# Concept: Time & Work Quick Revision

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**TIME and WORK**

While solving questions we may assume work to be done to be either 1 unit (Unitary Method) or the LCM of no. of days given in the question (LCM Method).

- Efficiency = Work done in 1 day = $\frac{\mathrm{Work\; done}}{\mathrm{Time\; taken}}$
- Work done = Efficiency × Time
- Time = $\frac{\mathrm{Work\; done}}{\mathrm{Efficiency}}$
- We can add individual efficiencies to get combined efficiency of 2 or more persons working together.
- We cannot add individual time to get combined time taken by 2 or more persons working together.

Here, we assume the total work to be done in 1 unit.

If X alone completes a work in ‘a’ days and Y alone completes the same work in ‘b’ days.

Hence, efficiency of X = $\frac{1}{a}$ and that of Y = $\frac{1}{b}$.

Now, if X and Y work together: their combined efficiency = sum of their individual efficiencies.

Hence, their combined efficiency = $\frac{1}{a}$+$\frac{1}{b}$

Time taken by them to complete the whole work together (n) = $\frac{\mathrm{Work\; to\; be\; done}}{\mathrm{combined\; efficiency}}$ = $\frac{1}{{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}}$

where n is the number of days taken by them complete the same work together.

This can also be written as: $\frac{1}{n}$ = $\frac{1}{a}$+$\frac{1}{b}$

Here, 1/n denotes the combined efficiency of X and Y.

When, there are only two people, this formula can be simplified to n = $\frac{ab}{a+b}$

Here, we assume the total work to be done as the LCM of the number of days given in the question. The rest of the process is exactly same as the Unitary method.

When it is given that X is twice as efficient Y, it means X does twice the work as Y in the same time, or X will require half the time required by Y to do the same work.

Hence, if

T_{X} = time taken by X

T_{Y} = time taken by Y

T_{Z} = time taken by Z

e_{X} = efficiency of X

e_{Y} = efficiency of Y

_{Z}= efficiency of Z

then, ${\mathrm{e}}_{\mathrm{X}}:{\mathrm{e}}_{\mathrm{Y}}:{\mathrm{e}}_{\mathrm{Z}}=\frac{1}{{\mathrm{T}}_{\mathrm{X}}}:\frac{1}{{\mathrm{T}}_{\mathrm{Y}}}:\frac{1}{{\mathrm{T}}_{\mathrm{Z}}}$

**WORKING IN SHIFTS**

In such type of questions, some people may or may not work for the entire duration. We can use either of the Unitary or LCM methods to solve these questions.

**Example**: A can do a work in 30 days. He worked for 5 days and the remaining work was
finished by B in 15 days. How long does it take for B alone to finish the work.

**Solution**:

Work done by A in 5 days = $\frac{1}{30}$ × 5 = $\frac{1}{6}$

Remaining work = 1 - $\frac{1}{6}$ = $\frac{5}{6}$

Now 5/6 of the work is done by B in 15 days.

∴ Whole work is done by B in 15 × $\frac{6}{5}$ = 18 days

The same question can be solved by considering the work done in difference periods of time.

**Example**: A can do a work in 30 days. He worked for 5 days and the remaining work was
finished by B in 15 days. How long does it take for B alone to finish the work.

**Solution**:

Let the time taken by B alone = B days.

Let us use the unitary method here. Hence, the total work to be done = 1 unit.

⇒ 1 = Work done in first 5 days + Work done in next 15 days.

⇒ 1 = e_{A} × 5 + e_{B} × 15

⇒ 1 = $\frac{1}{30}$ × 5 + $\frac{1}{\mathrm{B}}$ × 15

⇒ 1 = $\frac{1}{6}$ + $\frac{15}{\mathrm{B}}$

⇒ $\frac{5}{6}$ = $\frac{15}{\mathrm{B}}$

⇒ B = 18 days

Such type of questions can also be solved by considering the work done by different people.

Total work done = Sum of the work done by people involved.

**Example**: A can do a work in 20 days. He worked for 5 days and the remaining work was
finished by A and B together in 9 days. How long does it take for B alone to finish the work.

**Solution**:

Let the time taken by B alone = B days.

Let us use the unitary method here. Hence, the total work to be done = 1 unit.

⇒ 1 = Work done by A + Work done by B.

⇒ 1 = e_{A} × 14 + e_{B} × 9

⇒ 1 = $\frac{1}{20}$ × 14 + $\frac{1}{\mathrm{B}}$ × 9

⇒ 1 = $\frac{7}{10}$ + $\frac{9}{\mathrm{B}}$

⇒ $\frac{3}{10}$ = $\frac{9}{\mathrm{B}}$

⇒ B = 30 days

**PAYMENTS**

When payment is to be divided for a certain work, it should be divided in proportion of work done by each person.

Hence, ratio of payment = ratio of work done.

P_{A} : P_{B} : P_{C} = e_{A} × T_{A} : e_{B} × T_{B} : e_{C} × T_{C}

Here, P_{A}, e_{A} & T_{A} are Payment, efficiency and Time taken by A. [Same goes for B & C as well.]

When, everyone works for same amount of time, ratio of payment

P_{A} : P_{B} : P_{C} = e_{A} : e_{B} : e_{C}

**WORKING IN ROTATION**

When people are working on alternate/rotional basis, we calculate the work done per cycle and then calculate the number of cycles required to complete the work.

When number of cycles is an integer, we can multiply the number of cycles with days per cycle to calculate total days.

Even in this case, if one of the persons/pipes working do negative work, then we'll have to analyse the last cycle.

When the number of cycles is not an integer, we need to break them into complete and incomplete cycles. In such cases it matters who starts the work.

**MAN-DAYS**

The underlying assumption in such type of questions is that everyone works with same efficiency.

$\frac{{M}_{1}\times {H}_{1}\times {D}_{1}}{{W}_{1}}=\frac{{M}_{2}\times {H}_{2}\times {D}_{2}}{{W}_{2}}$

We can also have questions where people will have different efficiencies. In such cases we assume different efficiencies of people and compare the total work done by different group of people.

Example: If a men and b women complete a work in x days, while c men and d women complete twice work in y days, then

Work done by a men and b women in x days = (am + bw) × x [m and w are efficies of men and women resp.]

Work done by c men and d women in y days = (cm + dw) × y [m and w are efficies of men and women resp.]

Finally comparing work done, we get

⇒ (cm + dw) × y = 2 × (am + bw) × x

**PIPES & CISTERNS**

More or less the concepts used to solve problems on Pipes & Cisterns are same as in Time & Work. Here the work done is in terms of filling or emptying a cistern.

Usually filling the tank is taken as positive work done and emptying the tank is taken as negative work done.

Efficiency of a pipe ∝ cross-section of the pipe (i.e. Area)

Efficiency of a pipe ∝ speed of the flow of water

∴ Efficiency of a pipe ∝ Area × (speed of the flow of water)

For a circular cross-section pipe

Area ∝ (radius)^{2}

Hence, Efficiency of a pipe ∝ r^{2} × (speed of the flow of water)