\displaystyle \textbf{Question 1: } \text{If }a:b=c:d,\text{ prove that:}
\displaystyle \text{(i) }5a+7b:5a-7b=5c+7d:5c-7d
\displaystyle \text{(ii) }(9a+13b)(9c-13d)=(9c+13d)(9a-13b)
\displaystyle \text{(iii) }xa+yb:xc+yd=b:d
\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }\frac{a}{b}=\frac{c}{d},
\displaystyle \Rightarrow \frac{5a}{7b}=\frac{5c}{7d}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \Rightarrow \frac{5a+7b}{5a-7b}=\frac{5c+7d}{5c-7d}
\displaystyle \therefore 5a+7b:5a-7b=5c+7d:5c-7d
\displaystyle \text{(ii) Since }\frac{a}{b}=\frac{c}{d},
\displaystyle \Rightarrow \frac{9a}{13b}=\frac{9c}{13d}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \Rightarrow \frac{9a+13b}{9a-13b}=\frac{9c+13d}{9c-13d}
\displaystyle \Rightarrow (9a+13b)(9c-13d)=(9c+13d)(9a-13b)
\displaystyle \text{(iii) Since }\frac{a}{b}=\frac{c}{d},
\displaystyle \Rightarrow \frac{xa}{yb}=\frac{xc}{yd}
\displaystyle \text{Applying componendo,}
\displaystyle \Rightarrow \frac{xa+yb}{yb}=\frac{xc+yd}{yd}
\displaystyle \Rightarrow \frac{xa+yb}{xc+yd}=\frac{yb}{yd}=\frac{b}{d}
\displaystyle \therefore xa+yb:xc+yd=b:d
\\

\displaystyle \textbf{Question 2: } \text{If }a:b=c:d,\text{ prove that }(6a+7b)(3c-4d)=(6c+7d)(3a-4b).
\displaystyle \text{Answer:}
\displaystyle \text{Given }\frac{a}{b}=\frac{c}{d}
\displaystyle \Rightarrow \frac{6a}{7b}=\frac{6c}{7d}
\displaystyle \text{Applying componendo,}
\displaystyle \Rightarrow \frac{6a+7b}{7b}=\frac{6c+7d}{7d}
\displaystyle \Rightarrow \frac{6a+7b}{6c+7d}=\frac{b}{d}\qquad\cdots(1)
\displaystyle \text{Also, }\frac{a}{b}=\frac{c}{d}
\displaystyle \Rightarrow \frac{3a}{4b}=\frac{3c}{4d}
\displaystyle \text{Applying dividendo,}
\displaystyle \Rightarrow \frac{3a-4b}{4b}=\frac{3c-4d}{4d}
\displaystyle \Rightarrow \frac{3a-4b}{3c-4d}=\frac{b}{d}\qquad\cdots(2)
\displaystyle \text{From (1) and (2),}
\displaystyle \frac{6a+7b}{6c+7d}=\frac{3a-4b}{3c-4d}
\displaystyle \Rightarrow (6a+7b)(3c-4d)=(6c+7d)(3a-4b)
\\

\displaystyle \textbf{Question 3: } \text{Given }\frac{a}{b}=\frac{c}{d},\text{ prove that }\frac{3a-5b}{3a+5b}=\frac{3c-5d}{3c+5d}.\hfill\text{[2000]}
\displaystyle \text{Answer:}
\displaystyle \text{Given }\frac{a}{b}=\frac{c}{d}
\displaystyle \Rightarrow \frac{3a}{5b}=\frac{3c}{5d}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \Rightarrow \frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}
\displaystyle \text{Taking reciprocals of both sides,}
\displaystyle \Rightarrow \frac{3a-5b}{3a+5b}=\frac{3c-5d}{3c+5d}
\\

\displaystyle \textbf{Question 4: } \text{If }\frac{5x+6y}{5u+6v}=\frac{5x-6y}{5u-6v},\text{ prove that }x:y=u:v.
\displaystyle \text{Answer:}
\displaystyle \frac{5x+6y}{5u+6v}=\frac{5x-6y}{5u-6v}
\displaystyle \Rightarrow \frac{5x+6y}{5x-6y}=\frac{5u+6v}{5u-6v}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{(5x+6y)+(5x-6y)}{(5x+6y)-(5x-6y)}=\frac{(5u+6v)+(5u-6v)}{(5u+6v)-(5u-6v)}
\displaystyle \Rightarrow \frac{10x}{12y}=\frac{10u}{12v}
\displaystyle \Rightarrow \frac{x}{y}=\frac{u}{v}
\displaystyle \therefore x:y=u:v
\\

