10. Ratio and Proportion Previous Year Questions

By Dividendo: [ (6x² + 3x − 5) − (3x − 5) ] / (3x − 5) = [ (9x² + 2x + 5) − (2x + 5) ] / (2x + 5). [1 Mark]
6x² / (3x − 5) = 9x² / (2x + 5).
Dividing by 3x² (since x ≠ 0): 2 / (3x − 5) = 3 / (2x + 5). [1 Mark]
2(2x + 5) = 3(3x − 5) ⇒ 4x + 10 = 9x − 15 ⇒ 5x = 25 ⇒ x = 5. [1 Mark]

Correct Option: (d) x = 3, y = 150
Speed = 60/2 = 30 m/s.
90/x = 30 ⇒ x = 3.
y/5 = 30 ⇒ y = 150. [1 Mark]

(a) Cube root on both sides: (a + b)/(a − b) = ⁴/₃. [2 Marks]
(b) By Componendo & Dividendo: [ (a+b)+(a−b) ] / [ (a+b)−(a−b) ] = (4+3)/(4−3) ⇒ 2a/2b = 7/1 ⇒ a : b = 7 : 1. [2 Marks]

Mean proportional between $a$ and $b$ is given by $x = \sqrt{ab}$.
$x = \sqrt{4 \times 9} = \sqrt{36} = 6$.
Correct Option: (b) [1 Mark]

Step 1 Let the required number be $x$.
The resulting numbers $(4 + x), (6 + x), (8 + x), (11 + x)$ are in proportion:
$\frac{4 + x}{6 + x} = \frac{8 + x}{11 + x}$.1 Mark

Step 2 Cross-multiplying:
$(4 + x)(11 + x) = (6 + x)(8 + x)$
$44 + 15x + x^2 = 48 + 14x + x^2$.1 Mark

Step 3 Subtracting $x^2$ from both sides:
$15x - 14x = 48 - 44 \implies x = 4$.
The number to be added is 4.1 Mark

Step 1 Express RHS as $\frac{3}{1}$ and apply Componendo and Dividendo: $\frac{a + b}{a - b} = \frac{c + d}{c - d}$:
$$\frac{(\sqrt{2x+2} + \sqrt{2x-1}) + (\sqrt{2x+2} - \sqrt{2x-1})}{(\sqrt{2x+2} + \sqrt{2x-1}) - (\sqrt{2x+2} - \sqrt{2x-1})} = \frac{3 + 1}{3 - 1}$$1 Mark
Step 2 Simplify terms:
$$\frac{2\sqrt{2x+2}}{2\sqrt{2x-1}} = \frac{4}{2} \implies \frac{\sqrt{2x+2}}{\sqrt{2x-1}} = 2$$1 Mark
Step 3 Squaring both sides:
$$\frac{2x + 2}{2x - 1} = 4 \implies 2x + 2 = 4(2x - 1) \implies 2x + 2 = 8x - 4$$
$$6x = 6 \implies \mathbf{x = 1}$$.1 Mark

Method Factoring numerators and denominators:
$$\frac{x(x + 2)}{2(x + 2)} = \frac{y(y + 3)}{3(y + 3)}$$ [1.5 Marks]
Cancelling common non-zero terms $(x + 2)$ and $(y + 3)$:
$$\frac{x}{2} = \frac{y}{3} \implies \frac{x}{y} = \frac{2}{3} \implies \mathbf{x : y = 2 : 3}$$ [1.5 Marks]
(Alternative by Componendo & Dividendo also accepted).

Step 1 Express $x$ as $\frac{x}{1}$ and apply Componendo and Dividendo:
$$\frac{x + 1}{x - 1} = \frac{(\sqrt{2a + 1} + \sqrt{2a - 1}) + (\sqrt{2a + 1} - \sqrt{2a - 1})}{(\sqrt{2a + 1} + \sqrt{2a - 1}) - (\sqrt{2a + 1} - \sqrt{2a - 1})} = \frac{2\sqrt{2a + 1}}{2\sqrt{2a - 1}} = \frac{\sqrt{2a + 1}}{\sqrt{2a - 1}}$$ [1.5 Marks]
Step 2 Squaring both sides:
$$\frac{(x + 1)^2}{(x - 1)^2} = \frac{2a + 1}{2a - 1} \implies \frac{x^2 + 2x + 1}{x^2 - 2x + 1} = \frac{2a + 1}{2a - 1}$$ [1 Mark]
Step 3 Applying Componendo and Dividendo again:
$$\frac{(x^2 + 2x + 1) + (x^2 - 2x + 1)}{(x^2 + 2x + 1) - (x^2 - 2x + 1)} = \frac{(2a + 1) + (2a - 1)}{(2a + 1) - (2a - 1)}$$
$$\frac{2(x^2 + 1)}{4x} = \frac{4a}{2} \implies \frac{x^2 + 1}{2x} = 2a \implies x^2 + 1 = 4ax \implies \mathbf{x^2 - 4ax + 1 = 0}$$
Hence Proved. [1.5 Marks]

Step 1 Set up proportional ratio definition:
$\frac{K + 3}{K + 2} = \frac{3K - 7}{2K - 3}$. [1 Mark]
Step 2 Cross-multiply and expand terms:
$(K + 3)(2K - 3) = (K + 2)(3K - 7)$
$2K^2 - 3K + 6K - 9 = 3K^2 - 7K + 6K - 14$
$2K^2 + 3K - 9 = 3K^2 - K - 14$. [1 Mark]
Step 3 Collect all terms into standard quadratic form:
$3K^2 - 2K^2 - K - 3K - 14 + 9 = 0$
$K^2 - 4K - 5 = 0$
$(K - 5)(K + 1) = 0 \implies \mathbf{K = 5 \text{ or } K = -1}$. [1 Mark]

