## Capacity to do work or rate of work in unit time is of main interest

The problems on work and time deal with working agents, usually men, women or boys, who work alone or together on the same work. The rates of work or work capacities of the working agents are different so that combining the working agents together creates a bit of complexity. Usually in how many days a work can be completed is asked for.

In contrast, in work-wage problems we are interested in how much one working agent should earn after working on a job for certain number of days. Though in this type of problems also, we have working agents with different work capacities, time of completion of a work is not of interest, of interest is how much a working agent might earn from the work. The guiding principle here obviously is, the more is the work capacity of a working agent the more will be the agent's earning per unit time.

In pipes and cisterns problems, we don't have men, women or boys working and in the first look these problems seem to be different from the work-time problems. But if you look into the essentials of this type of problems, you will find little difference from work-time problems. In both work-time and pipes and cistern problems, * working agents do work at different rates*, nature of working agents do not change the nature of problems or solution methods.

In pipes and cisterns problems, the draining or filling pipes are the two types of working agents working at different rates in filling up or emptying the cistern. while in work-time problems the men, women or boys work at different rates to complete the same work. The difference between the two types lies in the work destruction of the draining pipes equivalence of which we don't find in work-time problems.

The basic roles of the working agents in work-time and pipes and cisterns problems being same along with work completion time and working agent inverse proportionality remaining same, the methods of solution are also primarily same for these two types of problems.

Let us take up the more important Work-Time problems first in details.

### Work-Time problems

As explained earlier, working agents do same work at different rates of work, alone or together. Generally we are to find work completion time. It is best to explain problems through examples.

#### First example of work and time problem

Javed can complete a work in 6 days, while Shalini completes the same work in 12 days and Javed and Ankur together can complete the work in 4 days. In how many days can Shalini and Ankur working together complete the work.

**Problem analysis of first example**

If you go about dissecting this commonly encountered form of work-time problem, you should identify the different elements step by step.

**Working agents**

Here we find three working agents with identifiable names, Javed, Shalini and Ankur. The names are not important though. The important information lies in the **individual capacities of their work.**

These individual work capacities are the* variables of interest in a work- time problem* and these only are related through the problem statements in forming simple linear equations. Here the problem solver's job are two:

- to form the linear equations, carefully following the rules of work-time problem domain, and
- to solve the simple equations in finding the answer to the question asked for.

#### The single most important concept in work and time problems

The * crucial concept* used in solving work-time problems is,

Convert each working agent's work capacity in terms of portion of work completed per day (we assume unit time to be a day, though it can be any unit of time), so that when two such agents work together on the same job simultaneously for a day, we are able to say at the end of the day how much portion of work has been completed by just adding the two rates of work as portions of work.

For example, if two agents A and B individually can complete a work in say, $D_A$ and $D_B$ days respectively, * working individually, the work portion* they complete in one day will be,

$\displaystyle\frac{1}{D_A}$ and $\displaystyle\frac{1}{D_B}$ portions of work respectively.

With this vital information obtained, we are now in a position to find out the amount of work completed by the two working together for a day as,

$\displaystyle\frac{1}{D_A} + \displaystyle\frac{1}{D_B}$ portion of work.

To be more specific and for ease of understanding, if we assume $D_A=3$ days and $D_B=6$ days, in one day the portion of work that would be completed by A and B working together will be,

$\displaystyle\frac{1}{3} + \displaystyle\frac{1}{6} = \displaystyle\frac{1}{2}$.

It means in one day, working together **A and B will complete half of the work.**

* This is the first part of problem solving*, namely,

- invert the number of days in which the agents complete the work individually alone and add the two fractions to form the
by the two working together.**portion of work completed in a day**

* We repeat,* the

*they complete the work individually, is the*

**result of adding the two inversions of work capacitiies of two agents expressed in days***This understanding is important.*

**PORTION of work they would complete in a day of working together.**In the * second part of problem solving *we need to find

**in how many days A and B would complete the work while working together.**In this speific case, as A and B could complete * half of the work while working together for a day,* they will

