$\displaystyle \text{Question 1: If } x, y \text{ and } z \text{ are in continued proportion, prove that: } \\ \\ \frac{(x+y)^2}{(y+z)^2} = \frac{x}{y} \text{. [2010]}$

If $\displaystyle x, y \text{ and } z$ are in continued proportion, then

$\displaystyle \frac{x}{y} = \frac{y}{z} \Rightarrow x = \frac{y^2}{z}$

Applying componendo and dividendo

$\displaystyle \frac{x+y}{x-y} = \frac{y+z}{y-z}$

$\displaystyle \Rightarrow \frac{x+y}{y+z} = \frac{x-y}{y-z}$

Squaring both sides

$\displaystyle \Rightarrow \frac{(x+y)^2}{(y+z)^2} = (\frac{x-y}{y-z})^2$

$\displaystyle \Rightarrow \frac{(x+y)^2}{(y+z)^2} = (\frac{x-y}{y-z})^2$

Substituting

$\displaystyle \Rightarrow \frac{(x+y)^2}{(y+z)^2} = (\frac{\frac{y^2}{z}-y}{y-z})^2$

$\displaystyle \Rightarrow \frac{(x+y)^2}{(y+z)^2} = (\frac{y^2-yz}{z(y-z)})^2 = \frac{y^2}{z^2} = \frac{zx}{z^2} = \frac{x}{z}$

$\displaystyle \\$

$\displaystyle \text{Question 2: Given } x= \frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 }} \\ \\ \text{ Use componendo and dividendo to prove that: } x^2= \frac{2a^2 x}{x^2+1} \text{ [2010 ] }$

$\displaystyle \text{Given } x= \frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 }}$

Applying componendo and dividendo

$\displaystyle \frac{x+1}{x-1} = \frac{(\sqrt{a^2+b^2}+\sqrt{a^2-b^2})+(\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 })}{(\sqrt{a^2+b^2}+\sqrt{a^2-b^2})-(\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 })}$

Simplifying

$\displaystyle \frac{x+1}{x-1} = \frac{\sqrt{a^2+b^2}}{\sqrt{a^2-b^2 }}$

Square both sides

$\displaystyle \frac{x^2+1+2x}{x^2-2x+1} = \frac{a^2+b^2}{a^2-b^2}$

Applying componendo and dividendo

$\displaystyle \frac{x^2+1+2x+x^2-2x+1}{x^2+1+2x-x^2+2x-1} = \frac{a^2+b^2+a^2-b^2}{a^2+b^2-a^2+b^2}$

$\displaystyle \frac{2(x^2+1)}{4x} = \frac{2a^2}{2b^2}$

$\displaystyle \frac{x^2+1}{2x} = \frac{a^2}{b^2}$

Simplifying

$\displaystyle b^2 = \frac{2a^2x}{x^2+1}$

$\displaystyle \\$

$\displaystyle \text{Question 3: If } \frac{x^2+y^2}{x^2-y^2} =2 \frac{1}{8} \text{ , find: }$

$\displaystyle \text{i) } \frac{x}{y}$           $\displaystyle \text{ii) } \frac{x^3+y^3}{x^3-y^3 } \text{ [2010] }$

$\displaystyle \text{i) } \text{Given } \frac{x^2+y^2}{x^2-y^2} =2 \frac{1}{8} = \frac{17}{8}$

Applying componendo and dividendo

$\displaystyle \frac{x^2+y^2+x^2-y^2}{x^2+y^2-x^2+y^2} = \frac{17+8}{17-8}$

$\displaystyle \frac{2x^2}{2y^2} = \frac{25}{9}$

Simplifying, we get

$\displaystyle \frac{x}{y} = \frac{5}{3}$

$\displaystyle \text{ii) } \frac{x^3+y^3}{x^3-y^3 }$

Applying componendo and dividendo

$\displaystyle \frac{x^3+y^3+x^3-y^3 }{x^3+y^3-x^3+y^3 }$

$\displaystyle \frac{2x^3}{2y^3} = \Big( \frac{x}{y} \Big)^3 = \Big( \frac{5}{3} \Big)^3 = \frac{125}{9}$

$\displaystyle \\$

Question 4: Using componendo and dividendo, find the value of $\displaystyle x$ : $\displaystyle \frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}} =9 \text{ [2010] }$

$\displaystyle \text{Given } \frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}} =9$

Applying componendo and dividendo

$\displaystyle \frac{(\sqrt{3x+4}+\sqrt{3x-5})+(\sqrt{3x+4}-\sqrt{3x-5})}{(\sqrt{3x+4}+\sqrt{3x-5})-(\sqrt{3x+4}-\sqrt{3x-5})} = \frac{9+1}{9-1}$

$\displaystyle \frac{2\sqrt{3x+4}}{2\sqrt{3x-5}} = \frac{10}{8}$

Simplifying

$\displaystyle \frac{\sqrt{3x+4}}{\sqrt{3x-5}} = \frac{5}{4}$

Square both sides

$\displaystyle \frac{3x+4}{3x-5} = \frac{25}{14}$

$\displaystyle 42x+56 = 75x-125$

Simplifying we get $\displaystyle x = 7$

$\displaystyle \\$

$\displaystyle \text{Question 5: If } x= \frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}} \\ \\ \text{using properties of proportion show that: } x^2-2ax+1 \text{ [2010] }$

