| Content On This Page | ||
|---|---|---|
| Time and Work: Basic Concepts (Efficiency, Total Work) | Problems involving Individual and Combined Work Rates | Pipes and Cisterns Problems |
| Solving Complex Problems on Time and Work | ||
Time and Work
Time and Work: Basic Concepts (Efficiency, Total Work)
The topic of Time and Work deals with problems concerning the time taken by individuals or groups to complete a piece of work, based on their working capabilities or rates. It establishes a relationship between the amount of work, the time taken, and the rate at which the work is done.
Basic Definitions:
- Work: This refers to the task or job that needs to be completed. Work can be anything from building a wall, digging a ditch, writing a book, filling a tank, etc. It can be measured in concrete units (like meters of wall, liters of water) or treated as a whole entity (e.g., "1 unit of work").
- Time: This is the duration taken to perform the work. It is typically measured in units like days, hours, minutes, or seconds.
- Efficiency (or Work Rate): This is the amount of work an individual or a group can complete per unit of time. It is a measure of how productive someone or something is. A higher efficiency means more work is done in the same amount of time, or the same amount of work is done in less time.
Fundamental Relationship: Work, Time, and Efficiency
The three quantities - Work, Time, and Efficiency - are related by a direct and simple formula:
$\boldsymbol{\text{Work} = \text{Efficiency} \times \text{Time}}$
... (i)
This formula can be rearranged to find Efficiency or Time if the other two quantities are known:
$\boldsymbol{\text{Efficiency} = \frac{\text{Work}}{\text{Time}}}$
... (ii)
$\boldsymbol{\text{Time} = \frac{\text{Work}}{\text{Efficiency}}}$
... (iii)
Inverse Relationship between Efficiency and Time
From equation (iii), if the Work to be done is constant (e.g., completing one specific task), Time is inversely proportional to Efficiency.
$\boldsymbol{\text{Time} \propto \frac{1}{\text{Efficiency}} \quad (\text{when Work is constant})}$
This is an intuitive relationship: if you increase your working rate (efficiency), you will take less time to finish the same job. For instance, if a worker doubles their efficiency, they will take half the time to complete the task.
Representing Total Work and Efficiency:
There are two common ways to represent the total work in problems:
1. Total Work as 1 Unit:
A simple approach is to consider the entire task or work to be completed as 1 unit. If a person can complete this 1 unit of work in $T$ units of time (e.g., $T$ days), then their efficiency (work done in 1 unit of time) is $\frac{\text{1 unit}}{\text{T units of time}} = \frac{1}{T}$ units/unit of time.
Example: If A can do a work in 8 days, then A's efficiency is $\frac{1}{8}$ of the work per day.
2. Total Work as the LCM of Times:
A more convenient approach, especially when dealing with multiple people or machines taking different amounts of time, is to assume the Total Work is equal to the Least Common Multiple (LCM) of the time periods given for each individual or group. This method makes it easier to calculate the efficiencies as integers.
Example: If A can do a work in 6 days and B can do the same work in 9 days.
Assume Total Work = LCM of 6 and 9 = 18 units.
$\begin{array}{c|cc} 2 & 6 \;, & 9 \\ \hline 3 & 3 \; , & 9 \\ \hline 3 & 1 \; , & 3 \\ \hline & 1 \; , & 1 \end{array}$
LCM $= 2 \times 3 \times 3 = 18$.
A's efficiency $= \frac{\text{Total Work}}{\text{Time taken by A}} = \frac{18 \text{ units}}{6 \text{ days}} = 3$ units/day.
B's efficiency $= \frac{\text{Total Work}}{\text{Time taken by B}} = \frac{18 \text{ units}}{9 \text{ days}} = 2$ units/day.
This LCM method is widely used as it avoids fractions in intermediate steps when calculating efficiencies.
Example 1. If a person can complete a task in 15 hours, what is their work rate per hour?
Answer:
Assume the Total Work is 1 unit.
Time taken $= 15$ hours.
Work rate (Efficiency) = $\frac{\text{Work}}{\text{Time}}$
Efficiency $= \frac{1 \text{ unit}}{15 \text{ hours}} = \boldsymbol{\frac{1}{15}}$ units/hour.
The person's work rate is $\boldsymbol{\frac{1}{15}}$ of the task per hour.
Competitive Exam Notes:
Time and Work problems are a staple in quantitative aptitude. Mastering the basic concepts and relationships is key.