*. This is simple application of*

**complete the whole work together in just two days***and*

**two-stage unitary method**

*inversion of the fractional portion of work completed in a day.*#### Problem modeling of the first example

Armed with this **conceptual base**, we can form the linear equations in our problem as,

$\displaystyle\frac{1}{D_J} + \displaystyle\frac{1}{D_A} = \displaystyle\frac{1}{4}$, where $D_J = 6$ is the number of days in which Javed completes the work alone and $D_A$ is the number of days Ankur takes to complete the work alone. As they complete the work working together in 4 days, in 1 day they complete one-fourth of the work. To form combined work linear equations, **we need always to convert all variables to 1 unit time span.**

So we have,

$\displaystyle\frac{1}{6} + \displaystyle\frac{1}{D_A} = \displaystyle\frac{1}{4}$,

Or, $\displaystyle\frac{1}{D_A} = \displaystyle\frac{1}{4} - \displaystyle\frac{1}{6} = \displaystyle\frac{1}{12}$.

Thus from a simple linear equation we obtained the value of the only unknown variable in our problem, Ankur's work capacity as, 12 days work completion.

#### Last step of the solution of the first example

So Shalini and Ankur both can complete the work individually in 12 days and hence working together they will complete the portion of the work in a day (note carefully the wording),

$\displaystyle\frac{1}{12} + \displaystyle\frac{1}{12} = \displaystyle\frac{1}{6}$.

It means they will take 6 days to complete the work while working together.

Not difficult at all, isn't it?

#### Second example of work and time problem

A boy completes a work in 24 days whereas a man completes the same work in 12 days. In how many days would 4 men and 4 boys working together would complete the work?

**The powerful concept of mandays or boydays**

Whenever we get the opportuniy we use the * mandays* concept. It is an elegant and powerful concept in time-work problem domain and will lead you to the solution in quick time. Let us explain.

A manday is the amount of work a man can do in a day.

From the statement that a man completes a work in 12 days we conclude the **amount of work is 12 mandays.**

The advantage of mandays concept is, it is not only used as the work capacity unit of a man, but also **helps to express the amount of work in the same unit.** It serves two purposes.

Because of this flexibility, when we know the anount of work as 12 mandays, we can say,

4 men can complete the work in 3 days, or,

1 man can complete the work in 12 days, or,

3 men can complete the work in 4 days, or,

2 men can complete the work in 6 days, or,

6 men can complete the work in 2 days, or for that matter,

12 men can complete the work in 1 day.

If the amount of work is 12 mandays, the product of number of men and number of days should be 12, that's all, the numbers may be any feasible. This concept thus gives a lot of flexibility in possibilities.

Observe, * if the number of men increases, the number of days to complete the work decreases proportionately.* This is one important hallmark of this work-time problem domain. We express this as,

$ \text{Number of men } \propto \displaystyle\frac{1}{\text{Number of days}} $

In wordy expression, it means **number of men is inversely proportional to the number of days to complete the work.** This makes sense. If we increase number of men they would finish the job in fewer days and vice versa.

#### Changing work amount

All through till now we have assumed that the work amount has remained fixed. But it may not necessarily be so. In a general situation the number of workers, the work completion time and the work amount all three can vary. Thus the general relationship is,

$ \text{Amount of Work}= \text{Number of men } $

$\hspace{20mm}\times{\text{Number of days}} $

#### Problem modeling and solution of second example on work-time problem

So by the problem statement,

Work amount = 24 boydays = 12 mandays.

It means,

1 manday = 2 boydays, that is, 1 man is work-equivalent to 2 boys.

So when 4 men and 4 boys work together, effectively, $2\times{4} + 4 = 12$ boys worked together.

As the work amount is 24 boydays, 12 boys will finish it in 2 days. That simple.

#### Alternate conventional solution of second example on Work-time problem

Working alone, in a day, the boy completes, $\displaystyle\frac{1}{24}$th part of the work and the man completes $\displaystyle\frac{1}{12}$th part of the work. So working together 4 men and 4 boys will complete in a day,

$ \displaystyle\frac{4}{24} + \displaystyle\frac{4}{12} = \displaystyle\frac{12}{24} = \displaystyle\frac{1}{2} \text{ of the work} $

Thus they will take 2 days to complete the work when working together.

