Problem: In a camp there were stored food of 48 soldiers for 7 weeks. if 8 more soldiers join the camp lets find for how many weeks it will be sufficient with the same food?
My approach:
48 soldiers: 7 weeks
1 soldier: $ {7\times 48}$ weeks
($48+8$) soldiers: $\dfrac{7\times 48}{56}=6$ weeks.
Is my approach correct?
Yes, your approach is good. Alternatively, using algebra instead of ratios:
The number of soldiers $s$ is inversely proportional to the number of weeks $w$ that the given food lasts. I.e., there is some constant $K$ such that $$sw=K.$$ So, $$s_1w_1=K=s_2w_2.$$
Since $s_1=48,\quad w_1=7,\quad s_2=48+8=56,$ $$w_2=6.$$
P.S. Such problems sometimes involve three related quantities (e.g., number of soldiers, number of weeks, number of combat rations) that are in direct and inverse proportion; in this case, a three-column table is a good way to keep track of the quantities as they vary with one another.
