A garrison had provisions for a certain number of days. After 10 days, one-fifth (1/5) of the men deserted. Under the new (reduced) strength, the remaining provisions will now last for exactly as many days as originally planned. For how many days was the garrison originally provisioned?

Introduction: Unitary-method and work–time questions often rely on the invariant “total resource” concept. Here the total food (in man-days) is fixed, while the consumption rate changes when some men desert. Translating the story into man-days lets us write and solve a simple equation for the original planned duration.

Given Data / Assumptions:

• The garrison was originally provisioned for T days with M men.
• After 10 days, one-fifth of the men desert → remaining men = (4/5)M.
• The remaining provisions will now last “just as long as before,” meaning the remaining number of days equals the original planned duration T (classic phrasing in aptitude problems).
• Total food is measured in man-days and is consumed at “men per day.”

Concept / Approach: Total food = M * T man-days. Food consumed in first 10 days = M * 10. Food left = M * (T − 10). After desertion, daily consumption = (4/5)M. Remaining days = [food left] / [new daily consumption] = T (by statement), which gives an equation in T alone.

Step-by-Step Solution:

Verification / Alternative check: If T = 50, then food left after 10 days is enough for (50 − 10) * (5/4) = 40 * 1.25 = 50 days at the reduced strength, matching the condition.

Why Other Options Are Wrong: 70, 48, 45, or 40 days fail the equation (T − 10) * 5/4 = T when substituted; only T = 50 satisfies it exactly.

Common Pitfalls: Equating the total time after desertion to T (i.e., 10 + remaining) rather than the remaining time alone to T. The wording standardly implies the latter in such problems.

Final Answer: 50 days