As we like to avoid fractions, we will always use the mandays concept when possible.

#### Third example on work-time problem

24 boys and 13 men can finish a work in 4 days, while 16 boys and 12 men can finish the same work in 5 days. Find the ratio of per day work capacities of the man to the boy.

**Solution 3:** The amount of work is same in both cases and it is,

$ \text{Amount of work } = 96 \text{ bd} + 52 \text{ md} $

$\hspace{23mm}= 80 \text{ bd} + 60 \text{ md}$

Or, $ 16 \text{ bd} = 8 \text{ md} $

Or, 1 manday = 2 boyday, that is, ratio of per day work capacities of a man to a boy is 2 : 1, the man works twice as much as the boy in one day.

**Key concept:** Amount of work in terms of $man\times{days}=mandays$. This remains same.

#### Fourth example on work-time problem

Working together Bittu and Piklu can finish a work in 4 days, Piklu and Chhotu can finish the same work in 6 days and Chhotu and Bittu can finish this work in 3 days. How many days would they take to finish the work if all three of them work together?

**Solution 4:**

One day's work of Bittu and Piklu,

$ \displaystyle\frac{1}{Bittu} + \displaystyle\frac{1}{Piklu} = \frac{1}{4} $

One day's work of Piklu and Chhotu,

$ \displaystyle\frac{1}{Piklu} + \displaystyle\frac{1}{Chhotu} = \displaystyle\frac{1}{6} $

One day's work of Chhotu and Bittu,

$ \displaystyle\frac{1}{Chhotu} + \displaystyle\frac{1}{Bittu} = \displaystyle\frac{1}{3} $

Adding them we get,

$ 2\left(\displaystyle\frac{1}{Bittu} + \displaystyle\frac{1}{Piklu} + \displaystyle\frac{1}{Chhotu}\right) $

$= \left(\displaystyle\frac{1}{4} + \displaystyle\frac{1}{6} + \displaystyle\frac{1}{3}\right) $

$= \displaystyle\frac{9}{12} $

So, in one day the three, working together can do,

$ \left(\displaystyle\frac{1}{Bittu} + \displaystyle\frac{1}{Piklu} + \displaystyle\frac{1}{Chhotu}\right) $

$= \displaystyle\frac{9}{24} $

$= \displaystyle\frac{3}{8} \text{ part of the work} $

Applying unitary method, we then have, required number of days in which all three working together will finish the job as inverse of 3/8,

$ \displaystyle\frac{8}{3}=2\displaystyle\frac{2}{3}, \quad \text{ a little less than 3 days} $

**Key concept:** Working together: In one day Bittu will do $1/Bittu$ part of the job if he takes $Bittu$ days to finish a work. This one day's work can be added to one day's work of any other person Piklu, to get one day's work of Bittu and Piklu working together. **Inversion is the key action.**

#### Fifth example on work-time problem

Bittu alone can do a work in 18 days and working with Piklu can finish the same work in 12 days. How many days Piklu will take to do this work alone?

**Solution 5:**

One day's work of Bittu and Piklu working together,

$ \displaystyle\frac{1}{Bittu} + \displaystyle\frac{1}{Piklu} = \displaystyle\frac{1}{18} + \displaystyle\frac{1}{Piklu}= \displaystyle\frac{1}{12}$

So, $ \displaystyle\displaystyle\frac{1}{Piklu} = \displaystyle\frac{1}{12} - \displaystyle\frac{1}{18} = \displaystyle\frac{1}{36} $

Hence Piklu will take a long 36 days to do the work alone.

**Key concept:** One day's work = inversion of number of days; adding one day's work of two persons working together adds inverted number of days for each person.

**Sixth example on work-time problem**

Bittu alone can do a work in 10 days and Piklu does the same work 25% faster than Bittu. How many days Piklu will take to do this work alone?

**Solution 6:**

Piklu is 25% faster than Bittu means:

One day's work of Piklu $= \displaystyle\frac{1}{Piklu} = 125$% of One day's work of Bittu

$= \displaystyle\frac{5}{4}\times{\frac{1}{10}}$

$=\displaystyle\frac{1}{8} $

So Piklu will take 8 days to finish the work.

