Question 1: If a:b=c:d , prove that:

i) 5a+7b:5a-7b=5c+7d:5c-7d

ii) (9a+13b)(9c-13d)=(9c+13d)(9a-13b)

iii)   xa+yb:xc+yd=b:d

Answer:

i) 5a+7b:5a-7b=5c+7d:5c-7d

Given \frac{a}{b} = \frac{c}{d}

 \Rightarrow \frac{5a}{7b} = \frac{5c}{7d} 

By componendo and dividendo

 \Rightarrow \frac{5a+7b}{5a-7b} = \frac{5c+7d}{5c-7d} 

ii) (9a+13b)(9c-13d)=(9c+13d)(9a-13b)

Given \frac{a}{b} = \frac{c}{d}

 \Rightarrow \frac{9a}{13b} = \frac{9c}{13d}

By componendo and dividendo

 \Rightarrow \frac{9a+13b}{9a-13b} = \frac{9c+13d}{9c-13d} 

or  (9a+13b)(9c-13d)=(9c+13d)(9a-13b)

iii)   xa+yb:xc+yd=b:d

Given \frac{a}{b} = \frac{c}{d}

 \Rightarrow \frac{xa}{yb} = \frac{xc}{yd}

By componendo

 \Rightarrow \frac{xa+yb}{yb} = \frac{xc+yd}{yd}

Cross multiplying

 \Rightarrow \frac{xa+yb}{xc+yd} = \frac{yb}{yd}

or  \frac{xa+yb}{xc+yd} = \frac{b}{d}

or   xa+yb:xc+yd=b:d

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Question 2: If a:b=c:d , prove that: (6a+7b)(3c-4d)=(6c+7d)(3a-4b) .

Answer:

Given \frac{a}{b} = \frac{c}{d}

\Rightarrow \frac{6a}{7b} = \frac{6c}{7d}

By Componendo

\frac{6a+7b}{7b} = \frac{6c+7d}{7d}

or   \frac{6a+7b}{6c+7d} = \frac{7b}{7d}

or   \frac{6a+7b}{6c+7d} = \frac{b}{d}

Again   \frac{a}{b} = \frac{c}{d}

\frac{3a}{4b} = \frac{3c}{4d}

By Dividendo

\frac{3a-4b}{4b} = \frac{3c-4d}{4d}

or   \frac{3a-4b}{3c-4d} = \frac{4b}{4d}

or   \frac{3a-4b}{3c-4d} = \frac{b}{d}

Hence (6a+7b)(3c-4d)=(6c+7d)(3a-4b)

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Question 3: 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 4: If \frac{5x+6y}{5u+6v} = \frac{5x-6y}{5u-6v} ; Then prove that x:y=u:v .

Answer:

\frac{5x+6y}{5u+6v} = \frac{5x-6y}{5u-6v}

\Rightarrow \frac{5x+6y}{5x-6y} = \frac{5u+6v}{5u-6v}

By componendo and dividendo

\Rightarrow \frac{(5x+6y) +( 5x-6y)}{(5x+6y) -(5x-6y} = \frac{(5u+6v)+(5u-6v)}{(5u+6v) - (5u-6v)}

\Rightarrow \frac{10x}{12y} = \frac{10u}{12v}

\Rightarrow x:y=u:v .

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Question 5: If  (7a+8b)(7c-8d)=(7a-8b)(7c+8d) ; prove that  a:b=c:d

Answer:

Given  (7a+8b)(7c-8d)=(7a-8b)(7c+8d)

 \frac{7a+8b}{7a-8b} = \frac{7c+8d}{7c-8d}

By componendo and dividendo

 \frac{7a+8b+7a-8b}{7a+8b -7a+8b} = \frac{7c+8d+7c-8d}{7c+8d -7c+8d}

\Rightarrow \frac{14a}{16b} = \frac{14c}{16d}

 a:b=c:d . Hence Proved.