- Core Relation: Work = Efficiency $\times$ Time. Understand how to derive E = W/T and T = W/E.
- Efficiency vs. Time: For a fixed amount of work, efficiency and time are inversely proportional. If efficiency increases, time decreases, and vice versa.
- Representing Work: Choose between assuming Total Work = 1 unit or taking Total Work = LCM of the given times. The LCM method is usually preferred for multiple workers as it simplifies calculations by dealing with integers.
- Efficiency Unit: Efficiency is always expressed as "work done per unit of time" (e.g., units/day, units/hour). If work is 1 unit, it's "fraction of work per unit of time".
Problems involving Individual and Combined Work Rates
A common type of Time and Work problem involves individuals or groups working either alone or together to complete a task. These problems utilize the concept of work rates (efficiency) and how they combine.
Working Together:
When multiple individuals or groups work together on the same task, their efficiencies (work rates) are typically additive, assuming they don't interfere with each other's work.
If Person A's efficiency is $E_A$ (work done by A per unit time) and Person B's efficiency is $E_B$ (work done by B per unit time), then their combined efficiency when working together on the same task is $E_{A+B} = E_A + E_B$.
The total time taken for them to complete the work together is given by:
$\text{Time taken together} = \frac{\text{Total Work}}{\text{Combined Efficiency}}$
Formula for Two Individuals Working Together:
If Person A can complete a certain work in $t_A$ days and Person B can complete the same work in $t_B$ days.
Let's assume the Total Work is 1 unit.
A's efficiency (work done by A in 1 day) $= \frac{1}{t_A}$ units/day.
B's efficiency (work done by B in 1 day) $= \frac{1}{t_B}$ units/day.
Combined efficiency (work done by A and B together in 1 day) $= E_A + E_B = \frac{1}{t_A} + \frac{1}{t_B}$.
Find the sum of efficiencies:
$\frac{1}{t_A} + \frac{1}{t_B} = \frac{t_B}{t_A t_B} + \frac{t_A}{t_A t_B} = \frac{t_A + t_B}{t_A t_B}$ units/day.
Time taken by A and B working together $= \frac{\text{Total Work}}{\text{Combined Efficiency}} = \frac{1 \text{ unit}}{\left(\frac{t_A + t_B}{t_A t_B}\right) \text{ units/day}}$
$\boldsymbol{\text{Time taken together} = \frac{t_A t_B}{t_A + t_B}}$ days
... (iv)
This formula directly gives the time taken when two individuals work together, given their individual times.
Formula for Three Individuals Working Together:
If A, B, and C can complete a work in $t_A, t_B,$ and $t_C$ days respectively, their individual efficiencies are $\frac{1}{t_A}, \frac{1}{t_B}, \frac{1}{t_C}$.
Their combined efficiency $= \frac{1}{t_A} + \frac{1}{t_B} + \frac{1}{t_C}$.
Time taken by A, B, and C working together $= \frac{\text{Total Work}}{\text{Combined Efficiency}} = \frac{1}{\left(\frac{1}{t_A} + \frac{1}{t_B} + \frac{1}{t_C}\right)}$ days.
$\boldsymbol{\text{Time taken together} = \frac{1}{\left(\frac{1}{t_A} + \frac{1}{t_B} + \frac{1}{t_C}\right)}}$ days
... (v)
Example 1. Ram can build a wall in 12 days and Shyam can build the same wall in 18 days. If they work together, in how many days can they build the wall?
Answer:
Time taken by Ram ($t_R$) $= 12$ days.
Time taken by Shyam ($t_S$) $= 18$ days.
Method 1: Using Unit of Work per Day (Efficiency).
Assume the Total Work is 1 unit (building the entire wall).
Ram's efficiency $= \frac{1}{12}$ of the wall per day.
Shyam's efficiency $= \frac{1}{18}$ of the wall per day.
Combined efficiency (work done by Ram and Shyam together in 1 day) $= \text{Ram's efficiency} + \text{Shyam's efficiency} = \frac{1}{12} + \frac{1}{18}$.
Find the sum:
$\frac{1}{12} + \frac{1}{18} = \frac{3}{36} + \frac{2}{36} = \frac{3+2}{36} = \frac{5}{36}$
So, Ram and Shyam together complete $\frac{5}{36}$ of the wall in 1 day.
Time taken to complete the total work (1 unit) $= \frac{\text{Total Work}}{\text{Combined Efficiency}} = \frac{1}{5/36}$ days.
