In this chapter, we will be studying about the concept of TIME AND WORK and some rules to solve the problems quickly.

**Time:** It is defined as a sequence of actions and events that take place in present, future, past and time is measured in seconds, minutes, hours, days, weeks, months and years.

**Work:** When a force acts upon an object to cause a displacement of the object, then it is said that a work has been done upon the object.

- Work = Distance
- Rate at which work is done = speed
- Number of days required to do the work = Time

- Worker [latex]\propto[/latex] Work ⇒ More workers can do more work and vice-versa.
- Time [latex]\propto[/latex] Work ⇒ In more time, more work can be done and vice-versa.
- Worker [latex]\propto[/latex] [latex]\frac{1}{Time}[/latex] ⇒ More workers can do more work in less time and vice-versa.

- e.g. If John can finish a work in 10 days, then he will finish [latex]\frac{1}{10}[/latex] part of the work in 1 day.
- e.g. If Mary can finishes [latex]\frac{1}{7}[/latex] part of a work in 1 days, then she will finish the complete work in7 days.

- [latex]\frac{M_{1}D_{1}}{W_{1}}[/latex] = [latex]\frac{M_{2}D_{2}}{W_{2}}[/latex]

- Here [latex]M_{1}[/latex] = 3, [latex]W_{1}[/latex] = [latex]\frac{1}{3}[/latex], [latex]D_{1}[/latex] = 6, and [latex]M_{2}[/latex] = 4, [latex]w_{2}[/latex] = [latex]\frac{1}{2}[/latex]. We have to find out [latex]D_{2}[/latex]. Using Rule 2, we have.

[latex]\frac{M_{1}D_{1}}{W_{1}}[/latex] = [latex]\frac{M_{2}D_{2}}{W_{2}}[/latex] = [latex]\frac{3 \times 6}{\frac{1}{3}}[/latex] = [latex]\frac{4 \times D_{2}}{\frac{1}{2}}[/latex] ⇒ [latex]D_{2}[/latex] = 6.75 Days.

- [latex]\frac{M_{1}D_{1}T_{1}}{W_{1}}[/latex] = [latex]\frac{M_{2}D_{2}T_{2}}{W_{2}}[/latex]

- Here [latex]M_{1}[/latex] = 4, [latex]W_{1}[/latex] = [latex]\frac{1}{5}[/latex], [latex]D_{1}[/latex] = 4, [latex]T_{1}[/latex] = 3.5, and [latex]M_{2}[/latex] = 2, [latex]W_{2}[/latex] = ?, [latex]T_{2}[/latex] = 9, [latex]D_{2}[/latex] = 5.

We have to find out [latex]W_{2}[/latex]. Usin

[latex]\frac{M_{1}D_{1}T_{1}}{W_{1}}[/latex] = [latex]\frac{M_{2}D_{2}T_{2}}{W_{2}}[/latex] = [latex]\frac{4 \times 4 \times 3.5}{\frac{1}{5}}[/latex] = [latex]\frac{4 \times 5 \times 9}{W_{2}}[/latex] ⇒ [latex]W_{2}[/latex] = 1.56 (or) 1 and [latex]\frac{5}{9}[/latex] part more of work.

1. Work done by A in 1 day = [latex]\frac{1}{x}[/latex], and work done by B in 1 day = [latex]\frac{1}{y}[/latex]

2. work done by A and B together in 1 day = [latex]\frac{1}{x}[/latex] + [latex]\frac{1}{y}[/latex]

3. Total days taken to complete the work by A and B working together = [latex]\frac{xy}{x + y}[/latex]

1. work done by A, B and C together in 1 day = [latex]\frac{1}{x}[/latex] + [latex]\frac{1}{y}[/latex] + [latex]\frac{1}{z}[/latex]

