Commercial Mathematics

Work and time

Work and time problems rely on inverse proportionality between the number of workers, time taken, and the total quantum of work completed.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Principle of Inverse Proportionality and Unit Work Rate

If $n$ persons can complete a task in $d$ days working $h$ hours per day with efficiency $\eta$, then the total work $W$ is invariant and governed by the relation \frac{n_1 d_1 h_1 \eta_1}{W_1} = \frac{n_2 d_2 h_2 \eta_2}{W_2}.

📐 Geometric Interpretation

Representing work as a rectangular area where length corresponds to time and height corresponds to rate of working. The total area remains constant for a fixed task.

⚙️ Algebraic & Structural Meaning

Establishes a joint variation model $W = k \cdot n \cdot d \cdot h \cdot \eta$, allowing linear and inverse proportion equations to be solved systematically.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Uniform Efficiency Assumption	`Constant across individuals or scaled proportionally`	Strictly Mandatory	If individual efficiencies vary non-linearly (e.g., fatigue or synergy factors), simple summation of 1-day work rates ($1/A + 1/B$) fails.
Independence of Workers	`Additive contribution domain`	Mandatory for joint work	Interference or bottlenecks among workers will violate the additive rate principle $R_{total} = \sum R_i$.
Continuous Operation / Fixed Hours	`Bounded time intervals $[0, T]`	Essential for time normalization	Varying daily working hours without proportional adjustment leads to incorrect daily rate calculations.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Two-Worker Combined Time

$$T = \frac{xy}{x + y}$$

Rule: If A can do work in $x$ days and B in $y$ days, together they finish it in the product divided by the sum.

Application: undefined

LCM Unit Work Method

$$Total Work = \text{LCM}(x, y, z), \quad Rate_A = \frac{W}{x}$$

Rule: Assume total work is the LCM of given days to avoid fractional arithmetic completely.

Application: Extensively used to solve 3-worker and pipe-cistern problems.

Men-Days-Hours Chain Rule

$$\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$$

Rule: Equate the man-hour-day product divided by respective work units when parameters change.

Alternate Days Work Completion

$$2k \text{ days} \implies 2k \times (R_A + R_B) \text{ work}$$

Rule: Calculate work done in a 2-cycle period, then multiply to approach total work before adding remainder days.

Efficiency Ratio to Time Ratio

$$Efficiency_A : Efficiency_B = b : a \implies Time_A : Time_B = a : b$$

Rule: Time taken is inversely proportional to efficiency; use proportional parts directly.

Application: Crucial for advanced PGT questions involving percentage efficiency comparisons.

Pipes and Cisterns Negative Work

$$Net Rate = \frac{1}{x} + \frac{1}{y} - \frac{1}{z}$$

Rule: Treat leakages or outlet pipes as negative rates of work ($-\frac{1}{z}$) in the LCM setup.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Reciprocal Addition for Alternate Days

Common Pitfall: Candidates simply take the combined rate and apply it directly when workers alternate days, ignoring the specific order of who starts.

Counterexample: A takes 10 days, B takes 15 days, working on alternate days starting with A.

Why it fails: The work does not distribute evenly in fractional days; discrete cycles of 2 days must be computed integer-wise.

🛡️ Exam Defense: Always map out complete 2-day cycles and calculate remaining work explicitly before assigning single-day increments.

⚠️ COMMON ERROR #02

Confusing 'More Efficient' with 'More Time'

Common Pitfall: Assuming a worker who is $50\%$ more efficient takes $50\%$ more time.

Counterexample: Efficiency $\eta_2 = 1.5 \eta_1$

Why it fails: Time is inversely proportional to efficiency: $T_2 = \frac{T_1}{1.5} = \frac{2}{3} T_1$, not $1.5 T_1$.