Time taken together $= 1 \times \frac{36}{5} = \frac{36}{5} = 7.2$ days.
They can build the wall together in $\boldsymbol{7.2}$ days.
Method 2: Using LCM Method for Total Work.
Assume Total Work = LCM of 12 and 18.
$\begin{array}{c|cc} 2 & 12 \;, & 18 \\ \hline 2 & 6 \; , & 9 \\ \hline 3 & 3 \; , & 9 \\ \hline 3 & 1 \; , & 3 \\ \hline & 1 \; , & 1 \end{array}$
LCM $= 2 \times 2 \times 3 \times 3 = 36$. Let Total Work $= 36$ units (e.g., 36 bricks if that simplifies visualization).
Ram's efficiency $= \frac{\text{Total Work}}{\text{Time taken by Ram}} = \frac{36 \text{ units}}{12 \text{ days}} = 3$ units/day.
Shyam's efficiency $= \frac{\text{Total Work}}{\text{Time taken by Shyam}} = \frac{36 \text{ units}}{18 \text{ days}} = 2$ units/day.
Combined efficiency = Ram's efficiency + Shyam's efficiency $= 3 + 2 = 5$ units/day.
Time taken together $= \frac{\text{Total Work}}{\text{Combined Efficiency}} = \frac{36 \text{ units}}{5 \text{ units/day}} = \frac{36}{5} = 7.2$ days.
They can build the wall together in $\boldsymbol{7.2}$ days.
Method 3: Using Formula (iv).
Given $t_A = 12$ and $t_B = 18$.
$\text{Time taken together} = \frac{t_A t_B}{t_A + t_B} = \frac{12 \times 18}{12 + 18}$
$= \frac{216}{30} = \frac{21.6}{3} = 7.2$ days.
All methods give the same answer.
Working Alternately:
Problems involving individuals working on alternate days or hours require calculating the work done in one full cycle of turns. A cycle usually consists of one turn for each individual involved in the alternate working pattern. Calculate the time taken for one cycle and the work done in one cycle. Then determine how many cycles are needed for the total work.
Part of Work Done / Remaining Work:
If an individual or a group works for a specific duration, the work done during that time can be calculated using the formula: Work Done = Efficiency $\times$ Time. The Remaining Work is then calculated as Total Work - Work Done. The time needed to complete the remaining work by others will depend on their efficiency and the remaining work amount.
Example 2. A can do a work in 15 days and B can do the same work in 25 days. They start working together, but after 5 days, A leaves. In how many days will B finish the remaining work?
Answer:
Time taken by A $= 15$ days.
Time taken by B $= 25$ days.
Assume Total Work = LCM of 15 and 25.
$\begin{array}{c|cc} 5 & 15 \;, & 25 \\ \hline 3 & 3 \; , & 5 \\ \hline 5 & 1 \; , & 5 \\ \hline & 1 \; , & 1 \end{array}$
LCM $= 5 \times 3 \times 5 = 75$. Let Total Work $= 75$ units.
A's efficiency $= \frac{\text{Total Work}}{\text{Time taken by A}} = \frac{75 \text{ units}}{15 \text{ days}} = 5$ units/day.
B's efficiency $= \frac{\text{Total Work}}{\text{Time taken by B}} = \frac{75 \text{ units}}{25 \text{ days}} = 3$ units/day.
A and B start working together. Their combined efficiency $= \text{A's efficiency} + \text{B's efficiency} = 5 + 3 = 8$ units/day.
They work together for 5 days.
Work done by A and B together in 5 days = Combined efficiency $\times$ Time $= 8 \text{ units/day} \times 5 \text{ days} = 40$ units.
Remaining work = Total Work - Work done together $= 75 - 40 = 35$ units.
After 5 days, A leaves. B finishes the remaining work alone.
Time taken by B to finish the remaining work = $\frac{\text{Remaining Work}}{\text{B's efficiency}}$
Time taken by B $= \frac{35 \text{ units}}{3 \text{ units/day}} = \frac{35}{3}$ days.
Convert the improper fraction to a mixed number:
$\frac{35}{3} = 11 \frac{2}{3}$
B will finish the remaining work in $\boldsymbol{11\frac{2}{3}}$ days.
Competitive Exam Notes:
Problems involving combined work rates are fundamental. The LCM method for setting total work is highly recommended for efficiency and clarity.
- Combined Efficiency: Add individual efficiencies when working together.
