Chain Rule - Important Formulas, Questions and Study Material

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}=?



 \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.


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