Chapter 10, Problem 11MPS

To determine

Find the time needed for each to do the job alone.

Answer to Problem 11MPS

The smaller plant take 12.4 hours and the larger plant takes 8.4 hours to do the job individually.

Explanation of Solution

Given:

Two water purification plants can process a day’s water supply in 5 h.Operating alone, the smaller plant would need 4 h more than the larger one would.

Calculation:

Let the smaller plant take x hours to do the job alone.

Let the larger plant take y hours to do the job alone.

Given , operating alone, the smaller plant would need 4 h more than the larger one would.

So, x = 4 + y

Job done by smaller plant in 1 hour = Efficiency of smaller plant = $\frac{1}{x}$

Job done by larger plant in 1 hour = Efficiency of larger plant = $\frac{1}{y}$

Efficiency of smaller and larger plant when work together = $\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$

Time taken to complete the work when both plants work together = $\begin{array}{l}\frac{1}{\left(\frac{x+y}{xy}\right)}\mathrm{..............................}[time\alpha \frac{1}{efficiency}]\\ =\frac{xy}{x+y}hr\end{array}$

Given time taken to complete the work when two work together is 5 h.

So,

  $\begin{array}{l}\frac{xy}{x+y}=5\\ \Rightarrow \frac{(y+4)y}{(y+4)+y}=\mathrm{5...........}[x=y+4]\\ \Rightarrow \frac{y^2+4y}{2y+4}=5\\ \Rightarrow y^2+4y=10y+\mathrm{20..............}[multiply\text{ 2y+4 each side]}\\ \Rightarrow y^2-6y-20=\mathrm{0...................}[\text{subtract 10y+20 from each side]}\\ \Rightarrow y^2-6y-20=0=(1)y^2+(-6)y+(-20)\\ \Rightarrow y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-6)\pm \sqrt{{(-6)}^2-4(1)(-20)}}{2(1)}\\ \Rightarrow y=\frac{6\pm \sqrt{36+80}}{2}=3\pm \frac{\sqrt{116}}{2}\\ \Rightarrow y=3+\mathrm{5.38.............................}[\text{time is non-negative]}\\ \Rightarrow y\approx 8.4\text{ hours}\\ \Rightarrow x=y+4=12.4\text{ hours}\end{array}$

So, the smaller plant take 12.4 hours and the larger plant takes 8.4 hours to do the job individually.