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(C) Using the formula $\frac{M_{1} D_{1} H_{1}}{W_{1}} = \frac{M_{2} D_{2} H_{2}}{W_{2}}$,where $M$ is the number of examiners,$D$ is the number of da

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$10$

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(C) Using the formula $\frac{M_{1} D_{1} H_{1}}{W_{1}} = \frac{M_{2} D_{2} H_{2}}{W_{2}}$,where $M$ is the number of examiners,$D$ is the number of days,$H$ is the number of hours per day,and $W$ is the amount of work (number of papers).Given:$M_{1} = 4, D_{1} = 10, H_{1} = 5, W_{1} = 1$$M_{2} = 2, D_{2} = 20, H_{2} = ?, W_{2} = 2$Substituting the values:$\frac{4 	imes 10 	imes 5}{1} = \frac{2 	imes 20 	imes H_{2}}{2}$$\Rightarrow 200 = 20 	imes H_{2}$$\Rightarrow H_{2} = \frac{200}{20} = 10$ hours.Therefore,$2$ examiners would have to work $10$ hours a day.

Four examiners can examine a certain number of answer papers in $10$ days by working for $5$ hours a day. For how many hours in a day would $2$ examiners have to work in order to examine twice the number of answer papers in $20$ days?

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