.

Total Time = Time

_{walk }+ Time_{Run =}(_{ }Distance_{walk }/_{ }Speed_{run }) +(_{ }Distance_{walk}/ Speed_{ run})**Step 2:**

By substituting the values in the formulas

Total Time = (14/5) + (14/10)

= 42/10

= 4.2 hours

Therefore, the man takes 4.2 hours to finish his journey.

**Example 3:**

**Question:**A man takes 6 hours in walking to a certain place and riding back. He would have taken 2 hours less by riding both ways. What will be the time required by him to walk both ways.

**Solution:**

**Step 1:**

A man takes 6 hours in walking to a certain place and
riding back.

So, time

_{walk}+ time_{ride}= 6 - Equation 1**Step 2:**

He will take 2 hours less if he rides both ways

time

_{ride}+ time_{ride}= 4 - Equation 2
2 x ride = 4

ride = 4/2

ride = 2

ride = 2

**Step 3:**

by substituting the ride value in the first formula we get,

time

_{walk }+ 2 = 6
time

_{walk }= 6 - 2
time

_{walk}= 4**Step 4:**

We need to find the time taken by the man to walk both
ways,

time

_{walk}+ time_{walk}
4 + 4 = 8

Therefore, man takes 8 hours to walk both ways.

**Module 2 :**How to solve Time and Distance Problems using Time and Distance Formulas when Distance remains constant.**Example**

**1:**

**Question:**A cyclist travels a certain distance in 6 hours at a uniform speed. In the return journey, he increases his speed too 2 km/hour and covers the same distance in 5 hours. What was his initial speed?

**Solution:**

**Step 1:**

In the question it is mentioned that the distance remains
constant

So, d

_{1}= d_{2...................................}1**Step 2:**

Distance = speed x time ..........................2

**Step 3:**

By substituting the equation 1 in equation 2 we get,

speed

_{1 }x time_{1}= speed_{2}x time_{2}_{}

**Step 4:**

By substituting the values in the above formula

s

_{1}x 6 = (s_{1 }+ 2) x 5 ( as the speed is increases 2km/hr
6s

_{1 }= 5s_{1}x 10
6s

_{1 }- 5s_{1}= 10
s

_{1}_{ = }10 km/hr
Therefore, the initial speed of the man was 10 km/hr

**Example 2:**

**Question:**A student walks to school at a rate of 2.5 km/hr and reached six min late. Next day he increases his speed and then reached 10 min early. What is the distance of the school from his home?

**Solution:**

**Regular Method:**

**Step1:**

Let us assume the right time to reach school as 't.'

Let us assume t

_{1 }- 6 to be the late time taken by the student to reach school.
Let us assume t

_{2}+ 10 as the time when the student reache school early.
Now we know can say that,

t

_{1}- 6 = t_{2}+ 10**Step 2:**

As we know, Time =
Distance / Time

Substitute the Time formula into above equation we get,

(d

_{1 }/ s_{1})_{ }- 6 = (d_{2 }/ s_{2}) + 10**Step 3:**

As the student travels from his home to school i.e.
distance remains constants.

d

_{1}= d_{2}can be assumed as 'd.'**Step 4:**

As the units are varying we need to convert them into,

The speed is given in
km/hr we need to convert 6 minutes and 10 minutes into hours.

So, 6 minutes = 6/60 hr and 10 minutes = 10/60 hr.

**Step 5:**

By substituting the values in the time and distance
formulas,

d/2.5 - 6/60 = d/2.5 + 2 + 10/60

d/2.5 - d/4.5 = 6/60 + 10/60

d [ 4.5 + 2.5] / [ 2.5 x 4.5] = 16/60

d = 16/60 x (2.5 x
4.5) / 2

d = 1.5 km

Therefore, the distance traveled by the student from
home to school is 1.5 km.

**Smart Method**

The same problem can be solved via using the formula-

Here, Î”t is the difference between the time from the actual time.

Here, Î”t is the difference between the time from the actual time.

i.e. 10 + 6 = 16 sec (This can be verified from the traditional method)

16/60 hours

s₁ and s₂ is the speed limits.

s 〜 s₂ is the difference between the speed.

By substituting the values in the formulas we get,

Therefore, the distance traveled by the student from home to school is 1.5 km.

The formula is used to solve these kinds of problems where d is constant. Be careful before you use the formula. This formula is the short-cut method to solve the time and distance problem in the minimum time taken.

s 〜 s₂ is the difference between the speed.

By substituting the values in the formulas we get,

d = 16/60 x (2.5 x 4.5) / 2

d = 1.5 kmTherefore, the distance traveled by the student from home to school is 1.5 km.

The formula is used to solve these kinds of problems where d is constant. Be careful before you use the formula. This formula is the short-cut method to solve the time and distance problem in the minimum time taken.

Do write down in the comment section how this blog helped you to solve Time and Distance Problems using Time and Distance Formulas.

Stay tuned for the next post.

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