Master the fundamental relationship between time spent and work completed, individual and combined unitary work rates, efficiency proportions, worker departure/arrival scenarios, and negative work in pipes and cisterns.
2.3 Time and Work: Fundamental Concepts
The physical or mental effort directed towards doing something refers to work. In simple mathematical language, work is an outcome of an effort applied for a particular time.
Direct Proportionality: Work is directly proportional to time ($\text{Work} \propto \text{Time}$). That is, more time given for work will increase the total work done.
If the amount of work (i.e. a task) remains constant, then two things play a vital role in determining the duration:
Number of persons working
Number of hours worked per day
Inverse Principle: In a task of constant magnitude, if the number of persons increases, the time required decreases, and vice versa.
Relation of Time and Work Done
To mathematically quantify time and work, we adopt the Unitary Rate Method where a complete job is treated as $1$ unit of work:
Let a complete piece of work be denoted by $W$.
If Ravi does work $W$ in $n$ days, then work done by Ravi in $1\text{ day} = \frac{1}{n}\text{ of } W = \frac{W}{n}$.
If Nitish does the same work $W$ in $m$ days, then work done by Nitish in $1\text{ day} = \frac{1}{m}\text{ of } W = \frac{W}{m}$.
If Ravi and Nitish work together and finish the entire work in $p$ days, then work finished by both in $1\text{ day} = \frac{1}{p}\text{ of } W$.
$$\text{Therefore, } \frac{W}{n} + \frac{W}{m} = \frac{W}{p}$$
Dividing both sides by $W$:
$$\frac{1}{n} + \frac{1}{m} = \frac{1}{p} \quad \ldots(1)$$
Hence, the total number of days $p$ taken by both of them working together is:
$$\mathbf{p = \frac{mn}{m + n}} \quad \ldots(2)$$
Generalization for Multiple Persons:
If $a$ persons work together for the same task taking $n_1, n_2, \dots, n_a$ days respectively, the combined rate becomes:
$$\frac{1}{n_1} + \frac{1}{n_2} + \dots + \frac{1}{n_a} = \frac{1}{p}$$
where $n_a$ represents the number of days taken by the $a^{\text{th}}$ person alone.
Example 6 (CBSE Support Material)
Reena completes her canvas painting in 4 days. Mihir finishes the same canvas painting in 5 days. If they both work together, find out the number of days taken by them to finish it.
Step 1: Identify individual time spans:
Time taken by Reena alone ($m$) = $4\text{ days}$.
Time taken by Mihir alone ($n$) = $5\text{ days}$.
Step 2: Apply combined time formula:
$$\text{Number of days taken together } (p) = \frac{m \times n}{m + n} = \frac{4 \times 5}{4 + 5} = \frac{20}{9}\text{ days}$$
$$\mathbf{p = 2\frac{2}{9}\text{ days}}$$
Negative Work and Pipes & Cisterns
In practical problems, not all agents contribute towards completion. Some agents work against progress.
Definition of Negative Work:
If two objects or persons are working against each other due to which a delay or incompletion in task occurs, it is referred to as negative work.
Key convention:
Inlet pipe (filling tank) $\implies$ Positive Work Rate ($+\frac{1}{A}$)
Outlet pipe or leakage (emptying tank) $\implies$ Negative Work Rate ($-\frac{1}{B}$)
Net filling rate per hour: $\mathbf{\frac{1}{\text{Net Time}} = \frac{1}{\text{Inlet}} - \frac{1}{\text{Outlet}}}$
Example 7 (CBSE Support Material)
Pipe A can fill a tank in 20 hours. But due to a leakage it is taking thrice the time to fill the tank. If the tank is full, how long will it take to drain out the tank, if the pipe A is closed?
Solution:
Let the initial time taken by pipe A alone $= x\text{ hours}$ (given $x = 20\text{ hours}$).
Water filled by Pipe A in 1 hour $= \frac{1}{x} = \frac{1}{20}$.
