ICSE Class 10 Mathematics • Chapter 05
Ratio: Comparison of two quantities of the same kind. Written as a:b or $\frac{a}{b}$
Proportion: Equality of two ratios. If $\frac{a}{b} = \frac{c}{d}$, then a, b, c, d are in proportion.
Continued Proportion: a:b = b:c, i.e., $b^2 = ac$
| Type | Condition | Formula |
|---|---|---|
| Mean Proportional | a:x = x:b | $x = \sqrt{ab}$ |
| Third Proportional | a:b = b:x | $x = \frac{b^2}{a}$ |
| Fourth Proportional | a:b = c:x | $x = \frac{bc}{a}$ |
Example 1: Find the mean proportional between 3 and 27.
Mean proportional = $\sqrt{3 \times 27} = \sqrt{81} = 9$
Example 2: Find third proportional to 4 and 12.
Let third proportional = x. Then 4:12 = 12:x
$x = \frac{12^2}{4} = \frac{144}{4} = 36$
If $\frac{a}{b} = \frac{c}{d}$, then:
| Property Name | Statement | Result |
|---|---|---|
| Invertendo | Invert both ratios | $\frac{b}{a} = \frac{d}{c}$ |
| Alternendo | Swap middle terms | $\frac{a}{c} = \frac{b}{d}$ |
| Componendo | Add 1 to both sides | $\frac{a+b}{b} = \frac{c+d}{d}$ |
| Dividendo | Subtract 1 from both sides | $\frac{a-b}{b} = \frac{c-d}{d}$ |
| Componendo-Dividendo | Combine both | $\frac{a+b}{a-b} = \frac{c+d}{c-d}$ |
Componendo-Dividendo is the MOST USED property in exams!
If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a+b}{a-b} = \frac{c+d}{c-d}$
Technique: If $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k$
Then: $a = bk$, $c = dk$, $e = fk$
This helps convert ratio problems into algebraic expressions!
Example 3: If $\frac{a}{3} = \frac{b}{4} = \frac{c}{5}$, find $\frac{a+b+c}{3a-b+c}$
Solution: Let each ratio = k
$a = 3k$, $b = 4k$, $c = 5k$
$\frac{a+b+c}{3a-b+c} = \frac{3k+4k+5k}{9k-4k+5k} = \frac{12k}{10k} = \frac{6}{5}$
Example 4: If $\frac{3x+4y}{3x-4y} = \frac{5}{3}$, find $\frac{x}{y}$
Solution: Apply Componendo-Dividendo
$\frac{(3x+4y)+(3x-4y)}{(3x+4y)-(3x-4y)} = \frac{5+3}{5-3}$
$\frac{6x}{8y} = \frac{8}{2}$
$\frac{x}{y} = \frac{8 \times 8}{2 \times 6} = \frac{64}{12} = \frac{16}{3}$
∴ x:y = 16:3
Example 5: If $\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} = \frac{4}{1}$, find x.
Solution: Apply Componendo-Dividendo
$\frac{2\sqrt{x+1}}{2\sqrt{x-1}} = \frac{4+1}{4-1} = \frac{5}{3}$
$\frac{\sqrt{x+1}}{\sqrt{x-1}} = \frac{5}{3}$
Squaring: $\frac{x+1}{x-1} = \frac{25}{9}$
$9(x+1) = 25(x-1)$
$9x + 9 = 25x - 25$
$34 = 16x$
$x = \frac{17}{8}$
| Given | Property | Apply |
|---|---|---|
| $\frac{a+b}{a-b} = \frac{m}{n}$ | Componendo-Dividendo | $\frac{a}{b} = \frac{m+n}{m-n}$ |
| a:b = b:c | Continued Proportion | $b^2 = ac$ |
| Find mean of a and b | Mean Proportional | $\sqrt{ab}$ |
| Find third after a, b | Third Proportional | $\frac{b^2}{a}$ |
BOARD If $\frac{8a-5b}{8c-5d} = \frac{8a+5b}{8c+5d}$, prove that $\frac{a}{b} = \frac{c}{d}$
BOARD Using properties of proportion, solve: $\frac{\sqrt{x+5}+\sqrt{x-16}}{\sqrt{x+5}-\sqrt{x-16}} = \frac{7}{3}$