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

Answer:

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]

Answer:

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]

Answer:

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} = (\frac{x}{y})^3 = (\frac{5}{3})^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]

Answer:

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]

Answer:

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]

Answer:

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]

Answer:

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]

Answer:

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]

Answer:

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]

Answer:

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]

Answer:

  \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]

Answer:

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]

Answer:

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]

Answer:

Let the number subtracted = x 

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

 \frac{23-x}{30-x}={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]

Answer:

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]

Answer:

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.

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