Official Syllabus Item: 23
Arithmetic & Applications
Time & Work, Efficiency & Pipes-Cisterns
Master unitary work rates, combined multi-person output, workers leaving/joining, alternate days, efficiency ratios, wages distribution, and pipes & cisterns for RIMC Entrance Exam.
Overview & Fundamental Concept
Time and Work is one of the most intellectually rigorous and reliable 10-mark questions in the RIMC Mathematics Paper (Paper I, 200 Marks, 90 Minutes). Prescribed under Item 23 of Chapter 8 in the official RIMC brochure, questions in this module appear regularly across both Part A (5 Marks) and Part B (10 Marks). The syllabus evaluates a candidate's mastery over fractional daily work rates, pairwise working pairs $(A+B, B+C, C+A)$, workers departing prior to completion or joining mid-task, alternate-day cyclic schedules, proportional division of wages based on work done, and fluid dynamics of inlet pipes and drainage leaks in cisterns. In the descriptive exam booklet, RIMC examiners demand formal declarations of daily rates ('Work done by A in 1 day = 1/a'), explicit residual work calculations ($W_{\text{remaining}} = 1 - W_{\text{done}}$), and boxed final answers with standard units (days, hours, rupees).
Official Syllabus Coverage Breakdown
| Syllabus Item | Prescribed Scope | Exam Relevance & Pedagogical Objective |
|---|---|---|
| Item 23 | Time and work: Daily work rates, combined work, efficiency, pipes and cisterns, and wage distribution | Core brochure syllabus item. Tests reciprocal proportional reasoning, multi-rate aggregation, dynamic schedule adjustments, and work-based compensation. |
1. The Unitary Work Rate Axiom & Reciprocal Principle
In the unitary method of Time and Work, Total Work is always defined as exactly $1$ complete unit (representing 100% of the project).
The Fundamental Reciprocal Principle: Time taken and the work rate per unit time are reciprocals of each other:
$$\text{If a person completes a piece of work in } n \text{ days, then in } 1 \text{ day he completes } \mathbf{\frac{1}{n}} \text{ part of the work.}}$$
$$\text{Conversely, if a person completes } \frac{1}{n} \text{ part of the work in } 1 \text{ day, he will take } \mathbf{n} \text{ days to finish the entire work.}}$$
Combined Work Rates of Multiple Independent Workers
- Two Workers (A and B): If A takes $a$ days and B takes $b$ days:
$$(A + B)\text{'s } 1\text{ day's work} = \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}$$
$$\text{Time taken together} = \frac{ab}{a + b}\text{ days}$$ - Three Workers (A, B, and C):
$$(A + B + C)\text{'s } 1\text{ day's work} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{bc + ca + ab}{abc}$$
$$\text{Time taken together} = \frac{abc}{ab + bc + ca}\text{ days}$$
2. Pairwise Working Systems (A+B, B+C, C+A)
A classic RIMC 10-mark problem gives the combined rates of pairs of workers: $(A+B)$ finishes in $x$ days, $(B+C)$ in $y$ days, and $(C+A)$ in $z$ days.
Summing all three pairs yields twice the combined capacity:
$$2(A + B + C)\text{'s } 1\text{ day's work} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$$
$$(A + B + C)\text{'s } 1\text{ day's work} = \frac{1}{2} \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)$$
To find the individual time taken by any single worker, subtract the appropriate pair from $(A + B + C)$:
• A's 1 day's work $= (A + B + C) - (B + C)$
• B's 1 day's work $= (A + B + C) - (C + A)$
• C's 1 day's work $= (A + B + C) - (A + B)$
3. Work Dynamics: Departures, Alternate Days & Wage Division
In advanced practical scenarios, the composition of the workforce alters dynamically during execution:
Workers Leaving Before Completion
When worker B departs $k$ days before the project is scheduled to finish, let the total project duration be $T$ days:
- Worker A works for the entire $T$ days, while B works for only $(T - k)$ days.
- Governing equation: $T \times \left(\frac{1}{a}\right) + (T - k) \times \left(\frac{1}{b}\right) = 1$.
Alternate Day Working Schedules
When A works on Day 1, B works on Day 2, A on Day 3, etc.:
- A full cycle consists of $2\text{ days}$.
- Work accomplished in one 2-day cycle $= \frac{1}{a} + \frac{1}{b}$.
- Determine the maximum number of complete 2-day cycles that can fit into $1$ unit of work, and evaluate the remaining fraction on Day $(2k + 1)$ with worker A.
Efficiency & Division of Wages
Wages are distributed strictly in proportion to the total work completed by each individual (or their daily work rate if all worked for the same number of days):
- Ratio of Wages $= (A\text{'s total work}) : (B\text{'s total work}) : (C\text{'s total work})$.
- Efficiency is inversely proportional to time taken: $\frac{\text{Efficiency}_A}{\text{Efficiency}_B} = \frac{\text{Time}_B}{\text{Time}_A}$.
4. The Fluid Mechanics of Pipes & Cisterns
Pipes and Cisterns problems follow the exact same mathematical laws as human workers, with one vital distinction: leakage or drainage pipes perform NEGATIVE work.
• Inlet Pipe: Fills the tank $\implies$ Work Rate is positive ($+\frac{1}{t_{\text{in}}}$).
• Outlet / Waste Pipe / Leak: Empties the tank $\implies$ Work Rate is negative ($-\frac{1}{t_{\text{out}}}$).
• Net Filling Rate:
$$\text{Net Rate per hour} = \left( \frac{1}{t_{\text{in}}} - \frac{1}{t_{\text{out}}} \right)$$
Illustrated Conceptual Diagrams & Visual Schematics
Work Rate Unit Model & Cistern Inflow/Outflow Dynamics
Unitary Work Principles: If person A completes a job in $n$ days, 1 day's work $= rac{1}{n}$. In cistern problems, filling pipes do positive work ($+ rac{1}{a}$) while drainage/emptying leaks do negative work ($- rac{1}{c}$). Net Rate $= rac{1}{a} + rac{1}{b} - rac{1}{c}$.