**Key concept:** One day's work = inversion of number of days; $y$ is 25% more than $x$ means, $y=125$% of $x$, or, $y=1.25x=\displaystyle\frac{5x}{4}$.

#### The powerful Work rate technique in solving Time and work problems faster

In this technique,

Instead of taking the variable as number of days required for a worker to complete a job,

This is the work rate for the particular worker per unit time.we directly assume the worker variable to be the portion of job the worker completes in a day.

The advantages with this approach are, for many time and work problems formation of linear equations are straightforward without involving fractions. Consequently solving the equations to arrive at the final solution is also easier and faster. This approach is also conceptually easier to grasp.

This is a new problem solver's approach which is nothing but a value addition to the classical techniques. Along with Mandays technique, we would usually adopt this direct and efficient approach in solving most of the time and work problems.

To explain the technique we will solve a new problem using this technique.

#### Seventh** example on work-time problem**

Two workers A and B working together complete a piece of work in 5 days. If A worked twice as efficiently as he actually did, and B worked one-third as efficiently as he actually did, the work would have been completed in 3 days. How many days A would then require to complete the piece of job when working alone?

- $6\frac{1}{4}$ days
- $8\frac{3}{4}$ days
- $5\frac{1}{5}$ days
- $7\frac{1}{2}$ days

#### Solution 7 using efficient Work rate technique

Assuming portion of work $W$ done by A and B individually in 1 day to be $A$ and $B$, as A and B take 5 days to complete the job working together,

$5(A+B)=W$.

Each day A and B together do $(A+B)$ portion of work $W$, so in 5 days the work done by them is, $5(A+B)$ which is equal to $W$.

If A works twice as efficiently as he actually did, his new wok rate becomes simply, $2A$.

Similarly the new work rate of B becomes, $\frac{1}{3}B$.

So from the second statement, as A and B working together at their new work rates finish the job in 3 days,

$3\left(2A+\frac{1}{3}B\right)=W$,

Or. $6A+B=W$.

From first equation and this new equation we need to eliminate B and get a relation between $A$ and $W$ which will determine in how many days A alone finishes the work.

First equation is,

$5(A+B)=W$,

Or, $5A+5B=W$.

From second equation we have,

$6A+B=W$,

Or, $30A+5B=5W$.

Subtracting the expanded first equation from the last equation,

$25A=4W$,

Or, $\displaystyle\frac{25}{4}A=W$.

It means, in $\displaystyle\frac{25}{4}=6\frac{1}{4}$ days A will complete the job when working alone.

**Answer:** Option a: $6\frac{1}{4}$ days.

Observe how naturally the relationships are formed and solution reached with this new approach.

While solving most of the work time problems we will use this * Work rate technique* and the

*as required by the problem.*

**Mandays technique**### Work and wages Problems, basic principle

- Wage is the money earned by a worker in a day. It is
It is directly proportional to amount of work done in a day, but remains usually fixed for a particular worker.**daily and variable.**

$ \text{Wage} \propto \text{Amount of work done} $

$\hspace{14mm}\propto \text{Amount of work done per day} \times {\text{Number of days}} $

- Daily wage of a person is inversely proportional to number of days he takes to finish a work. If Pintu finishes a work in 2 days which Bonku takes 4 days to finish, Pintu will obviously earn double the wage per day compared to that of Bonku.

$ \text{Wage} \propto \displaystyle\frac{1}{\text{Number of days to finish the work}} $

Thus ratio of wage earned by Pintu and Bonku working together in a day will be in ratio of the amount of work done by them. In this case,

$\displaystyle\frac{\text{Bonku's wage per day}}{\text{Pintu's wage per day}} = \frac{1}{2} $

#### First example of Work-wage problem

Pintu, Bittu and Bonku undertake to do a work for Rs.6400. Pintu and Bittu worked together to finish 25% of the work. Bonku then took up the work and finished the rest of it alone. How much will Bonku get?