After the leakage develops, the effective time taken to fill the tank becomes thrice $= 3x\text{ hours}$.
$\implies$ Net water filled in tank per hour with leakage $= \frac{1}{3x} = \frac{1}{60}$.
Let the full tank get emptied by the leak in $p$ hours.
Water drained out by leak in 1 hour $= \frac{1}{p}$.
$$\text{Net filling rate} = \text{Filling rate of Pipe A} - \text{Emptying rate of Leak}$$
$$\frac{1}{3x} = \frac{1}{x} - \frac{1}{p} \implies \frac{1}{p} = \frac{1}{x} - \frac{1}{3x} = \frac{2}{3x}$$
$$p = \frac{3x}{2}\text{ hours} = \frac{3}{2} \times 20 = \mathbf{30\text{ hours}}$$
Conclusion: The tank is completely drained out in 30 hours.
Work Efficiency and Man-Day Principles
1. Efficiency - Time Reciprocal Relationship:
Efficiency ($\eta$) is the rate of doing work per unit time.
$$\text{Efficiency} \propto \frac{1}{\text{Time taken}}$$
If Person A is $k$ times as efficient as Person B:
$$\frac{\text{Efficiency of A}}{\text{Efficiency of B}} = \frac{k}{1} \implies \frac{\text{Time taken by A}}{\text{Time taken by B}} = \frac{1}{k}$$
$$\text{Time taken by A} = \frac{1}{k} \times \text{Time taken by B}$$
2. Chain Rule (Compound Proportion):
For a group of $M_1$ persons working $D_1$ days for $H_1$ hours each to produce $W_1$ work, and $M_2$ persons working $D_2$ days for $H_2$ hours to produce $W_2$ work:
$$\mathbf{\frac{M_1 \times D_1 \times H_1}{W_1} = \frac{M_2 \times D_2 \times H_2}{W_2}}$$
When the work is identical ($W_1 = W_2$), the total person-hours are invariant:
$$M_1 D_1 H_1 = M_2 D_2 H_2$$
Check Your Progress 2.3
Official CBSE Applied Mathematics Support Material Questions with Full Step-by-Step Solutions
Question 1Efficiency & Combined Work
Navya is twice as efficient as Nitti. If they take 10 days to finish a certain job together, how much time will they take individually to finish the same job?
Step-by-step Solution: 1. Establish efficiency and rates:
Let Nitti's rate of work per day $= x$.
Since Navya is twice as efficient as Nitti, Navya's rate of work per day $= 2x$.
2. Combined daily work:
Working together, in 1 day they complete:
$$\text{Combined Rate} = x + 2x = 3x$$
3. Equate to given combined duration:
They take 10 days to complete the entire job together $\implies \text{Combined Rate} = \frac{1}{10}$.
$$3x = \frac{1}{10} \implies x = \frac{1}{30}$$
4. Find individual durations:
Final Answer: Navya takes 15 days; Nitti takes 30 days.
Question 2Multiple Worker Proportions
A piece of work is finished in 30 days by A. Since C is thrice as good as A and A is twice as good as B. If A, B, C work together, then how many days will they take to finish the work?
Step-by-step Solution: 1. Determine individual daily work rates:
A finishes the work in 30 days $\implies \text{Rate of A} = \frac{1}{30}$ work/day.
C is thrice as good as A $\implies \text{Rate of C} = 3 \times \frac{1}{30} = \frac{3}{30} = \frac{1}{10}$ work/day.
A is twice as good as B $\implies \text{Rate of A} = 2 \times \text{Rate of B} \implies \text{Rate of B} = \frac{1}{2} \times \frac{1}{30} = \frac{1}{60}$ work/day.