\displaystyle \textbf{Question 5: } \text{If }(7a+8b)(7c-8d)=(7a-8b)(7c+8d),\text{ prove that }a:b=c:d.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }(7a+8b)(7c-8d)=(7a-8b)(7c+8d)
\displaystyle \Rightarrow \frac{7a+8b}{7a-8b}=\frac{7c+8d}{7c-8d}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{(7a+8b)+(7a-8b)}{(7a+8b)-(7a-8b)}=\frac{(7c+8d)+(7c-8d)}{(7c+8d)-(7c-8d)}
\displaystyle \Rightarrow \frac{14a}{16b}=\frac{14c}{16d}
\displaystyle \Rightarrow \frac{a}{b}=\frac{c}{d}
\displaystyle \therefore a:b=c:d
\\

\displaystyle \textbf{Question 6: }
\displaystyle \text{(i) If }x=\frac{6ab}{a+b},\text{ find }\frac{x+3a}{x-3a}+\frac{x+3b}{x-3b}.
\displaystyle \text{(ii) If }a=\frac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}},\text{ find }\frac{a+2\sqrt{2}}{a-2\sqrt{2}}+\frac{a+2\sqrt{3}}{a-2\sqrt{3}}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }x=\frac{6ab}{a+b}
\displaystyle \Rightarrow \frac{x}{3a}=\frac{2b}{a+b}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{x+3a}{x-3a}=\frac{2b+a+b}{2b-a-b}=\frac{a+3b}{b-a}\qquad\cdots(1)
\displaystyle \text{Similarly, }\frac{x}{3b}=\frac{2a}{a+b}
\displaystyle \Rightarrow \frac{x+3b}{x-3b}=\frac{2a+a+b}{2a-a-b}=\frac{3a+b}{a-b}\qquad\cdots(2)
\displaystyle \text{Adding (1) and (2),}
\displaystyle \frac{x+3a}{x-3a}+\frac{x+3b}{x-3b}=\frac{a+3b}{b-a}+\frac{3a+b}{a-b}
\displaystyle =-\frac{a+3b}{a-b}+\frac{3a+b}{a-b}
\displaystyle =\frac{2a-2b}{a-b}=2
\displaystyle \text{(ii) Given, }a=\frac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}}
\displaystyle \Rightarrow \frac{a}{2\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}+\sqrt{3}}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{a+2\sqrt{2}}{a-2\sqrt{2}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}}{2\sqrt{3}-\sqrt{2}-\sqrt{3}}
\displaystyle =\frac{\sqrt{2}+3\sqrt{3}}{\sqrt{3}-\sqrt{2}}\qquad\cdots(1)
\displaystyle \text{Similarly, }\frac{a}{2\sqrt{3}}=\frac{2\sqrt{2}}{\sqrt{2}+\sqrt{3}}
\displaystyle \Rightarrow \frac{a+2\sqrt{3}}{a-2\sqrt{3}}=\frac{2\sqrt{2}+\sqrt{2}+\sqrt{3}}{2\sqrt{2}-\sqrt{2}-\sqrt{3}}
\displaystyle =\frac{3\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}\qquad\cdots(2)
\displaystyle \text{Adding (1) and (2),}
\displaystyle \frac{a+2\sqrt{2}}{a-2\sqrt{2}}+\frac{a+2\sqrt{3}}{a-2\sqrt{3}}
\displaystyle =\frac{\sqrt{2}+3\sqrt{3}}{\sqrt{3}-\sqrt{2}}+\frac{3\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}
\displaystyle =-\frac{\sqrt{2}+3\sqrt{3}}{\sqrt{2}-\sqrt{3}}+\frac{3\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}
\displaystyle =\frac{2\sqrt{2}-2\sqrt{3}}{\sqrt{2}-\sqrt{3}}=2
\\

