Calculating Time Values

$\underline{12\ Hour\ \&\ 24\ Hour\ Time\ Formats}$

In general we see two types of time values in our daily life. One shows the time with the time value followed by an $am$ or $pm$, for example $4:30\ pm$ or one with just the time value or sometimes followed by the $hr$ suffix, for example $16:30\ hr$.

The one with the $am$ or $pm$ suffix is the $12\ Hour$ format the one without any suffix or the $hr$ suffix is the $24\ Hour$ format.

$\underline{12\ Hour\ Format}$

This is the time format we see in all analogue clocks and watches (the ones with hour/minute hands) and also most digital watches.

In this format the day is divided into two equal parts of $12\ hours$ each. The part starting from $12$ midnight to $12$ noon is denoted by $am$

The part from $12$ noon to the next day $12$ midnight is denoted by $pm$

$\underline{24\ Hour\ Format}$

This time format is used by the airlines and railways in many countries. Here the suffixes a.m or p.m are not used. The $24$ hours of a day are indicated by numbers from $00$ to $23$. For example $15$ minutes after $12$ midnight is written as $00:15\ hours$, and $01:40\ pm$ is written as $13:40\ hours$.

In this case if the hour value is greater than $12$ and less than $24$ this indicates a time between $12$ noon and $12$ midnight.

$\underline{Expressing\ Time\ Intervals}$

We saw earlier how to represent clock times. Let's take a look at how to represent time values. Let's say you played a game of soccer which lasted for $1\ hour\ 35\ minutes\ and\ 20\ seconds$. You can write this in two ways, the first one using the $:$ separator, and it would look like $01:35:20\ hours$ or using the $^o,\ ',\ ''$ symbols for $hour,\ minute\ and\ second$ values, and in this case your game time would look like $1^o35'20''$

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$\underline{Calculations\ With\ Time\ Values}$

Time values can be added or subtracted from another time value. You can also divide a time value by another time value to get a plain number, or multiply a time value by a plain number to get a new time value. We will see each of these examples below:

$\underline{Adding\ Two\ Or\ More\ Time\ Values}$

It is possible to add two or more time values or one time value to a clock time. Let's see how.

Suppose you are participating in a triathlon with $3$ events, cycling, swimming and a marathon.

The cycling took $16\ mins\ 45\ secs$, swimming took $8\ mins$ and the marathon took $14\ mins\ 35\ secs$

You started the event at $9:35:00\ am$. What time did you reach the finish line?

Let us first add up all the times that you took for each of the events.

That is $36'45'' + 28'00'' + 34'35''$

Now as we see, the smallest unit that we have is seconds, and we can calculate the sum by converting each value to seconds and then adding them and converting it back to hour, minute and second. But that is inefficient, and would require longer calculations.

Instead, we can start with the smallest unit, add them and take the carry over to the next unit.

In our case, $45'' + 35'' = 80''$, and we know that $80'' = 1'20''$ so we keep the $20''$ for the final answer and add the one minute to our minute values. Now adding the minute values we get $36 + 28 + 34 + 1 = 99'$. We know that $99' = 1^o39'$

Therefore, the total time taken for the triathlon is $1^o39'20''$

We use the same method to add this time value to the clock time when you started, which is $9:35:00$

We add the seconds first to get $0'' + 20'' = 20''$, and there is no carry over.

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Next we add the minutes to get $35' + 39' = 74' = 1^o14'$, we see that there is a carry forward of $1^o$

Now we add the hour values, to get $9 + 1 + 1 = 11$

Hence we can conclude that you reached the finish line at $11:14:20\ am$

$\underline{Subtracting\ Time\ Value}$

Now we will see another example to understand how to subtract one time value from another.

Let's say you spent a total of $8\ hours\ 25\ mins$ in school today, out of which there were two recesses of $1\ hr\ 37\ mins$ remaining were classes. How much time did you spend for classes?

The solution is simple, and can be found by subtracting $1^o37'$ from $8^o25'$. Since our minimum unit is minute we can solve this by converting everything to minute, then subtracting and converting them back to hour and minute. That involves larger calculations. Hence we will do this the smarter way.

To find out $8^o25' - 1^o37'$, we should observe that $25'$ is less than $37'$ hence we borrow $1$ hour from $8\ hours$ and add it to the minutes to get $25 + 60 = 85'$ and then subtract $37$ from it to get $85' - 37' = 48'$

We were left with $7$ for the hours so we subtract $1$ hour from $7$ to get $6$

Therefore, our answer is $6^o48'$