Class 10: Ratio and Proportion – ICSE Board Problems

$\displaystyle \textbf{Question 1: If }x,\ y\text{ and }z\text{ are in continued proportion, prove that }$
$\displaystyle \frac{(x+y)^2}{(y+z)^2}=\frac{x}{y}.\hfill \text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }x,\ y\text{ and }z\text{ are in continued proportion,}$
$\displaystyle \frac{x}{y}=\frac{y}{z}$
$\displaystyle \Rightarrow y^2=xz$
$\displaystyle \text{Now, }\frac{(x+y)^2}{(y+z)^2}=\frac{x}{y}\iff y(x+y)^2=x(y+z)^2$
$\displaystyle \text{L.H.S.}=y(x+y)^2=x^2y+2xy^2+y^3$
$\displaystyle =x^2y+2x(xz)+y(xz)$
$\displaystyle =x^2y+2x^2z+xyz$
$\displaystyle =x(x^2+2xz+yz)$
$\displaystyle \text{Also, }\text{R.H.S.}=x(y+z)^2=x(y^2+2yz+z^2)$
$\displaystyle =x(xz+2yz+z^2)$
$\displaystyle =x(xz+2yz+z^2)$
$\displaystyle \text{Since }y^2=xz,\ \ x^2z=xy^2\text{ and }yz^2=y(y^2/x)=y^3/x,$
$\displaystyle \text{or more directly, substituting }z=\frac{y^2}{x},$
$\displaystyle \text{L.H.S.}=y\left(x+y\right)^2=\text{R.H.S.}=x\left(y+\frac{y^2}{x}\right)^2$
$\displaystyle \therefore \frac{(x+y)^2}{(y+z)^2}=\frac{x}{y}$
$\\$

$\displaystyle \textbf{Question 2: Given }x=\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}-\sqrt{a^2-b^2}},$
$\displaystyle \text{use componendo and dividendo to prove that }b^2=\frac{2a^2x}{x^2+1}.\hfill \text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle x=\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}-\sqrt{a^2-b^2}}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{x+1}{x-1}=\frac{(\sqrt{a^2+b^2}+\sqrt{a^2-b^2})+(\sqrt{a^2+b^2}-\sqrt{a^2-b^2})}{(\sqrt{a^2+b^2}+\sqrt{a^2-b^2})-(\sqrt{a^2+b^2}-\sqrt{a^2-b^2})}$
$\displaystyle \frac{x+1}{x-1}=\frac{2\sqrt{a^2+b^2}}{2\sqrt{a^2-b^2}}$
$\displaystyle \frac{x+1}{x-1}=\frac{\sqrt{a^2+b^2}}{\sqrt{a^2-b^2}}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle \frac{(x+1)^2}{(x-1)^2}=\frac{a^2+b^2}{a^2-b^2}$
$\displaystyle \frac{x^2+2x+1}{x^2-2x+1}=\frac{a^2+b^2}{a^2-b^2}$
$\displaystyle \text{Again applying componendo and dividendo,}$
$\displaystyle \frac{(x^2+2x+1)+(x^2-2x+1)}{(x^2+2x+1)-(x^2-2x+1)}=\frac{(a^2+b^2)+(a^2-b^2)}{(a^2+b^2)-(a^2-b^2)}$
$\displaystyle \frac{2x^2+2}{4x}=\frac{2a^2}{2b^2}$
$\displaystyle \frac{x^2+1}{2x}=\frac{a^2}{b^2}$
$\displaystyle b^2(x^2+1)=2a^2x$
$\displaystyle \therefore b^2=\frac{2a^2x}{x^2+1}$
$\\$

