When quantities of different kinds are connected to one another so that we know how much of one quantity is equivalent to a given quantity of a second, etc. We can determine how much of the last kind is equivalent to a given of first kind by the chain rule.

Proportion and Indirect Proportion

If increase or decrease of a quantity $Q_{1}$ causes increase or decrease of another quantity $Q_{2}$ in the same extent then, $Q_{1}$ is directly proportional to $Q_{2}\Rightarrow Q_{1}\propto Q_{2}$

1) Number of persons $\propto$ Amount of work done, i.e. more persons, more work

2) Number of days $\propto$ Amount of work, i.e. more days, more work

3) Working rate $\propto$ Amount of work, i.e. more working rate, more work

4) Efficiency of man $\propto$ Amount of work, i.e. more efficiency of man, more work
Combining I,II,III,and IV, $(Man\times Days\times Work \,rate\times Efficiency) \propto$ Amount of work. If increase of a quantity $Q_{1} is\, indirectly \,proportional\, to \,Q_{2}\Rightarrow Q_{1}\propto \frac{1}{Q_{2}}$

5) Number of men $\propto\frac{1} {No.\,of \,days}$, i.e. more the men, less the no. of days required

Chain Rule Important Formulas

1) $\frac{Man_{1}\times Days_{1}\times Work\, rate_{1}}{Amount \,of\, work\,done_{1}}=\frac{Man_{2}\times Days_{2}\times Work\,rate_{2}}{Amount \,of \,work\,done_{2}}$

Remember, “Man days” required per unit work is always same. In fact, $Man \times Days$ specify the volume of job or work.

2) If in place of men there are engines burning coal for certain number of hours, then, the above equation changes to

$\frac{Number\, of \,engine_{1}\times Hours_{1}\times Consumption \,Rate_{1}}{Amount\, of\, coal \,burnt_{1}}=\frac{Engine_{2}\times Hours_{2}\times Consumption \,Rate_{2}}{Amount\, of\, coal\, burnt_{2}}$

Because, here the job of engine is to burn the coal.

3) If number of examiners examining a number of answer books in number of days by working a number of hours or day, since the job of examiner is to check the answer books,

then, $\frac{Number\, of\, examiner_{1}\times Days_{1}\times Work\, Rate_{1}}{No.\,of \,answer\, books checked_{1}}=\frac{Examiner_{2}\times Days_{2}\times Work\, Rate_{2}}{No.\,of\, answer\, books\, checked_{2}}$

Chain Rule Solved Examples and Solutions

Question 1) A garrison of 3000 men has provisions for 25 days, when given at the rate of 900 g per head. At the end of 11 days, a reinforcement arrives and it was found that now the provision will last 10 days more, when given at the rate of 840 g per head. What is the strength of reinforcement ? (MBA, ’82)

Solution : Let strength of reinforcement be x
Remaining food provisions after 11 days =$3000\times (25-11)\times 900$

Total men after 11 days=(3000+x)

$\therefore 3000\times 14\times 900=(3000+x)\times 10\times 840$

$\Rightarrow$ x=1,500.

$\therefore$ The reinforcement had 1,500 men.

Question 2) Six diesel engines consume 900 litres of diesel, when each one is running for 5 h a day. How much diesel will be required by 9 engines, each running 8 h a day when 5 diesel engines of former type consume as much diesel as 8 diesel engines of the latter type.

Solution : Using the formula,

$N_{1}\times D_{1}\times \frac{R_{1}}{W_{1}} = N_{2}\times D_{2}\times \frac{R_{2}}{W_{2}}$

Since 5 diesel engines of I type=8 diesel engines of II type

$R_{1}=\frac{1}{5}and R_{2}=\frac{1}{8}$

$\therefore W_{1}=900 litres, W_{2}$ = ?

(Since Amount of work = Diesel consumption)

$\Rightarrow \frac{6\times 5\times \frac{1}{5}}{900}=\frac{9\times 8\times \frac{1}{8}}{W_{2}}$

$\Rightarrow {W_{2}}$=1,350 litres.

$\therefore$ The diesel required is 1,350 litres

Question 3) Two coal loading machines each working 12 hours per day for 8 days handles 9,000 tonnes of coal with an efficiency of 90%. While 3 other coal loading machines at an efficiency of 80% set to handle 12,000 tonnes of coal in 6 days. Find how many hours per day each should work.

Solution: Here $\frac{N_{1}\times D_{1}\times R_{1}\times E_{1}}{W_{1}}$

=$\frac{N_{2}\times D_{2}\times R_{2}\times E_{2}}{W_{2}}$

$N_{1}=2 R_{1}=12\frac{h}{day};N_{2}=3 R_{2}=?$

$E_{1}=\frac{90}{100}W_{1}=9,000;$

$E_{2}=\frac{80}{100}W_{2}=12,000$

$\Rightarrow \frac{2\times 8\times 12\times 90}{9,000\times 100}=\frac{3\times 6\times R_{2}\times 80}{12,000\times 100}$

$Rightarrow\, R_{2}=16 \frac{d}{day}$

$\therefore$ Each machine should work $16\frac{d}{day}$

Question 4) ’A’ can do a piece of work in $2\frac{1}{2}$ days which ‘B’ can do in $3\frac{1}{2}$ days. If ‘A’ wages are Rs.50 per week and ‘B’ wages are Rs.42.50 per week, what ‘A’ would have charged for doing a piece of work for which B received Rs 340? (ITI, ’90)
Solution : Total wage = Total Amount of work; Wage rate = work Rate

$\therefore \frac{N_{1}\times D_{1}\times R_{1} }{W_{1}}=\frac{N_{2}\times D_{2}\times R_{2}}{W}$

$\frac{1\times \frac{5}{2}\times 50}{W_{1}}=\frac{1\times \frac{10}{3}\times 42.50}{340}$

$\Rightarrow W_{1}$ = 300

$\therefore$ ‘A’ should charge Rs 300 for the job.

Question 5) Four lorries carrying 4 tons each move 128 tons in 8 days. In how many days will 6 tons in 8 days. In how many days will 6 lorries carrying 3 tons each move 540 tons ?

Solution : $\frac{N_{1}\times D_{1}\times R_{1}}{W_{1}} =\frac{N_{2}\times D_{2}\times R_{2}}{W_{2}}$

$\frac{4\times 8\times 4}{128}=\frac{6\times D_{2}\times 3}{540}$

$\Rightarrow D_{2}$ = 30 days.

Chain Rule Questions from previous year exams

Chain Rule Important Videos - Must Watch