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Question 6:

i) If  x= \frac{6ab}{a+b} , find the value of:  \frac{x+3a}{x-3a} + \frac{x+3b}{x-3b}

ii) If  a= \frac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}} , find the value of:  \frac{a+2\sqrt{2}}{a-2\sqrt{2}} + \frac{a+2\sqrt{3}}{a-2\sqrt{3}}

Answer:

i) Given  x= \frac{6ab}{a+b}

 \frac{x}{3a} = \frac{2b}{a+b}

By componendo and dividendo

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

 \frac{x+3a}{x-3a} = \frac{3b+a}{b - a} … … … … … … i)

Similarly

 \frac{x}{3b} = \frac{2a}{a+b}

By componendo and dividendo

 \frac{x+3b}{x-3b} = \frac{3a+b}{a-b}  … … … … … … ii)

Adding i) and ii)

 \frac{x+3a}{x-3a}+ \frac{x+3b}{x-3b} = \frac{3b+a}{b - a} + \frac{3a+b}{a-b} 

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

ii) Given  a= \frac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}}

 \frac{a}{2\sqrt{2}} = \frac{2\sqrt{3}}{\sqrt{2}+\sqrt{3}}

By componendo and dividendo

 \frac{a+2\sqrt{2}}{a-2\sqrt{2}} = \frac{2\sqrt{3}+\sqrt{2}+\sqrt{3} }{2\sqrt{3} - \sqrt{2}-\sqrt{3}}

 \frac{a+2\sqrt{2}}{a-2\sqrt{2}} = \frac{3\sqrt{3}+\sqrt{2} }{\sqrt{3} - \sqrt{2}}   … … … … … … i)

Similarly

 \frac{a}{2\sqrt{3}} = \frac{2\sqrt{2}}{\sqrt{2}+\sqrt{3}}

By componendo and dividendo

 \frac{a+2\sqrt{3}}{a-2\sqrt{3}} = \frac{2\sqrt{2}+ \sqrt{2}+\sqrt{3}}{2\sqrt{2} - \sqrt{2}-\sqrt{3}}

 \frac{a+2\sqrt{3}}{a-2\sqrt{3}} = \frac{3\sqrt{2}+\sqrt{3}}{\sqrt{2} -\sqrt{3}}    … … … … … … ii)

Adding i) and ii) we get

 \frac{a+2\sqrt{2}}{a-2\sqrt{2}}+ \frac{a+2\sqrt{3}}{a-2\sqrt{3}} = \frac{3\sqrt{3}+\sqrt{2} }{\sqrt{3} - \sqrt{2}} + \frac{3\sqrt{2}+\sqrt{3}}{\sqrt{2} -\sqrt{3}} 

 \frac{a+2\sqrt{2}}{a-2\sqrt{2}}+ \frac{a+2\sqrt{3}}{a-2\sqrt{3}} = \frac{2\sqrt{2}-2\sqrt{3} }{\sqrt{2} - \sqrt{3}} = 2

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Question 7: If  (a+b+c+d)(a-b-c+d)=(a+b-c-d)(a-b+c-d) ; prove that  a:b=c:d

Answer:

Given  (a+b+c+d)(a-b-c+d)=(a+b-c-d)(a-b+c-d)

 \frac{a+b+c+d}{a+b-c-d} = \frac{a-b+c-d}{a-b-c+d}

By componendo and dividendo

 \frac{(a+b+c+d)+(a+b-c-d)}{(a+b+c+d)-(a+b-c-d)} = \frac{(a-b+c-d)+(a-b-c+d)}{(a-b+c-d)-(a-b-c+d)}

\frac{a+b}{a-d} = \frac{c+d}{c-d}

Applying again componendo and dividendo

\frac{a+b+a-b}{a+b-a+d} = \frac{c+d+c-d}{c+d -c+d}

Simplifying

 \frac{a}{b} = \frac{c}{d} Hence Proved.

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Question 8: If   \frac{a-2b-3c+4d}{a+2b-3c-4d} = \frac{a-2b-3c-4d}{a+2b+3c+4d} , show that  2ab=3bc .