- LCM Method: Assume Total Work = LCM of times taken by individuals. This gives integer efficiencies. Calculate efficiencies = Total Work / Time. Calculate combined efficiency by adding. Calculate time together = Total Work / Combined Efficiency.
- Formula for Two: $\text{Time together} = \frac{t_A t_B}{t_A + t_B}$. Quick shortcut for two individuals.
- Part Work: Work Done = Efficiency $\times$ Time. Remaining Work = Total Work - Work Done.
- Working in Shifts/Alternately: Calculate work done in one full cycle of turns. Find the number of cycles to complete most of the work. Calculate the time for the remaining work based on whose turn it is.
Pipes and Cisterns Problems
Problems involving pipes and cisterns (tanks) are a direct application of the concepts from Time and Work. In these problems, the 'work' is typically filling or emptying a tank or cistern, and the 'workers' are the pipes or taps connected to the tank. Pipes that fill the tank are analogous to workers who do positive work, while pipes that empty the tank are analogous to workers who do negative work.
Filling and Emptying Pipes:
- Filling Pipe (or Inlet Pipe): A pipe that adds water (or liquid) to fill the tank. Its work rate (efficiency) is considered positive because it contributes towards filling the tank. If a filling pipe can fill a tank completely in $t_{fill}$ units of time (e.g., hours), the fraction of the tank filled by this pipe in one unit of time is $\frac{1}{t_{fill}}$. So, its efficiency is $+\frac{1}{t_{fill}}$ tank/unit time.
- Emptying Pipe (or Outlet Pipe): A pipe that removes water (or liquid) from the tank, thus emptying it. Its work rate (efficiency) is considered negative because it works against filling the tank. If an emptying pipe can empty a full tank completely in $t_{empty}$ units of time, the fraction of the tank emptied by this pipe in one unit of time is $\frac{1}{t_{empty}}$. So, its efficiency is $-\frac{1}{t_{empty}}$ tank/unit time.
The Total Work is usually considered as filling the entire tank, which can be taken as 1 unit of work.
Combined Efficiency of Pipes:
When multiple pipes are operating simultaneously on the same tank, their efficiencies (work rates) are combined to find the net rate at which the tank is being filled or emptied. The efficiencies of filling pipes are added, and the efficiencies of emptying pipes are subtracted.
$\boldsymbol{\text{Net Combined Efficiency} = (\text{Sum of Efficiencies of Filling Pipes}) - (\text{Sum of Efficiencies of Emptying Pipes})}$
... (i)
Using the individual efficiencies:
$\boldsymbol{\text{Net Efficiency} = E_{\text{fill},1} + E_{\text{fill},2} + \dots - E_{\text{empty},1} - E_{\text{empty},2} - \dots}$
If Total Work is 1 unit, and times are $t_{f1}, t_{f2}, \dots$ for filling pipes and $t_{e1}, t_{e2}, \dots$ for emptying pipes:
$\boldsymbol{\text{Net Efficiency} = \left(\frac{1}{t_{f1}} + \frac{1}{t_{f2}} + \dots\right) - \left(\frac{1}{t_{e1}} + \frac{1}{t_{e2}} + \dots\right)}$
... (ii)
The time taken to fill or empty the tank when all pipes are open together is given by:
$\boldsymbol{\text{Time taken} = \frac{\text{Total Work}}{\text{Net Combined Efficiency}}}$
... (iii)
If the Net Combined Efficiency is positive, the tank will be filled, and the time calculated is the time to fill it. If the Net Combined Efficiency is negative, the tank will be emptied (assuming it was full or partially full), and the absolute value of the time calculated is the time to empty it (or empty the filled portion).
Similar to Time and Work, the LCM of the times taken by individual pipes is often used as the Total Work (Capacity of the tank) to simplify calculations and get integer efficiencies.
Example 1. Pipe A can fill a tank in 10 hours and Pipe B can fill it in 15 hours. If both pipes are opened together, in how many hours will the tank be filled?
Answer:
Time for Pipe A (filling) $= 10$ hours.
Time for Pipe B (filling) $= 15$ hours.
Assume Total Work (Capacity of tank) = LCM of 10 and 15 $= 30$ units.
$\begin{array}{c|cc} 2 & 10 \;, & 15 \\ \hline 5 & 5 \; , & 15 \\ \hline 3 & 1 \; , & 3 \\ \hline & 1 \; , & 1 \end{array}$
LCM $= 2 \times 5 \times 3 = 30$.
Efficiency of Pipe A $= \frac{\text{Total Work}}{\text{Time taken by A}} = \frac{30 \text{ units}}{10 \text{ hours}} = +3$ units/hour (filling).
