The Hidden Power of Componendo-Dividendo in Solving Ratio and Proportion Problems

Published by Ganit Charcha | Category - High School Math Topics | 2025-08-19 01:09:09

Introduction:

In the world of algebraic shortcuts, Componendo and Dividendo stands out as a powerful tool for simplifying complex ratios and solving problems efficiently. Whether you're preparing for competitive exams or just brushing up your math skills, mastering this concept can give you a real edge. To refresh your understanding before diving into the solved problems below, we highly recommend watching this quick and clear explainer video: https://www.youtube.com/watch?v=uj81S90mrsI. It’s a great way to revisit the fundamentals and build confidence before tackling the examples!

Selected Problems on Ratio-Proportion with Solutions

Problem 1
If $ a : b = c : d $, then show that
(i) $ \frac{a^2 + b^2}{a^2 - b^2} = \frac{ac + bd}{ac - bd} $
(ii) $ \frac{a^2 + ab + b^2}{a^2 - ab + b^2} = \frac{c^2 + cd + d^2}{c^2 - cd + d^2} $
(iii) $ \sqrt{a^2 + c^2} = \sqrt{b^2 + d^2} = \frac{pa + qc}{pb + qd} $

Solutions:
(i) Let $ \frac{a}{b} = \frac{c}{d} $.
Then $ \frac{a^2}{b^2} = \frac{ac}{bd} $. [Multiplying both sides by $\frac{a}{b}$]

Using componendo-dividendo, we then get

$\frac{a^2 + b^2}{a^2 - b^2} = \frac{ac + bd}{ac - bd}$

(ii) Again, $ \frac{a}{b} = \frac{c}{d} \Rightarrow \frac{a^2}{b^2} = \frac{c^2}{d^2} $

Componendo gives,
$\frac{a^2 + b^2}{b^2} = \frac{c^2 + d^2}{d^2}$ (1)

Also, $ \frac{a}{b} = \frac{c}{d} \Rightarrow \frac{b}{a} = \frac{d}{c} \Rightarrow \frac{b^2}{ab} = \frac{d^2}{cd} \quad \text{(2)} $

Multiplying (1) and (2), we get
$\frac{a^2 + b^2}{ab} = \frac{c^2 + d^2}{cd}$

Thus, by componendo and dividendo, we have
$\frac{a^2 + ab + b^2}{a^2 - ab + b^2} = \frac{c^2 + cd + d^2}{c^2 - cd + d^2}$

(iii) Let $ \frac{a}{b} = \frac{c}{d} = k \Rightarrow a = bk,\, c = dk $

Then,
$\frac{\sqrt{a^2 + c^2}}{\sqrt{b^2 + d^2}} = \frac{\sqrt{b^2k^2 + d^2k^2}}{\sqrt{b^2 + d^2}} = \frac{k \sqrt{b^2 + d^2}}{\sqrt{b^2 + d^2}} = k $

And,
$\frac{pa + qc}{pb +qd} = \frac{p(bk) + q(dk)}{pb + qd} = k$

Hence proven.

Problem 2
If $ \frac{x}{a} = \frac{y}{b} = \frac{z}{c} $, then prove that,}
(i) $ \frac{x^3}{a^3} + \frac{y^3}{b^3} + \frac{z^3}{c^3} = \left( \frac{x + y + z}{a + b + c} \right)^3 $
(ii) $ \frac{x^3 + y^3 + z^3}{a^3 + b^3 + c^3} = \frac{xyz}{abc} $
(iii) $ (a + b + c)(x^2 + y^2 + z^2) = (ax + by + cz)^2 $

Solutions:
Let $ \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k \Rightarrow x = ak,\ y = bk,\ z = ck $
(i)
$\frac{x^3}{a^3} + \frac{y^3}{b^3} + \frac{z^3}{c^3} = \frac{a^3k^3}{a^3} + \frac{b^3k^3}{b^3} + \frac{c^3k^3}{c^3} = k^3$
$\left( \frac{x + y + z}{a + b + c} \right)^3 = \left( \frac{ak + bk + ck}{a + b + c} \right)^3 = \left( \frac{k(a + b + c)}{a + b + c} \right)^3 = k^3$
Hence, LHS = RHS.

(ii)
$\frac{x^3 + y^3 + z^3}{a^3 + b^3 + c^3} = \frac{k^3(a^3 + b^3 + c^3)}{a^3 + b^3 + c^3} = k^3$
$\frac{xyz}{abc} = \frac{(ak)(bk)(ck)}{abc} = \frac{k^3 abc}{abc} = k^3$
Hence, LHS = RHS.

