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

If $x, y \ and \ z$ are in continued proportion, then

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

Applying componendo and dividendo

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

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

Squaring both sides

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

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

Substituting

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

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

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

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

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

Simplifying

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

Square both sides

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

Applying componendo and dividendo

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

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

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

Simplifying

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

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

i) $\frac{x}{y}$   ii) $\frac{x^3+y^3}{x^3-y^3 }$  [2014]

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

Applying componendo and dividendo

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

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

Simplifying, we get

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

ii) $\frac{x^3+y^3}{x^3-y^3 }$

Applying componendo and dividendo

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

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

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Question 4: Using componendo and dividendo, find the value of $x$$\frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}}$ $=9$    [2011]

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

Applying componendo and dividendo

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

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

Simplifying

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

Square both sides

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

$42x+56 = 75x-125$

Simplifying we get $x = 7$

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Question 5: If $x=$ $\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}$ , using properties of proportion show that:  $x^2-2ax+1$   [2012]

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

Applying componendo and dividendo

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

Simplify

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

Now square both sides

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

Simplifying

$x^2+1 = 2ax$

or $x^2-2ax+1 = 0$

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

Given $\frac{a}{b}$ $=$ $\frac{c}{d}$

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

By componendo and dividendo

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

By Alternendo

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

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

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

Applying componendo and dividendo

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

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

Squaring both sides

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

Applying componendo and dividendo once again

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

simplifying

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

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

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

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

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

Applying componendo and dividendo

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

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

Taking the square root of both sides

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

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

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

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

Let the number added be $x$

Therefore $(6+x): (15+x) = (20+x): (43+x)$

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

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

$\Rightarrow x = 3$

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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 $x and y$ respectively.

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

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

Substituting

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

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

Substituting

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

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

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

$81x-729=48x+96$

$x=25$

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

Given $a:b=5:3$

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

Now substituting

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

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

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

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

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

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

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

Simplifying

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

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

Therefore $x = 6$

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

Let the number subtracted $= x$

Therefore $(23-x):(30-x)=(57-x):(78-x)$

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

Simplifying

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

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

Given  $6$ is the mean proportion between two numbers $x \ and \ y$

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

Also given  $48$  is the third proportion to $x \ and \ y$

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

Solving i) and ii)

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

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

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

Hence the numbers are $3 \ and \ 12$.

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

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

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

Applying Componendo and Dividendo

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

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

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

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