2. Combined daily work rate of A, B, and C:
$$\text{Combined Rate} = \text{Rate of A} + \text{Rate of B} + \text{Rate of C}$$
$$\text{Combined Rate} = \frac{1}{30} + \frac{1}{60} + \frac{1}{10} = \frac{2 + 1 + 6}{60} = \frac{9}{60} = \frac{3}{20}\text{ work/day}$$
3. Total time taken together:
$$\text{Days taken} = \frac{1}{3/20} = \frac{20}{3}\text{ days} = \mathbf{6\frac{2}{3}\text{ days}} \quad (\approx 6\text{ days } 16\text{ hours})$$
Question 3Worker Leaving Before Completion
X can do a piece of work in 60 days, whereas Y can do the same work in 40 days. Both started the work together, but X left 10 days before the completion of work. Find how many days will it take to complete the work?
Step-by-step Solution: 1. Individual 1-day rates:
Rate of X $= \frac{1}{60}$; Rate of Y $= \frac{1}{40}$.
2. Let total duration to complete the project be $T$ days:
Y worked for the entire project duration, i.e., $T$ days.
X left 10 days before completion, so X worked for $(T - 10)$ days.
3. Formulate total work equation:
$$\text{Work done by X} + \text{Work done by Y} = 1$$
$$(T - 10) \times \frac{1}{60} + T \times \frac{1}{40} = 1$$
LCM of 60 and 40 is 120. Multiply both sides by 120:
$$2(T - 10) + 3T = 120$$
$$2T - 20 + 3T = 120$$
$$5T = 140 \implies \mathbf{T = 28\text{ days}}$$
Verification: In 28 days, Y did $\frac{28}{40} = \frac{7}{10}$. X worked for $28 - 10 = 18$ days, doing $\frac{18}{60} = \frac{3}{10}$. Total $= \frac{7}{10} + \frac{3}{10} = 1$. Matches perfectly! Final Answer: The entire work will take 28 days to complete.
Question 4Project Workload & Departures
100 persons begin to work together on a project which was expected to be completed in 40 days. But after a few days 40 persons left. As a result, the project got delayed by 10 days. How many days after the commencement of the project did the 40 persons leave?
Step-by-step Solution: 1. Compute total planned person-days:
$$\text{Total Work} = 100 \text{ persons} \times 40 \text{ days} = 4000 \text{ person-days}$$
2. Determine actual duration and remaining workforce:
Due to the departure of 40 persons, the project was delayed by 10 days.
$$\text{Actual Total Time} = 40 + 10 = 50\text{ days}$$
After 40 persons left, the remaining workforce was $100 - 40 = 60\text{ persons}$.
3. Set up the linear person-days balance:
Let the 40 persons leave after $x$ days from commencement.
For the first $x$ days, all 100 persons worked $\implies 100x$ person-days.
For the remaining $(50 - x)$ days, only 60 persons worked $\implies 60(50 - x)$ person-days.
$$100x + 60(50 - x) = 4000$$
$$100x + 3000 - 60x = 4000$$
$$40x = 1000 \implies \mathbf{x = \frac{1000}{40} = 25\text{ days}}$$
Final Answer: The 40 persons left 25 days after the commencement of the project.
Question 5Three-Variable Efficiency Ratio
The efficiency of $x, y, z$ are in the ratio of $3:2:6$ to finish a task. If they work together, they can finish it in 2 hours; find the time taken by them if they do the task individually?
Step-by-step Solution: 1. Define hourly output units using common ratio $k$:
Let hourly efficiency of $x = 3k$ units/hr.
Hourly efficiency of $y = 2k$ units/hr.
Hourly efficiency of $z = 6k$ units/hr.
2. Combined hourly efficiency:
$$\text{Combined Efficiency} = 3k + 2k + 6k = 11k \text{ units/hr}$$
3. Total work required:
Working together, they complete the entire task in 2 hours:
$$\text{Total Work} = 11k \times 2 = 22k \text{ units}$$
4. Individual time taken:
Time for $x$: $\frac{22k}{3k} = \frac{22}{3}\text{ hours} = \mathbf{7\frac{1}{3}\text{ hours}}$ (7 hours 20 minutes).
Time for $y$: $\frac{22k}{2k} = \mathbf{11\text{ hours}}$.