$\displaystyle \text{Given } x= \frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}$

Applying componendo and dividendo

$\displaystyle \frac{x+1}{x-1} = \frac{(\sqrt{a+1}+\sqrt{a-1})+(\sqrt{a+1}-\sqrt{a-1})}{(\sqrt{a+1}+\sqrt{a-1})-(\sqrt{a+1}-\sqrt{a-1})}$

Simplify

$\displaystyle \frac{x+1}{x-1} = \frac{\sqrt{a+1}}{\sqrt{a-1}}$

Now square both sides

$\displaystyle \frac{x^2+1+2x}{x^2-2x+1} = \frac{a+1}{a-1}$

Simplifying

$\displaystyle x^2+1 = 2ax$

$\displaystyle \text{or } x^2-2ax+1 = 0$

$\displaystyle \\$

$\displaystyle \text{Question 6: Given, } \frac{a}{b} = \frac{c}{d} \text{, prove that: } \frac{3a-5b}{3a+5b} = \frac{3c-5d}{3c+5d} \text{ [2000] }$

$\displaystyle \text{Given } \frac{a}{b} = \frac{c}{d}$

$\displaystyle \Rightarrow \frac{3a}{5b} = \frac{3c}{5d}$

By componendo and dividendo

$\displaystyle \frac{3a+5b}{3a-5b} = \frac{3c+5d}{3c-5d}$

By Alternendo

$\displaystyle \frac{3a-5b}{3a+5b} = \frac{3c-5d}{3c+5d}$

$\displaystyle \\$

$\displaystyle \text{Question 7: If } x=\frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}} \text{, prove that: } 3bx^2-2ax+3b=0 \text{ [2007] }$

$\displaystyle \text{Given } x= \frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}}$

Applying componendo and dividendo

$\displaystyle \frac{x+1}{x-1} = \frac{(\sqrt{a+3b}+\sqrt{a-3b})-(\sqrt{a+3b}-\sqrt{a-3b})}{(\sqrt{a+3b}+\sqrt{a-3b})-(\sqrt{a+3b}-\sqrt{a-3b})}$

$\displaystyle \frac{x+1}{x-1} = \frac{2\sqrt{a+3b}}{-2\sqrt{a-3b}}$

Squaring both sides

$\displaystyle \frac{x^2+2x+1}{x^2-2x+1} = \frac{a+3b}{a-3b}$

Applying componendo and dividendo once again

$\displaystyle \frac{(x^2+2x+1)-(x^2-2x+1)}{(x^2+2x+1)-(x^2-2x+1)} = \frac{(a+3b)-(a-3b)}{(a+3b)-(a-3b)}$

simplifying

$\displaystyle \frac{x^2+1}{2x} = \frac{a}{3b}$

$\displaystyle 3b(x^2+1) = 2ax$

$\displaystyle 3bx^2-2ax+3b=0$ Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 8: Using the properties of proportion, solve for x Given: } \\ \\ \frac{(x^4+1)}{2x^2} = \frac{17}{8} \text{ [2013] }$

$\displaystyle \text{Given } \frac{(x^4+1)}{2x^2} = \frac{17}{8}$

Applying componendo and dividendo

$\displaystyle \frac{(x^4+1)+2x^2}{(x^4+1)-2x^2} = \frac{17+8}{17-8}$

$\displaystyle \frac{(x^2+1)^2}{(x^2-1)^2} = \frac{25}{9}$

Taking the square root of both sides

$\displaystyle \frac{x^2+1}{x^2-1} = \frac{5}{3}$

$\displaystyle 3x^2+3=5x^2-5$

$\displaystyle x^2 = 4 or x = \pm 2$

$\displaystyle \\$

Question 9: What least number must be added to each of the numbers $\displaystyle 6, 15, 20, \text{ and } 43$ to make them proportional. [2005, 2013]

Let the number added be $\displaystyle x$

$\displaystyle \text{Therefore } (6+x): (15+x) = (20+x): (43+x)$

$\displaystyle \Rightarrow (6+x) \times (43+x) = (20+x) \times (15+x)$

$\displaystyle \Rightarrow x^2+49x+258 = x^2+ 35x +300$

$\displaystyle \Rightarrow x = 3$

$\displaystyle \\$

Question 10: The monthly pocket money of Ravi and Sanjeev are in the ratio of 5:7 Their expenditures are in the ratio of 3:5. If each saves Rs. 80 per month, find their monthly pocket money. [2012]

Let monthly pocket of Rave and Sanjeev by $\displaystyle x \text{ and } y$ respectively.