Topper Strategy & Descriptive Step Breakdown
Mastering RIMC Mathematics Standard
In RIMC, clear step-by-step layout is essential. Write down all given values, declare formulas before substitution, and conclude with full sentences with proper units.
Official Descriptive Marking Scheme & Step Allocation
In RIMC Stage-I, answers without step justification receive only 20-30% of total marks even if the numeric answer is correct. Follow this strict 4-tier layout on every question:
Step 1: One-Day Fractional Work Rate Declarations
1 - 2 Marks
State the individual time taken and declare the reciprocal 1-day (or 1-hour) work rates for each person/pipe.
Booklet Layout:
Given:\nTime taken by A alone $= 10\text{ days} \implies \text{A's 1 day's work} = \frac{1}{10}$\nTime taken by B alone $= 15\text{ days} \implies \text{B's 1 day's work} = \frac{1}{15}$
Given:\nTime taken by A alone $= 10\text{ days} \implies \text{A's 1 day's work} = \frac{1}{10}$\nTime taken by B alone $= 15\text{ days} \implies \text{B's 1 day's work} = \frac{1}{15}$
Step 2: Combined Group Work Rate & Phase-1 Evaluation
2 - 3 Marks
Sum the fractional rates, find LCM of denominators, and compute the work completed during the joint phase.
Booklet Layout:
$$(A + B)\text{'s 1 day's work} = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6}$$
$$\text{Work done by (A + B) in 3 days} = 3 \times \frac{1}{6} = \frac{1}{2}$$
$$(A + B)\text{'s 1 day's work} = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6}$$
$$\text{Work done by (A + B) in 3 days} = 3 \times \frac{1}{6} = \frac{1}{2}$$
Step 3: Residual Work & Phase-2 Completion Time
2 - 4 Marks
Subtract Phase-1 work from 1 to obtain remaining work, then divide by the surviving worker's daily rate.
Booklet Layout:
$$\text{Remaining Work} = 1 - \frac{1}{2} = \frac{1}{2}$$
$$\text{Time taken by A to finish} = \frac{\text{Remaining Work}}{\text{A's daily rate}} = \frac{1/2}{1/10} = \frac{1}{2} \times 10 = 5\text{ days}$$
$$\text{Remaining Work} = 1 - \frac{1}{2} = \frac{1}{2}$$
$$\text{Time taken by A to finish} = \frac{\text{Remaining Work}}{\text{A's daily rate}} = \frac{1/2}{1/10} = \frac{1}{2} \times 10 = 5\text{ days}$$
Step 4: Boxed Final Value with Proper Units
1 Mark
State the final time clearly in words and enclose the answer in a neat box.
Booklet Layout:
$$\mathbf{\text{Answer: A will finish the remaining work in } 5\text{ days.}}$$
$$\mathbf{\text{Answer: A will finish the remaining work in } 5\text{ days.}}$$
Examiner Traps & Common Blunders (Marks Penalties)
Adding Days Directly Instead of Work Rates -4 to -5 Marks
Common Mistake: Computing time taken by A and B together as $10 + 15 = 25\text{ days}$ or $\frac{10 + 15}{2} = 12.5\text{ days}$.
Cadet Antidote: Two people working together MUST take LESS time than either person alone. Never add days directly; always add their fractional rates: $\frac{1}{10} + \frac{1}{15}$.
Forgetting to Subtract Drainage Pipe Work -3 Marks
Common Mistake: Adding the rates of inlet and outlet pipes: $\frac{1}{8} + \frac{1}{12}$.
Cadet Antidote: Outlet pipes empty the cistern, undoing work. Their rate is strictly negative: $\frac{1}{8} - \frac{1}{12}$.
Dividing Wages According to Days Rather Than Work Done -3 Marks
Common Mistake: Dividing a contract fee equally when one worker is twice as fast as the other.
Cadet Antidote: Wages MUST be divided in the exact ratio of the work accomplished by each person.
Pairwise System Factor of 2 Omission -2 Marks
Common Mistake: Adding $(A+B) + (B+C) + (C+A)$ and failing to divide by $2$ to find $(A+B+C)$.
Cadet Antidote: Adding the three pairs counts each worker twice ($2A + 2B + 2C$). You must multiply the sum of fractions by $\frac{1}{2}$.
Cadet Topper Mental Verification Shortcuts
The LCM Total Units Method for Mental Arithmetic
Assume total work = $\text{LCM}(a, b, c)$ units instead of 1. Daily rates become simple integers.
A takes 10 days, B takes 15 days. $\text{LCM} = 30\text{ units}$. A does $3\text{ units/day}$, B does $2\text{ units/day}$. Together they do $5\text{ units/day}$. Time $= 30 / 5 = 6\text{ days}$ in 2 seconds.
Direct Two-Person Product-over-Sum Formula
For two independent workers: $\text{Time together} = \frac{a \times b}{a + b}$.
A in 12 days, B in 24 days: $\text{Time} = \frac{12 \times 24}{12 + 24} = \frac{288}{36} = 8\text{ days}$ instantly.
Efficiency-to-Time Inversion Shortcut
If A is $n$ times as efficient as B, then A takes $\frac{1}{n}$ of the time B takes: $T_B = n \times T_A$.
A is 3 times as efficient as B and takes 60 days less than B: $3x - x = 60 \implies 2x = 60 \implies x = 30$. A takes 30 days, B takes 90 days.
Key Formulas & Conversion Cheat Sheet
Unitary Work Rate Formula
$$\text{Daily Work Rate} = \frac{1}{\text{Total Days } (n)}$$
Fraction of project completed in a single unit of time.
Combined Rate (Two Workers)
$$(A + B)\text{'s 1 day's work} = \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}$$
Joint work rate of two individuals.
Time Taken Together
$$T_{\text{together}} = \frac{ab}{a + b}$$
Product over sum formula for two workers.
Three Workers Combined Rate
$$(A + B + C)\text{'s rate} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$$
Sum of individual fractional rates.