Time for $z$: $\frac{22k}{6k} = \frac{11}{3}\text{ hours} = \mathbf{3\frac{2}{3}\text{ hours}}$ (3 hours 40 minutes).
Final Answer: Individually, $x$ takes $7\frac{1}{3}$ hours (7h 20m), $y$ takes 11 hours, and $z$ takes $3\frac{2}{3}$ hours (3h 40m).
Question 6Relative Efficiency & Time Difference
A is three times as efficient as B. Also, A takes 30 days less than B for doing a piece of work. Find the time taken by them if they work: (a) individually (b) together
Step-by-step Solution: 1. Formulate time variables based on efficiency:
Since efficiency is inversely proportional to time:
$$\frac{\text{Efficiency of A}}{\text{Efficiency of B}} = \frac{3}{1} \implies \frac{\text{Time taken by A}}{\text{Time taken by B}} = \frac{1}{3}$$
Let the time taken by B alone be $t$ days. Then the time taken by A alone is $\frac{t}{3}$ days.
2. Use the time difference condition:
A takes 30 days less than B:
$$t - \frac{t}{3} = 30 \implies \frac{2t}{3} = 30 \implies 2t = 90 \implies t = 45\text{ days}$$
(a) Individual Times:
Time taken by B alone: $t = \mathbf{45\text{ days}}$.
Time taken by A alone: $\frac{45}{3} = \mathbf{15\text{ days}}$ (which is $45 - 30 = 15$ days).
(b) Working Together:
$$\text{Combined Time } (p) = \frac{A \times B}{A + B} = \frac{15 \times 45}{15 + 45} = \frac{675}{60} = \frac{45}{4}\text{ days} = \mathbf{11.25\text{ days}} = \mathbf{11\frac{1}{4}\text{ days}}$$
Final Answer: (a) Individually: A takes 15 days, B takes 45 days. (b) Together: $11\frac{1}{4}$ days (11 days 6 hours).
Question 7Percentage Efficiency & Production Order
Machine P is 40% more efficient than Machine Q. Machine P can make 100 bags alone in 30 hours. Find the time taken to complete the order if both the machines work together?
Step-by-step Solution: 1. Relate the efficiencies of Machine P and Machine Q:
Efficiency of Machine P is $40\%$ higher than Machine Q:
$$\text{Efficiency of P} = (1 + 0.40) \times \text{Efficiency of Q} = 1.4 \times \text{Efficiency of Q} = \frac{7}{5} \times \text{Efficiency of Q}$$
Since time is inversely proportional to efficiency:
$$\frac{\text{Time taken by Q}}{\text{Time taken by P}} = 1.4 \implies \text{Time taken by Q} = 1.4 \times \text{Time taken by P}$$
$$\text{Time taken by Q} = 1.4 \times 30 = 42\text{ hours}$$
2. Compute combined rate for 100 bags:
In 1 hour:
Machine P completes $\frac{1}{30}$ of the 100-bag order.
Machine Q completes $\frac{1}{42}$ of the 100-bag order.
$$\text{Combined Rate} = \frac{1}{30} + \frac{1}{42} = \frac{7 + 5}{210} = \frac{12}{210} = \frac{2}{35}\text{ of order per hour}$$
3. Time taken working together:
$$\text{Time} = \frac{1}{2/35} = \frac{35}{2}\text{ hours} = \mathbf{17.5\text{ hours}} = \mathbf{17\text{ hours } 30\text{ minutes}}$$
Final Answer: Both machines together will take 17.5 hours (17 hours 30 minutes) to complete the 100-bag order.
Targeted Practice Worksheet
Comprehensive multi-section drill problems structured according to CBSE board exam patterns.
Section A: Unitary Rate & Basic Combinations
WS.1: Three Worker Team
Anil, Bimal, and Chetan can complete an accounting audit in 12 days, 15 days, and 20 days respectively. How many days will they take to complete the audit if all three work together?