\displaystyle \textbf{Question 7: } \text{If }(a+b+c+d)(a-b-c+d)=(a+b-c-d)(a-b+c-d),
\displaystyle \text{prove that }a:b=c:d.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }(a+b+c+d)(a-b-c+d)=(a+b-c-d)(a-b+c-d)
\displaystyle \Rightarrow \frac{a+b+c+d}{a+b-c-d}=\frac{a-b+c-d}{a-b-c+d}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{(a+b+c+d)+(a+b-c-d)}{(a+b+c+d)-(a+b-c-d)}
\displaystyle =\frac{(a-b+c-d)+(a-b-c+d)}{(a-b+c-d)-(a-b-c+d)}
\displaystyle \Rightarrow \frac{2a+2b}{2c+2d}=\frac{2a-2b}{2c-2d}
\displaystyle \Rightarrow \frac{a+b}{c+d}=\frac{a-b}{c-d}
\displaystyle \text{Again applying componendo and dividendo,}
\displaystyle \frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{(c+d)+(c-d)}{(c+d)-(c-d)}
\displaystyle \Rightarrow \frac{2a}{2b}=\frac{2c}{2d}
\displaystyle \Rightarrow \frac{a}{b}=\frac{c}{d}
\displaystyle \therefore a:b=c:d
\\

\displaystyle \textbf{Question 8: } \text{If }\frac{a-2b-3c+4d}{a+2b-3c-4d}=\frac{a-2b-3c-4d}{a+2b+3c+4d},
\displaystyle \text{show that }2ad=3bc.
\displaystyle \text{Answer:}
\displaystyle \frac{a-2b-3c+4d}{a+2b-3c-4d}=\frac{a-2b-3c-4d}{a+2b+3c+4d}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{(a-2b-3c+4d)+(a+2b-3c-4d)}{(a-2b-3c+4d)-(a+2b-3c-4d)}
\displaystyle =\frac{(a-2b-3c-4d)+(a+2b+3c+4d)}{(a-2b-3c-4d)-(a+2b+3c+4d)}
\displaystyle \Rightarrow \frac{2a-6c}{-4b+8d}=\frac{2a}{-4b-6c-8d}
\displaystyle \Rightarrow \frac{a-3c}{-2b+4d}=\frac{a}{-2b-3c-4d}
\displaystyle \Rightarrow (a-3c)(-2b-3c-4d)=a(-2b+4d)
\displaystyle \Rightarrow -2ab-3ac-4ad+6bc+9c^2+12cd=-2ab+4ad
\displaystyle \Rightarrow -3ac-8ad+6bc+9c^2+12cd=0
\displaystyle \text{This does not simplify to }2ab=3bc\text{ or }2ad=3bc\text{ in general.}
\\

\displaystyle \textbf{Question 9: } \text{If }(a^2+b^2)(x^2+y^2)=(ax+by)^2,\text{ prove that }\frac{a}{x}=\frac{b}{y}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }(a^2+b^2)(x^2+y^2)=(ax+by)^2
\displaystyle \Rightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2abxy+b^2y^2
\displaystyle \Rightarrow a^2y^2+b^2x^2-2abxy=0
\displaystyle \Rightarrow (ay-bx)^2=0
\displaystyle \Rightarrow ay=bx
\displaystyle \therefore \frac{a}{x}=\frac{b}{y}
\\

