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A contract is to be completed in 52 days and 125 identical robots were employed, each operational for 7 hours a day. After 39 days, five-seventh of the work was completed. How many additional robots would be required to complete the work on time, if each robot is now operational for 8 hours a day?

  1. 50
  2. 89
  3. 146
  4. 175
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Migrated from GO Mechanical 4 years ago by Arjun

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Let $y$ be the speed of work of a robot per hour and $S$ be the total amount of work done.

$\frac{5}{7}$ of work is completed in $39$ days with $7$ hours a day. So,

$\frac{5}{7}S = 125 \times 39 \times y \times 7  \to(1)$

Remaining work = $\frac{2}{7}S$ and now we have $13$ more days and $8$ hours per day. So,

$\frac{2}{7}S = x \times 13\times y \times 8 \to (2)$

where $x$ is the number of robots required.

$(1) /(2) \implies \frac{5}{2} x \times 8=125\times 3\times7  .$

$x = \frac{1050}{8} = 131.25.$

So number of robots required $=\lceil 131.25 \rceil = 132$. We have $125$ already and so the no. of additional robots required $= 132-125 = 7.$
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Total work to be completed in $52$ days &, there are $125$ identical robots , each works for $7 \hspace{0.1cm } hrs/day$

Assuming, Total work is $W$

As each robot is working for $7 \hspace{0.1cm } hrs/day$

each day $125$ robots completed a total of $125 \times 7\hspace{0.1cm }W$

In 39 days $\dfrac{5}{7}\hspace{0.1cm }W$ is completed

∴ $125 \times 7 \times 39 = \dfrac{5}{7}\hspace{0.1cm }W$

Or, $W = \dfrac{125 \times 7 \times 7 \times 39 }{5}$

∴ Total Work = $47775$

∴ $\dfrac{5}{7}W = \dfrac{5}{7} \times 47775$

$\qquad= 34125\hspace{0.1cm } W $ is completed already in $39$ days

Remaining Days = $(52-39)$ = $13$

∴ In $13$ days $(47775 - 34125)$ = $13650\hspace{0.1cm } W$ has to be done

In $13$ days $125$ Robots can complete

$13 \times 125 \times 8 = 13000\hspace{0.1cm } W$ $\qquad[\text{Now each robot works for 8 hrs/day]}$

∴ Remaining = $(13650-13000)$ = $650\hspace{0.1cm } W$

The remaining $650\hspace{0.1cm } W$ can be completed by $\dfrac{650}{13 \times 8} = 6.25 \approx 7$ robots in $13$ days
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