Efficiency of Pipe B $= \frac{\text{Total Work}}{\text{Time taken by B}} = \frac{30 \text{ units}}{15 \text{ hours}} = +2$ units/hour (filling).
Net combined efficiency of A and B (when both are open) $=$ Efficiency of A $+$ Efficiency of B
Net Efficiency $= 3 + 2 = +5$ units/hour.
Since the net efficiency is positive, the tank will be filled.
Time taken to fill the tank $= \frac{\text{Total Work}}{\text{Net Efficiency}}$
Time $= \frac{30 \text{ units}}{5 \text{ units/hour}} = 6$ hours.
The tank will be filled in $\boldsymbol{6}$ hours.
Alternative Method (Using Formula from Time & Work for two working together):
This is equivalent to two workers doing a job together. Time taken by A $= t_A = 10$ hours, time taken by B $= t_B = 15$ hours.
Time taken together $= \frac{t_A t_B}{t_A + t_B} = \frac{10 \times 15}{10 + 15}$ hours.
Time $= \frac{150}{25} = 6$ hours.
Both methods give the same answer.
Example 2. Pipe A can fill a tank in 20 hours and Pipe B can empty the same tank in 30 hours. If both pipes are opened together when the tank is empty, in how many hours will it be filled?
Answer:
Time for Pipe A (filling) $= 20$ hours.
Time for Pipe B (emptying) $= 30$ hours.
Assume Total Work (Capacity of tank) = LCM of 20 and 30 $= 60$ units.
$\begin{array}{c|cc} 2 & 20 \;, & 30 \\ \hline 2 & 10 \; , & 15 \\ \hline 5 & 5 \; , & 15 \\ \hline 3 & 1 \; , & 3 \\ \hline & 1 \; , & 1 \end{array}$
LCM $= 2^2 \times 5 \times 3 = 60$.
Efficiency of Pipe A $= \frac{60 \text{ units}}{20 \text{ hours}} = +3$ units/hour (filling).
Efficiency of Pipe B $= \frac{60 \text{ units}}{30 \text{ hours}} = -2$ units/hour (emptying).
Net combined efficiency of A and B $=$ Efficiency of A $+$ Efficiency of B (algebraic sum)
Net Efficiency $= (+3) + (-2) = 3 - 2 = +1$ unit/hour.
Since the net efficiency is positive, the tank will be filled.
Time taken to fill the tank $= \frac{\text{Total Work}}{\text{Net Efficiency}}$
Time $= \frac{60 \text{ units}}{1 \text{ unit/hour}} = 60$ hours.
The tank will be filled in $\boldsymbol{60}$ hours.
Example 3. Pipe P can fill a tank in 15 minutes, and Pipe Q can fill it in 20 minutes. Pipe R can empty the same tank in 30 minutes. If all three pipes are opened together, in how many minutes will the tank be filled?
Answer:
Time for Pipe P (filling) $= 15$ minutes.
Time for Pipe Q (filling) $= 20$ minutes.
Time for Pipe R (emptying) $= 30$ minutes.
Assume Total Work (Capacity of tank) = LCM of 15, 20, and 30.
$\begin{array}{c|ccc} 2 & 15 \;, & 20 \;, & 30 \\ \hline 2 & 15 \; , & 10 \; , & 15 \\ \hline 3 & 15 \; , & 5 \; , & 15 \\ \hline 5 & 5 \; , & 5 \; , & 5 \\ \hline & 1 \; , & 1 \; , & 1 \end{array}$
LCM $= 2 \times 2 \times 3 \times 5 = 60$.
Assume Total Work $= 60$ units.
Efficiency of Pipe P $= \frac{60 \text{ units}}{15 \text{ minutes}} = +4$ units/minute.
Efficiency of Pipe Q $= \frac{60 \text{ units}}{20 \text{ minutes}} = +3$ units/minute.
Efficiency of Pipe R $= \frac{60 \text{ units}}{30 \text{ minutes}} = -2$ units/minute.
Net combined efficiency of P, Q, and R = Efficiency of P + Efficiency of Q + Efficiency of R
Net Efficiency $= (+4) + (+3) + (-2) = 4 + 3 - 2 = 7 - 2 = +5$ units/minute.
Since the net efficiency is positive, the tank will be filled.
Time taken to fill the tank $= \frac{\text{Total Work}}{\text{Net Efficiency}}$
Time $= \frac{60 \text{ units}}{5 \text{ units/minute}} = 12$ minutes.
