**GETTING STARTED**

☛ If a person completes a work in **X days, then his /her one day work will be **** 1/X**

Meaning, if his total work has X parts, he would have completed 1 out of those X parts, where X is the number of days.

Eg. If a person can complete a work in 12 days, he will complete 1/12th work in 1 day.

**Conversely, if a day’s work of a person is 1/X , he will complete the work in X days**.

**Two People Working on a Job Together**

Let us say, **Ram completes a job in A days** and **Shyam takes B days to complete the same job**.

The total time taken by them is calculated as follows:

Ram’s one day work = 1/A

Shyam’s one day work = 1/B

**Total one day work = 1/A + 1/B = (A+B)/AB**

Thus the **work will be completed in AB/(A+B)** days.

**Eg. Bheem completes building a bridge in 4 days, and Hanuman completes building a bridge in 2 days. How many bridges will the complete together in 8 days?**

**Soln.**

Bheem’s one day work = 1/4

Hanuman’s one day work = 1/2

**Their one day work together = 1/4 + 1/2 = 3/4**

Therefore,** they take 4/3 days to construct one bridge together.**

Considering **that they have 8 days,**

**They will build 8/(4/3) = 6 bridges.**

**Concept of Man-Days**

The product of men working on a job and the days taken to finish the job is known as **‘man-days’. **

**SIGNIFICANCE:**

This is the same concept as **Inverse Proportion in Unitary Method**. It is used to find **the number of days 1 person will take to finish the job, given A people finish the job in X days**.

If A people can complete a work in X no. of days, then **one person will complete the work in AX** of days.

**Eg. Food supplies in fort guarded by 10 people will last for 10 days. If the no. of soldiers are increased to 25, how long will the supplies last?**

**Soln:.**

The food supplies are available for **10 soldiers for 10 days. **

Let **1 soldier eat 1 unit of food every day**, then there must be **10 x 10 =100 units of food in the fort**.

**If 1 soldier were to finish it, the food would last for 100 days.**

Given that the no. of soldiers is increased to 25. **Every day 25 units of food will be consumed in the fort. **

Thus, the **food will last for only 100/25 = 4 days.**

**Eg. If 10 persons can complete 2/3 rd task in 20 days, how long will the work be completed if 2 men were to complete the remaining task?**

Soln:. Given that **2/3rd task is completed in 20 days, the task will be completed in**

** 20x 3/2 = 30 days.**

10 men together can complete the task in 30 days.

If **1 man completes 1 unit of work a day, 300 units of work** together form the task.

Already 2/3 x 300 units = 200 units of work are completed.

**Remaining work = 100 units**

**2 persons will complete 2 units of work per day. **

Thus, time taken to complete remaining work = 100/2

= **50 days**

**PIPES & CISTERNS**

The pipes and cisterns problems **are very similar to t****he time and work problems**

☛If **a tap can fill a tank alone in X hours**, in **1 hour it will fill 1/X of the tank**.

**Eg**. **If a tank is filled by a tap in 2 hours**, then in first hour, **1/2 of the tank will be filled**.

The other half of the tank will be filled in the second hour.

☛ If **two taps are opened at the same time **and **one tap fills the tank in X hours while the other tap fills the tank in Y hours**. **Time taken **to fill the tank will be

= **XY/(X+Y) hours**.

Here both the taps are doing the work of filling up the tank.

**Eg. If one tap fills the tank in 8 hours while the other tank fills the tap in 10 hours. Find the total time to fill the tank completely.**

**Soln**:.

Tank filled in 1 hr = 1/8 + 1/10 = **18/80**

Thus, the tank will be completely filled in **80/18** hours.

☛ If one cistern fills the tank in X hours while the other empties the tank in Y hours, the **tank will be filled or emptied depending on the values of X and Y.**

**Eg. A tank is filled by a pipe in 3 hours while another tap, empties it in 5 hours. Will the tank be emptied or filled?**

**Soln:.**

**Here one tap is doing the work of filling the tank while the other tap is doing the work of emptying the tank.**

Work done by **filling tap in 1 hour** =**1/3**

Work done by **emptying tap in 1 hour** = **1/5**

**Work done by filling tap > Work done by emptying tank in 1 hour. So, the tank will be filled**** **

Had the time of emptying tap < Time to fill the tank, tank will be emptied.

In the example, **total work done by both taps = 1/3 – 1/5** (Since one tap is filling and the other is working against it, ie. emptying)

Work done in 1 hour= 1/3 – 1/5 = **2/15**

Therefore, **work will be completed (Tank will be filled) in 15/2 hrs = 7.5 hours.**