\displaystyle \textbf{Question 10: } \text{If }a,\ b\text{ and }c\text{ are in continued proportion, prove that:}
\displaystyle \text{(i) }\frac{a^2+ab+b^2}{b^2+bc+c^2}=\frac{a}{c}
\displaystyle \text{(ii) }\frac{a^2+b^2+c^2}{(a+b+c)^2}=\frac{a-b+c}{a+b+c}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\frac{a}{b}=\frac{b}{c}=k
\displaystyle \Rightarrow b=ck\text{ and }a=bk=ck^2
\displaystyle \text{(i) LHS}=\frac{a^2+ab+b^2}{b^2+bc+c^2}
\displaystyle =\frac{(ck^2)^2+(ck^2)(ck)+(ck)^2}{(ck)^2+(ck)c+c^2}
\displaystyle =\frac{c^2k^4+c^2k^3+c^2k^2}{c^2k^2+c^2k+c^2}
\displaystyle =\frac{c^2k^2(k^2+k+1)}{c^2(k^2+k+1)}=k^2
\displaystyle \text{RHS}=\frac{a}{c}=\frac{ck^2}{c}=k^2
\displaystyle \therefore \frac{a^2+ab+b^2}{b^2+bc+c^2}=\frac{a}{c}
\displaystyle \text{(ii) LHS}=\frac{a^2+b^2+c^2}{(a+b+c)^2}
\displaystyle =\frac{(ck^2)^2+(ck)^2+c^2}{(ck^2+ck+c)^2}
\displaystyle =\frac{c^2(k^4+k^2+1)}{c^2(k^2+k+1)^2}
\displaystyle =\frac{k^4+k^2+1}{(k^2+k+1)^2}
\displaystyle \text{RHS}=\frac{a-b+c}{a+b+c}
\displaystyle =\frac{ck^2-ck+c}{ck^2+ck+c}
\displaystyle =\frac{k^2-k+1}{k^2+k+1}
\displaystyle =\frac{(k^2-k+1)(k^2+k+1)}{(k^2+k+1)^2}
\displaystyle =\frac{k^4+k^2+1}{(k^2+k+1)^2}
\displaystyle \therefore \frac{a^2+b^2+c^2}{(a+b+c)^2}=\frac{a-b+c}{a+b+c}
\\

\displaystyle \textbf{Question 11: } \text{Using properties of proportion, solve for }x:
\displaystyle \text{(i) }\frac{\sqrt{x+5}+\sqrt{x-16}}{\sqrt{x+5}-\sqrt{x-16}}=\frac{7}{3}
\displaystyle \text{(ii) }\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\frac{4x-1}{2}
\displaystyle \text{(iii) }\frac{3x+\sqrt{9x^2-5}}{3x-\sqrt{9x^2-5}}=5
\displaystyle \text{Answer:}
\displaystyle \text{(i) Applying componendo and dividendo,}
\displaystyle \frac{2\sqrt{x+5}}{2\sqrt{x-16}}=\frac{7+3}{7-3}
\displaystyle \Rightarrow \frac{\sqrt{x+5}}{\sqrt{x-16}}=\frac{5}{2}
\displaystyle \Rightarrow \frac{x+5}{x-16}=\frac{25}{4}
\displaystyle \Rightarrow 4x+20=25x-400
\displaystyle \Rightarrow x=20
\displaystyle \text{(ii) Applying componendo and dividendo,}
\displaystyle \frac{2\sqrt{x+1}}{2\sqrt{x-1}}=\frac{4x-1+2}{4x-1-2}
\displaystyle \Rightarrow \frac{\sqrt{x+1}}{\sqrt{x-1}}=\frac{4x+1}{4x-3}
\displaystyle \Rightarrow \frac{x+1}{x-1}=\frac{(4x+1)^2}{(4x-3)^2}
\displaystyle \Rightarrow (x+1)(4x-3)^2=(x-1)(4x+1)^2
\displaystyle \Rightarrow 4x=5
\displaystyle \Rightarrow x=\frac{5}{4}
\displaystyle \text{(iii) Applying componendo and dividendo,}
\displaystyle \frac{6x}{2\sqrt{9x^2-5}}=\frac{5+1}{5-1}
\displaystyle \Rightarrow \frac{x}{\sqrt{9x^2-5}}=\frac{1}{2}
\displaystyle \Rightarrow \frac{x^2}{9x^2-5}=\frac{1}{4}
\displaystyle \Rightarrow 4x^2=9x^2-5
\displaystyle \Rightarrow 5x^2=5
\displaystyle \Rightarrow x=\pm1
\displaystyle \text{On checking in the given equation, }x=1\text{ is valid and }x=-1\text{ is not valid.}
\displaystyle \therefore x=1
\\

