# How to Solve Quickly Speed Time And Distance Questions

## Solving Speed, Distance and Time Question Quickly

Speed Time and Distance is a part of our study since school time but still, students get confused while solving the questions. This page will provide you the easiest way to How to Solve Speed Distance and Time questions quickly along with the help of some Solved Examples.

### How to Solve Speed Distance and Time Questions Quickly:-

Before solving the question, we have to know about the exact definitions of Speed Time and Distance.

### Definition of Speed, Distance and Time:

• Speed:

Speed can be defined as how quickly an object moves from one place to another.

Formula for speed = $\frac{distance}{time} \$

Or it can be written as, speed = $\frac{d}{t} \$

• Time:

Time is defined as distance divided by speed.

Formula for time = $\frac{distance}{speed} \$

Or it can be written as, $\frac{d}{s} \$

• Distance:

Area covered by an object from moving one place to another in a uniform speed and time is known as Distance.

Formula for distance = speed x time

Or it can be written as , s × t

### How to Solve Problems Quickly

Question 1 A dog runs from one side of a road to the other. The road is 80.0 meters across. The dog takes 16.0 seconds to cross the road. What is the speed of the dog?

Options:

A. 2.0 meters/second

B. 3.5 meters/second

C. 5.1 meters/second

D. 5.0 meters/second

Solution:    S = $\frac{d}{t} \$

S = $\frac{80.0}{16.0} \$

⇒ 5.0 meters/second

Question 2 An Old man and a Young man are working together in an office and stay together in a nearby apartment. The Old Man takes 30 minutes and the Young 20 minutes to walk from apartment to office. If one day the old man started at 10:00 AM and the young man at 10:05 AM from the apartment to office, when will they meet?

Options:

A. 10: 40 AM

B. 10: 15 AM

C. 10:00 AM

D. 10:05 am

Solution:    Ratio of old man speed to young man speed = 3:2

The distance covered by the old man in 5 min = 10units

The 10 unit is covered with relative speed=$\frac{10}{(3-2)} \$=10 min

so, they will meet at 10:15 am.

Question 3 A motorboat covers a certain distance downstream in 30 minutes, while it comes back in 45 minutes. If the speed of the stream is 5 km/h what is the speed of the boat in still water?

Options:

A. 10 kmph

B. 5 kmph

C. 20 kmph

D. 25 kmph

Solution:   Let distance between two points = d kms
Let speed of boat in still water = s
Let speed of river = r

Now :

Downstream speed = s + r = s + 5

= $\frac{d}{30}$ x 60 = 2d kmph

Upstream speed = s – r = s – 5

= d/45 x 60 = $\frac{4d}{30}$kmph

Solving s + 5 = 2d & s – 5 = 4d/3

2d – 5 = $\frac{4d}{30}$ + 5

=> 6d – 15 = 4d + 15

=> d = 15 km

=> s = 2d – 5 = 25 kmph

Question 4 Walking$\frac{6}{7^{th}} \$ of his usual speed, a man is 12 minutes late. What is the usual time taken by him to cover the distance?

Options:

A. 2 hrs

B. 1 hr 30 min

C. 1 hr 12 min

D. 39 min

Solution:    New speed of man = $\frac{6}{7} \$of usual speed

As we know speed & time are inversely proportional

Hence, new time= $\frac{7}{6} \$of usual time

Hence $\frac{7}{6} \$ of usual time- usual time= 12 minutes
$\frac{1}{6} \$ of usual time= 12 minutes

Therefore, usual time = 12×6 = 72 minutes

It means 1 hr 12 minutes

Question 5 A kid takes 6 hours for 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?

Options:

A. 6 hrs

B. 1 hr

C. 3 hrs

D. 8 hrs

Solution:    Time taken by the kid in walking to a certain place & riding back is 6 hrs

Time walk+ Time ride = 6

2 hrs of the time is reduced if he rides both the ways.

Time ride+ Time ride = 4

2× ride= 4

ride = $\frac{4}{2} \$

ride= 2

Now by putting the value of ride, we get

time walk + 2= 6

timewalk = 6-2 =4

Here we are finding the time taken by the kid if he walk both the ways

4+4=8

Therefore, he will take 8 hrs to walk both the ways.

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