🛡️ Exam Defense: Invert efficiency ratios immediately when converting to time domains.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Combined Work	Constant individual rates	$T = \frac{xy}{x+y}$	Combining two rectangular rate blocks	'A and B together can do...'
Chain Rule	Proportional scaling	$\frac{M_1D_1}{W_1} = \frac{M_2D_2}{W_2}$	Scaling rectangular volume of work	'If 15 men in 10 days...'
Alternate Days	Strict sequence alternation	Cycle-based integer accumulation	Stepped staircase work progression	'Working on alternate days starting with...'
Pipes & Cisterns	Constant inflow/outflow rates	Net Rate = Inflow - Outflow	Net flux across a container boundary	'Tap A fills, Tap B empties...'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

A can do a piece of work in 12 days and B in 15 days. They work together for 3 days, then A leaves. How long will B take to finish the remaining work?

(A) 7 days(B) 8 \frac{1}{4} days(C) 9 days(D) 8 days

📚 Method A: Formal Textbook Derivation

Step 1: Let total work = LCM(12, 15) = 60 units. Step 2: Rate of A = 5 units/day, Rate of B = 4 units/day. Step 3: Combined work in 3 days = $3(5+4) = 27$ units. Step 4: Remaining work = $60 - 27 = 33$ units. Step 5: Time taken by B = $\frac{33}{4} = 8\frac{1}{4}$ days.

⚡ Method B: 25-Second Exam Shortcut

LCM method instantly yields integer rates (5 and 4), avoiding fractions until the final division.

💡 Tip: Pay close attention to whether the question asks for 'remaining work time' or 'total time elapsed'.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If 12 men or 18 women can reap a field in 14 days, in how many days can 8 men and 16 women reap the same field?

(A) 9 days(B) 10 days(C) 9 \frac{9}{11} days(D) 12 days

📚 Method A: Formal Textbook Derivation

Step 1: Equate capacities: $12M = 18W \implies 2M = 3W$. Step 2: Convert total work into man-days: Total Work = $12 \times 14$ man-days = 168 man-days. Step 3: Convert target group ($8M + 16W$) into equivalent men: $16W = \frac{16}{3}M$. Total men = $8 + \frac{16}{3} = \frac{40}{3}$ men. Step 4: Time = $\frac{168}{40/3} = \frac{504}{40} = \frac{63}{5} = 12.6$ or check options. Wait, recalculate: $8M + 16W = 8M + 16(2/3M) = (24+32)/3 = 56/3 M$. Days = $12 \times 14 / (56/3) = \frac{12 \times 14 \times 3}{56} = \frac{3 \times 14 \times 3}{14} = 9$ days? Wait, $12 \times 14 = 168$. Equivalent men = $8 + 16 \times (2/3)$? Wait, $12M = 18W \implies M/W = 18/12 = 3/2 \implies W = (2/3)M$. Target: $8M + 16W = 8M + 16(2/3)M = (24+32)/3 M = (56/3)M$. Total Work = $12 \times 14 = 168$. Days = $168 \times 3 / 56 = 3 \times 3 = 9$ days.

⚡ Method B: 25-Second Exam Shortcut

Use the equivalence formula: Days = $\frac{\text{Given Men} \times \text{Given Days}}{\text{Target Men} + \text{Target Women} \times \frac{M}{W}}$.

💡 Tip: Always reduce the 'or' condition to a single gender unit before evaluating the 'and' condition.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If A and B together can complete a work in 15 days and B alone in 20 days, in how many days can A alone complete the work?

A 45 days

B 60 days

C 35 days

D 50 days

Derivation & Analysis:

Rate of A = $\frac{1}{15} - \frac{1}{20} = \frac{4-3}{60} = \frac{1}{60}$. Thus, A takes 60 days.

MCQ 2: A is twice as efficient as B, and together they finish a work in 14 days. How many days will A alone take?

A 21 days

B 42 days

C 28 days

D 14 days

Derivation & Analysis:

Efficiencies are $2:1$. Combined units = $2+1 = 3$. Total work = $3 \times 14 = 42$ units. A's time = $\frac{42}{2} = 21$ days.

MCQ 3: 10 men working 8 hours a day can complete a work in 18 days. How many hours a day must 15 men work to complete the same work in 12 days?

A 6 hours

B 8 hours

C 10 hours

D 12 hours

Derivation & Analysis:

Using $\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$: $10 \times 18 \times 8 = 15 \times 12 \times H_2 \implies 1440 = 180 H_2 \implies H_2 = 8$ hours.