Step 1 Write RHS as fraction $\frac{4}{1}$ and apply Componendo and Dividendo:
$\frac{(\sqrt{5x} + \sqrt{2x - 6}) + (\sqrt{5x} - \sqrt{2x - 6})}{(\sqrt{5x} + \sqrt{2x - 6}) - (\sqrt{5x} - \sqrt{2x - 6})} = \frac{4 + 1}{4 - 1}$. [1 Mark]
Step 2 Simplify both sides:
$\frac{2\sqrt{5x}}{2\sqrt{2x - 6}} = \frac{5}{3} \implies \frac{\sqrt{5x}}{\sqrt{2x - 6}} = \frac{5}{3}$. [1 Mark]
Step 3 Square both sides and solve for $x$:
$\frac{5x}{2x - 6} = \frac{25}{9} \implies \frac{x}{2x - 6} = \frac{5}{9}$
$9x = 5(2x - 6) \implies 9x = 10x - 30 \implies 10x - 9x = 30 \implies \mathbf{x = 30}$.
Answer: $x = 30$. [1 Mark]

Solution For $a, b, c$ in continued proportion, $b^2 = ac$.
$(x + 6)^2 = (x + 3)(x + 10)$
$x^2 + 12x + 36 = x^2 + 13x + 30 \implies 13x - 12x = 36 - 30 \implies \mathbf{x = 6}$.
Correct Answer: (b) 6 [1 Mark]

Step 1 Write RHS as $\frac{3}{1}$ and apply Componendo and Dividendo:
$\frac{(\sqrt{2x + 3} + \sqrt{2x - 3}) + (\sqrt{2x + 3} - \sqrt{2x - 3})}{(\sqrt{2x + 3} + \sqrt{2x - 3}) - (\sqrt{2x + 3} - \sqrt{2x - 3})} = \frac{3 + 1}{3 - 1}$. [1 Mark]
Step 2 Simplify both sides:
$\frac{2\sqrt{2x + 3}}{2\sqrt{2x - 3}} = \frac{4}{2} \implies \frac{\sqrt{2x + 3}}{\sqrt{2x - 3}} = 2$. [1 Mark]
Step 3 Square both sides and solve for $x$:
$\frac{2x + 3}{2x - 3} = 4 \implies 2x + 3 = 4(2x - 3) = 8x - 12$
$8x - 2x = 3 + 12 \implies 6x = 15 \implies x = \frac{15}{6} = \mathbf{2.5}$ (or $\frac{5}{2}$). [1 Mark]

Step 1 Express as ratio and apply Componendo & Dividendo:
$$\frac{2x + \sqrt{4x^2 - 1}}{2x - \sqrt{4x^2 - 1}} = \frac{4}{1}$$ $$\frac{(2x + \sqrt{4x^2 - 1}) + (2x - \sqrt{4x^2 - 1})}{(2x + \sqrt{4x^2 - 1}) - (2x - \sqrt{4x^2 - 1})} = \frac{4 + 1}{4 - 1}$$ [1 Mark]
Step 2 Simplify LHS and RHS:
$$\frac{4x}{2\sqrt{4x^2 - 1}} = \frac{5}{3} \implies \frac{2x}{\sqrt{4x^2 - 1}} = \frac{5}{3}$$ [1 Mark]
Step 3 Square both sides:
$$\frac{4x^2}{4x^2 - 1} = \frac{25}{9} \implies 36x^2 = 25(4x^2 - 1) = 100x^2 - 25$$ $$100x^2 - 36x^2 = 25 \implies 64x^2 = 25 \implies x^2 = \frac{25}{64}$$ Since $x$ is positive: $x = \sqrt{\frac{25}{64}} = \mathbf{\frac{5}{8}}$ [1 Mark]
Answer: $x = \frac{5}{8}$.

Step 1 Since $b$ is the mean proportional between $a$ and $c$:
$$b^2 = ac \implies b^4 = a^2 c^2$$ [1 Mark]
Step 2 Substitute $b^2 = ac$ and $b^4 = a^2 c^2$ into LHS:
$$\text{Numerator} = a^4 + a^2(ac) + a^2 c^2 = a^4 + a^3 c + a^2 c^2 = a^2(a^2 + ac + c^2)$$ [1 Mark]
$$\text{Denominator} = a^2 c^2 + (ac)c^2 + c^4 = a^2 c^2 + ac^3 + c^4 = c^2(a^2 + ac + c^2)$$
Step 3 Cancel common terms:
$$\text{LHS} = \frac{a^2(a^2 + ac + c^2)}{c^2(a^2 + ac + c^2)} = \mathbf{\frac{a^2}{c^2}} = \text{RHS}$$ [1 Mark]
Hence Proved.

(i) Find m : n Apply Componendo and Dividendo:
$$\frac{(7m + 2n) + (7m - 2n)}{(7m + 2n) - (7m - 2n)} = \frac{5 + 3}{5 - 3}$$ $$\frac{14m}{4n} = \frac{8}{2} = 4 \implies \frac{7m}{2n} = 4 \implies \frac{m}{n} = \frac{8}{7}$$ $$\mathbf{m : n = 8 : 7}$$ [1.5 Marks]
(ii) Find (m² + n²) / (m² - n²) Square both sides of $\frac{m}{n} = \frac{8}{7}$:
$$\frac{m^2}{n^2} = \frac{64}{49}$$ Apply Componendo and Dividendo:
$$\frac{m^2 + n^2}{m^2 - n^2} = \frac{64 + 49}{64 - 49} = \mathbf{\frac{113}{15}}$$ [1.5 Marks]
Answers: (i) $m : n = 8 : 7$, (ii) $\frac{113}{15}$.