**Solution 1:**

Though this seems to be a problem of Work and wages, actually it is solved by simple arithmetic principles.

Pintu and Bittu finished 25% of the work and so received 25% of the contracted amount of money, which is Rs.1600. As Bonku finished the rest of the work alone, he must take away rest of the contracted amount, that is,

$ \text{Rs.}6400 - \text{Rs.}1600 = \text{Rs.}4800 $.

#### Second example of Work-wage problem

Venky can do a work in 6 days and Himu can do the same work in 7 days working alone. If the total amount given for this work is Rs.780 when they worked together, what will be the share of Venky?

**Solution 2:**

Here Venky and Himu's share of the total will be in the ratio of inverse of their number of days of completion of the work, that is, in the ratio of 1/6 : 1/7 which is, 7 : 6.

Let us assume the ratio to be $7x : 6x$. So total is $13x=Rs.780$.

Thus, $x=Rs.60$, and Venky's share, $7x = Rs.420$.

#### Third example of Work-wage problem

This time Chandra, Venky and Himu take Rs. 535 for doing a work together. If each takes 5, 6 and 7 days respectively to complete the work when working alone, what will be the share of Chandra?

**Solution 3:**

The share of the three will be in the share of inverse of their number of days of completion of work, that is in the ratio of 1/5 : 1/6 : 1/7.

Multiplying each of the member of the ratio by 210, the LCM of 5, 6, and 7, we get the ratio of the share as, 42 : 35 : 30. Let as assume the ratio to be, $42x : 35x : 30x$.

So, total share is,

$(42x + 35x + 30x) = 107x = \text{Rs. }535$.

Thus,

$x=\text{Rs. }5$

and

Chandra's share, $42x=\text{Rs. }210$.

#### Fourth example of Work-wage problem

25 men worked together for 16 days to get an wage of Rs.11500. How many women must work together for 48 days to receive an wage of Rs.31050, if daily a woman receives half the wage of a man?

**Solution 4:**

Daily wage of a man,

$ \text{Wage of a man per day }=\text{Rs.}\displaystyle\frac{11500}{25\times{16}}$

$\hspace{50mm}=\text{Rs.}\displaystyle\frac{115}{4} $

Daily wage of a woman being half that of a man, it is, Rs.$\displaystyle\frac{115}{8}$.

So to receive Rs.31050 in 48 days, the number of women working would be,

$ \text{Number of women }=\displaystyle\frac{31050}{48\times{\displaystyle\frac{115}{8}}}$

$\hspace{41mm}=\displaystyle\frac{31050}{6\times{115}}$

$\hspace{41mm}=\displaystyle\frac{5175}{115}$

$\hspace{41mm}=\displaystyle\frac{1035}{23}$

$\hspace{41mm}=45 $.

#### Fifth example of Work-wage problem

Manish and Manav together do a work in 15 days to receive a total amount of Rs.400. If Manish can do the same work alone in 20 days, how much did Manav receive for the work?

**Solution 5:**

Applying the per day work principle, we have,

$ \displaystyle\frac{1}{20} + \displaystyle\frac{1}{\text{Manav}} = \displaystyle\frac{1}{15}$,

So,

$\displaystyle\frac{1}{\text{Manav}}=\frac{1}{15} - \displaystyle\frac{1}{20}=\displaystyle\frac{1}{60} $

Thus Manav does the work in 60 days and will receive the share in the ratio of, 1/60 : 1/20, that is, 20 : 60 or 1 : 3.

As the total is Rs.400, Manav will receive 1 out of 4 share of Rs.400, that is, Rs.100.

### Pipes and Cisterns Problems, basic principles

A cistern is to be filled or emptied by pipes connected to it.

The pipes are of two types:

**Filling pipe:** One or more than one pipe fills the cistern with specific rate of filling. The rate of filling of each filling pipe is stated in terms of how much time it takes to fill the whole cistern. For example, "A cistern has one filling pipe that can fill the cistern in 6 hours."

**Emptying pipe:** one or more than one emptying pipe empties the water in the cistern. Capacity of an emptying pipe is stated in terms of how much time it takes to empty the whole cistern. For example, "An emptying pipe can empty the cistern in 8 hours."

