Here's a question I've had a bit of trouble answering:

A military member is assigned $5$ years to serve abroad. For each month they serve in which they have exemplary behavior though, $10$ days is taken off their $5$ year contract. Assuming they have exemplary behavior every month, how much time do they have to serve altogether? How many days get taken off their $5$ year contract?

You can't just take $60\text{ months} \times 10 \text{ days}$ and then deduct it from the $5$ years because they won't be serving all $5$ years.


Observe that after 36 months, that is 3 years, 360 days are taken off the $5$ year contract. Now, after 3 years, the contract remains of $4$ years and 5 days, which means $12$ months and $5$ days more to go. But for every month, $10$ days get cut off from the contract. So, after $1$st month, the contractual time remaining is $10$ months $25$ days. After $2$nd month, we are left with $9$ months $15$ days. After the $3$rd, we are left with $8$ months $5$ days. After the $4$th, we are left with $6$ months $25$ days. After the $5$th month, time left would be $5$ months $15$ days. After $6$th month, we have $4$ months $5$ days. After $7$th month, we are left with $2$ months $25$ days which will be served completely.

Hence, the total time of service would be $3$ years, $9$ months and $5$ days.

If the leap year concept is taken into consideration, the calculation might change a bit. But the process is the same as described above.

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    $\begingroup$ How would you do this with equations? Say if there was a problem with many years and not dealing with just integers. $\endgroup$ – Chan C Dec 9 '18 at 7:20
  • $\begingroup$ Also some months have 31 days vs 30. $\endgroup$ – Chan C Dec 9 '18 at 7:23
  • $\begingroup$ Why will the $2$ month$- 25$ day period be served completely? Shouldn't it be $3$ years, $9$ months, $5$ days of service? $\endgroup$ – Shubham Johri Dec 9 '18 at 7:23
  • $\begingroup$ @ShubhamJohri You are correct. The time of service will be what you say $\endgroup$ – Aniruddha Deshmukh Dec 9 '18 at 7:45
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    $\begingroup$ Of course! But then the total reduction will not cause decrease in the number of months. Only decrease in number of days. So, I did it in my mind and did not write it here which is why I made a mistake while writing. I am correcting it now. $\endgroup$ – Aniruddha Deshmukh Dec 10 '18 at 5:48

To do this with equations, measure the time in days, and ignore leap years for now. There are 365 days in a year and 12 months, so every $\frac {365}{12}$ days, they earn $10$ days credit. Let $d$ be the number of days that they actually serve. Then they earn $$\frac {10}{\frac {365}{12}}d = \frac{120}{365}d$$days of credit total.

The actual days on duty plus the credited days will add up to the full 5 years of service:

$$d + \frac{120}{365}d = 5\times 365\\d= \frac{5\cdot 365}{\frac{485}{365}} = 1373.5 \text{ days}$$ which rounds up to 1374 days, or 3 years, 9 months, 5 days.

But of course, somewhere in the 5 year term, we are going to have a leap year day. Possibly two, but likely just one. So the calculation is one or two days short. Thus the actual duty would be 3 years, 9 months, 6 days (or 7 days).

Note that this "solution by equation" makes a number of simplifying assumptions. The credit days accrue steadily, instead of in 10-day lumps at the beginning of each month. The months are all of even length, just under 30.5 days.

The equation method only estimates the complexity of the real situation, and thus, its result is less accurate. In order to model the situation more accuratebrly, you need to break it down into details, as Aniruddha Deshmukh has done. Even that solution is only approximate, as an exact calculation would require more information than is given in the question:

  • On what year, month and day does the tour of duty start? (The year is only needed to figure out where the leap year days fall.)
  • How exactly are the credit days awarded? (For example, on the first day of the following month? Or maybe a month later, or possibly continuously as my equation calculates.)
  • $\begingroup$ When Aniruddha stopped at month 7 " After 7th month, we are left with 2 months 25 days which will be served completely." couldn't he of went on for 1 more period or 2 depending on which day the 10 day reduction happens on. In that case it would match the equation model that you did. I don't understand why he stopped on month 7. Thanks I appreciate your help. $\endgroup$ – Chan C Dec 10 '18 at 1:57
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    $\begingroup$ @ChanC - I hadn't double-checked Aniruddha's numbers, so I didn't notice the error. I was just aware of the extra error in setting it up as I had, and assumed that accounted for the difference. I've adjusted my remarks now. But the warning about the innaccuracies of this method still holds. $\endgroup$ – Paul Sinclair Dec 10 '18 at 23:46

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