(iii)
Since $ x = ak,\, y = bk,\, z = ck $, we have:
$x^2 = a^2k^2,\quad y^2 = b^2k^2,\quad z^2 = c^2k^2$
So,
$x^2 + y^2 + z^2 = k^2(a^2 + b^2 + c^2)$
Also,
$ax + by + cz = a(ak) + b(bk) + c(ck) = k(a^2 + b^2 + c^2)$
Thus,
$(ax + by + cz)^2 = k^2(a^2 + b^2 + c^2)^2$
And,
$(a^2 + b^2 + c^2)(x^2 + y^2 + z^2) = k^2(a^2 + b^2 + c^2)^2$
Hence, LHS = RHS.

Problem 3
If $ \frac{a}{b} = \frac{b}{c} $, prove the following:
(i) $ \left( \frac{a + b}{b + c} \right)^2 = \frac{a^2 + b^2}{b^2 + c^2} $
(ii) $ a^2 b^2 c^2 \left( \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} \right) = a^3 + b^3 + c^3 $
(iii) $ \frac{abc(a + b + c)^3}{(ab + bc + ca)^3} = 1 $

Solutions:
Let $ \frac{a}{b} = \frac{b}{c} = k \Rightarrow a = bk = ck^2,\ b = ck,\ c = c $

(i)
LHS = $\left( \frac{a + b}{b + c} \right)^2 = \left( \frac{bk + ck}{ck + c} \right)^2 = \left( \frac{(b + c)k}{b + c} \right)^2 = k^2$

RHS = $\frac{a^2 + b^2}{b^2 + c^2} = \frac{(ck^2)^2 + (ck)^2}{(ck)^2 + c^2} = \frac{c^2k^2(k^2 + 1)}{c^2(k^2 + 1)} = k^2$

Hence, LHS = RHS.

(ii)}
LHS = $ a^2 b^2 c^2 ( \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3}) = (ck^2)^2 (ck)^2 c^2 (\frac{1}{(ck^2)^3} + \frac{1}{(ck)^3} + \frac{1}{c^3} ) $

= $c^6k^6 \left(\frac{1}{c^3k^6} + \frac{1}{c^3k^3} + \frac{1}{c^3} \right) = c^3 + c^3k^3 + c^3k^6 = (ck^2)^3 + (ck)^3 + c^3$

And this simplifies to:
$ = a^3 + b^3 + c^3 $

(iii)
Given $ \frac{a}{b} = \frac{b}{c} = k \Rightarrow a = bk,\, b = ck $

Then, we have
$ abc = bk \cdot ck \cdot c = c^3 k^3 $
$ a + b + c = bk + ck + c = c (k^2 + k + 1)$ $\Rightarrow$ $(a + b + c)^3 = c^3(k^2 + k + 1)^3 $

$ ab + bc + ca = bk \cdot ck + ck \cdot c + c \cdot bk = c^2k^3 + c^2k + c^2k = c^2k(k^2 + 1 + k) = c^2k(k^2 + k + 1)$
$\Rightarrow$ $(ab + bc + ca)^3 = c^6k^3(k^2 + k + 1)^3 $

Hence,
$\frac{abc( a + b + c)^3}{ab + bc + ca)^3} = \frac{c^3k^3 \cdot c^3((k^2 + k + 1)^3}{c^6k^3(k^2 + k + 1)^3} = 1$

Problem 4
If $ \frac{x}{lm - n^2} = \frac{y}{mn - l^2} = \frac{z}{nl - m^2} $, then prove that $lx + my + nz = 0$.

Solution:
Since, $\frac{x}{lm - n^2} = \frac{y}{mn - l^2} = \frac{z}{nl - m^2}$ then,
$$ \frac{lx}{l^2m - ln^2} = \frac{my}{m^2n - ml^2} = \frac{nz}{n^2l - nm^2} = \frac{lx + my + nz}{l^2m - ln^2 +m^2n - ml^2 + n^2l - nm^2} = \frac{lx + my + nz}{0} $$
Denominator of the last ratio is 0, so, multiplying this denominator with any other ratio, we have
$lx + my + nz = 0\cdot \frac{nz}{n^2l - nm^2}$ $\Rightarrow$ $lx + my + nz = 0$.

Problem 5
If $ \frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c} $, then prove that $(b - c)x + (c - a)y + (a - b)z = 0$.

Solution:
Let the common ratio be $ k $. This implies, $x = k(b + c - a),\ y = k(c + a - b),\ z = k(a + b - c) $

Then, we have
$ (b - c)x = k(b - c)(b + c - a) = k(b^2 - c^2)- k(ab - ac) $
$ (c - a)y = k(c - a)(c + a - b) = k (c^2 - a^2) - k(bc - ab) $
$ (a - b)z = k(a - b)(a + b - c) = k(a^2 - b^2) - k(ac - bc) $

Add all the three eualities, we get
$ (b - c)x + (c - a)y + (a - b)z = k(b^2 - c^2 + c^2 - a^2 + a^2 - b^2) - k(ab - ac +bc - ab + ac - bc) = 0 $

Problem 6
If $ \frac{x}{a} = \frac{y}{b} = \frac{z}{c} $, then prove that
$ \frac{x^2 - yz}{a^2 - bc} = \frac{y^2 - zx}{b^2 - ca} = \frac{z^2 - xy}{c^2 - ab}$.

