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Let the distance covered by that person be ‘d’.

s*t = (1/3)s*(t+30) → t = t/3 + 10 → t = 15.

Problems involving Time, Distance and Speed are solved based on one simple formula.

Distance = Speed * Time

Which implies →

Speed = Distance / Time and

Time = Distance / Speed

Let us take a look at some simple examples of distance, time and speed problems.

Example 1. A boy walks at a speed of 4 kmph. How much time does he take to walk a distance of 20 km?

Example 1. A boy walks at a speed of 4 kmph. How much time does he take to walk a distance of 20 km?

Solution

Time = Distance / speed = 20/4 = 5 hours.

Example 2. A cyclist covers a distance of 15 miles in 2 hours. Calculate his speed.

Example 2. A cyclist covers a distance of 15 miles in 2 hours. Calculate his speed.

Solution

Speed = Distance/time = 15/2 = 7.5 miles per hour.

Example 3. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. What should be its speed to cover the same distance in 1.5 hours?

Example 3. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. What should be its speed to cover the same distance in 1.5 hours?

Solution

Distance covered = 4*40 = 160 miles

Speed required to cover the same distance in 1.5 hours = 160/1.5 = 106.66 mph

Now, take a look at the following example:

Now, take a look at the following example:

Example 4. If a person walks at 4 mph, he covers a certain distance. If he walks at 9 mph, he covers 7.5 miles more. How much distance did he actually cover?

Now we can see that the direct application of our usual formula Distance = Speed * Time or its variations cannot be done in this case and we need to put in extra effort to calculate the given parameters.

Let us see how this question can be solved.

Solution

For these kinds of questions, a table like this might make it easier to solve.

Distance | Speed | Time |

d | 4 | t |

d+7.5 | 9 | t |

Let the distance covered by that person be ‘d’.

Walking at 4 mph and covering a distance ‘d’ is done in a time of ‘d/4’

IF he walks at 9 mph, he covers 7.5 miles more than the actual distance d, which is ‘d+7.5’.

He does this in a time of (d+7.5)/9.

Since the time is same in both the cases →

d/4 = (d+7.5)/9 → 9d = 4(d+7.5) → 9d=4d+30 → d = 6.

So, he covered a distance of 6 miles in 1.5 hours.

Example 5. A train is going at 1/3 of its usual speed and it takes an extra 30 minutes to reach its destination. Find its usual time to cover the same distance.

Example 5. A train is going at 1/3 of its usual speed and it takes an extra 30 minutes to reach its destination. Find its usual time to cover the same distance.

Solution

Here, we see that the distance is same.

Let us assume that its usual speed is ‘s’ and time is ‘t’, then

Distance | Speed | Time |

d | s | t min |

d | S+1/3 | t+30 min |

s*t = (1/3)s*(t+30) → t = t/3 + 10 → t = 15.

So the actual time taken to cover the distance is 15 minutes.

*Note:*Note the time is expressed in terms of ‘minutes’. When we express distance in terms of miles or kilometers, time is expressed in terms of hours and has to be converted into appropriate units of measurement.