$\displaystyle \textbf{Question 3: If }\frac{x^2+y^2}{x^2-y^2}=2\frac{1}{8},\text{ find:}$
$\displaystyle \text{(i) }\frac{x}{y}\qquad\qquad\text{(ii) }\frac{x^3+y^3}{x^3-y^3}.\hfill \text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given }\frac{x^2+y^2}{x^2-y^2}=2\frac{1}{8}=\frac{17}{8}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{(x^2+y^2)+(x^2-y^2)}{(x^2+y^2)-(x^2-y^2)}=\frac{17+8}{17-8}$
$\displaystyle \frac{2x^2}{2y^2}=\frac{25}{9}$
$\displaystyle \frac{x^2}{y^2}=\frac{25}{9}$
$\displaystyle \therefore \frac{x}{y}=\frac{5}{3}$
$\displaystyle \text{(ii) }\frac{x^3}{y^3}=\left(\frac{x}{y}\right)^3=\left(\frac{5}{3}\right)^3=\frac{125}{27}$
$\displaystyle \therefore \frac{x^3+y^3}{x^3-y^3}=\frac{125+27}{125-27}$
$\displaystyle =\frac{152}{98}=\frac{76}{49}$
$\\$

$\displaystyle \textbf{Question 4: Using componendo and dividendo, find the value of }x:$
$\displaystyle \frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}}=9.\hfill \text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }\frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}}=9$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{(\sqrt{3x+4}+\sqrt{3x-5})+(\sqrt{3x+4}-\sqrt{3x-5})}{(\sqrt{3x+4}+\sqrt{3x-5})-(\sqrt{3x+4}-\sqrt{3x-5})}=\frac{9+1}{9-1}$
$\displaystyle \frac{2\sqrt{3x+4}}{2\sqrt{3x-5}}=\frac{10}{8}$
$\displaystyle \frac{\sqrt{3x+4}}{\sqrt{3x-5}}=\frac{5}{4}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle \frac{3x+4}{3x-5}=\frac{25}{16}$
$\displaystyle 16(3x+4)=25(3x-5)$
$\displaystyle 48x+64=75x-125$
$\displaystyle 27x=189$
$\displaystyle x=7$
$\\$

$\displaystyle \textbf{Question 5: If }x=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}},$
$\displaystyle \text{using properties of proportion, show that }x^2-2ax+1=0.\hfill \text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }x=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{x+1}{x-1}=\frac{(\sqrt{a+1}+\sqrt{a-1})+(\sqrt{a+1}-\sqrt{a-1})}{(\sqrt{a+1}+\sqrt{a-1})-(\sqrt{a+1}-\sqrt{a-1})}$
$\displaystyle \frac{x+1}{x-1}=\frac{2\sqrt{a+1}}{2\sqrt{a-1}}$
$\displaystyle \frac{x+1}{x-1}=\frac{\sqrt{a+1}}{\sqrt{a-1}}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle \frac{(x+1)^2}{(x-1)^2}=\frac{a+1}{a-1}$
$\displaystyle \frac{x^2+2x+1}{x^2-2x+1}=\frac{a+1}{a-1}$
$\displaystyle (a-1)(x^2+2x+1)=(a+1)(x^2-2x+1)$
$\displaystyle ax^2+2ax+a-x^2-2x-1=ax^2-2ax+a+x^2-2x+1$
$\displaystyle 4ax-2x^2-2=0$
$\displaystyle x^2-2ax+1=0$
$\\$

$\displaystyle \textbf{Question 6: Given }\frac{a}{b}=\frac{c}{d},\text{ prove that }\frac{3a-5b}{3a+5b}=\frac{3c-5d}{3c+5d}.\hfill \text{[ICSE 2000]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }\frac{a}{b}=\frac{c}{d}$
$\displaystyle \Rightarrow \frac{3a}{5b}=\frac{3c}{5d}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}$
$\displaystyle \text{Taking reciprocals of both sides,}$
$\displaystyle \frac{3a-5b}{3a+5b}=\frac{3c-5d}{3c+5d}$
$\displaystyle \therefore \frac{3a-5b}{3a+5b}=\frac{3c-5d}{3c+5d}$
$\\$