Solution:
Given,
$ \frac{x}{a} = \frac{y}{b} = \frac{z}{c} $
$\Rightarrow \frac{x^2}{a^2} = \frac{y^2}{b^2} = \frac{z^2}{c^2} $

$\Rightarrow \frac{x^2}{a^2} = \frac{y^2}{b^2} = \frac{z^2}{c^2} = \frac{yz}{bc} = \frac{zx}{ca} = \frac{xy}{ab} $
[This happens because,
$ \frac{x^2}{a^2} = \frac{x}{a}.\frac{x}{a} = \frac{y}{b}.\frac{z}{c} = \frac{yz}{bc} $]

$\Rightarrow \frac{x^2}{a^2} = \frac{yz}{bc} = \frac{y^2}{b^2} = \frac{zx}{ca} = \frac{z^2}{c^2} = \frac{xy}{ab} $

$\Rightarrow \frac{x^2}{a^2} = \frac{-yz}{-bc} = \frac{y^2}{b^2} = \frac{-zx}{-ca} = \frac{z^2}{c^2} = \frac{-xy}{-ab} $

$\Rightarrow \frac{x^2 - yz}{a^2 - bc} = \frac{y^2 - zx}{b^2 - ca} = \frac{z^2 - xy}{c^2 - ab}$ [Applying Addendo successively on the pairs (1st, 2nd)-terms, then (3rd, 4th)-terms and (5th, 6th)-terms]

Problem 7
If $ x = \frac{8ab}{a + b} $, then find the value of $ \frac{x + 4a}{x - 4a} + \frac{x + 4b}{x - 4b}$.

Solution:
Since, $ x = \frac{8ab}{a + b} $, this implies $ \frac{x}{4a} = \frac{2b}{a + b}$.

Applying, componendo-dividendo, we get
$ \frac{x + 4a}{x - 4a} = \frac{2b + a + b}{2b - a - b} = \frac{a + 3b}{b - a}$.

Similarly, $ x = \frac{8ab}{a + b} $, implies $ \frac{x}{4b} = \frac{2a}{a + b}$.

Applying, componendo-dividendo, we get
$ \frac{x + 4b}{x - 4b} = \frac{2a + a + b}{2a - a - b} = \frac{3a + b}{a - b}$.

Adding, we get $ \frac{x + 4a}{x - 4a} + \frac{x + 4b}{x - 4b} = \frac{a + 3b}{b - a} - \frac{3a + b}{b - a} = \frac{2(b-a)}{b-a} = 2$

Problem 8
If $ \frac{x + y}{3a -b} = \frac{y+z}{3b-c} = \frac{z+x}{3c-a} $, then prove that $\frac{x + y + z}{a + b + c} = \frac{ax + by + cz}{a^2 + b^2 + c^2} $.

Solution:
Given, $\frac{x + y}{3a -b} = \frac{y+z}{3b-c} = \frac{z+x}{3c-a}$
Implies $\frac{x + y}{3a -b} = \frac{y+z}{3b-c} = \frac{z+x}{3c-a}$ = $\frac{x+y+z}{a+b+c}$ [By Applying Addendo]

Similarly, $\frac{x + y}{3a -b} = \frac{y+z}{3b-c} = \frac{z+x}{3c-a}$ implies
$\frac{x + y}{3a -b} = \frac{y+z}{3b-c} = \frac{z+x}{3c-a} = \frac{(x+y)+(y+z) -(z+x)}{(3a-b)+(3b-c)-(-3c-a)} = \frac{(y+z)+(z+x) -(x+y)}{(3b-c)+(3c-a)-(-3a-b)} = \frac{(z+x)+(x+y) -(y+z)}{(3c-a)+(3a-b)-(-3b-c)} $

This further implies,
$\frac{x + y}{3a -b} = \frac{y+z}{3b-c} = \frac{z+x}{3c-a} = \frac{y}{2a + b -2c} = \frac{z}{2b + c - 2a} = \frac{x}{2c + a -2b} $

=> $\frac{x + y}{3a -b} = \frac{y+z}{3b-c} = \frac{z+x}{3c-a} = \frac{by}{2ab + b^2 -2bc} = \frac{cz}{2bc + c^2 - 2ac} = \frac{ax}{2ac + a^2 -2ab} = \frac{ax + by + cz}{a^2 + b^2 + c^2} $ [Applying Addendo on on last 3 ratios]

Therefore, each ratio given in the problem is equal to both $\frac{x+y+z}{a+b+c}$ and $\frac{ax+by+cz}{a^2+b^2+c^2}$.
Therfore, $\frac{x + y + z}{a + b + c} = \frac{ax + by + cz}{a^2 + b^2 + c^2}$.

The Hidden Power of Componendo-Dividendo in Solving Ratio and Proportion Problems

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