2. Total days taken to complete the work by A, B and C working together = [latex]\frac{xyz}{xy + yz + zx}[/latex]

- Work of 1 day = [latex]\frac{Total \ work}{Total \ no.of \ working \ days}[/latex]
- Total work = (Work of 1 day) × (Total no. of working days)
- Remaining work = 1 - (Work done)
- Work done by A = (Work done in 1 day by A) × (Total no. of days worked by A)

- A left the work ‘m’ days before completion, then total days taken to complete the work is [latex]\frac{(x + m)y}{x + y}[/latex]
- B left the work ‘m’ days before completion, then total days taken to complete the work is [latex]\frac{(y + m)x}{x + y}[/latex]

Let's see some examples for applying these rules:

- A’s share : B’s share : C’s share

= (B’s time × C’s time) : (A’s time × C’s time) : (A’s time × B’s time)

= 96 : 72 : 48

= 4 : 3 : 2

∴ B’s share = [latex]\frac{1550}{9} \times 3[/latex]= Rs. 516.66.

- Use

- Use

- Total work of X and Y: [latex]\frac{1}{8}[/latex] + [latex]\frac{1}{24}[/latex] = [latex]\frac{1}{6}[/latex], Therefore, Working together, machines X and Y can produce 1,000 bolts in 6 hours.

- Ratio of work done by A and B = 3 : 1.

Ratio of time taken by A and B to finish a work = 1 : 3

- a) If [latex]M_{1}[/latex] persons can do [latex]W_{1}[/latex] work in [latex]D_{1}[/latex] days and [latex]M_{2}[/latex] persons can
do [latex]W_{2}[/latex] work in [latex]D_{2}[/latex] days then, [latex]M_{1}W_{2}D_{1}[/latex]= [latex]M_{2}W_{1}D_{2}[/latex]

b) If the persons work for [latex]T_{1}[/latex] and [latex]T_{2}[/latex] hours per day respectively then, [latex]M_{1}W_{2}D_{1}T_{1}[/latex]= [latex]M_{2}W_{1}D_{2}T_{2}[/latex]

c)If the persons have efficiency of [latex]E_{1}[/latex] and [latex]E_{2}[/latex] respectively then, [latex]M_{1}W_{2}D_{1}T_{1}E_{1}[/latex]= [latex]M_{2}W_{1}D_{2}T_{2}E_{2}[/latex]

- By A alone in [latex]\frac{2XYZ}{XY + YZ - ZX}[/latex]days

By B alone in [latex]\frac{2XYZ}{YZ + ZX - XY}[/latex]days

By C alone in [latex]\frac{2XYZ}{ZX + XY - YZ}[/latex]days

By A, B, and C together in [latex]\frac{2XYZ}{XY + YZ + ZX}[/latex]days

- [latex]\frac{D_1 * D_2 * D_3} {D_1D_2 + D_2D_3 + D_3D_1}[/latex]

- Given that,

A can do the work in 6 hours ⇒ A's 1 hour work = [latex]\frac{1}{6}[/latex]

B can do the work in 12 hours ⇒ B's hour work = [latex]\frac{1}{12}[/latex]

Then, A's an B's work in 1 hour is

(A + B)'s work = [latex]\frac{1}{6}[/latex] + [latex]\frac{1}{12}[/latex] = [latex]\frac{1}{2}[/latex]

Therefore, both can do the work in [latex]\frac{2}{1}[/latex] days = 2 days

- P can do the work in 8 days of 10 hours ⇒ 8 x 10 ⇒ 80 hours

Q can do the work in 6 days of 8 hours ⇒ 6 x 8 ⇒ 48 hours

Now, P's 1 hour work = [latex]\frac{1}{80}[/latex]

Q's 1 hour work = [latex]\frac{1}{48}[/latex]

(P + Q)'s 1 hour work = [latex]\frac{1}{80}[/latex] + [latex]\frac{1}{48}[/latex] = [latex]\frac{3 + 5}{240}

[/latex] = [latex]\frac{8}{240}[/latex] = [latex]\frac{1}{30}[/latex]

