Question

A fort is provided with food for 80 soldiers to last for 60 days. Find how long would the food last if 20 additional soldiers join after 15 days?

Answer 1

Hint: In this type of question we have to use the concept of general mathematics. Here, we can see that when the number of soldiers increases, the food would last for fewer days thus there is an inverse proportion between soldiers and number of days. To solve this question we have to formulate an equation in the form of variables and get the answer.

Complete step by step answer:
Now we have to find the number of days for which the food lasts if in addition to 80 soldiers 20 soldiers join the ford.
As given that the fort is provided with food for 80 soldiers to last for 60 days we can write,
\[\Rightarrow \text{Total Food Available = Total soldiers }\times \text{ Total no}\text{. of days }\]
By substituting number of soldiers and number of days we get,
\[\begin{align}
  & \Rightarrow \text{Total Food Available = }80\times 60 \\
 & \Rightarrow \text{Total Food Available = }4800 \\
\end{align}\]
After 15 days 20 additional soldiers join the fort. So let us suppose that the remaining food lasts for \[x\] days now. Thus we get,
\[\Rightarrow \text{Total Food Available = Food used for 15 days }\left( 80\text{ soldiers} \right)\text{ + Food available for x days }\left( 80+20\text{ soldiers} \right)\]
\[\Rightarrow 4800=\left( \text{15}\times 80 \right)+\left( 100\times x \right)\]
On simplification, we can write,
\[\begin{align}
  & \Rightarrow 4800=1200+100x \\
 & \Rightarrow 4800-1200=100x \\
 & \Rightarrow 3600=100x \\
 & \Rightarrow \dfrac{3600}{100}=x \\
 & \Rightarrow x=36 \\
\end{align}\]
Hence, after adding 20 soldiers the remaining food lasts for 36 days.
Thus the food will last for in all, \[36+15=51\text{ days}\].

Note: In this type of question students have to take care during the formation of the equation. They have to make sure that equal entities must be present on both sides. Also students have to understand the question properly as it is a word problem.