Pairwise Aggregation Formula
$$(A + B + C)\text{'s rate} = \frac{1}{2} \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)$$
Derives 3-worker rate from pair combinations.
Residual Work Formula
$$W_{\text{remaining}} = 1 - W_{\text{completed}}$$
Work left to be completed by surviving workers.
Net Cistern Filling Rate
$$\text{Net Rate} = \frac{1}{t_{\text{fill}}} - \frac{1}{t_{\text{empty}}}$$
Combines filling inlets with draining outlets.
Universal Man-Day-Hour Chain Rule
$$\frac{M_1 \times D_1 \times H_1}{W_1} = \frac{M_2 \times D_2 \times H_2}{W_2}$$
Multi-variable workforce, time, and output balance.
Solved Model Examples (Descriptive Format with Step Mark Distribution)
Model 1: Worker Departure After Few Days
5 Marks
A can do a piece of work in $10\text{ days}$, and B can do it in $15\text{ days}$. They work together for $3\text{ days}$ and then B leaves the job. In how many days will A finish the remaining work?
Step-by-Step Model Solution:
Step 1: Declare One-Day Work Rates
1.5 Marks
Work done by A in $1\text{ day} = \frac{1}{10}$.
Work done by B in $1\text{ day} = \frac{1}{15}$.
Work done by B in $1\text{ day} = \frac{1}{15}$.
Step 2: Compute Combined Work in 3 Days
1.5 Marks
$$(A + B)\text{'s 1 day's work} = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6}$$
Work done by $(A + B)$ in $3\text{ days}$:
$$W_3 = 3 \times \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$$
Work done by $(A + B)$ in $3\text{ days}$:
$$W_3 = 3 \times \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$$
Step 3: Calculate Remaining Work & Time for A to Finish
2 Marks
$$\text{Remaining Work} = 1 - \frac{1}{2} = \frac{1}{2}$$
Time taken by A to complete $\frac{1}{2}$ work:
$$t = \frac{\text{Remaining Work}}{\text{A's 1 day's rate}} = \frac{1/2}{1/10} = \frac{1}{2} \times 10 = \mathbf{5\text{ days}}$$
$$\mathbf{\text{Answer: A will finish the remaining work in } 5\text{ days.}}$$
Time taken by A to complete $\frac{1}{2}$ work:
$$t = \frac{\text{Remaining Work}}{\text{A's 1 day's rate}} = \frac{1/2}{1/10} = \frac{1}{2} \times 10 = \mathbf{5\text{ days}}$$
$$\mathbf{\text{Answer: A will finish the remaining work in } 5\text{ days.}}$$
Examiner Scoring Key: Setting up 1/10 + 1/15 = 1/6 carries 1.5 marks. Computing remaining work = 1/2 carries 1.5 marks. Concluding with 5 days earns full 5 marks.
Model 2: Pairwise Workers System (A+B, B+C, C+A)
5 Marks
A and B can do a piece of work in $12\text{ days}$, B and C in $15\text{ days}$, and C and A in $20\text{ days}$. In how many days can all three together finish the work? Also find in how many days A alone could do it.
Step-by-Step Model Solution:
Step 1: Express Daily Work of Pairs & Sum Them
2 Marks
$$(A + B)\text{'s 1 day's work} = \frac{1}{12}$$
$$(B + C)\text{'s 1 day's work} = \frac{1}{15}$$
$$(C + A)\text{'s 1 day's work} = \frac{1}{20}$$
Adding all three equations:
$$2(A + B + C)\text{'s 1 day's work} = \frac{1}{12} + \frac{1}{15} + \frac{1}{20}$$
$\text{LCM}(12, 15, 20) = 60$:
$$= \frac{5 + 4 + 3}{60} = \frac{12}{60} = \frac{1}{5}$$
$$(B + C)\text{'s 1 day's work} = \frac{1}{15}$$
$$(C + A)\text{'s 1 day's work} = \frac{1}{20}$$
Adding all three equations:
$$2(A + B + C)\text{'s 1 day's work} = \frac{1}{12} + \frac{1}{15} + \frac{1}{20}$$
$\text{LCM}(12, 15, 20) = 60$:
$$= \frac{5 + 4 + 3}{60} = \frac{12}{60} = \frac{1}{5}$$
Step 2: Compute Combined Rate of (A + B + C)
1.5 Marks
$$(A + B + C)\text{'s 1 day's work} = \frac{1}{2} \times \frac{1}{5} = \frac{1}{10}$$
$$\text{Time taken by all three together} = 10\text{ days}$$
$$\text{Time taken by all three together} = 10\text{ days}$$
Step 3: Compute A's Individual Rate and Time
1.5 Marks
A's 1 day's work $= (A + B + C) - (B + C)$:
$$= \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30}$$
$$\text{Time taken by A alone} = \mathbf{30\text{ days}}$$
$$\mathbf{\text{Answer: Together = } 10\text{ days}; \quad \text{A alone = } 30\text{ days}}$$
$$= \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30}$$
$$\text{Time taken by A alone} = \mathbf{30\text{ days}}$$
$$\mathbf{\text{Answer: Together = } 10\text{ days}; \quad \text{A alone = } 30\text{ days}}$$
Examiner Scoring Key: Dividing by 2 to get (A+B+C) = 1/10 carries 1.5 marks. Subtracting (B+C) to isolate A = 1/30 carries 1.5 marks.
Model 3: Alternate Days Cyclic Working Schedule
7 Marks
A can complete a project in $16\text{ days}$ and B can complete it in $12\text{ days}$. They work on alternate days, with A starting on the first day. In how many days will the entire project be completed?
Step-by-Step Model Solution:
Step 1: Compute Work in One 2-Day Cycle
2.5 Marks
Day 1: A works and does $\frac{1}{16}$ of the work.
Day 2: B works and does $\frac{1}{12}$ of the work.