Both will finish the work in 30 hours

Therefore, number of days of 6[latex]\frac{2}{5}[/latex] hours each = 30 x [latex]\frac{32}{5}[/latex] = 6 x 32

= 192 days

- Given

S and T can do a work in 15 days = (S + T)'s work in 1 hour = [latex]\frac{1}{15}[/latex]

T and U can do a work in 20 days = (T + U)'s work in 1 hour = [latex]\frac{1}{20}[/latex]

U and S can do a work in 30 days = (U + S)'s work in 1 hour = [latex]\frac{1}{30}[/latex]

By adding them all,

(S + T + T + U + U + S) = [latex]\frac{1}{15}[/latex] + [latex]\frac{1}{20}[/latex] + [latex]\frac{1}{30}[/latex]

⇒ 2(S + T + U) = [latex]\frac{4 + 3 + 2}{60}[/latex]

⇒ 2(S + T + U) = [latex]\frac{9}{60}[/latex]

⇒ (S + T + U) = [latex]\frac{9}{60}[/latex] x [latex]\frac{1}{2}[/latex]

⇒ (S + T + U) = [latex]\frac{3}{20}[/latex] x [latex]\frac{1}{2}[/latex]

⇒ (S + T + U) = [latex]\frac{3}{40}[/latex]

Thus, S, T and U together can do the work in [latex]\frac{40}{3}[/latex] days

Now, S's 1 day work = [((S + T + U)'s work in 1 hour] -[(T + U)'s work in 1 hour]

⇒ [latex]\frac{3}{40}[/latex] - [latex]\frac{1}{20}[/latex]

⇒ [latex]\frac{3 - 2}{40}[/latex]

⇒ [latex]\frac{1}{40}[/latex]

Therefore, S alone can do the work in 40 days

Similarly, T's 1 day work = [latex]\frac{3}{40}[/latex] - [latex]\frac{1}{30}[/latex]

⇒ [latex]\frac{9 - 4}{120}[/latex]

⇒ [latex]\frac{5}{120}[/latex]

⇒ [latex]\frac{1}{24}[/latex]

Therefore, T alone can do the work in 24 days

U's 1 day work = [latex]\frac{3}{40}[/latex] - [latex]\frac{1}{15}[/latex]

⇒ [latex]\frac{9 - 8}{120}[/latex]

⇒ [latex]\frac{1}{120}[/latex]

Therefore, U alone can do the work in 120 days

- Let,

Number of men [latex]M_1[/latex] = 5

Number of cycles [latex]W_1[/latex] = 10

Number of days [latex]D_1[/latex] = 6

Number of hours [latex]T_1[/latex] = 6

Number of men [latex]M_2[/latex] = 12

Number of cycles [latex]W_2[/latex] = 16

Number of hours [latex]T_2[/latex] = 8

Number of days [latex]D_2[/latex] = ?

Now, consider

[latex]M_{1}W_{2}D_{1}T_{1}[/latex]= [latex]M_{2}W_{1}D_{2}T_{2}[/latex]

substitute the given values, i.e.

5 x 6 x 6 x 16 = [latex]D_{2}[/latex] x 12 x 10 x 8

⇒ [latex]D_{2}[/latex] x 12 x 10 x 8 = 2880

⇒ [latex]D_{2}[/latex] = [latex]\frac{2880}{960}[/latex]

⇒ [latex]D_{2}[/latex] = 3 days

Therefore, Required number of days = 3 days.

- Given

Number of work days of A = [latex]D_1[/latex] = 10 days

Number of work days of B = [latex]D_2[/latex] = 15 days

Now, consider

[latex]\frac{D_1 * D_2}{D_1 + D_2}[/latex]

⇒ [latex]\frac{10 * 15}{10 + 15}[/latex]

⇒ [latex]\frac{150}{25}[/latex]

⇒ 6

Therefore, A and B can do the work together in 6 days.