$\displaystyle \textbf{Question 7: If }x=\frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}},\text{ prove that }$
$\displaystyle 3bx^2-2ax+3b=0.\hfill \text{[ICSE 2007]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }x=\frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{x+1}{x-1}=\frac{(\sqrt{a+3b}+\sqrt{a-3b})+(\sqrt{a+3b}-\sqrt{a-3b})}{(\sqrt{a+3b}+\sqrt{a-3b})-(\sqrt{a+3b}-\sqrt{a-3b})}$
$\displaystyle \frac{x+1}{x-1}=\frac{2\sqrt{a+3b}}{2\sqrt{a-3b}}$
$\displaystyle \frac{x+1}{x-1}=\frac{\sqrt{a+3b}}{\sqrt{a-3b}}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle \frac{(x+1)^2}{(x-1)^2}=\frac{a+3b}{a-3b}$
$\displaystyle \frac{x^2+2x+1}{x^2-2x+1}=\frac{a+3b}{a-3b}$
$\displaystyle \text{Applying componendo and dividendo again,}$
$\displaystyle \frac{(x^2+2x+1)+(x^2-2x+1)}{(x^2+2x+1)-(x^2-2x+1)}=\frac{(a+3b)+(a-3b)}{(a+3b)-(a-3b)}$
$\displaystyle \frac{2x^2+2}{4x}=\frac{2a}{6b}$
$\displaystyle \frac{x^2+1}{2x}=\frac{a}{3b}$
$\displaystyle 3b(x^2+1)=2ax$
$\displaystyle 3bx^2+3b=2ax$
$\displaystyle 3bx^2-2ax+3b=0$
$\displaystyle \therefore 3bx^2-2ax+3b=0$
$\\$

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

$\displaystyle \textbf{Question 9: What least number must be added to each of the numbers }6,\ 15,\ 20\text{ and }43$
$\displaystyle \text{to make them proportional?}\hfill \text{[ICSE 2005, 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required number be }x$
$\displaystyle (6+x):(15+x)=(20+x):(43+x)$
$\displaystyle (6+x)(43+x)=(20+x)(15+x)$
$\displaystyle x^2+49x+258=x^2+35x+300$
$\displaystyle 49x-35x=300-258$
$\displaystyle 14x=42$
$\displaystyle x=3$
$\displaystyle \therefore \text{The least number to be added is }3$
$\\$

$\displaystyle \textbf{Question 10: The monthly pocket money of Ravi and Sanjeev are in the ratio }5:7.\text{ Their expenditures}$
$\displaystyle \text{are in the ratio }3:5.\text{ If each saves Rs. }80\text{ per month, find their monthly pocket money.}\hfill \text{[ICSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the monthly pocket money of Ravi and Sanjeev be Rs. }5x\text{ and Rs. }7x\text{ respectively}$
$\displaystyle \text{Ravi's expenditure }=\text{Rs. }(5x-80)$
$\displaystyle \text{Sanjeev's expenditure }=\text{Rs. }(7x-80)$
$\displaystyle \frac{5x-80}{7x-80}=\frac{3}{5}$
$\displaystyle 5(5x-80)=3(7x-80)$
$\displaystyle 25x-400=21x-240$
$\displaystyle 4x=160$
$\displaystyle x=40$
$\displaystyle \therefore \text{Ravi's monthly pocket money }=5x=5(40)=\text{Rs. }200$
$\displaystyle \text{Sanjeev's monthly pocket money }=7x=7(40)=\text{Rs. }280$
$\\$

$\displaystyle \textbf{Question 11: If }(x-9):(3x+6)\text{ is the triplicate ratio of }4:9,\text{ find }x.\hfill \text{[ICSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Triplicate ratio of }4:9=4^3:9^3=64:729$
$\displaystyle \frac{x-9}{3x+6}=\frac{64}{729}$
$\displaystyle 729(x-9)=64(3x+6)$
$\displaystyle 729x-6561=192x+384$
$\displaystyle 537x=6945$
$\displaystyle x=\frac{6945}{537}=\frac{2315}{179}$
$\\$

