For how many days will the food last in garrison?

Question

A garrison has sufficient food for $75$ soldiers for a period of $90$ days. After $10$ days, one third of the soldiers leave. After another $10$ days, $5$ soldiers return, From this day on, how many days will the food last ?

I did

$F \propto s\times d \\ F_{1} =k\times75\times 90 \\ F_{2} =k\times50\times 80 \\ F_{3} =k\times55\times 70 $

Now I don't know how to proceed .

I look for a short and simple way.

I have studied maths up to $12$th grade.

Answer 1

As mentioned it the comments, imagine every soldier needs a meal a day. You start with $90\cdot75$ meals.

After 10 days, you have $75\cdot10$ meals less.

25 soldies leave, so in the following 10 days, only $50\cdot10$ meals are eaten.

Than 5 soldiers return, so you have 55 soldiers. All you have to do is evaluate in how many days will the soldiers eat the remaining meals.

Answer 2

$F_{0} = 75*90$ (food for 90 days and 75)

$F_{1} = F_{0}-75*10$ (75 soldiers eat for 10 days)

$F_{2} = F_{1}-50*10$ (50 soldiers eat for 10 days)

$F_{3} = F_{2}-t*55$ (55 soldiers eat for $t$ days)

I guess you can figure out what $F_{3}$ is.

Then calculate $t$, the remaining days.

Answer 3

Let us say each soldier eats $1$ meal per day . You have $75*90$ meals.First $10$ days you use $75*10$ meals. Second $10$ days, you use $50*10$ meals. You now have $M= 75*90 - 75*10 - 50*10$ meals = $5500$ meals. You now have $55$ soldiers. Therefore, no. of days = $M/55 =100$.