In every competitive exam problems related to Time, Work and wages concepts are usually asked, As there is a limited number of types in this time and work concepts you can ensure your marks with minimum efforts. Let Time taken by a group or one/two persons in doing certain works or number of persons required to complete a certain work are commonly asked. Questions based on the comparison of men, women, children works, efficiency, strength are also asked.
Relation Between Time, Work and Capacity
\begin{align*}
Capacity \ =\dfrac { Work }{ Time }
\end{align*}
also written as
\begin{align*}
\begin{pmatrix} Capacity \\ or \\ strength \\ or \\efficiency \end{pmatrix}=\dfrac {Work}{Time}
\end{align*}
Time allotted and men engaged in a project are inversely proportional to one another.
Here the terms capacity, strength and efficiency are same and are used according to situations.
.·. We have
\begin{align*}
\begin{pmatrix} Capacity \\ or \\ strength \\ or \\efficiency \end{pmatrix}=\dfrac {W}{M\times D\times H}\end{align*}
Here W=Work, M=number of persons, D=Days, H=Hours/day.
Formulas Of Time And Work Concepts
1.If A can do a piece of work in n days then at a uniform rate of working
\begin{align*}A \ will \ finish \ \left( 1/n\right) ^{th} \ work \ in \ one \ day.
\end{align*}
2. Let A, B, C can do a piece of work in T1, T2, T3 days respectively. If they have worked for D1, D2, D3 days respectively then,
\begin{align*}
Amount \ of \ work \ done \ by \ A \ is \left( \dfrac {D_{1}}{T_{1}}\right)
\end{align*} \begin{align*}
Amount \ of \ work \ done \ by \ B \ is \left( \dfrac {D_{2}}{T_{2}}\right)
\end{align*} \begin{align*}
Amount \ of \ work \ done \ by \ C \ is \left( \dfrac {D_{3}}{T_{3}}\right)
\end{align*}
The amount of work done by A, B, C together will be equal to 1 if the work is completed. \begin{align*}i, e \
\left[ \dfrac {D_{1}}{T_{1}}+\dfrac {D_{2}}{T_{2}}+\dfrac {D_{3}}{T_{3}}\right] =1 \end{align*}
3. Let M1 men can finish W1 work in D1 days and M2 men can finish W2 work in D2 days, then\begin{align*} \dfrac {M_{1}\times D_{1}}{W_{1}}=\dfrac {M_{2}\times D_{2}}{W_{2}} \end{align*}
4. Let M1 men can finish W1 work in D1 days working T1 time each day and M2 men can finish W2 work in D2 days working T2 time each day, then \begin{align*} \dfrac {M_{1}\times D_{1}\times T_{1}}{W_{1}}=\dfrac {M_{2}\times D_{2}\times T_{2}}{W_{2}} \end{align*}
5. Let A can do a piece of work in X days and B can do the same work in Y days,
then both of them working together can complete the same work in \begin{align*} \dfrac {XY}{X+Y} \ days. \end{align*}
6. Let A, B, and C working alone can do a piece of work in X, Y, and Z days then
all of them working together can complete the same work in \begin{align*} \dfrac {XYZ}{XY+YZ+ZX} \ days. \end{align*}
7. Let A and B working together can finish a piece of work in X days, B and C in Y days, C and A in Z days, then
- A, B and C operating along will complete the work in \begin{align*} \dfrac {2XYZ}{XY+YZ+ZX} \ days. \end{align*}
- A alone can complete the work in \begin{align*} \dfrac {2XYZ}{XY+YZ-ZX} \ days. \end{align*}
- B alone can complete the work in \begin{align*} \dfrac {2XYZ}{YZ+ZX-XY} \ days. \end{align*}
- C alone can complete the work in \begin{align*} \dfrac {2XYZ}{XY+ZX-YZ} \ days. \end{align*}
8. Two persons A and B working together can complete a piece of work in X days. If A operating alone will complete the work in Y days, then
B working alone can complete the work in \begin{align*} \dfrac {XY}{X-Y} \ days \end{align*}
9. Let A can do a piece of work in X days and B can do the same work in Y days and when they started working together
- A has left the work m days before completion of the work then The
time taken to complete work is \begin{align*} \left[ \dfrac {\left( X+m\right) Y}{X+Y}\right] \ days. \end{align*} - B left the work m days before completion of the work then
The time taken to complete work is\begin{align*} \left[ \dfrac {\left( Y+m\right) X}{X+Y}\right] \ days. \end{align*}
10. Let A and B working together will complete a certain work in X days. They worked together for Y days and so B left the work. A finished the rest of the work in D days. Then
total time taken by A alone to complete the work is \begin{align*} \dfrac {Xd}{X-Y} \ days \end{align*}
11. Consider a certain amount of food is available for X days for A men at a certain place and after Y days, B men join then
the remaining food will serve men for \begin{align*} \left[ \dfrac {A\left( X-Y\right) }{A+B}\right] \ days. \end{align*}
12. Let the certain amount of food is available for X days for A men at a certain place and after Y days, B men leave then
the remaining food will serve men for \begin{align*} \left[ \dfrac {A\left( X-Y\right) }{A-B}\right] \ days. \end{align*}
13. Let some individuals complete a certain work in X days. if there were less people then the work would be completed in Y days more
the number of people present initially was \begin{align*} \left[ \dfrac {a\left( x-y\right) }{y}\right] \ people. \end{align*}