$\displaystyle \textbf{Question 12: If }a:b=5:3,\text{ find }(5a+8b):(6a-7b).\hfill \text{[ICSE 2002]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }a:b=5:3$
$\displaystyle \frac{a}{b}=\frac{5}{3}$
$\displaystyle a=\frac{5b}{3}$
$\displaystyle \frac{5a+8b}{6a-7b}=\frac{5\left(\frac{5b}{3}\right)+8b}{6\left(\frac{5b}{3}\right)-7b}$
$\displaystyle =\frac{\frac{25b}{3}+\frac{24b}{3}}{10b-7b}$
$\displaystyle =\frac{\frac{49b}{3}}{3b}$
$\displaystyle =\frac{49}{9}$
$\displaystyle \therefore (5a+8b):(6a-7b)=49:9$
$\\$

$\displaystyle \textbf{Question 13: The work done by }(x-3)\text{ men in }(2x+1)\text{ days and the work done by }$
$\displaystyle (2x+1)\text{ men in }(x+4)\text{ days are in the ratio }3:10.\text{ Find the value of }x.\hfill \text{[ICSE 2003]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Work done by }(x-3)\text{ men in }(2x+1)\text{ days }=(x-3)(2x+1)$
$\displaystyle \text{Work done by }(2x+1)\text{ men in }(x+4)\text{ days }=(2x+1)(x+4)$
$\displaystyle \frac{(x-3)(2x+1)}{(2x+1)(x+4)}=\frac{3}{10}$
$\displaystyle \frac{x-3}{x+4}=\frac{3}{10}$
$\displaystyle 10(x-3)=3(x+4)$
$\displaystyle 10x-30=3x+12$
$\displaystyle 7x=42$
$\displaystyle x=6$
$\\$

$\displaystyle \textbf{Question 14: What number should be subtracted from each of the numbers }23,\ 30,\ 57\text{ and }78$
$\displaystyle \text{so that the resulting numbers are in proportion?}\hfill \text{[ICSE 2004]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number to be subtracted be }x$
$\displaystyle (23-x):(30-x)=(57-x):(78-x)$
$\displaystyle \frac{23-x}{30-x}=\frac{57-x}{78-x}$
$\displaystyle (23-x)(78-x)=(30-x)(57-x)$
$\displaystyle 1794-101x+x^2=1710-87x+x^2$
$\displaystyle 1794-1710=101x-87x$
$\displaystyle 84=14x$
$\displaystyle x=6$
$\displaystyle \therefore \text{The required number is }6$
$\\$

$\displaystyle \textbf{Question 15: }6\text{ is the mean proportion between two numbers }x\text{ and }y,\text{ and }48\text{ is the third}$
$\displaystyle \text{proportion to }x\text{ and }y.\text{ Find the numbers.}\hfill \text{[ICSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }6\text{ is the mean proportion between }x\text{ and }y$
$\displaystyle \frac{x}{6}=\frac{6}{y}$
$\displaystyle xy=36$
$\displaystyle x=\frac{36}{y}\qquad \cdots\text{(i)}$
$\displaystyle \text{Also, }48\text{ is the third proportion to }x\text{ and }y$
$\displaystyle \frac{x}{y}=\frac{y}{48}$
$\displaystyle y^2=48x\qquad \cdots\text{(ii)}$
$\displaystyle \text{Substituting (i) in (ii),}$
$\displaystyle y^2=48\left(\frac{36}{y}\right)$
$\displaystyle y^3=48\times36$
$\displaystyle y^3=1728$
$\displaystyle y=12$
$\displaystyle x=\frac{36}{12}=3$
$\displaystyle \therefore \text{The required numbers are }3\text{ and }12$
$\\$

$\displaystyle \textbf{Question 16: If }\frac{8a-5b}{8c-5d}=\frac{8a+5b}{8c+5d},\text{ prove that }\frac{a}{b}=\frac{c}{d}.\hfill \text{[ICSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }\frac{8a-5b}{8c-5d}=\frac{8a+5b}{8c+5d}$
$\displaystyle \frac{8c+5d}{8c-5d}=\frac{8a+5b}{8a-5b}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{(8c+5d)+(8c-5d)}{(8c+5d)-(8c-5d)}=\frac{(8a+5b)+(8a-5b)}{(8a+5b)-(8a-5b)}$
$\displaystyle \frac{16c}{10d}=\frac{16a}{10b}$
$\displaystyle \frac{c}{d}=\frac{a}{b}$
$\displaystyle \therefore \frac{a}{b}=\frac{c}{d}$
$\\$

