Question

A garrison had provisions for a certain number of days. After 10 days, 1/5 of the men leave and it is found that the provisions will now last just as long as before. How long was that ?

Answer 1

Answer:

50 days

Step-by-step explanation:

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\\\;:-}}}$

So, garrison had provisions for a certain number of days. After 10 days, 1/5 of the men leave and it is found that the provisions will now last just as long as before.

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Initially, let x men had provisions for y days.

Then, x men had provisions for (y-10) days.

$(x-\frac{x}{5})$ men had provisions for y days.

★ Solution :-

Therefore, x(y−10)=$\frac{4x}{5}$

5xy−50x=4xy

xy−50x=0

x(y−50)=0

y=50

That was answer to your question

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Answer 2

Answer:

The provision was for 50 days.

Step-by-step explanation:

Let's assume that, the garrison had provision for x days for y number of men.

Therefore, in x days, y men consume the whole provision.

=> In x days, each men consume $\frac{1}{y}$ part of the provision.

=> In each day, each men consume $\frac{1}{xy}$ part of the provision.

=> In 10 days, y men consumed $\frac{10}{x}$ part of the provision.

And $1 - \frac{10}{x} = \frac{x - 10}{x}$ part of the provision left.

After 10 days, $\frac{y}{5}$ people left and $y - \frac{y}{5} = \frac{4y}{5}$ men remained.

According to the problem,

Remaining provision is just sufficient for $\frac{4y}{5}$ men for x days.

For the remaining period, each day, each man will consume:

$= \frac{x - 10}{x} . \frac{1}{x} .\frac{1}{\frac{4y}{5}} = \frac{5(x - 10)}{4x^2y}$

Therefore, $\frac{1}{xy} = \frac{5(x - 10)}{4x^2y}$

$=>1 = \frac{5(x - 10)}{4x}$

$=>4x = 5x - 50$

∴ $x = 50$

The provision was for 50 days.