Work completed in one $2\text{-day cycle}$:
$$W_{\text{cycle}} = \frac{1}{16} + \frac{1}{12} = \frac{3 + 4}{48} = \frac{7}{48}\text{ of total work}$$
Day 2: B works and does $\frac{1}{12}$ of the work.
Work completed in one $2\text{-day cycle}$:
$$W_{\text{cycle}} = \frac{1}{16} + \frac{1}{12} = \frac{3 + 4}{48} = \frac{7}{48}\text{ of total work}$$
Step 2: Fit Maximum Complete Cycles
2.5 Marks
We need $\frac{7}{48} \times k \le 1$.
Since $7 \times 6 = 42 < 48$ and $7 \times 7 = 49 > 48$, take $k = 6\text{ full cycles}$.
In $6\text{ cycles}$ (which is $6 \times 2 = 12\text{ days}$):
$$\text{Work completed in 12 days} = 6 \times \frac{7}{48} = \frac{42}{48} = \frac{7}{8}$$
$$\text{Remaining Work} = 1 - \frac{42}{48} = \frac{6}{48} = \frac{1}{8}$$
Since $7 \times 6 = 42 < 48$ and $7 \times 7 = 49 > 48$, take $k = 6\text{ full cycles}$.
In $6\text{ cycles}$ (which is $6 \times 2 = 12\text{ days}$):
$$\text{Work completed in 12 days} = 6 \times \frac{7}{48} = \frac{42}{48} = \frac{7}{8}$$
$$\text{Remaining Work} = 1 - \frac{42}{48} = \frac{6}{48} = \frac{1}{8}$$
Step 3: Allocate Remaining Work on Day 13
2 Marks
On Day $13$, it is A's turn to work.
A's daily capacity $= \frac{1}{16} = \frac{3}{48}$.
Since remaining work $\frac{6}{48} > \frac{3}{48}$, A works the entire 13th day and completes $\frac{3}{48}$ work.
$$\text{Work left at end of Day 13} = \frac{6}{48} - \frac{3}{48} = \frac{3}{48} = \frac{1}{16}$$
On Day $14$, it is B's turn.
B's capacity $= \frac{1}{12} = \frac{4}{48}$.
Time taken by B to finish $\frac{3}{48}$ work:
$$t_B = \frac{3/48}{4/48} = \frac{3}{4}\text{ day}$$
$$\text{Total Time Taken} = 12 + 1 + \frac{3}{4} = \mathbf{13\frac{3}{4}\text{ days}}$$
$$\mathbf{\text{Answer: The project will be completed in } 13\frac{3}{4}\text{ days.}}$$
A's daily capacity $= \frac{1}{16} = \frac{3}{48}$.
Since remaining work $\frac{6}{48} > \frac{3}{48}$, A works the entire 13th day and completes $\frac{3}{48}$ work.
$$\text{Work left at end of Day 13} = \frac{6}{48} - \frac{3}{48} = \frac{3}{48} = \frac{1}{16}$$
On Day $14$, it is B's turn.
B's capacity $= \frac{1}{12} = \frac{4}{48}$.
Time taken by B to finish $\frac{3}{48}$ work:
$$t_B = \frac{3/48}{4/48} = \frac{3}{4}\text{ day}$$
$$\text{Total Time Taken} = 12 + 1 + \frac{3}{4} = \mathbf{13\frac{3}{4}\text{ days}}$$
$$\mathbf{\text{Answer: The project will be completed in } 13\frac{3}{4}\text{ days.}}$$
Examiner Scoring Key: Stepwise cycle allocation (12 days for 42/48 work, full 13th day for A, 3/4 day for B) is rigorously checked. Final answer 13 3/4 days earns 7 marks.
Model 4: Division of Wages Based on Work Contribution
7 Marks
A, B, and C undertake to complete an engineering construction contract for ₹ $5,290$. A and B together do $\frac{19}{23}$ of the total work, while B and C together do $\frac{8}{23}$ of the work. Determine how much money should be paid to each person.
Step-by-Step Model Solution:
Step 1: Express Total Work as 1 Unit and Deduce C's Share
2.5 Marks
Total work $= 1$.
Work done by $(A + B) = \frac{19}{23}$.
Since $(A + B + C) = 1$, work done by C alone is:
$$\text{Work of C} = 1 - (A + B) = 1 - \frac{19}{23} = \mathbf{\frac{4}{23}}$$
Work done by $(A + B) = \frac{19}{23}$.
Since $(A + B + C) = 1$, work done by C alone is:
$$\text{Work of C} = 1 - (A + B) = 1 - \frac{19}{23} = \mathbf{\frac{4}{23}}$$
Step 2: Deduce A's Share and B's Share
2.5 Marks
Given work done by $(B + C) = \frac{8}{23}$.
Work done by A alone is:
$$\text{Work of A} = 1 - (B + C) = 1 - \frac{8}{23} = \mathbf{\frac{15}{23}}$$
Now compute B's share:
$$\text{Work of B} = 1 - (\text{Work of A} + \text{Work of C}) = 1 - \left(\frac{15}{23} + \frac{4}{23}\right) = 1 - \frac{19}{23} = \mathbf{\frac{4}{23}}$$
Check: $\frac{15}{23} + \frac{4}{23} + \frac{4}{23} = \frac{23}{23} = 1$ (Verified!).
Work done by A alone is:
$$\text{Work of A} = 1 - (B + C) = 1 - \frac{8}{23} = \mathbf{\frac{15}{23}}$$
Now compute B's share:
$$\text{Work of B} = 1 - (\text{Work of A} + \text{Work of C}) = 1 - \left(\frac{15}{23} + \frac{4}{23}\right) = 1 - \frac{19}{23} = \mathbf{\frac{4}{23}}$$
Check: $\frac{15}{23} + \frac{4}{23} + \frac{4}{23} = \frac{23}{23} = 1$ (Verified!).