$\displaystyle \textbf{Question 17: If }x=\frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}},\text{ then prove that}$
$\displaystyle 3bx^2-2ax+3b=0.\hfill \text{[ICSE 2007]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{x}{1}=\frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{x+1}{x-1}=\frac{(\sqrt{a+3b}+\sqrt{a-3b})+(\sqrt{a+3b}-\sqrt{a-3b})}{(\sqrt{a+3b}+\sqrt{a-3b})-(\sqrt{a+3b}-\sqrt{a-3b})}$
$\displaystyle \frac{x+1}{x-1}=\frac{2\sqrt{a+3b}}{2\sqrt{a-3b}}$
$\displaystyle \frac{x+1}{x-1}=\frac{\sqrt{a+3b}}{\sqrt{a-3b}}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle \frac{(x+1)^2}{(x-1)^2}=\frac{a+3b}{a-3b}$
$\displaystyle \frac{x^2+2x+1}{x^2-2x+1}=\frac{a+3b}{a-3b}$
$\displaystyle \text{Again applying componendo and dividendo,}$
$\displaystyle \frac{(x^2+2x+1)+(x^2-2x+1)}{(x^2+2x+1)-(x^2-2x+1)}=\frac{(a+3b)+(a-3b)}{(a+3b)-(a-3b)}$
$\displaystyle \frac{2x^2+2}{4x}=\frac{2a}{6b}$
$\displaystyle \frac{x^2+1}{2x}=\frac{a}{3b}$
$\displaystyle 3b(x^2+1)=2ax$
$\displaystyle 3bx^2+3b=2ax$
$\displaystyle 3bx^2-2ax+3b=0$
$\displaystyle \therefore 3bx^2-2ax+3b=0$
$\\$

$\displaystyle \textbf{Question 18: Given that }\frac{a^3+3ab^2}{b^3+3a^2b}=\frac{63}{62}.$
$\displaystyle \text{Using componendo and dividendo, find }a:b.\hfill \text{[ICSE 2009]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{a^3+3ab^2}{b^3+3a^2b}=\frac{63}{62}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{(a^3+3ab^2)+(b^3+3a^2b)}{(a^3+3ab^2)-(b^3+3a^2b)}=\frac{63+62}{63-62}$
$\displaystyle \frac{a^3+b^3+3a^2b+3ab^2}{a^3-b^3-3a^2b+3ab^2}=\frac{125}{1}$
$\displaystyle \frac{(a+b)^3}{(a-b)^3}=\frac{5^3}{1^3}$
$\displaystyle \text{Taking cube root of both sides,}$
$\displaystyle \frac{a+b}{a-b}=\frac{5}{1}$
$\displaystyle \text{Again applying componendo and dividendo,}$
$\displaystyle \frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{5+1}{5-1}$
$\displaystyle \frac{2a}{2b}=\frac{6}{4}$
$\displaystyle \frac{a}{b}=\frac{3}{2}$
$\displaystyle \therefore a:b=3:2$
$\\$

$\displaystyle \textbf{Question 19: If }x,\ y\text{ and }z\text{ are in continued proportion, then prove that}$
$\displaystyle \frac{(x+y)^2}{(y+z)^2}=\frac{x}{z}.\hfill \text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }x,\ y\text{ and }z\text{ are in continued proportion}$
$\displaystyle \frac{x}{y}=\frac{y}{z}$
$\displaystyle y^2=xz$
$\displaystyle \text{L.H.S.}=\frac{(x+y)^2}{(y+z)^2}$
$\displaystyle =\frac{x^2+y^2+2xy}{y^2+z^2+2yz}$
$\displaystyle =\frac{x^2+xz+2xy}{xz+z^2+2yz}$
$\displaystyle =\frac{x(x+z+2y)}{z(x+z+2y)}$
$\displaystyle =\frac{x}{z}$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \frac{(x+y)^2}{(y+z)^2}=\frac{x}{z}$
$\\$