\displaystyle \textbf{Question 12: } \text{If }x=\frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}},\text{ prove that } \\ 3bx^2-2ax+3b=0.\hfill\text{[2007]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\sqrt{a+3b}\text{ and }B=\sqrt{a-3b}.
\displaystyle \text{Given, }x=\frac{A+B}{A-B}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{x+1}{x-1}=\frac{(A+B)+(A-B)}{(A+B)-(A-B)}
\displaystyle \Rightarrow \frac{x+1}{x-1}=\frac{2A}{2B}=\frac{A}{B}
\displaystyle \Rightarrow \frac{x+1}{x-1}=\frac{\sqrt{a+3b}}{\sqrt{a-3b}}
\displaystyle \text{Squaring both sides,}
\displaystyle \frac{(x+1)^2}{(x-1)^2}=\frac{a+3b}{a-3b}
\displaystyle \Rightarrow \frac{x^2+2x+1}{x^2-2x+1}=\frac{a+3b}{a-3b}
\displaystyle \text{Applying componendo and dividendo again,}
\displaystyle \frac{(x^2+2x+1)+(x^2-2x+1)}{(x^2+2x+1)-(x^2-2x+1)}
\displaystyle =\frac{(a+3b)+(a-3b)}{(a+3b)-(a-3b)}
\displaystyle \Rightarrow \frac{2x^2+2}{4x}=\frac{2a}{6b}
\displaystyle \Rightarrow \frac{x^2+1}{2x}=\frac{a}{3b}
\displaystyle \Rightarrow 3b(x^2+1)=2ax
\displaystyle \Rightarrow 3bx^2-2ax+3b=0
\\

\displaystyle \textbf{Question 13: } \text{Using the properties of proportion, solve for }x:\frac{x^4+1}{2x^2}=\frac{17}{8}.\hfill\text{[2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\frac{x^4+1}{2x^2}=\frac{17}{8}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{x^4+1+2x^2}{x^4+1-2x^2}=\frac{17+8}{17-8}
\displaystyle \Rightarrow \frac{(x^2+1)^2}{(x^2-1)^2}=\frac{25}{9}
\displaystyle \text{Taking square root of both sides,}
\displaystyle \frac{x^2+1}{x^2-1}=\frac{5}{3}
\displaystyle \Rightarrow 3x^2+3=5x^2-5
\displaystyle \Rightarrow 2x^2=8
\displaystyle \Rightarrow x^2=4
\displaystyle \Rightarrow x=\pm2
\\

\displaystyle \textbf{Question 14: } \text{If }x=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}},\text{ express }n\text{ in terms of }x\text{ and }m.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \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})}
\displaystyle \Rightarrow \frac{x+1}{x-1}=\frac{\sqrt{m+n}}{\sqrt{m-n}}
\displaystyle \text{Squaring both sides,}
\displaystyle \frac{x^2+2x+1}{x^2-2x+1}=\frac{m+n}{m-n}
\displaystyle \text{Applying componendo and dividendo again,}
\displaystyle \frac{(x^2+2x+1)+(x^2-2x+1)}{(x^2+2x+1)-(x^2-2x+1)}=\frac{(m+n)+(m-n)}{(m+n)-(m-n)}
\displaystyle \Rightarrow \frac{x^2+1}{2x}=\frac{m}{n}
\displaystyle \therefore n=\frac{2mx}{x^2+1}
\\

\displaystyle \textbf{Question 15: } \text{If }\frac{x^3+3xy^2}{3x^2y+y^3}=\frac{m^3+3mn^2}{3m^2n+n^3},\text{ show that }nx=my.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\frac{x^3+3xy^2}{3x^2y+y^3}=\frac{m^3+3mn^2}{3m^2n+n^3}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{(x^3+3xy^2)+(3x^2y+y^3)}{(x^3+3xy^2)-(3x^2y+y^3)}=\frac{(m^3+3mn^2)+(3m^2n+n^3)}{(m^3+3mn^2)-(3m^2n+n^3)}
\displaystyle \Rightarrow \frac{(x+y)^3}{(x-y)^3}=\frac{(m+n)^3}{(m-n)^3}
\displaystyle \Rightarrow \frac{x+y}{x-y}=\frac{m+n}{m-n}
\displaystyle \text{Applying componendo and dividendo again,}
\displaystyle \frac{(x+y)+(x-y)}{(x+y)-(x-y)}=\frac{(m+n)+(m-n)}{(m+n)-(m-n)}
\displaystyle \Rightarrow \frac{2x}{2y}=\frac{2m}{2n}
\displaystyle \Rightarrow \frac{x}{y}=\frac{m}{n}
\displaystyle \therefore nx=my
\\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.