Solution:
Combined 1-day work $= \frac{1}{12} + \frac{1}{15} + \frac{1}{20}$.
LCM of 12, 15, 20 is 60.
Combined Rate $= \frac{5 + 4 + 3}{60} = \frac{12}{60} = \frac{1}{5}$.
Total days taken $= 5\text{ days}$.
WS.2: Finding Missing Partner's Time
P and Q together can finish a coding project in 8 days. If P alone can finish it in 12 days, in how many days can Q alone finish the same project?
A and B undertake to do a piece of work for ₹4,800. A alone can do it in 6 days while B alone can do it in 8 days. With the help of C, they finish it in 3 days. What is C's share of the wage?
Solution:
Total work in 1 day by A, B, C $= \frac{1}{3}$.
Rate of C $= \frac{1}{3} - \left(\frac{1}{6} + \frac{1}{8}
ight) = \frac{1}{3} - \frac{7}{24} = \frac{8 - 7}{24} = \frac{1}{24}$.
Ratio of 1-day work (A : B : C) $= \frac{1}{6} : \frac{1}{8} : \frac{1}{24} = 4 : 3 : 1$.
Total ratio parts $= 4 + 3 + 1 = 8$.
C's share $= \frac{1}{8} \times 4800 = \mathbf{₹600}$.
Section C: Work on Alternate Days
WS.4: Alternate Day Working
A can do a piece of work in 9 days and B in 12 days. If they work on alternate days with A beginning on the first day, in how many days will the work be completed?
Solution:
LCM of 9 and 12 = 36 units (Total Work).
A's daily work $= 36/9 = 4$ units. B's daily work $= 36/12 = 3$ units.
In a 2-day cycle (A on Day 1, B on Day 2), work completed $= 4 + 3 = 7$ units.
In 5 cycles (10 days), work completed $= 5 \times 7 = 35$ units.
Remaining work $= 36 - 35 = 1$ unit.
On Day 11, it is A's turn (efficiency 4 units/day). Time taken $= \frac{1}{4}$ day.
Total days $= 10 + \frac{1}{4} = \mathbf{10\frac{1}{4}\text{ days}}$.
Section D: Pipes and Cisterns Inlets & Outlets
WS.5: Two Inlets and One Waste Pipe
Two pipes A and B can fill a cistern in 15 minutes and 20 minutes respectively. A waste pipe C can empty it in 30 minutes. If all three pipes are opened together, in how many minutes will the cistern be full?
Solution:
Net rate in 1 minute $= \frac{1}{15} + \frac{1}{20} - \frac{1}{30}$.
LCM of 15, 20, 30 is 60.
Net rate $= \frac{4 + 3 - 2}{60} = \frac{5}{60} = \frac{1}{12}$ cistern/min.
Total time $= \mathbf{12\text{ minutes}}$.
High-Yield Revision Sheet & Formula Cheat-Sheet
Quick-reference formulas, inverse laws, and key examination traps to review before tests:
Concept
Mathematical Formula
Key Notes
Unitary Rate
$$\text{Rate} = \frac{1}{n} \text{ work/day}$$
If a person finishes a job in $n$ days, their 1-day output is reciprocal of $n$.
Two-Person Combined Time
$$p = \frac{mn}{m + n}$$
Harmonic product-over-sum rule for two agents working simultaneously.
Multi-Person Combined Time
$$\frac{1}{p} = \sum_{i=1}^k \frac{1}{n_i}$$
Sum individual daily rates to find the combined daily rate, then invert.
Efficiency & Time
$$\frac{\eta_A}{\eta_B} = \frac{t_B}{t_A}$$
Higher efficiency means strictly shorter completion time (inverse relation).
Treat emptying pipes and leaks with negative sign in summation.
Common Examination Pitfalls to Avoid:
Never add days directly! If A takes 4 days and B takes 5 days, together they DO NOT take $4 + 5 = 9$ days. Always add their rates ($\frac{1}{4} + \frac{1}{5}$).
