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A customer asks for the production of x number of goods. The company produces y number of goods daily. Out of which z% are units for sale. The order will be completed in :

1[ x /{100y *(1 -z)}] days
2[(100yz) /x] days
3[100x /{y(100 -z)}] days
4100 / {y*(z -1)} days

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Daily supply, \\ = (100 - z) % of y \\ = \frac{{(100 - z) y}}{100} \\ Thus, required number of days \\ = [\frac{(100 x)}{(100 - z) y}] \end{aligned}

$\eqalign{ & {\text{Daily}}\,{\text{supply}}, \cr & = \left( {100 - z} \right)\% \,{\text{of}}\,y \cr & = \frac{{\left\{ {\left( {100 - z} \right)y} \right\}}}{{100}} \cr & {\text{Thus,}}\,{\text{required}}\,{\text{number}}\,{\text{of}}\,{\text{days}} \cr & = \left[ {\frac{{\left( {100x} \right)}}{{\left( {100 - z} \right)y}}} \right] \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Daily supply, \\ = (100 - z) % of y \\ = \frac{{(100 - z) y}}{100} \\ Thus, required number of days \\ = [\frac{(100 x)}{(100 - z) y}] \end{aligned}

$\eqalign{ & {\text{Daily}}\,{\text{supply}}, \cr & = \left( {100 - z} \right)\% \,{\text{of}}\,y \cr & = \frac{{\left\{ {\left( {100 - z} \right)y} \right\}}}{{100}} \cr & {\text{Thus,}}\,{\text{required}}\,{\text{number}}\,{\text{of}}\,{\text{days}} \cr & = \left[ {\frac{{\left( {100x} \right)}}{{\left( {100 - z} \right)y}}} \right] \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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