The tank will be filled in $\boldsymbol{12}$ minutes.
Competitive Exam Notes:
Pipes and Cisterns problems are a specific type of Time and Work problem. Apply the same principles but distinguish between filling (positive efficiency) and emptying (negative efficiency).
- Filling = Positive Work: A pipe filling a tank contributes positively to the 'work' of filling.
- Emptying = Negative Work: A pipe emptying a tank contributes negatively, working against the filling process.
- Efficiency Sign Convention: Use + for filling pipes' efficiencies and - for emptying pipes' efficiencies.
- Net Efficiency: Add all efficiencies algebraically. Net Efficiency $= \sum E_{\text{filling}} - \sum E_{\text{emptying}}$.
- Total Work (Capacity): Assume the capacity of the tank is the LCM of the times taken by individual pipes to fill or empty it.
- Time Taken: Time $= \frac{\text{Total Work}}{\text{Net Efficiency}}$. If Net Efficiency is positive, it's time to fill. If negative, it's time to empty (if tank was full or partially full).
- Partial Filling/Emptying: If the tank is initially partially full or empty, calculate the remaining capacity that needs to be filled or emptied.
- Leakage: A leak in a tank acts as an emptying pipe.
Solving Complex Problems on Time and Work
Beyond basic problems involving individuals working together, Time and Work questions can become more complex by introducing varying efficiencies, scenarios where workers join or leave a project mid-way, or problems involving different types of workers (like men, women, children) with different working capacities. These require careful analysis and systematic application of the fundamental concepts.
Problems involving Men, Women, and Children:
These problems provide information relating the efficiencies of different categories of workers. The first step is to establish a numerical relationship between the efficiencies of men, women, and/or children. This allows you to convert a group of mixed workers into an equivalent number of workers of a single category (often men) to solve the problem.
The relationship between efficiencies is usually given indirectly, e.g., "X men can do a work in D days, and Y women can do the same work in D days". This implies that the total work done by X men in D days is equal to the total work done by Y women in D days. Total Work = (Number of Men $\times$ Efficiency of 1 Man $\times$ Days) = (Number of Women $\times$ Efficiency of 1 Woman $\times$ Days).
If X men do a job in $D_1$ days and Y women do the same job in $D_2$ days, and Z children do the same job in $D_3$ days:
Total Work $= X \times E_{man} \times D_1 = Y \times E_{woman} \times D_2 = Z \times E_{child} \times D_3$
By equating these, you can find the ratio of efficiencies (e.g., $\frac{E_{man}}{E_{woman}} = \frac{Y \times D_2}{X \times D_1}$). Once the ratio is known, you can assume convenient integer values for the efficiencies and proceed using the LCM method or convert the entire mixed group into an equivalent number of workers of one type.
Example 1. 4 men or 6 women can do a piece of work in 30 days. In how many days will 8 men and 10 women finish the same work?
Answer:
Given: 4 men can do the work in 30 days.
Total Work $= 4 \text{ men} \times 30 \text{ days} = 120$ man-days.
Also, 6 women can do the same work in 30 days.
Total Work $= 6 \text{ women} \times 30 \text{ days} = 180$ woman-days.
Since the work is the same, we can equate the total work units:
$\boldsymbol{120 \text{ man-days} = 180 \text{ woman-days}}$
... (i)
This equation gives the relationship between the working capacities of men and women. From (i), we can say that the work done by 120 men in 1 day is equal to the work done by 180 women in 1 day.
120 men $= 180$ women
Divide both sides by their GCD, 60:
$\boldsymbol{2 \text{ men} = 3 \text{ women}}$
... (ii)
This relation (ii) means that the efficiency of 2 men is equal to the efficiency of 3 women.
We need to find the time taken by a group of 8 men and 10 women to finish the same work.
Let's convert this group into an equivalent number of men using the relation (ii).
From (ii), 3 women $= 2$ men.
To find the equivalent number of men for 10 women, use proportionality: $\frac{\text{Men}}{\text{Women}} = \frac{2}{3}$. So, Men $= \frac{2}{3} \times \text{Women}$.
Equivalent men for 10 women $= \frac{2}{3} \times 10 = \frac{20}{3}$ men.
Total number of workers in the new group, expressed as men:
Equivalent men $= 8 \text{ men} + 10 \text{ women} = 8 \text{ men} + \frac{20}{3} \text{ men} = \left(8 + \frac{20}{3}\right) \text{ men}$.