Answer:

Given  \frac{(a-2b-3c+4d)}{(a+2b-3c-4d)} = \frac{(a-2b-3c-4d)}{(a+2b+3c+4d)}

By componendo and dividendo

 \frac{(a-2b-3c+4d)+ (a+2b-3c-4d)}{(a-2b-3c+4d)-(a+2b-3c-4d)} = \frac{(a-2b-3c-4d)+(a+2b+3c+4d)}{(a-2b-3c-4d)-(a+2b+3c+4d)}

Simplifying

 \frac{(a-3c)}{(a+3c)} = \frac{(4d-2b)}{(-4d-2b)}

Applying again componendo and dividendo

 \frac{(a-3c)+(a+3c)}{(a-3c)-(a+3c)} = \frac{(4d-2b)+(-4d-2b)}{(4d-2b)-(-4d-2b)}

Simplifying

\frac{a}{-3c} = \frac{-b}{2d}

2ad=3bc

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Question 9: If  (a^2+b^2 )(x^2+y^2 )=(ax+by)^2 ; prove that:  \frac{a}{x} = \frac{b}{y} .

Answer:

Given  (a^2+b^2 )(x^2+y^2 )=(ax+by)^2

a^2x^2+b^2x^2+a^2y^2+b^2y^2 = a^2x^2+b^2y^2+2abxy

a^2y^2+b^2x^2-2abxy = 0

\Rightarrow (ay-bx)^2 = 0 \Rightarrow ay-bx=0

Therefore ay = bx \Rightarrow \frac{a}{x} = \frac{b}{y} Hence Proved.

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Question 10: If  a, b \ and \ c are in continued proportion, prove that:

i) \frac{a^2+ab+b^2}{b^2+bc+c^2} = \frac{a}{c}

ii) \frac{a^2+b^2+c^2}{(a+b+c)^2} = \frac{a-b+c}{a+b+c}

Answer:

Given \frac{a}{b} = \frac{b}{c} = k

 a = bk \ and \ b = ck

or  a = ck^2

i)  \frac{a^2+ab+b^2}{b^2+bc+c^2} = \frac{a}{c}

LHS  = \frac{{ck^2}^2+{ck^2}{ck}+{ck}^2}{{ck}^2+{ck}c+c^2}

 = \frac{c^2k^4+c^2k^3+c^2k^2}{c^2k^2+c^2k+ c^2}

 =  \frac{c^2k^2(k^2+k+1)}{c^2(k^2+k+1)}

 =k^2

RHS  = \frac{a}{c} = \frac{ck^2}{c} = k^2

Hence LHS = RHS. Hence Proved.

ii)  \frac{a^2+b^2+c^2}{(a+b+c)^2} = \frac{a-b+c}{a+b+c}

LHS  = \frac{a^2+b^2+c^2}{(a+b+c)^2}

 = \frac{{(ck^2)}^2+{ck}^2+c^2}{({ck^2}+ck+c)^2}

 = \frac{c^2(k^4+k^2+1)}{(k^2+k+1)^2}

 = \frac{k^4+k^2+1}{(k^2+k+1)^2}

RHS  = \frac{a-b+c}{a+b+c}

 = \frac{ck^2-ck+c}{ck^2+ck+c}

 = \frac{c(k^2-k+1)}{c(k^2+k+1)}

 = \frac{k^2-k+1}{k^2+k+1}

 = \frac{(k^2-k+1)(k^2+k+1)}{(k^2+k+1)^2}

 = \frac{k^4+k^2+1}{(k^2+k+1)^2}

Hence LHS = RHS. Hence Proved.

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Question 11: Using Properties of proportion, solve for  x :

i) \frac{\sqrt{x+5}+\sqrt{x-16}}{\sqrt{x+5}-\sqrt{x-16}} = \frac{7}{3}

ii) \frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} = \frac{4x-1}{2}

iii) \frac{3x+\sqrt{9x^2-5}}{3x-\sqrt{9x^2-5}} =5

Answer:

i)  \frac{\sqrt{x+5}+\sqrt{x-16}}{\sqrt{x+5}-\sqrt{x-16}} = \frac{7}{3}

Applying componendo and dividendo

 \frac{\sqrt{x+5}+\sqrt{x-16}+ \sqrt{x+5}-\sqrt{x-16}}{ \sqrt{x+5}+\sqrt{x-16}- \sqrt{x+5}+\sqrt{x-16}} = \frac{7+3}{7-3}