**Pipes working together:** Different type of pipes may work together. To get the effect of two or more pipes working together,

**first:**the combined effect of the pipes in 1 hour is found out by summing up the per hour filling or emptying capacities of the pipes. The per hour emptying capacities are subtracted and filling capacities are added.

**second:**the result of adding and subtracting is inverted to get the time that the pipes will take to fill up or empty the cistern working together.

#### First example of pipes and cisterns problem

Two pipes fill up a cistern in 3 hours and 6 hours respectively working independently. A third pipe empties the cistern in 4 hours working independently. How long will it take to fill or empty the cistern if all three pipes work together?

Portion of cistern filled or emptied by the effect of three pipes working together in 1 hour is,

$ \displaystyle\frac{1}{3} + \displaystyle\frac{1}{6} - \displaystyle\frac{1}{4}=\displaystyle\frac{1}{2} - \displaystyle\frac{1}{4}=\displaystyle\frac{1}{4} $

Thus the three pipes working together will take 4 hours to fill it up.

The mechanism of "Pipes and Cisterns" is thus similar to the mechanism of "Work and Time" except that here we have emptying pipe while, in work environment there is no work destroying person usually.

#### Second example of pipes and cisterns problem

In a cistern one pipe fills it in 6 hours and another empties it in 4 hours. Both are opened when the cistern is half full. How long will it take to empty the cistern?

**Solution 2:**

In 1 hour the part of the cistern that will be emptied is,

$ \displaystyle\frac{1}{4} - \displaystyle\frac{1}{6} = \displaystyle\frac{1}{12} $

So in 12 hours the full cistern will be emptied and in 6 hours the half-filled cistern will be emptied.

#### Third example of pipes and cisterns problem

A pipe that can fill a cistern in 5 hours actually fills it in 10 hours due to a leak at the bottom of the cistern. If the pipe is closed, how much time will the leak take to empty the cistern?

**Solution 3:**

In 1 hour, the pipe and the leak working together fills up the portion of the cistern,

$ \displaystyle\frac{1}{5} - \displaystyle\frac{1}{\text{leak}} = \displaystyle\frac{1}{10}$

So, $ \displaystyle\frac{1}{\text{leak}} = \displaystyle\frac{1}{5} - \displaystyle\frac{1}{10} = \displaystyle\frac{1}{10} $.

So the leak, if left alone, will empty the filled up tank in 10 hours.

#### Fourth example of pipes and cisterns problem

A leak in a cistern empties the full cistern in 8 mins. If a tap that fills water in the cistern at the rate of 8 litres per minute fills it in 8 mins find the capacity of the cistern.

**Solution 4:**

In a minute the tap and the leak working together fills up the portion of cistern,

$ \displaystyle\frac{1}{\text{tap}} - \displaystyle\frac{1}{8} = \displaystyle\frac{1}{8}$

So,

$ \displaystyle\frac{1}{\text{tap}} = \displaystyle\frac{1}{8} + \displaystyle\frac{1}{8} = \displaystyle\frac{1}{4} $.

So the tap fills up the cistern with the leak closed in 4 mins. Thus the capacity of the cistern is, $4\times{8}=32$ litres.

**Note:** As Pipes and Cisterns problem are very similar to Work time problems, we will apply * the direct Work rate technique for these problems also.* For a filling pipe A, filling rate, the portion of tank filled in 1 hour will be $A$ and for a draining pipe B, draining rate, the portion of tank drained in 1 hour will be $B$. If the two pipes A and B are open simultaneously, the portion of tank filled up in 1 hour will simply be, $A-B$, the effective filling rate of the two pipes working together.

*Usually in the area of Work time problems, there is no work agent*

*equivalent*to a draining pipe, unless an work agent destroys a portion of work per unit time.We will leave you now with 10 related problems as exercise. The answers are at the end.

### Exercise problems

#### Problems exercise 1.

If 3 weavers can weave 14 cotton sarees in 14 days, how many more weavers are required to weave the same 14 cotton sarees in 6 days?

