Time and Work Scenarios Simplified: Tips for Faster Solutions
Time and Work problems can seem tricky, but recognizing common scenarios can make solving them much easier. Below is a table of frequently encountered scenarios and their corresponding quick methods to help you navigate these problems with ease.
| Concept/Scenario | Formula/Shortcut | Example |
|---|---|---|
| A works 2× faster than B | A’s Time = $\frac{1}{2}$ $\times$ B’s Time, or A’s Efficiency = 2 × B’s Efficiency | A finishes in 10 days, so B finishes in 2 × 10 = 20 days. |
| A and B together, B leaves | Total Work = Combined Work + Individual Work of A (after B leaves) | A and B together do a job in 4 days; B leaves after 2 days. A completes the rest in 3 days. |
| Partial Jobs | Work Done = Efficiency × Time | A’s efficiency is $\frac{1}{4}$. Work done in 2 days = $\frac{1}{4}$ × 2 = $\frac{1}{2}$. |
| Inverse Proportionality | Time ∝ $\frac{1}{\text{Efficiency}}$ | A is 2× more efficient than B, so A will take half the time of B to finish the same job. |
| Pipes and Cisterns | Rate of Filling/Emptying = Filling Rate − Emptying Rate | Pipe A fills a tank in 5 hours, and Pipe B empties it in 10 hours. Net rate = $\frac{1}{5}$ – $\frac{1}{10}$. |
| Reverse Pipes and Cisterns | Net Rate of Filling = 1 − $\frac{\text{Filling Rate of Leaks}}{\text{Total Filling Rate}}$ | A leak empties $ \frac{1}{4}$ tank. Effective rate = 1 – $\frac{1}{4}$ = $\frac{3}{4}$. |
| Alternate Working Days | Total Work = Days A works × A’s Efficiency + Days B works × B’s Efficiency | A and B alternate, working 1 day each. If A = $\frac{1}{3}$, and B = $\frac{1}{6}$. Total Work Done in 4 Days (2 Cycles) =2×($\frac{1}{3}$ + $\frac{1}{6}$)=1 job. |
By mastering these methods, you can tackle Time and Work problems confidently and efficiently. Practice these scenarios to make problem-solving second nature!
You can practice questions on this topic at Time and Work.