$\displaystyle \textbf{Question 20: Using componendo and dividendo, find the value of }x:$
$\displaystyle \frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}}=9.\hfill \text{[ICSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}}=9$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{(\sqrt{3x+4}+\sqrt{3x-5})+(\sqrt{3x+4}-\sqrt{3x-5})}{(\sqrt{3x+4}+\sqrt{3x-5})-(\sqrt{3x+4}-\sqrt{3x-5})}=\frac{9+1}{9-1}$
$\displaystyle \frac{2\sqrt{3x+4}}{2\sqrt{3x-5}}=\frac{10}{8}$
$\displaystyle \frac{\sqrt{3x+4}}{\sqrt{3x-5}}=\frac{5}{4}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle \frac{3x+4}{3x-5}=\frac{25}{16}$
$\displaystyle \text{Again applying componendo and dividendo,}$
$\displaystyle \frac{(3x+4)+(3x-5)}{(3x+4)-(3x-5)}=\frac{25+16}{25-16}$
$\displaystyle \frac{6x-1}{9}=\frac{41}{9}$
$\displaystyle 6x-1=41$
$\displaystyle 6x=42$
$\displaystyle x=7$
$\\$

$\displaystyle \textbf{Question 21: Given, }\frac{a}{b}=\frac{c}{d},\text{ prove that }\frac{3a-5b}{3a+5b}=\frac{3c-5d}{3c+5d}.\hfill \text{[ICSE 2000]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{a}{b}=\frac{c}{d}$
$\displaystyle \frac{3a}{5b}=\frac{3c}{5d}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}$
$\displaystyle \text{Applying invertendo,}$
$\displaystyle \frac{3a-5b}{3a+5b}=\frac{3c-5d}{3c+5d}$
$\displaystyle \therefore \frac{3a-5b}{3a+5b}=\frac{3c-5d}{3c+5d}$
$\\$

$\displaystyle \textbf{Question 22: Two numbers are in the ratio }3:5.\text{ If }8\text{ is added to each number,}$
$\displaystyle \text{then the ratio becomes }2:3.\text{ Find the numbers.}\hfill \text{[ICSE 2001]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two numbers be }3x\text{ and }5x$
$\displaystyle \frac{3x+8}{5x+8}=\frac{2}{3}$
$\displaystyle 3(3x+8)=2(5x+8)$
$\displaystyle 9x+24=10x+16$
$\displaystyle x=8$
$\displaystyle \therefore \text{The two numbers are }3x=24\text{ and }5x=40$
$\\$

$\displaystyle \textbf{Question 23: If }a:b=5:3,\text{ then find }(5a+8b):(6a-7b).\hfill \text{[ICSE 2002]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }a:b=5:3$
$\displaystyle \frac{a}{b}=\frac{5}{3}$
$\displaystyle \frac{5a+8b}{6a-7b}=\frac{\frac{5a+8b}{b}}{\frac{6a-7b}{b}}$
$\displaystyle =\frac{5\left(\frac{a}{b}\right)+8}{6\left(\frac{a}{b}\right)-7}$
$\displaystyle =\frac{5\left(\frac{5}{3}\right)+8}{6\left(\frac{5}{3}\right)-7}$
$\displaystyle =\frac{\frac{25}{3}+8}{\frac{30}{3}-7}$
$\displaystyle =\frac{\frac{25+24}{3}}{\frac{30-21}{3}}$
$\displaystyle =\frac{\frac{49}{3}}{\frac{9}{3}}$
$\displaystyle =\frac{49}{9}$
$\displaystyle \therefore (5a+8b):(6a-7b)=49:9$
$\\$