Equivalent men $= \left(\frac{24+20}{3}\right) \text{ men} = \frac{44}{3}$ men.
Now we know that 4 men can do the work in 30 days, and we need to find the time taken by $\frac{44}{3}$ men.
This is an inverse proportion problem: More men take less time to complete the same work. Using the formula $M_1 D_1 = M_2 D_2$ (for constant Work):
$M_1 = 4$ men, $D_1 = 30$ days.
$M_2 = \frac{44}{3}$ men, $D_2 = ?$ days.
$\boldsymbol{4 \times 30 = \frac{44}{3} \times D_2}$
... (iii)
$\boldsymbol{120 = \frac{44}{3} \times D_2}$
Solve for $D_2$:
$\boldsymbol{D_2 = 120 \times \frac{3}{44}}$
Simplify the expression:
$D_2 = \cancel{120}^{\normalsize 30} \times \frac{3}{\cancel{44}^{\normalsize 11}} = \frac{30 \times 3}{11} = \frac{90}{11}$ days.
Convert the improper fraction to a mixed number:
$\frac{90}{11} = 8 \frac{2}{11}$
$D_2 = 8\frac{2}{11}$ days.
8 men and 10 women will finish the work in $\boldsymbol{8\frac{2}{11}}$ days.
Alternative Method (Using LCM and Efficiencies directly):
From the given information, 4 men's 1-day work = $\frac{1}{30}$ of total work. 6 women's 1-day work = $\frac{1}{30}$ of total work.
Let 1 man's 1-day work (efficiency) $= m$ and 1 woman's 1-day work (efficiency) $= w$.
$4m \times 30 = \text{Total Work}$
$6w \times 30 = \text{Total Work}$
Equating work: $120m = 180w \implies 2m = 3w \implies \frac{m}{w} = \frac{3}{2}$.
Assume 1 man's efficiency $m=3k$ and 1 woman's efficiency $w=2k$. For simplicity, let $k=1$. So, $m=3$ units/day, $w=2$ units/day.
Total Work $= 4m \times 30 = 4 \times 3 \times 30 = 12 \times 30 = 360$ units.
(Check with women: Total Work $= 6w \times 30 = 6 \times 2 \times 30 = 12 \times 30 = 360$ units. Consistent).
Now consider 8 men and 10 women. Their combined efficiency:
Combined efficiency $=(8 \times m) + (10 \times w) = (8 \times 3) + (10 \times 2) = 24 + 20 = 44$ units/day.
Time taken by 8 men and 10 women = $\frac{\text{Total Work}}{\text{Combined Efficiency}}$
Time $= \frac{360 \text{ units}}{44 \text{ units/day}} = \frac{360}{44}$ days.
Simplify the fraction:
Time $= \frac{\cancel{360}^{\normalsize 90}}{\cancel{44}^{\normalsize 11}} = \frac{90}{11}$ days $= 8\frac{2}{11}$ days.
Both methods provide the same result.
Problems involving Wages:
When a group of individuals or groups are paid for completing a task, the total wages are distributed among them based on the amount of work each has contributed. The fundamental principle is that Wages are proportional to the Work Done.
$\boldsymbol{\text{Ratio of Wages} = \text{Ratio of Work Done}}$
... (iv)
If all individuals or groups work for the same amount of time (e.g., they work together until the job is finished), then the amount of work done by each is proportional to their efficiency (Work = Efficiency $\times$ Time, and Time is constant). In this specific case:
$\boldsymbol{\text{Ratio of Wages} = \text{Ratio of Efficiencies}}$
... (v)
If they work for different amounts of time, calculate the fraction of work done by each person/group and distribute the wages in that ratio.
Example 2. A, B, and C can complete a work in 10, 15, and 20 days respectively. They started working together and finished the work. If they receive $\textsf{₹ } 12000$ for the total work, find the share of each.
Answer:
Time taken by A $= 10$ days.
Time taken by B $= 15$ days.
Time taken by C $= 20$ days.
Since A, B, and C worked together until the work was finished, they all worked for the same duration. Therefore, the total wages will be distributed among them in the ratio of their efficiencies.
Assume Total Work = LCM of 10, 15, and 20 $= 60$ units.
