Positive and Negative Work Aptitude Formulas, Definitions, & Examples:

Overview:

Concept of Positive and Negative work:

If there are two groups of workers \(x\) and \(y\). The group \(x\) is working to complete a task, where as group \(y\) is woking to destroy the task.

Here group \(x\) is doing positive work and group \(y\) is doing Negative work, then task will be completed $$ \left[Work \ of \ group \ x - Work \ of \ group \ y\right] $$

Note: We can also use this concept in the cases of Pipes and Cistern (Tank), discussed in the next chapter.

Example (1): \(P\) can finish a task in \(5 \ days\), where as \(Q\) can destroy the same task in \(10 \ days\). If they start working at the same time then how many days it will take to complete the task.

Solution: \(P\) can finish a task in \(5 \ days\), so in one day \(P\) can finish \(\frac{1}{5}\) part of the task.

\(Q\) can destroy the task in \(10 \ days\), so in one day \(Q\) can destroy \(\frac{1}{10}\) part of the task, then $$ \left[Work \ of \ P - Work \ of \ Q\right] $$ $$ \left[\frac{1}{5} - \frac{1}{10}\right] $$ $$ \left[\frac{2 - 1}{10}\right] = \frac{1}{10} $$

Hence task will be finish in \(10 \ days\).

Example (2): \(Jack\) can finish a task in \(20 \ days\), where as \(Johnson\) can destroy the same task in \(25 \ days\). If they start working at the same time then how many days it will take to complete the task.

Solution: \(Jack\) can finish a task in \(20 \ days\), so in one day \(Jack\) can finish \(\frac{1}{20}\) part of the task.

\(Johnson\) can destroy the task in \(25 \ days\), so in one day \(Johnson\) can destroy \(\frac{1}{25}\) part of the task, then $$ \left[Work \ of \ Jack - Work \ of \ Johnson\right] $$ $$ \left[\frac{1}{20} - \frac{1}{25}\right] $$ $$ \left[\frac{5 - 4}{100}\right] = \frac{1}{100} $$

Hence task will be finish in \(100 \ days\).