Left $x$ days before completion: Express the worker's working time as $(T - x)$ where $T$ is the total project duration.
Leakage delays: In leakage problems, remember that the pipe continues to supply water while the leak drains it; net filling rate is $\text{Inlet} - \text{Leak}$.
CBSE Board-Level 10-Question Mock Test
Section 2.3 Time & Work • Maximum Marks: 25 • Standard Time: 45 Minutes
CBSE Blueprint Aligned
Q1 • MCQ (1 Mark)
A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is:
A. 7/15
B. 8/15
C. 1/10
D. 2/5
Correct Answer: B (8/15)
In 1 day, A + B do $\frac{1}{15} + \frac{1}{20} = \frac{4 + 3}{60} = \frac{7}{60}$.
In 4 days, work done $= 4 \times \frac{7}{60} = \frac{7}{15}$.
Fraction of work left $= 1 - \frac{7}{15} = \frac{8}{15}$.
Q2 • MCQ (1 Mark)
If 12 men or 18 women can reap a field in 14 days, then the number of days that 8 men and 16 women will take to reap it is:
A cistern can be filled by two pipes in 30 minutes and 40 minutes respectively. Both pipes are opened together. When should the first pipe be turned off so that the cistern may be just full in 24 minutes?
A. After 12 minutes
B. After 15 minutes
C. After 18 minutes
D. After 20 minutes
Correct Answer: A (After 12 minutes)
Second pipe worked for the entire 24 minutes: Work done $= 24 \times \frac{1}{40} = \frac{3}{5}$.
Remaining work for the first pipe $= 1 - \frac{3}{5} = \frac{2}{5}$.
Time taken by first pipe $= \frac{2/5}{1/30} = \frac{2}{5} \times 30 = 12\text{ minutes}$.
Q4 • Assertion & Reason (1 Mark)
Assertion (A): If A is twice as efficient as B, then for doing a piece of work, A takes half the time taken by B. Reason (R): The efficiency of a worker is directly proportional to the time taken to finish the work.
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 of (A).
C. (A) is true but (R) is false.
D. (A) is false but (R) is true.
Correct Answer: C ((A) is true but (R) is false)
Efficiency is inversely proportional to time taken ($\eta \propto \frac{1}{T}$), not directly proportional. Thus Assertion (A) is true while Reason (R) is false.
Q5 • Very Short Answer (2 Marks)
A and B can complete a work in 12 days and 18 days respectively. They worked together for 4 days, after which B was replaced by C. If the remaining work was completed in 4 days, in how many days can C alone complete the entire work?
Click to View Model Solution
Work done by A and B in 4 days $= 4 \times \left(\frac{1}{12} + \frac{1}{18}
ight) = 4 \times \frac{5}{36} = \frac{5}{9}$.
Remaining work $= 1 - \frac{5}{9} = \frac{4}{9}$.
In the next 4 days, A and C worked: Work done by A $= 4 \times \frac{1}{12} = \frac{1}{3} = \frac{3}{9}$.
Work done by C in 4 days $= \frac{4}{9} - \frac{3}{9} = \frac{1}{9}$.
In 4 days C does $\frac{1}{9}$ of the work $\implies$ C alone takes $4 \times 9 = \mathbf{36\text{ days}}$.
Q6 • Very Short Answer (2 Marks)
A water tank has an inlet pipe that can fill it in 6 hours and a leak that can empty it in 10 hours. If the tank is currently half full and both the inlet and leak operate together, find the time required to fill the remaining half.
Click to View Model Solution
Net filling rate per hour $= \frac{1}{6} - \frac{1}{10} = \frac{5 - 3}{30} = \frac{2}{30} = \frac{1}{15}$.
Total time to fill the entire empty tank $= 15\text{ hours}$.
Since only half of the tank needs to be filled, time required $= \frac{1}{2} \times 15 = \mathbf{7.5\text{ hours}}$ (7 hours 30 minutes).
