# Question on work done by x machines in some days?

May 27, 2017

Option C

#### Explanation:

Given that the initial estimated count of machines be $x$
Given that the adjusted count of machines is $x + 3$
Given that the new count of days to complete the task is 12 days.

Let the amount of work be measured in 'machine days'.

$\textcolor{b l u e}{\text{Initial condition}}$

So the amount of work to complete the work is:

$12 \left(x + 3\right)$ machine days.

Set $12 \left(x + 3\right) \text{ machine days } =$ the amount of work for 1 job

$12 \left(x + 3\right) = 1 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(1\right)$

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$\textcolor{b l u e}{\text{Changed condition}}$

Not 3 but only 1 additional machine is used.

The total amount of work still needs to be done to complete the task.

Let the new count of days be $d$

$d \left(x + 1\right) = 1 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(2\right)$

Using $E q u a t i o n \left(1\right)$ substitute for 1 in $E q u a t i o n \left(2\right)$

$d \left(x + 1\right) = 12 \left(x + 3\right) \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left({2}_{a}\right)$

Divide both sides by $\left(x + 1\right)$

$d = \frac{12 \left(x + 3\right)}{x + 1}$
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$\textcolor{b l u e}{\text{Determine the number of days more than originally estimated}}$

$d - 10 \text{ "=" } \frac{12 \left(x + 3\right)}{x + 1} - 10$

Set the value of $1 \equiv \frac{x + 1}{x + 1} \text{ }$so $\text{ } 10 \equiv \frac{10 \left(x + 1\right)}{x + 1}$

Thus we have:

$d - 10 \text{ "=" } \frac{12 \left(x + 3\right)}{x + 1} - 10$

$\text{ "=" } \frac{12 \left(x + 3\right)}{x + 1} - \frac{10 \left(x + 1\right)}{x + 1}$

$\text{ "=" } \frac{12 x + 36 - 10 x - 10}{x + 1}$

$\text{ "=" } \frac{2 x + 26}{x + 1}$

This matches option C