$\begin{array}{c|ccc} 2 & 10 \;, & 15 \;, & 20 \\ \hline 2 & 5 \; , & 15 \; , & 10 \\ \hline 3 & 5 \; , & 15 \; , & 5 \\ \hline 5 & 5 \; , & 5 \; , & 5 \\ \hline & 1 \; , & 1 \; , & 1 \end{array}$
LCM $= 2 \times 2 \times 3 \times 5 = 60$.
A's efficiency $= \frac{\text{Total Work}}{\text{Time taken by A}} = \frac{60 \text{ units}}{10 \text{ days}} = 6$ units/day.
B's efficiency $= \frac{\text{Total Work}}{\text{Time taken by B}} = \frac{60 \text{ units}}{15 \text{ days}} = 4$ units/day.
C's efficiency $= \frac{\text{Total Work}}{\text{Time taken by C}} = \frac{60 \text{ units}}{20 \text{ days}} = 3$ units/day.
Ratio of efficiencies = A : B : C $= 6 : 4 : 3$.
The total wages $\textsf{₹ } 12000$ will be divided in this ratio.
Sum of the terms in the ratio $= 6 + 4 + 3 = 13$.
Calculate A's share:
A's share $= \frac{\text{A's ratio term}}{\text{Sum of ratio terms}} \times \text{Total Wages} = \frac{6}{13} \times \textsf{₹ } 12000$
Calculate the amount:
A's share $= \frac{6 \times 12000}{13} = \frac{72000}{13}$
$\begin{array}{r} 5538.46... \\ 13{\overline{\smash{\big)}\,72000.00}} \\ \underline{-~\phantom{(}65\phantom{000.00)}} \\ 70\phantom{00.00)} \\ \underline{-~\phantom{()}\,65\phantom{00.00)}} \\ 50\phantom{0.00)} \\ \underline{-~\phantom{()}\,39\phantom{0.00)}} \\ 110\phantom{.00)} \\ \underline{-~\phantom{()}\,104\phantom{.00)}} \\ 60\phantom{0)} \\ \underline{-~\phantom{()}\,52\phantom{0)}} \\ 80 \\ \underline{-~\phantom{()}\,78} \\ 2 \end{array}$
A's share $\approx \textsf{₹ } 5538.46$
Calculate B's share:
B's share $= \frac{\text{B's ratio term}}{\text{Sum of ratio terms}} \times \text{Total Wages} = \frac{4}{13} \times \textsf{₹ } 12000$
B's share $= \frac{4 \times 12000}{13} = \frac{48000}{13} \approx \textsf{₹ } 3692.31$
Calculate C's share:
C's share $= \frac{\text{C's ratio term}}{\text{Sum of ratio terms}} \times \text{Total Wages} = \frac{3}{13} \times \textsf{₹ } 12000$
C's share $= \frac{3 \times 12000}{13} = \frac{36000}{13} \approx \textsf{₹ } 2769.23$
Check (with approximate values): $5538.46 + 3692.31 + 2769.23 = 12000$. (Matches total wages).
A's share is $\boldsymbol{\textsf{₹ } \frac{72000}{13}}$ (or $\approx \textsf{₹ } 5538.46$), B's share is $\boldsymbol{\textsf{₹ } \frac{48000}{13}}$ (or $\approx \textsf{₹ } 3692.31$), and C's share is $\boldsymbol{\textsf{₹ } \frac{36000}{13}}$ (or $\approx \textsf{₹ } 2769.23$). It is best to leave answers as fractions unless asked to round.
Competitive Exam Notes:
Complex Time and Work problems require a solid grasp of the basic concepts and the ability to apply them logically in multi-step scenarios.
- Establish Efficiency Relations: In problems with different types of workers, use the given information (e.g., X men = Y women for the same work/time) to find the ratio of their efficiencies. Convert all workers to a single category or use their relative efficiencies with the LCM method.
- Variable Time/Workers: If workers join or leave, calculate the work done during the periods when the group composition is constant. Subtract this from the total work to find the remaining work. Calculate the time needed to complete the remaining work by the new group.
- Wages Distribution: Wages are directly proportional to the work done. If time is the same, wages are proportional to efficiency. If time is different, calculate the exact fraction/amount of work done by each person and distribute wages in that ratio. Work done by person $i$ = Efficiency of $i \times$ Time worked by $i$.
- MDH/W Formula: Remember $\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$ for comparing two scenarios with potentially different numbers of Men (M), Days (D), Hours per day (H), and Work Done (W). Omit terms that are constant or not mentioned.
- LCM Method Benefits: Using LCM for total work simplifies calculations involving fractions, especially when dealing with remaining work or mixed groups.