#### Problem exercise 2.

Out of 120 trees 40 labourers cut 40 trees in 4 hours and after that 8 labourers fell ill and left the job. How many hours more will be required to finish cutting all the trees?

#### Problem exercise 3.

A team of 16 men is required to complete an assignment in 10 days. How much should the team strength be reduced so that it takes 16 days to complete the assigment?

#### Problem exercise 4.

Runu can do a piece of work in 6 days and Biju in 8 days. They begin together and with the help of a boy finish the work in 3 days. If they get Rs.2400 as wages, how much should Runu get?

#### Problem exercise 5.

A cistern can be filled up by two pipes A and B in 30 minutes and 20 minutes respectively. Both the pipes were opened together when the cistern was empty. After some time pipe B was stopped and the cistern was filled up in a total time of 18 minutes. When was the pipe B stopped?

#### Problem exercise 6.

A piece of work is done by B in 12 days while A does it in 20 days. A worked for a few days and then B started working in place of A. If the work was completed in a total of 14 days when did A leave?

#### Problem exercise 7.

A, B and C can do a piece of work in 20, 30 and 12 days respectively. A and B begin to do the work and taking help of C every 3rd day complete the work. If they receive Rs.6000 as wages how much did C receive?

#### Problem exercise 8.

If 30 men work, they take 6 more days than 40 men would have taken to complete the work. How many days would 60 men take to complete the work?

#### Problem exercise 9.

Ravi is thrice as fast as Mintu and is able to finish a work in 60 days less than Mintu. Find how long would they take to complete the work together.

#### Problem exercise 10.

7 men and 7 boys can do a work in 2 days. How much time will 1 man and 1 boy take to do the work?

### Answers to problem exercises

Problem 1: 4 more weavers will be required.

Problem 2: 10 hours more and total 14 hours will be required.

Problem 3: Team strength to be reduced by 6 from 16 to 10.

Problem 4: Runu gets Rs.1200 as share.

Problem 5: At 8 minutes from the start.

Problem 6: After 5 days from start.

Problem 7: C receives Rs. 1500 as his share of wage.

Problem 8: In 12 days.

Problem 9: 22.5 days.

Problem 10: 14 days.

### Useful resources to refer to

**7 steps for sure success in SSC CGL Tier 1 and Tier 2 competitive tests**

**How to solve Arithmetic problems on Work-time, Work-wages and Pipes-cisterns**

**Basic concepts on Arithmetic problems on Speed-time-distance Train-running Boat-rivers**

**How to solve a hard CAT level Time and Work problem in a few confident steps 3**

**How to solve a hard CAT level Time and Work problem in a few confident steps 2**

**How to solve a hard CAT level Time and Work problem in few confident steps 1**

**How to solve Work-time problems in simpler steps type 1**

**How to solve Work-time problem in simpler steps type 2 **

**SSC CGL Tier II level Solution Set 10 on Time-work Work-wages Pipes-cisterns 1**

**SSC CGL Tier II level Question Set 10 on Time-work Work-wages Pipes-cisterns 1**

**SSC CGL level Solution Set 67 on Time-work Work-wages Pipes-cisterns 6 **

**SSC CGL level Question Set 67 on Time-work Work-wages Pipes-cisterns 6**

**SSC CGL level Solution Set 66 on Time-Work Work-Wages Pipes-Cisterns 5**

**SSC CGL level Question Set 66 on Time-Work Work-Wages Pipes-Cisterns 5**

**SSC CGL level Solution Set 49 on Time and work in simpler steps 4**

**SSC CGL level Question Set 49 on Time and work in simpler steps 4**

**SSC CGL level Solution Set 48 on Time and work in simpler steps 3**

**SSC CGL level Question Set 48 on Time and work in simpler steps 3**

**SSC CGL level Solution Set 44 on Work-time Pipes-cisterns Speed-time-distance**

**SSC CGL level Question Set 44 on Work-time Pipes-cisterns Speed-time-distance**

**SSC CGL level Solution Set 32 on work-time, work-wage, pipes-cisterns**

**SSC CGL level Question Set 32 on work-time, work-wages, pipes-cisterns**