\frac{\sqrt{x+5}}{\sqrt{x-16}} = \frac{5}{2}

Squaring both sides

\frac{(x+5)}{(x-16)} = \frac{25}{4}  \Rightarrow 21x = 420

x = 20

ii)  \frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} = \frac{4x-1}{2}

Applying componendo and dividendo

 \frac{\sqrt{x+1}+\sqrt{x-1}+\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}-\sqrt{x+1}+\sqrt{x-1}} = \frac{4x-1+2}{4x-1-2}

 \frac{2\sqrt{x+1}}{2\sqrt{x-1}} = \frac{4x+1}{4x-3}

Squaring both sides

 \frac{x+1}{x-1} = \frac{16x^1+1+8x}{16x^2+9-24x}

Applying componendo and dividendo

 \frac{x+1+x-1}{x+1-x+1} = \frac{16x^2+1+8x+16x^2+9-24x}{16x^1+1+8x-16x^2-9+24x}

 \frac{2x}{2} = \frac{32x^2+10-16x}{32-8x}

or 4x=5 \Rightarrow x = \frac{5}{4}

iii)  \frac{3x+\sqrt{9x^2-5}}{3x-\sqrt{9x^2-5}} =5

Applying componendo and dividendo

 \frac{3x+\sqrt{9x^2-5}+3x-\sqrt{9x^2-5}}{3x+\sqrt{9x^2-5}-3x+\sqrt{9x^2-5}} =5

\frac{6x}{2\sqrt{9x^2-5}} = \frac{6}{4} 

\frac{x}{\sqrt{9x^2-5}} = \frac{1}{2} 

Squaring both sides

\frac{x^2}{9x^2-5} = \frac{1}{4} 

5x^2 = 5 \ or\  x = 1 

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Question 12: 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 13: 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 14: If  x= \frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}} , express  n in terms of  x \ and \ m .

Answer:

Given  x= \frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}

Applying componendo and dividendo

 \frac{x+1}{x-1} = \frac{(\sqrt{m+n}+\sqrt{m-n})+(\sqrt{m+n}-\sqrt{m-n})}{(\sqrt{m+n}+\sqrt{m-n})-(\sqrt{m+n}-\sqrt{m-n})}

Simplifying

 \frac{x+1}{x-1} = \frac{\sqrt{m+n}}{\sqrt{m-n}}

Squaring both sides

 \frac{x^2+1+2x}{x^2+1-2x} = \frac{m+n}{m-n}

Applying componendo and dividendo again

 \frac{(x^2+1+2x)+(x^2+1-2x)}{(x^2+1+2x)-(x^2+1-2x)} = \frac{m+n+m-n}{m+n-m+n}

Simplifying

\frac{x^2+1}{2x} = \frac{m}{n} 

or n = \frac{2mx}{x^2+1}

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Question 15: If   \frac{x^3+3xy^2}{3x^2 y+y^3} = \frac{m^3+3mn^2}{3m^2 n+n^3 } , show that:  nx=my .

Answer:

Given  \frac{x^3+3xy^2}{3x^2 y+y^3} = \frac{m^3+3mn^2}{3m^2 n+n^3 }

Applying componendo and dividendo

 \frac{(x^3+3xy^2)+(3x^2 y+y^3)}{(x^3+3xy^2)-(3x^2 y+y^3)} = \frac{(m^3+3mn^2)-(3m^2 n+n^3)}{(m^3+3mn^2)-(3m^2 n+n^3) }

\frac{(x+y)^3}{(x-y)^3} = \frac{(m+n)^3}{(m-n)^3}

or \frac{(x+y)}{(x-y)} = \frac{(m+n)}{(m-n)}

Applying componendo and dividendo again

\frac{(x+y)+(x+y)}{(x+y)-(x-y)} = \frac{(m+n)+(m-n)}{(m+n)-(m-n)}

Simplifying

\frac{x}{y} = \frac{m}{n}

or nx = my