Step 3: Calculate Monetary Share for Each Worker
2 Marks
Total Contract Amount $= ₹ 5,290$. Notice $5290 \div 23 = 230$.
$$\text{Share of A} = \frac{15}{23} \times 5290 = 15 \times 230 = \mathbf{₹ 3,450}$$
$$\text{Share of B} = \frac{4}{23} \times 5290 = 4 \times 230 = \mathbf{₹ 920}$$
$$\text{Share of C} = \frac{4}{23} \times 5290 = 4 \times 230 = \mathbf{₹ 920}$$
Check: $3450 + 920 + 920 = ₹ 5,290$.
$$\mathbf{\text{Answer: A gets } ₹\;3,450; \quad \text{B gets } ₹\;920; \quad \text{C gets } ₹\;920}$$
$$\text{Share of A} = \frac{15}{23} \times 5290 = 15 \times 230 = \mathbf{₹ 3,450}$$
$$\text{Share of B} = \frac{4}{23} \times 5290 = 4 \times 230 = \mathbf{₹ 920}$$
$$\text{Share of C} = \frac{4}{23} \times 5290 = 4 \times 230 = \mathbf{₹ 920}$$
Check: $3450 + 920 + 920 = ₹ 5,290$.
$$\mathbf{\text{Answer: A gets } ₹\;3,450; \quad \text{B gets } ₹\;920; \quad \text{C gets } ₹\;920}$$
Examiner Scoring Key: Finding C = 4/23 and A = 15/23 directly from 1 - (A+B) and 1 - (B+C) carries 5 marks. Computing exact rupee values carries 2 marks.
Model 5: Worker Leaving Before Scheduled Completion
10 Marks
A can complete a piece of work in $20\text{ days}$, and B can complete it in $30\text{ days}$. They start the work together, but B leaves $5\text{ days}$ before the work is finished. In how many total days was the entire work completed?
Step-by-Step Model Solution:
Step 1: Declare Variable for Total Duration
2.5 Marks
Let the total number of days taken to complete the entire work be $T\text{ days}$.
A worked for the entire duration of $T\text{ days}$.
B left $5\text{ days}$ before completion, so B worked for $(T - 5)\text{ days}$.
A worked for the entire duration of $T\text{ days}$.
B left $5\text{ days}$ before completion, so B worked for $(T - 5)\text{ days}$.
Step 2: Express Daily Rates and Formulate Work Equation (ATQ)
3 Marks
A's 1 day's work $= \frac{1}{20}$.
B's 1 day's work $= \frac{1}{30}$.
Total Work $= 1$:
$$\text{Work done by A} + \text{Work done by B} = 1$$
$$T \times \left(\frac{1}{20}\right) + (T - 5) \times \left(\frac{1}{30}\right) = 1$$
B's 1 day's work $= \frac{1}{30}$.
Total Work $= 1$:
$$\text{Work done by A} + \text{Work done by B} = 1$$
$$T \times \left(\frac{1}{20}\right) + (T - 5) \times \left(\frac{1}{30}\right) = 1$$
Step 3: Clear Denominators & Solve for T
2.5 Marks
Multiply the entire equation by $\text{LCM}(20, 30) = 60$:
$$60 \times \left(\frac{T}{20}\right) + 60 \times \left(\frac{T - 5}{30}\right) = 60 \times 1$$
$$3T + 2(T - 5) = 60$$
$$3T + 2T - 10 = 60$$
$$5T = 60 + 10 = 70 \implies T = \frac{70}{5} = \mathbf{14\text{ days}}$$
$$60 \times \left(\frac{T}{20}\right) + 60 \times \left(\frac{T - 5}{30}\right) = 60 \times 1$$
$$3T + 2(T - 5) = 60$$
$$3T + 2T - 10 = 60$$
$$5T = 60 + 10 = 70 \implies T = \frac{70}{5} = \mathbf{14\text{ days}}$$
Step 4: Verification & Conclude
2 Marks
A worked for $14\text{ days}$: $14 \times \frac{1}{20} = \frac{7}{10}$ work.
B worked for $14 - 5 = 9\text{ days}$: $9 \times \frac{1}{30} = \frac{3}{10}$ work.
Total work $= \frac{7}{10} + \frac{3}{10} = \frac{10}{10} = 1$ (Verified!).
$$\mathbf{\text{Answer: The entire work was completed in } 14\text{ days.}}$$
B worked for $14 - 5 = 9\text{ days}$: $9 \times \frac{1}{30} = \frac{3}{10}$ work.
Total work $= \frac{7}{10} + \frac{3}{10} = \frac{10}{10} = 1$ (Verified!).
$$\mathbf{\text{Answer: The entire work was completed in } 14\text{ days.}}$$
Examiner Scoring Key: Setting up T/20 + (T-5)/30 = 1 is the cleanest algebraic approach carrying 5.5 marks. Solving for T = 14 days earns full 10 marks.
Model 6: Dual Inlets with an Emptying Drainage Pipe
10 Marks
Two pipes A and B can fill a military reservoir in $12\text{ hours}$ and $16\text{ hours}$ respectively. A waste pipe C can empty the completely filled reservoir in $8\text{ hours}$.
(a) If all three pipes are opened simultaneously when the reservoir is empty, how long will it take to fill the reservoir completely?
(b) If Pipe C is kept open for the first $4\text{ hours}$ and then closed, in how much additional time will Pipes A and B fill the remaining reservoir?
(a) If all three pipes are opened simultaneously when the reservoir is empty, how long will it take to fill the reservoir completely?
(b) If Pipe C is kept open for the first $4\text{ hours}$ and then closed, in how much additional time will Pipes A and B fill the remaining reservoir?
Step-by-Step Model Solution:
Step 1: Compute Hourly Rates of All Three Pipes
2.5 Marks
Pipe A: fills $\frac{1}{12}$ reservoir/hour ($+1/12$).
Pipe B: fills $\frac{1}{16}$ reservoir/hour ($+1/16$).
Pipe C: empties $\frac{1}{8}$ reservoir/hour ($-1/8$).
Pipe B: fills $\frac{1}{16}$ reservoir/hour ($+1/16$).
Pipe C: empties $\frac{1}{8}$ reservoir/hour ($-1/8$).
