Topic Included: | Formulas, Definitions & Exmaples. |
Main Topic: | Quantitative Aptitude. |
Sub-topic: | Componendo and Dividendo Rule. |
Questions for practice: | 10 Questions. |
The Componendo and Dividendo Rule is used to ease the long calculation and solve the equations in a quick way, when dealing with equations of fractions and rational functions.
If $$ \frac{A}{B} = \frac{C}{D} $$ then, $$ \frac{A + B}{B} = \frac{C + D}{D} $$
If $$ \frac{A}{B} = \frac{C}{D} $$
Then by adding 1 to both side we get $$ \left(\frac{A}{B} + 1\right) = \left(\frac{A}{B} + 1\right) $$ $$ \bbox[5px,border:2px solid #800000] {\frac{A + B}{B} = \frac{C + D}{D}} \qquad (1) $$
If $$ \frac{A}{B} = \frac{C}{D} $$ then, $$ \frac{A - B}{B} = \frac{C - D}{D} $$
If $$ \frac{A}{B} = \frac{C}{D} $$
Then by subtracting 1 from both side we get $$ \left(\frac{A}{B} - 1\right) = \left(\frac{A}{B} - 1\right) $$ $$ \bbox[5px,border:2px solid #800000] {\frac{A - B}{B} = \frac{C - D}{D}} \qquad (2) $$
If $$ \frac{A}{B} = \frac{C}{D} $$ then, $$ \frac{A + B}{A - B} = \frac{C + D}{C - D} $$
If $$ \frac{A}{B} = \frac{C}{D} $$
Then by dividing equation \((1)\) from equation \((2)\) we get $$ \left[\frac{\frac{A + B}{B}}{\frac{A - B}{B}} = \frac{\frac{C + D}{D}}{\frac{C - D}{D}}\right] $$ $$ \bbox[5px,border:2px solid #800000] {\frac{A + B}{A - B} = \frac{C + D}{C - D}} \qquad (3) $$ The final equation is called componendo and dividendo.
Example: If \(\frac{a}{b} = \frac{4}{3}\), then find the value of \(\frac{a + b}{a - b}\)?
Solution: Given values, \(a = 4\) and \(b = 3\), we know, according to componendo dividendo \(\frac{a}{b} = \frac{c}{d}\), then from the equation \((3)\), $$ \frac{A + B}{A - B} = \frac{C + D}{C - D} $$ $$ \frac{a + b}{a - b} = \frac{4 + 3}{4 - 3} = \frac{7}{1} = 7 $$ Here we can see that the componendo and dividendo rule makes it easy to solve rational functions.