Machine K and Machine N, working simultaneously and independently at

Machine K and Machine N, working simultaneously and independently at their respective constant rates, process 2/3 of the shipment of a certain chemical product in 1.6 hours.

Inference: To process \(\frac{2}{3}^{\text{rd}}\) of the shipment, Machine K and Machine N, working simultaneously and independently at their respective constant rates, take \(1.6\) hours. Hence, to process the entire shipment the machines would have taken \(\frac{3}{2}^{\text{rd}}\) the time taken to process \(\frac{2}{3}^{\text{rd}}\) of the shipment.

To process \(\frac{2}{3}^{\text{rd}}\) unit of the shipment ----------------- \(1.6\) hours

To process \(1\) unit of the shipment ----------------- \(1.6 * \frac{3}{2} = 2.4\) hours

Then machine K stopped working, and Machine N, working alone at it's constant rate, processed the rest of the shipment in 2 hours

Inference: Hence, machine N processed the remaining \(\frac{1}{3}^{\text{rd}}\) shipment in 2 hours. Therefore, to process the entire shipment alone, machine K would have taken thrice the time taken to process \(\frac{1}{3}^{\text{rd}}\) shipment.

To process \(\frac{1}{3}^{\text{rd}}\) unit of the shipment Machine K, working alone, took ----------------- \(2\) hours

To process \(1\) unit of the shipment Machine K, working alone, will take ----------------- \(2 * 3 = 6\) hours

If Machine N, working alone at its constant rate, takes \(T_1\) hours to complete one unit of task and Machine K, working alone at its constant rate, takes \(T_2\) hours to complete one unit of the same task, the time taken by Machine K and Machine N, working simultaneously and independently at their respective constant rates, to complete one unit of task is given by -

\(T_{\text{together}} = \frac{T_1 * T_2}{T_1 + T_2}\)

In this question -

• \(T_{\text{together}} = 2.4\) hours
  
• \(T_1 = 6\) hours
  
• \(T_2 =\) ?

Substituting the values in the above formula

\(2.4 = \frac{6T_2}{6 + T_2}\)

\(14.4 + 2.4T_2 = 6T_2\)

\(3.6T_2 = 14.4\)

\(T_2 = \frac{14.4}{3.6} = 4.0\)

Option B