Step 2: Solve Part (a) - All Three Pipes Open
3 Marks
Net rate when all three pipes are open:
$$\text{Net Rate} = \frac{1}{12} + \frac{1}{16} - \frac{1}{8}$$
$\text{LCM}(12, 16, 8) = 48$:
$$\text{Net Rate} = \frac{4 + 3 - 6}{48} = \frac{7 - 6}{48} = \frac{1}{48}\text{ reservoir per hour}$$
$$\text{Total Time to fill (Part a)} = \frac{1}{1/48} = \mathbf{48\text{ hours}}$$
$$\text{Net Rate} = \frac{1}{12} + \frac{1}{16} - \frac{1}{8}$$
$\text{LCM}(12, 16, 8) = 48$:
$$\text{Net Rate} = \frac{4 + 3 - 6}{48} = \frac{7 - 6}{48} = \frac{1}{48}\text{ reservoir per hour}$$
$$\text{Total Time to fill (Part a)} = \frac{1}{1/48} = \mathbf{48\text{ hours}}$$
Step 3: Solve Part (b) - First 4 Hours with Pipe C
2.5 Marks
In the first $4\text{ hours}$, all three pipes are open at net rate $\frac{1}{48}$:
$$\text{Water filled in 4 hours} = 4 \times \frac{1}{48} = \frac{4}{48} = \frac{1}{12}$$
$$\text{Remaining reservoir to be filled} = 1 - \frac{1}{12} = \frac{11}{12}$$
$$\text{Water filled in 4 hours} = 4 \times \frac{1}{48} = \frac{4}{48} = \frac{1}{12}$$
$$\text{Remaining reservoir to be filled} = 1 - \frac{1}{12} = \frac{11}{12}$$
Step 4: Additional Time with Only Pipes A and B Operating
2 Marks
When Pipe C is closed, only A and B fill:
$$\text{Combined Rate of (A + B)} = \frac{1}{12} + \frac{1}{16} = \frac{4 + 3}{48} = \frac{7}{48}$$
$$\text{Additional Time} = \frac{\text{Remaining Volume}}{\text{Rate of (A + B)}} = \frac{11/12}{7/48} = \frac{11}{12} \times \frac{48}{7} = \frac{11 \times 4}{7} = \frac{44}{7} = \mathbf{6\frac{2}{7}\text{ hours}}$$
$$\mathbf{\text{Answer: (a) } 48\text{ hours}; \quad \text{(b) Additional time = } 6\frac{2}{7}\text{ hours (or } 6\text{ h } 17\text{ m } 9\text{ s})}}$$
$$\text{Combined Rate of (A + B)} = \frac{1}{12} + \frac{1}{16} = \frac{4 + 3}{48} = \frac{7}{48}$$
$$\text{Additional Time} = \frac{\text{Remaining Volume}}{\text{Rate of (A + B)}} = \frac{11/12}{7/48} = \frac{11}{12} \times \frac{48}{7} = \frac{11 \times 4}{7} = \frac{44}{7} = \mathbf{6\frac{2}{7}\text{ hours}}$$
$$\mathbf{\text{Answer: (a) } 48\text{ hours}; \quad \text{(b) Additional time = } 6\frac{2}{7}\text{ hours (or } 6\text{ h } 17\text{ m } 9\text{ s})}}$$
Examiner Scoring Key: In part (a), net rate 1/48 yields 48 hours. In part (b), remaining volume 11/12 divided by 7/48 yields 44/7 = 6 2/7 hours. Award 5 marks per part.
RIMC Practice & Drill Exercises
Attempt these questions on paper following proper step-wise conventions, then click Reveal Solution to verify your answers.
Question 1
5 Marks
A and B together can do a piece of work in $12\text{ days}$, while B alone can finish it in $30\text{ days}$. In how many days can A alone finish the work?
Model Descriptive Solution:
Step 1: State Combined & B's Work Rate
2.5 Marks
$$(A + B)\text{'s 1 day's work} = \frac{1}{12}, \quad \text{B's 1 day's work} = \frac{1}{30}$$
Step 2: Isolate A's Rate
2.5 Marks
$$\text{A's 1 day's work} = \frac{1}{12} - \frac{1}{30} = \frac{5 - 2}{60} = \frac{3}{60} = \frac{1}{20}$$
A alone finishes in $20\text{ days}$.
A alone finishes in $20\text{ days}$.
Final Answer: $20\text{ days}$
Question 2
5 Marks
A pipe can fill a water tank in $8\text{ hours}$ and another pipe can empty it in $12\text{ hours}$. If both pipes are opened simultaneously in an empty tank, in how many hours will the tank be full?
Model Descriptive Solution:
Step 1: Net Hourly Rate
2.5 Marks
$$\text{Net Rate} = \frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24}$$
Step 2: Total Time
2.5 Marks
$$\text{Time} = \frac{1}{1/24} = 24\text{ hours}$$
Final Answer: $24\text{ hours}$
Question 3
5 Marks
A is twice as good a workman as B, and together they finish a piece of work in $14\text{ days}$. In how many days can A alone finish the work?
Model Descriptive Solution:
Step 1: Set Up Efficiency Ratio
2.5 Marks
Let B's 1 day's work be $x$. A's 1 day's work $= 2x$.
Together: $2x + x = 3x = \frac{1}{14} \implies x = \frac{1}{42}$.
Together: $2x + x = 3x = \frac{1}{14} \implies x = \frac{1}{42}$.
Step 2: Compute A's Time
2.5 Marks
A's 1 day's work $= 2x = 2 \times \frac{1}{42} = \frac{1}{21}$.
A alone takes $21\text{ days}$.
A alone takes $21\text{ days}$.
Final Answer: $21\text{ days}$
Question 4
7 Marks
A can do a work in $15\text{ days}$ and B in $20\text{ days}$. If they work on it together for $4\text{ days}$, what fraction of the work is left unfinished?
