Time & Work – Advanced Quantitative Aptitude Notes
(MIT‑Level, First Principles & Rigorous Reasoning)
Audience – Advanced undergraduate students, graduate students, or high‑level competitive‑exam aspirants.
Goal – Understand time & work from a mathematical perspective, derive all classic formulas rigorously, explore edge cases, and extend the model to continuous‑time and stochastic settings.
1. Foundations
1.1 The Basic Model
Let
- (W) be a finite, positive amount of work (e.g., “painting a wall”).
- (v_i>0) be the constant rate of worker (i) measured in units of (W) per unit time.
- (t_i) be the time worker (i) works.
The total work done by worker (i) is (v_i t_i).
If workers work simultaneously without interference, the total work is additive:
[ \sum_{i=1}^{n} v_i t_i = W \quad\text{(Equation 1)} ]
Assumptions
- Work is continuous and divisible (no indivisibility).
- Workers do not impede one another (no bottlenecks).
- Rates (v_i) are time‑independent (constant).
These are the first principles from which all classical “time & work” problems arise.
1.2 Units & Dimensional Analysis
- Rate (v_i): ([v_i]=\frac{[W]}{[t]}).
- Work (W): ([W]) (e.g., “one wall”).
- Time (t): ([t]).
Dimensional consistency is a quick check for errors.
2. Classical Formulas – Derivation from First Principles
2.1 Single Worker
With one worker (v) working until (t):
[ v t = W ;;\Longrightarrow;; t = \frac{W}{v} ]
2.2 Two Workers Working Together
Let workers A and B have rates (v_A) and (v_B). Working together for time (t):
[ (v_A+v_B)t = W ;;\Longrightarrow;; t = \frac{W}{v_A+v_B} ]
Interpretation – Their rates add because work is additive.
2.3 Work Done by One Worker after Others Finish
Suppose A works alone for (t_A) and then B joins. Let total time be (T).
Work by A alone: (v_A t_A).
Remaining work: (W - v_A t_A).
Work done by B (for time (T - t_A)): (v_B (T - t_A)).
Equation:
[ v_A t_A + v_B (T - t_A) = W ]
Solve for (T) or (t_A) as needed.
2.4 General (n) Workers
With (n) workers and a common finish time (T):
[ \sum_{i=1}^{n} v_i T = W ;;\Longrightarrow;; T = \frac{W}{\sum_{i=1}^{n} v_i} ]
This is the linear superposition principle for rates.
3. Edge Cases & Non‑Standard Situations
| Scenario | What Happens? | Key Takeaway |
|---|---|---|
| Zero Rate ((v_i = 0)) | Worker contributes nothing; effectively removed. | Always safe to drop. |
| Negative Rate | Implies undoing work (e.g., a repairer removing paint). | Can be modeled, but most aptitude problems forbid it. |
| Infinite Workers | (\sum v_i \to \infty) → (T \to 0). | Theoretically instantaneous, but practically impossible due to interference. |
| Time‑Dependent Rate (e.g., fatigue) | (v_i(t)) changes; integral needed: (\int_0^T v_i(t)dt=W). | Requires calculus. |
| Work Not Divisible | Work is atomic (e.g., building a bridge). | Discrete combinatorial models apply. |
| Interference / Bottleneck | Workers cannot all work simultaneously. | Requires network flow or queuing theory. |
4. Continuous‑Time Extension
When a worker’s rate varies with time, we replace the product (v t) by an integral.
Let (v_i(t)) be continuous on ([0, T]). Then
[ \int_{0}^{T} v_i(t),dt = W_i \quad\text{(work done by worker } i) ]
For multiple workers:
[ \sum_{i=1}^{n} \int_{0}^{T} v_i(t),dt = W ]
If (v_i(t) = v_i (1-\alpha t)) (linear fatigue with (\alpha>0)):
[ \int_{0}^{T} v_i(1-\alpha t),dt = v_i \left(T - \frac{\alpha}{2}T^2\right) = W_i ]
Solve for (T) implicitly, often requiring quadratic equations.
5. Optimisation Problems
5.1 Minimising Time with Variable Work Allocation
Suppose workers A and B have rates (v_A, v_B), and we may choose how long each works before they exchange roles. Let (t_A) be time A works alone, then B works alone for (T-t_A):
[ v_A t_A + v_B (T - t_A) = W ]
We want to minimise (T) under constraints (t_A \ge 0),