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A garrison had provision for a certain number of days. After 10 days, of the men desert and it is found that the provision will now last as long as before. How long was that ?

115 days
225 days
335 days
450 days

Answer:- 1

Explanation:-

Solution:

Initially, Let there be x men having food for y days
After 10 days, x men had food for days
Also,

(x - \frac{x}{5})

$\left( {x - \frac{x}{5}} \right)$ men had food for y days

\begin{aligned} ∴ x (y - 10) = \frac{4 x}{5} \times y \\ \Leftrightarrow 5 x y - 50 x = 4 x y \\ \Leftrightarrow x y - 50 x = 0 \\ \Leftrightarrow x (y - 50) = 0 \\ \Leftrightarrow y - 50 = 0 \\ \Leftrightarrow y = 50 \end{aligned}

$\eqalign{ & \therefore \,x\left( {y - 10} \right) = \frac{{4x}}{5} \times y \cr & \Leftrightarrow 5xy - 50x = 4xy \cr & \Leftrightarrow xy - 50x = 0 \cr & \Leftrightarrow x\left( {y - 50} \right) = 0 \cr & \Leftrightarrow y - 50 = 0 \cr & \Leftrightarrow y = 50 \cr}$

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men had food for y days

\begin{aligned} ∴ x (y - 10) = \frac{4 x}{5} \times y \\ \Leftrightarrow 5 x y - 50 x = 4 x y \\ \Leftrightarrow x y - 50 x = 0 \\ \Leftrightarrow x (y - 50) = 0 \\ \Leftrightarrow y - 50 = 0 \\ \Leftrightarrow y = 50 \end{aligned}

$\eqalign{ & \therefore \,x\left( {y - 10} \right) = \frac{{4x}}{5} \times y \cr & \Leftrightarrow 5xy - 50x = 4xy \cr & \Leftrightarrow xy - 50x = 0 \cr & \Leftrightarrow x\left( {y - 50} \right) = 0 \cr & \Leftrightarrow y - 50 = 0 \cr & \Leftrightarrow y = 50 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

Initially, Let there be x men having food for y days
After 10 days, x men had food for days
Also,

(x - \frac{x}{5})

$\left( {x - \frac{x}{5}} \right)$ men had food for y days

\begin{aligned} ∴ x (y - 10) = \frac{4 x}{5} \times y \\ \Leftrightarrow 5 x y - 50 x = 4 x y \\ \Leftrightarrow x y - 50 x = 0 \\ \Leftrightarrow x (y - 50) = 0 \\ \Leftrightarrow y - 50 = 0 \\ \Leftrightarrow y = 50 \end{aligned}

$\eqalign{ & \therefore \,x\left( {y - 10} \right) = \frac{{4x}}{5} \times y \cr & \Leftrightarrow 5xy - 50x = 4xy \cr & \Leftrightarrow xy - 50x = 0 \cr & \Leftrightarrow x\left( {y - 50} \right) = 0 \cr & \Leftrightarrow y - 50 = 0 \cr & \Leftrightarrow y = 50 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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