Chain Rule

When two or more elements are given in a scenario than chain rule can be applied. There are two figures in each of these two elements. While one figure is missing in one of the elements. To find out this missing part of the figure we use the chain rule.  Suppose in the question, ‘days’ part is missing. Thus, you will compare ‘days’ with other elements independently to find out the element and ignore the other elements. Let us see chain rule examples. 

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Principles of the Chain Rule

When you do the comparison there are mainly two principles that have to be followed:

1. If the missing part is not greater than the given part than the numerator should also be small than the denominator. That is, it should be a/b < 1. Here a and b are the part given in the other elements.
2. If the given part is smaller than the missing part, then the denominator should also be kept smaller than the numerator. Thus, it should be a/b > 1. Here also a and b are the given parts of the other element.

Chain Rule Examples

Some of the types of chain rule problems that are asked in the exam.

Question 1

In school, there are some chocolates for 240 adults and 400 children. If the chocolates are taken away by 300 children, then how many adults will be provided with the remaining chocolates?

Here it is clearly given that there are chocolates for 400 children and 300 of them has already taken the chocolates. So, the total remaining children that need to be provided with chocolates is 400 – 300, which is 100.

It is also given in the question that 400 children are equal to 240 adults. So, 1 child = 240/400 adults. Therefore, 100 children will be the same as 240/400 x 100 adults. This equals up to 60 adults. So, the required answer over here is 60 adults.

Learn more about Geometric Series here in detail.

Question 2

It takes 48 laborers to construct a house in 24 days. Find out the number of days that 36 laborers will take to build a house.

This is a case of indirect proportion. This means that less the number of people more will be the days taken by them. Suppose, the required number of days is ‘x’. So, the laborers and days here will be in ratio.

This will be 36: 48:: 24: x. Also, note that because of indirect proportion, 48/36 will be written as 36/48. So, x = 48 x 24/36 => 32. So the required number of days is 32.

Question 3

Suppose that a bike travels ‘a’ km in ‘b’ liters. Find the amount of petrol the bike requires for 7/2 of ‘a’ km.

Suppose the required amount of petrol to be found out is ‘X’ liters. The bike, in this case, uses ‘b’ liters to travel ‘a’ km. More the amount of traveling (km), more will be the petrol required (liters).

So, the ratio of petrol: the ratio of km. Thus, a : 7a/2 :: b : X. Therefore, 7ab/2a = 7b/2. So, the answer required is 7/2 liters of petrol.

Practice Questions

Q. A bus travels 11.25 km of the journey with 750 ml of fuel. Find the cost to travel 63.75 km of 1 liter of petrol is priced at Rs. 61.

A. 260                    B. 259.25                  C. 275.25                D. 260.25

Answer: B. 259.25

Q. It takes 6 days to complete 2/5th of the work. How many more days will it take to complete the work at this rate?

A. 6 days               B. 9 days                C. 12 days                 D. 15 days

Answer: B. 9 days