Model Descriptive Solution:
Step 1: Combined Daily Rate
3 Marks
$$(A + B)\text{'s 1 day} = \frac{1}{15} + \frac{1}{20} = \frac{4 + 3}{60} = \frac{7}{60}$$
Step 2: Work Done in 4 Days and Remaining Fraction
4 Marks
$$\text{Work in 4 days} = 4 \times \frac{7}{60} = \frac{7}{15}$$
$$\text{Fraction Left} = 1 - \frac{7}{15} = \frac{8}{15}$$
$$\text{Fraction Left} = 1 - \frac{7}{15} = \frac{8}{15}$$
Final Answer: $\frac{8}{15}\text{ of the work}$
Question 5
7 Marks
A and B can complete a work in $18\text{ days}$ and $12\text{ days}$ respectively. A started the work alone and then after $6\text{ days}$, B joined him till the completion of the work. How many total days did the work take?
Model Descriptive Solution:
Step 1: Work Done by A in 6 Days
2.5 Marks
A's 6 days' work $= 6 \times \frac{1}{18} = \frac{1}{3}$.
Remaining work $= 1 - \frac{1}{3} = \frac{2}{3}$.
Remaining work $= 1 - \frac{1}{3} = \frac{2}{3}$.
Step 2: Time for (A + B) to Finish Remaining Work
3 Marks
$$(A + B)\text{'s 1 day} = \frac{1}{18} + \frac{1}{12} = \frac{2 + 3}{36} = \frac{5}{36}$$
$$\text{Time} = \frac{2/3}{5/36} = \frac{2}{3} \times \frac{36}{5} = \frac{24}{5} = 4.8\text{ days}$$
$$\text{Time} = \frac{2/3}{5/36} = \frac{2}{3} \times \frac{36}{5} = \frac{24}{5} = 4.8\text{ days}$$
Step 3: Total Duration
1.5 Marks
$$\text{Total Time} = 6 + 4.8 = 10.8\text{ days} = 10\frac{4}{5}\text{ days}$$
Final Answer: $10\frac{4}{5}\text{ days}$ (or $10.8\text{ days}$)
Question 6
7 Marks
Three taps A, B, and C can fill an overhead tank in $12, 15,$ and $20\text{ hours}$ respectively. If tap A is open all the time and B and C are open for one hour each alternately, in how many hours will the tank be full?
Model Descriptive Solution:
Step 1: Rates of the Two Alternating Pairs
3 Marks
Hour 1: $(A + B)$ open $\implies \frac{1}{12} + \frac{1}{15} = \frac{5 + 4}{60} = \frac{9}{60} = \frac{3}{20}$.
Hour 2: $(A + C)$ open $\implies \frac{1}{12} + \frac{1}{20} = \frac{5 + 3}{60} = \frac{8}{60} = \frac{2}{15}$.
Hour 2: $(A + C)$ open $\implies \frac{1}{12} + \frac{1}{20} = \frac{5 + 3}{60} = \frac{8}{60} = \frac{2}{15}$.
Step 2: Work in One 2-Hour Cycle
2 Marks
$$\text{Work in 2 hours} = \frac{9}{60} + \frac{8}{60} = \frac{17}{60}$$
Step 3: Solve for Total Time
2 Marks
In 3 cycles ($6\text{ hours}$): $3 \times \frac{17}{60} = \frac{51}{60}$.
Remaining $= 1 - \frac{51}{60} = \frac{9}{60}$.
In 7th hour, $(A + B)$ operates and fills exactly $\frac{9}{60}$.
Total time $= 6 + 1 = 7\text{ hours}$.
Remaining $= 1 - \frac{51}{60} = \frac{9}{60}$.
In 7th hour, $(A + B)$ operates and fills exactly $\frac{9}{60}$.
Total time $= 6 + 1 = 7\text{ hours}$.
Final Answer: $7\text{ hours}$
Question 7
7 Marks
A, B, and C can complete a task in $10, 20,$ and $30\text{ days}$ respectively. They worked together and earned ₹ $1,100$. Find the share of each.
Model Descriptive Solution:
Step 1: Ratio of Daily Work Rates
3.5 Marks
$$\text{Ratio} = \frac{1}{10} : \frac{1}{20} : \frac{1}{30}$$
Multiply by $\text{LCM}(10, 20, 30) = 60$:
$$\text{Ratio} = 6 : 3 : 2$$
Sum of terms $= 6 + 3 + 2 = 11$.
Multiply by $\text{LCM}(10, 20, 30) = 60$:
$$\text{Ratio} = 6 : 3 : 2$$
Sum of terms $= 6 + 3 + 2 = 11$.
Step 2: Allocate Wages
3.5 Marks
A $= \frac{6}{11} \times 1100 = ₹ 600$.
B $= \frac{3}{11} \times 1100 = ₹ 300$.
C $= \frac{2}{11} \times 1100 = ₹ 200$.
B $= \frac{3}{11} \times 1100 = ₹ 300$.
C $= \frac{2}{11} \times 1100 = ₹ 200$.
Final Answer: A: ₹ $600$; B: ₹ $300$; C: ₹ $200$
Question 8
10 Marks
A can do a piece of work in $24\text{ days}$, B in $32\text{ days}$, and C in $64\text{ days}$. All begin to do it together, but A leaves after $6\text{ days}$ and B leaves $6\text{ days}$ before the completion of the work. How many days did the work last?
Model Descriptive Solution:
Step 1: Declare Variable for Total Duration
3 Marks
Let total days be $T$.
A works for $6\text{ days}$.
B works for $(T - 6)\text{ days}$.
C works for the entire $T\text{ days}$.
A works for $6\text{ days}$.
B works for $(T - 6)\text{ days}$.
C works for the entire $T\text{ days}$.