MCQ 4: Two pipes A and B can fill a tank in 20 minutes and 30 minutes respectively. If both pipes are opened simultaneously, find the time taken to fill the tank.

A 12 minutes

B 15 minutes

C 10 minutes

D 25 minutes

Derivation & Analysis:

Using $\frac{xy}{x+y} = \frac{20 \times 30}{20 + 30} = \frac{600}{50} = 12$ minutes.

MCQ 5: Pipe A fills a tank in 12 hours, while pipe B empties it in 15 hours. If both are opened together in an empty tank, when will the tank be full?

A 50 hours

B 60 hours

C 45 hours

D 55 hours

Derivation & Analysis:

Net rate = $\frac{1}{12} - \frac{1}{15} = \frac{5-4}{60} = \frac{1}{60}$. Time = 60 hours.

MCQ 6: A contractor undertakes to finish a road in 40 days and employs 100 men. After 24 days, he finds that only $\frac{1}{3}$ of the work has been completed. How many additional men must he employ to finish the work on time?

A 150

B 200

C 100

D 250

Derivation & Analysis:

Remaining work = $\frac{2}{3}$, remaining time = $40 - 24 = 16$ days. Using $\frac{100 \times 24}{1/3} = \frac{M_2 \times 16}{2/3} \implies M_2 = 300$. Additional men = $300 - 100 = 200$.

MCQ 7: A can do a work in 20 days and B in 30 days. They start together, but A leaves 5 days before the work is completed. In how many days is the total work finished?

A 12 days

B 14 days

C 15 days

D 16 days

Derivation & Analysis:

Let total days be $t$. Work done by A in $(t-5)$ days + Work done by B in $t$ days = 1. $\frac{t-5}{20} + \frac{t}{30} = 1 \implies 3(t-5) + 2t = 60 \implies 5t = 75 \implies t = 15$ days.

MCQ 8: Assertion (A): If 5 persons complete a task in 10 days, 10 persons will complete it in 5 days. Reason (R): Number of persons and time taken are inversely proportional when work is constant.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

By inverse variation $M_1 D_1 = M_2 D_2 \implies 5 \times 10 = 10 \times 5 = 50$. Both assertion and reason are mathematically sound.

MCQ 9: Three men A, B, C can complete a work in 10, 12, and 15 days respectively. They start working together, but A leaves 2 days after the start and B leaves 3 days before the completion of work. How long does the work last?

A 6 days

B 7 days

C 8 days

D 5 days

Derivation & Analysis:

Let total days be $x$. A works for 2 days, B works for $x-3$ days, C works for $x$ days. $\frac{2}{10} + \frac{x-3}{12} + \frac{x}{15} = 1 \implies \frac{1}{5} + \frac{x-3}{12} + \frac{x}{15} = 1$. LCM = 60: $12 + 5(x-3) + 4x = 60 \implies 12 + 5x - 15 + 4x = 60 \implies 9x - 3 = 60 \implies 9x = 63 \implies x = 7$ days.

Q10

MCQ 10: A pump can be used to fill a tank with water in 2 hours. Because of a leak, it took 2 hours and 20 minutes to fill the tank. What is the time taken by the leak to empty the full tank?

A 12 hours

B 14 hours

C 10 hours

D 15 hours

Derivation & Analysis:

Pump rate = $\frac{1}{2}$. Net rate with leak = $\frac{1}{7/3} = \frac{3}{7}$. Leak rate = $\frac{1}{2} - \frac{3}{7} = \frac{7-6}{14} = \frac{1}{14}$. Thus, leak empties in 14 hours.

Common questions

Fractions introduce calculation fatigue and higher probability of arithmetic errors under exam stress. Assuming total work as the LCM of given days converts all rates into clean integers, speeding up problem-solving by at least 40%.

Students mistakenly average the rates or assume continuous fractional distribution. Because work is completed in discrete blocks (cycles of 2 or more days), remainder work must always be calculated step-by-step.

Convert all 'OR' conditions into an equivalence ratio (e.g., $2M = 3W$), express all individuals in terms of a single base unit (men or women), and then apply the standard man-days invariant product.