$M$ = Men or women or any person
$E$ = Efficiency of the person
$D$ = Days
$H$ = Hours
$W$ = Work
Note: In general $Inputs$ are on the left side,
$Outputs$ are on the right side of the equation. Here, Men, efficiency etc are
inputs. Work is output. Solved Example:
If 12 carpenters working 6 hours a day can make 30 chairs in 24 days, how
many chairs will 18 carpenters make in 36 days, each working 8 hours a
day?
Solution:
Let us prepare small table to understand the problem.
$\begin{array}{*{20}{c}}
{Men}&{Hours}&{Days}&{Chairs} \\ \hline
{12}&6&{24}&{30} \\
{18}&8&{36}&? \\
\end{array}$
Formula: $\dfrac{{{M_1}{E_1}{D_1}{H_1}}}{{{M_2}{E_2}{D_2}{H_2}}} =
\dfrac{{{W_1}}}{{{W_2}}}$
$\therefore \dfrac{{12 \times 6 \times 24}}{{18 \times 8 \times 36}} =
\dfrac{{30}}{x}$
($\because $ E is omitted as the efficicney is same.)
$\Rightarrow \require{cancel}\dfrac{{{12} \times 6 \times
\cancel{24}^2}}{{{18}\times 8 \times \cancel{36}^3}} = \dfrac{{30}}{x}$
$\Rightarrow \require{cancel}\dfrac{{{12} \times \cancel6 \times
\cancel{24}^2}}{{\cancel{18}_3 \times 8 \times \cancel{36}^3}} =
\dfrac{{30}}{x}$
$\Rightarrow \require{cancel}\dfrac{{\cancel{12}^4 \times \cancel6 \times
\cancel{24}^2}}{{\cancel{18}_\cancel3 \times 8 \times \cancel{36}^3}} =
\dfrac{{30}}{x}$
$\Rightarrow x = 90$
Exercise
136 men can complete a piece of work in 18 days
working 6 hours a day. In how many days will 27 men complete the same work
working 8 hours a day?
3If five cats can kill five mice in five minutes,
how long will it take 100 cats to kill 100 mice?
A5 B1 C100 D500
Answer: A
Explanation:
$\begin{array}{*{20}{c}}
{cats}&{Time}&{Work} \\ \hline
{5}&5&{5} \\
{100}&?&{100} \\
\end{array}$
Formula: $\dfrac{{{M_1}{E_1}{D_1}{H_1}}}{{{M_2}{E_2}{D_2}{H_2}}} =
\dfrac{{{W_1}}}{{{W_2}}}$
Work = mice. Also, Time in given in minutues and days are not given.
$\therefore \dfrac{{5 \times 5}}{{100 \times x}} = \dfrac{5}{{100}}$
$\Rightarrow \require{cancel} \dfrac{{\cancel5 \times 5}}{\cancel{100}
\times {x}} = \dfrac {\cancel{5}}{\cancel{100}}$
$\Rightarrow x = 5$
436 men can complete a piece of work in 18 days
working 6 hours a day. In how many days will 27 men with half the efficiency
complete "double" the work working 8 hours a day?
A18 B36 C72 D108
Answer: C
Explanation:
If the efficiciency of the former men is $2$ units then later men is $1$
unit.
$\begin{array}{*{20}{c}}
{Men}&{Effi.}&{Days}&{Hrs}&{Work} \\ \hline
{36}&2&{18}&{6}&{1} \\
{27}&1&{?}&{8}&{2} \\
\end{array}$
Formula: $\dfrac{{{M_1}{E_1}{D_1}{H_1}}}{{{M_2}{E_2}{D_2}{H_2}}} =
\dfrac{{{W_1}}}{{{W_2}}}$
$\require{cancel}\therefore \dfrac{{{36} \times 2 \times 18 \times 6}}{{27
\times 1 \times x \times 8}} = \dfrac{{1}}{2}$
$\Rightarrow x = 72$
5If 18 pumps of 25 watts can raise 250 tonnes of
water to a height of 15 mts in 10 days working 7 hours a day, how many pumps
of 40 wtts will be required to raise 200 tonnes of water to a height of 12
mts in 14 days, working 9 hours a day?
A24 B8 C12 D4
Answer: D
Explanation:
Here, Height is output as it requires inputs to pump.
$\begin{array}{*{20}{c}}
{Pumps}&{Watts}&{Days}&{Hrs}&{Work}&{Ht.}\\ \hline
{36}&25&{10}&{7}&{250}&{15} \\
{x}&40&{14}&{9}&{200}&{12} \\
\end{array}$
$\require{cancel}\therefore \dfrac{{{18} \times 25 \times 10 \times 7 }}{{x
\times 40 \times 14 \times 9}} = \dfrac{{250} \times 15}{200 \times 12}$
$\Rightarrow x = 4$
6If 30 men working 7 hours a day can make 12
tables in 18 days, how many days will 45 women working 9 hours a day take to
make 32 chairs? Given, 4 men can make 3 tables in the same time as 3 women
can make 4 chairs.
A42 B21 C14 DCannot be determined
Answer: C
Explanation:
We cannot directly apply chain rule here as work is different in both cases
as well as on the first job men are working and on the second women are
working.
Let the efficiency of a man = $m$, Woman = $w$
Also, a Table= $t$ units, chair = $c$ units.
$\therefore \dfrac{{30 \times m \times 18 \times 7}}{{45 \times w \times d
\times 9}} = \dfrac{{12t}}{{32c}} \tag{1}$
It is given, 4 men can make 3 tables in the same time as 3 women can make 4
chairs.
$\therefore \dfrac{{4 \times m}}{{3 \times w}} = \dfrac{{3t}}{{4c}}$
$\therefore \dfrac{m}{w} = \dfrac{{9t}}{{16c}} \tag{2}$
Substituting equation $(2)$ in equation $(1)$,
$\therefore \dfrac{{30 \times 9t \times 18 \times 7}}{{45 \times 16c \times
d \times 9}} = \dfrac{{12t}}{{32c}}$
$\therefore d = 14$
7If 9 engines consume 24 metric tonnes of coal,
when each is working 8 hours per day, how much coal will be required for 8
engines, each running 13 hours a day, it is given that 3 engines of former
type consume as much as 4 engines of latter type?
A13 B26 C32 DCannot be determined
Answer: B
Explanation:
Let the engine one capacity = $E_1$ and engine two capacity = $E_2$
Given,
$\therefore \dfrac{{9 \times {E_1} \times 8}}{{8 \times {E_2} \times 13}} =
\dfrac{{24}}{x} \tag{1}$
Given, 3 engines of former type consume as much as 4 engines of latter
type
$\therefore \dfrac{{3 \times {E_1}}}{{4 \times {E_2}}} = \dfrac{1}{1}$
$\therefore \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{4}{3} \tag{2} $
Substituting equation $(2)$ in $(1)$,
$\therefore \dfrac{{9 \times 4 \times 8}}{{8 \times 3 \times 13}} =
\dfrac{{24}}{x}$
$\therefore x = 26$