$\displaystyle \frac{x}{y} = \frac{5}{7} \Rightarrow x = \frac{5}{7} y$

$\displaystyle \frac{x-80}{y-80} = \frac{3}{5}$

Substituting

$\displaystyle \frac{ \frac{5}{7} y-80}{y-80} = \frac{3}{5}$

$\displaystyle \frac{25}{7} x-400=3x-240 \Rightarrow x=280$

Substituting

$\displaystyle y = \frac{5}{7} \times 280 = 200$

$\displaystyle \\$

Question 11: If $\displaystyle (x-9):(3x+6)$ is the triplicate ratio of $\displaystyle 4:9$ , find $\displaystyle x$ . [2014]

$\displaystyle \frac{x-9}{3x+6} = \frac{4^2}{9^2} = \frac{16}{81}$

$\displaystyle 81x-729=48x+96$

$\displaystyle x=25$

$\displaystyle \\$

Question 12: If $\displaystyle a:b=5:3$ , find $\displaystyle (5a+8b):(6a-7b)$ . [2002]

$\displaystyle \text{Given } a:b=5:3$

$\displaystyle \text{or } \frac{a}{b} = \frac{5}{3} \Rightarrow a = b \frac{5}{3}$

Now substituting

$\displaystyle \frac{5a+8b}{6a-7b} = \frac{5 \times b\frac{5}{3}+8b}{6 \times b\frac{5}{3}-7b} = \frac{25+24}{30-21} = \frac{49}{9}$

Hence $\displaystyle (5a+8b):(6a-7b) = \frac{49}{9}$

$\displaystyle \\$

Question 13: The work done by $\displaystyle (x-3)$ men in $\displaystyle (2x+1)$ days and the work done by $\displaystyle (2x+1)$ men in $\displaystyle (x+4)$ days are in the ratio $\displaystyle 3:10$ . Find the value of $\displaystyle x$ . [2003]

Amount of work done by $\displaystyle (x-3)$ men in $\displaystyle (2x+1)$ days $\displaystyle = (x-3)(2x+1)$

Similarly, amount of work done by $\displaystyle (2x+1)$ men in $\displaystyle (x+4)$ days $\displaystyle = (2x+1)(x+4)$

$\displaystyle \text{Given } \frac{(x-3)(2x+1)}{(2x+1)(x+4)} = \frac{3}{10}$

$\displaystyle 10(2x^2+x-6x-3)=3(2x^2+8x+x+4)$

Simplifying

$\displaystyle 2x^2-11x-6=0$

$\displaystyle (x-6)(2x+1) = 0 \Rightarrow x = 6 or x=- \frac{1}{2} (not possible)$

$\displaystyle \text{Therefore } x = 6$

$\displaystyle \\$

Question 14: What number should be subtracted from each of the numbers $\displaystyle 23, 30, 57 \text{ and } 78$ ; so that the ratios are in proportion. [2004]

Let the number subtracted $\displaystyle = x$

$\displaystyle \text{Therefore } (23-x):(30-x)=(57-x):(78-x)$

$\displaystyle \frac{23-x}{30-x} = \frac{57-x}{78-x}$

Simplifying

$\displaystyle x^2-101x+1794 = x^2-87x+1710 \Rightarrow x =6$

$\displaystyle \\$

Question 15: $\displaystyle 6$ is the mean proportion between two numbers $\displaystyle x \text{ and } y \text{ and } 48$ is the third proportion to $\displaystyle x \text{ and } y$ . Find the numbers. [2011]

$\displaystyle \text{Given } 6$ is the mean proportion between two numbers $\displaystyle x \text{ and } y$

$\displaystyle \text{Therefore } \frac{x}{6} ={6}{y} \Rightarrow xy=36 \Rightarrow x = \frac{36}{y}$ … … … … … … i)

Also $\displaystyle \text{Given } 48$ is the third proportion to $\displaystyle x \text{ and } y$

$\displaystyle \text{Therefore } \frac{x}{y} = \frac{y}{48} \Rightarrow y^2=48x$ … … … … … … ii)

Solving i) and ii)

$\displaystyle y^2 = 48 \frac{36}{y}$

$\displaystyle y^3 = 2^3 \times 6^3 \Rightarrow y = 12$

$\displaystyle \text{Hence } x = \frac{36}{12} = 3$

Hence the numbers are $\displaystyle 3 \text{ and } 12$ .

$\displaystyle \\$

$\displaystyle \text{Question 16: If } \frac{8a-5b}{8c-5d} = \frac{8a+5b}{8c+5d} \text{, prove that } \frac{a}{b} = \frac{c}{d} \text{ [2008] }$

$\displaystyle \text{Given } \frac{8a-5b}{8c-5d} = \frac{8a+5b}{8c+5d}$

$\displaystyle \text{or } \frac{8c+5d}{8c-5d} = \frac{8a+5b}{8a-5b}$

Applying Componendo and Dividendo

$\displaystyle \frac{8c+5d+8c-5d}{8c+5d-8c+5d} = \frac{8a+5b+8a-5b}{8a+5b-8a+5b}$

$\displaystyle \frac{16c}{10d} = \frac{16a}{10b}$

$\displaystyle \frac{c}{d} = \frac{a}{b}$

$\displaystyle \text{or } \frac{a}{b} = \frac{c}{d} \text{ Hence proved.}$