Q7 • Short Answer (3 Marks)
A, B, and C can do a piece of work in 20, 30, and 60 days respectively. In how many days can A finish the work if he is assisted by B and C on every third day?
Click to View Model Solution
Total work $= \text{LCM}(20, 30, 60) = 60\text{ units}$.
Daily efficiencies: A $= 3$ units, B $= 2$ units, C $= 1$ unit.
Day 1: A works alone $\implies 3$ units.
Day 2: A works alone $\implies 3$ units.
Day 3: A + B + C work together $\implies 3 + 2 + 1 = 6$ units.
Total work in 1 three-day cycle $= 3 + 3 + 6 = 12\text{ units}$.
Number of cycles to complete 60 units $= \frac{60}{12} = 5\text{ cycles}$.
Total days $= 5 \times 3 = \mathbf{15\text{ days}}$.
Q8 • Short Answer (3 Marks)
25 men were employed to complete a bridge construction project in 36 days, working 8 hours a day. After 20 days, only $\frac{5}{12}$ of the work was completed. How many additional men must be employed so that the work may be completed on time, each person now working 9 hours a day?
Pipe A can fill a swimming pool in 10 hours, and Pipe B in 15 hours. Both pipes are opened together, but after 2 hours Pipe A is closed. Pipe B continues to fill the pool for 4 hours, after which an outlet Pipe C is opened which empties the pool at a constant rate. If the pool is emptied in 10 hours after C was opened, find how long Pipe C alone would take to empty the full pool.
Click to View Model Solution
Let total pool capacity $= 30\text{ units}$.
Rate of A $= 30/10 = 3$ units/hr. Rate of B $= 30/15 = 2$ units/hr.
First 2 hours: Both A and B run $\implies 2 \times (3 + 2) = 10$ units filled.
Next 4 hours: B alone runs $\implies 4 \times 2 = 8$ units filled.
Total filled after 6 hours $= 10 + 8 = 18$ units.
Now, Pipe C is opened with Pipe B still running. The pool is emptied in 10 hours.
In these 10 hours, B adds $10 \times 2 = 20$ units. Total water C must remove $= 18 + 20 = 38$ units in 10 hours.
Rate of Pipe C $= \frac{38}{10} = 3.8$ units/hr.
Time for C alone to empty full pool (30 units) $= \frac{30}{3.8} = \frac{300}{38} = \frac{150}{19} = \mathbf{7\frac{17}{19}\text{ hours}} \quad (\approx 7.89\text{ hours})$.
Q10 • Case-Based Integrated Problem (4 Marks)
Case Study: Solar Rooftop Installation Project
A municipal corporation invites tenders for installing solar photovoltaic panels across a district hospital. Contractor Apex can complete the installation in 30 days with a standard team. Contractor Bright can complete the same project in 20 days. To finish ahead of the monsoon, the municipal engineer awards the contract to both contractors jointly. However, after 8 days of joint work, Contractor Apex has to withdraw its technicians due to an emergency at another hospital.
What fraction of the total installation was completed jointly during the first 8 days? (1 Mark)
How many more days will Contractor Bright require to complete the remaining installation alone? (2 Marks)
What is the total duration taken from start to completion of the solar project? (1 Mark)
Click to View Model Solution
Solution:
1. Joint 1-day rate of Apex and Bright $= \frac{1}{30} + \frac{1}{20} = \frac{2 + 3}{60} = \frac{5}{60} = \frac{1}{12}$.
In 8 days, fraction completed $= 8 \times \frac{1}{12} = \mathbf{\frac{2}{3}}$ of the project.
2. Remaining work after 8 days $= 1 - \frac{2}{3} = \frac{1}{3}$.
Bright's rate $= \frac{1}{20}$ project/day.
Days required by Bright alone $= \frac{1/3}{1/20} = \frac{20}{3} = \mathbf{6\frac{2}{3}\text{ days}}$ (6 days 16 hours).
3. Total duration from start to finish $= 8 + 6\frac{2}{3} = \mathbf{14\frac{2}{3}\text{ days}}$.