Step 2: Set Up Work Sum Equation
4 Marks
$$\frac{6}{24} + \frac{T - 6}{32} + \frac{T}{64} = 1$$
Multiply by $\text{LCM}(24, 32, 64) = 192$:
$$8(6) + 6(T - 6) + 3T = 192$$
$$48 + 6T - 36 + 3T = 192 \implies 9T + 12 = 192$$
Multiply by $\text{LCM}(24, 32, 64) = 192$:
$$8(6) + 6(T - 6) + 3T = 192$$
$$48 + 6T - 36 + 3T = 192 \implies 9T + 12 = 192$$
Step 3: Solve for T
3 Marks
$$9T = 192 - 12 = 180 \implies T = \frac{180}{9} = 20\text{ days}$$
Final Answer: $20\text{ days}$
Question 9
10 Marks
Two pipes A and B can fill a cistern in $37\frac{1}{2}\text{ minutes}$ and $45\text{ minutes}$ respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the pipe B is turned off after how many minutes?
Model Descriptive Solution:
Step 1: Compute Rates
3 Marks
Pipe A: $37\frac{1}{2} = \frac{75}{2}\text{ mins} \implies \text{Rate} = \frac{2}{75}\text{ per min}$.
Pipe B: $45\text{ mins} \implies \text{Rate} = \frac{1}{45}\text{ per min}$.
Pipe B: $45\text{ mins} \implies \text{Rate} = \frac{1}{45}\text{ per min}$.
Step 2: Tank Filled in 30 Mins
4 Marks
Pipe A runs for all $30\text{ minutes}$. Let B run for $t\text{ minutes}$.
$$30 \times \frac{2}{75} + t \times \frac{1}{45} = 1$$
$$\frac{60}{75} + \frac{t}{45} = 1 \implies \frac{4}{5} + \frac{t}{45} = 1$$
$$30 \times \frac{2}{75} + t \times \frac{1}{45} = 1$$
$$\frac{60}{75} + \frac{t}{45} = 1 \implies \frac{4}{5} + \frac{t}{45} = 1$$
Step 3: Solve for t
3 Marks
$$\frac{t}{45} = 1 - \frac{4}{5} = \frac{1}{5} \implies t = \frac{45}{5} = 9\text{ minutes}$$
Final Answer: $9\text{ minutes}$
Question 10
10 Marks
A cistern has a leak which would empty it in $8\text{ hours}$. A tap is turned on which admits $6\text{ litres}$ of water a minute into the cistern, and it is now emptied in $12\text{ hours}$. How many litres does the cistern hold?
Model Descriptive Solution:
Step 1: Compute Hourly Rate of the Inlet Tap
4 Marks
Leak alone empties in $8\text{ hours} \implies -\frac{1}{8}\text{ per hour}$.
Leak + Inlet empties in $12\text{ hours} \implies -\frac{1}{12}\text{ per hour}$.
$$\text{Inlet Rate} = \left(-\frac{1}{12}\right) - \left(-\frac{1}{8}\right) = \frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24}\text{ per hour}$$
Inlet tap fills the empty tank in $24\text{ hours}$.
Leak + Inlet empties in $12\text{ hours} \implies -\frac{1}{12}\text{ per hour}$.
$$\text{Inlet Rate} = \left(-\frac{1}{12}\right) - \left(-\frac{1}{8}\right) = \frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24}\text{ per hour}$$
Inlet tap fills the empty tank in $24\text{ hours}$.
Step 2: Convert Inflow Rate to Litres per Hour
3 Marks
Inflow $= 6\text{ litres/minute} = 6 \times 60 = 360\text{ litres/hour}$.
Step 3: Calculate Capacity
3 Marks
$$\text{Capacity} = 24\text{ hours} \times 360\text{ litres/hour} = 8,640\text{ litres}$$
Final Answer: $8,640\text{ litres}$
High-Yield FAQs & Revision Notes
Why is total work assumed to be 1?
1 represents 100% of the completed job. The work done in a single day is expressed as a proper fraction of this total (e.g., 1/10).
What is the relationship between efficiency and time?
Efficiency and time are inversely proportional. A person who is twice as efficient takes half as much time to do the same task.
How are wages divided among workers?
Strictly in the ratio of the total work completed by each person during the project.
How do you handle alternate day problems?
Group days into 2-day cycles, find the work completed per cycle, fit as many whole cycles as possible, and calculate the fraction remaining on subsequent days.
Why do drainage pipes have a negative rate?
Because they subtract water from the cistern, working against the inlet filling pipes.
How do you solve pairwise problems (A+B, B+C, C+A)?
Add all three equations to get $2(A+B+C)$, divide by 2 to get $(A+B+C)$, then subtract the appropriate pair to isolate any individual.
What is the LCM method for time and work?
Assuming total work equals the LCM of individual days converts fractional daily work rates into easy integer daily units.
What is the formula when worker B leaves k days before completion?
Let total time be $T$. Then $T(1/a) + (T - k)(1/b) = 1$.
How do you find the capacity of a tank from filling rate in litres/min?
Find the time $t$ (in hours) the inlet tap needs to fill the tank, convert the flow rate to litres/hour, and multiply by $t$.
How does this module map to the official RIMC syllabus?
Directly covers Serial No. 23 ('Time and work') of Chapter 8 broader syllabus in the official RIMC brochure.
How does this topic relate to the official RIMC syllabus?
This module directly covers Serial No. 23 of the broader syllabus published in Chapter 8 of the official RIMC admission brochure, ensuring 100% syllabus alignment.
Last-Minute Cadet High-Yield Revision Matrix
| Topic / Concept | High-Yield Exam Rules & Memory Anchor |
|---|---|
| Work Rate | If completed in $n$ days, 1 day's work $= 1/n$. Never add days directly. |
| Product/Sum | Two workers together: $\text{Time} = \frac{ab}{a+b}$. |
| Pairwise System | $2(A+B+C) = (A+B) + (B+C) + (C+A)$. Divide by 2! |
| Departure Before End | $T(1/a) + (T - k)(1/b) = 1$ where $T$ is total project duration. |
| Pipes & Cisterns | Inlet is positive ($+1/t_1$); Leak is negative ($-1/t_2$). |
| Wages Distribution | Divide money in the exact ratio of the work done by each person. |