Componendo-Dividendo: Concept, Solved Examples & Practice Questions (2025)

Componendo-Dividendo

Componendo and dividendo is a theorem on proportions that allows for a quick way to perform calculations and reduce the number of expansions needed. It is particularly useful when dealing with equations involving fractions or rational functions in mathematical Olympiads, especially when you see fractions.

Hence, if the ratio of any two numbers is equal to the ratio of another two numbers, then the ratios of the sum of numerator and denominator to the difference of numerator and denominator of both rational numbers are equal. First, let us start by defining some important terms. Let us see:

Browse more Topics under Ratios And Proportions

• Propotion Of Quantities
• Proportionals (Third, Fourth and Mean)
• Comparison of Ratios
• Invertendo, and Alternendo
• Componendo, and Dividendo
• Duplicate Ratios
• Variations
• Ratios and Proportion Practice Questions

Direct Proportion

Two quantities are said to be directly proportional, if on the increase (or decrease) of the one, the other increases (or decreases) to the same extent.

Example 1:Cost is directly proportional to the number of articles. (More Articles, More Cost)
Example 2: Work done is directly proportional to the number of men work (More Men, More Work).

Indirect Proportion

Two quantities are said to be indirectly proportional if on the increase of the one, the other decreases to the same extent and vice-versa.

Example 1:The time taken by a car is covering a certain distance is inversely proportional to the speed of the car. (More speed, Less is the time taken to cover a distance.)

Example 2: The time taken to finish a work is inversely proportional to the number of persons working at it. (More persons, Less is the time taken to finish a job).
Remarks: In solving problems by the chain rule, we compare every item with the term to be found out.

Key Points

If p : q :: r : s, then:

The method of finding the 4 th term of a proportion when three are given is known as the rule of three as above. Three or more quantities are said to be in compound proportion if one quantity depends on the other remaining quantities. If p,q,r,s are four quantities & if p : q :: r : s then:

Componendo

(p+q)/q = (r+s)/s

Dividendo

(p-q)/p = (r-s)/s

Componendo & Dividendo

(p+q)/(p-q) = (r+s)/(r-s)

Invertendo

q/p = s/r

Alternendo

p/r = q/s

Solved Examples For You

Part I

Example 3: Prove the componendo and dividendo formula.

Answer: Let a, b, c and d be four quantities such that, a : b = c : d

In other words, we may write a/b = c/d

We can write b/a = d/c [By invertendo]

(a+b)/b = (c+d)/d [By componendo].

Thus we have, (a-b)/b = (c-d)/d [By dividendo]

Hence, we can write (a+b)/(a-b) = (c+d)/(c-d) [On dividing (i) by (ii)]

In other words we may write, (a+b):(a-b) = (c+d):(c-d).

Hence we can write a:b = c:d and (a+b):(a-b) = (c+d):(c-d).Therefore, it is proved that if the ratio of a to b is equal to the ratio of c to d, then the ratio of a + b toa − b is equal to the ratio of c+ dtoc − d. This property is called the componendo and dividendo rule.

Example 4: If p = 4xy/(x+y), then the value of (p+2x)/(p-2x) + (p+2x)/(p-2y) will be:

A) px – xy B) P(x+y) C) xyp D) 2

Answer: Here we have: p = 4xy/(x+y) which will imply that p/2x = 2y/(x+y). In other words we can say that:

(p+2x)/(p-2x) = (2y + x + y)/(2y – (x + y)) [By componendo and dividendo]

(p + 2x)/(p-2x) = (3y + x)/(y – x) ….. (i)

Again, we can say that p = 4xy/(x+y) and we can say that p/2y = 2x/(x + y)

Also, this means thta (p + 2y)/(p – 2y) = (2x + x + y)/(2x – (x +y)) [By componendo and dividendo]

Therefore, we have: (p + 2y)/(p – 2y) = (3x + y)/(x – y) ….(ii)

Adding (i) and (ii), we get:

(p + 2x)/(p – 2x) + (p + 2y)/(p – 2y) = (3y + x)/(y – x) + (3x + y)/(x – y)

In other words, this can be written as: (3y + x)/(y -x) – (3x + y)/(y – x) = (3y + x – 3x – y)/(y -x) = (2y – 2x)/(y – x)

Hence, we have: 2(y – x)/(y – x) = 2. Hence the correct option here isD) 2.

Part II

Let us see a few more examples to get ourselves well acquainted with the concept of componendo and dividendo.

Example 5: If (4a + 9b) x (4c – 9d) = (4a – 9b) (4c + 9d), then the value of a:b can be equal to:

A) c:d B) 2 C) ab:c D) bc:a

Answer: Here we have: (4a + 9b) x (4c – 9d) = (4a – 9b) (4c + 9d). In other words we can say that: (4a + 9b)/(4a – 9b) = (4c + 9d)/(4c – 9d)

Hence we have: {(4a + 9b) + (4a – 9b)}/{(4a + 9b) – (4a – 9b)} = {(4c + 9d) + (4c – 9d)}/{(4c + 9d) – (4c – 9d)} [By componendo and dividendo]

In other words we have: 8a/18b = 8c/18d.

Or we may write: a/b = c/d i.e. a:b :: c : d.

Hence a:b = c: d and the correct option here isA) c:d. Let us see another example below.

Practice Question:

Q 1: If (x : y) = 2 : 1, then (x² – y²) : (x² + y²) is:

A) 3: 5 B) 5 : 3 C) 1:3 D) 3:1

Ans:A) 3: 5