$\displaystyle \textbf{Question 24: If }x=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}},\text{ then using}$
$\displaystyle \text{the properties of proportion, show that }x^2-2ax+1=0.\hfill \text{[ICSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{x}{1}=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{x+1}{x-1}=\frac{(\sqrt{a+1}+\sqrt{a-1})+(\sqrt{a+1}-\sqrt{a-1})}{(\sqrt{a+1}+\sqrt{a-1})-(\sqrt{a+1}-\sqrt{a-1})}$
$\displaystyle \frac{x+1}{x-1}=\frac{2\sqrt{a+1}}{2\sqrt{a-1}}$
$\displaystyle \frac{x+1}{x-1}=\frac{\sqrt{a+1}}{\sqrt{a-1}}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle \frac{(x+1)^2}{(x-1)^2}=\frac{a+1}{a-1}$
$\displaystyle \frac{x^2+2x+1}{x^2-2x+1}=\frac{a+1}{a-1}$
$\displaystyle \text{Again applying componendo and dividendo,}$
$\displaystyle \frac{(x^2+2x+1)+(x^2-2x+1)}{(x^2+2x+1)-(x^2-2x+1)}=\frac{(a+1)+(a-1)}{(a+1)-(a-1)}$
$\displaystyle \frac{2x^2+2}{4x}=\frac{2a}{2}$
$\displaystyle \frac{x^2+1}{2x}=a$
$\displaystyle x^2+1=2ax$
$\displaystyle x^2-2ax+1=0$
$\displaystyle \therefore x^2-2ax+1=0$
$\\$

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

$\displaystyle \textbf{Question 26: If }\frac{x^2+y^2}{x^2-y^2}=\frac{17}{8},\text{ then find:}$
$\displaystyle \text{(i) }x:y\qquad\text{(ii) }\frac{x^3+y^3}{x^3-y^3}.\hfill \text{[ICSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{x^2+y^2}{x^2-y^2}=\frac{17}{8}$
$\displaystyle \text{(i) Applying componendo and dividendo,}$
$\displaystyle \frac{(x^2+y^2)+(x^2-y^2)}{(x^2+y^2)-(x^2-y^2)}=\frac{17+8}{17-8}$
$\displaystyle \frac{2x^2}{2y^2}=\frac{25}{9}$
$\displaystyle \frac{x^2}{y^2}=\left(\frac{5}{3}\right)^2$
$\displaystyle \text{Taking square root of both sides,}$
$\displaystyle \frac{x}{y}=\frac{5}{3}$
$\displaystyle \therefore x:y=5:3$
$\displaystyle \text{(ii) From part (i), }\frac{x}{y}=\frac{5}{3}$
$\displaystyle \text{Cubing both sides,}$
$\displaystyle \frac{x^3}{y^3}=\left(\frac{5}{3}\right)^3=\frac{125}{27}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{x^3+y^3}{x^3-y^3}=\frac{125+27}{125-27}$
$\displaystyle \frac{x^3+y^3}{x^3-y^3}=\frac{152}{98}=\frac{76}{49}$
$\\$

$\displaystyle \textbf{Question 27: If }(x-9):(3x+6)\text{ is the duplicate ratio of }4:9,\text{ then find the}$
$\displaystyle \text{value of }x.\hfill \text{[ICSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Duplicate ratio of }4:9=4^2:9^2=16:81$
$\displaystyle \frac{x-9}{3x+6}=\frac{16}{81}$
$\displaystyle 81(x-9)=16(3x+6)$
$\displaystyle 81x-729=48x+96$
$\displaystyle 33x=825$
$\displaystyle x=25$
$\\$

$\displaystyle \textbf{Question 28: If }(3a+2b):(5a+3b)=18:29,\text{ then find }a:b.\hfill \text{[ICSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{3a+2b}{5a+3b}=\frac{18}{29}$
$\displaystyle \frac{\frac{3a}{b}+2}{\frac{5a}{b}+3}=\frac{18}{29}$
$\displaystyle \text{Let }\frac{a}{b}=k$
$\displaystyle \frac{3k+2}{5k+3}=\frac{18}{29}$
$\displaystyle 29(3k+2)=18(5k+3)$
$\displaystyle 87k+58=90k+54$
$\displaystyle 4=3k$
$\displaystyle k=\frac{4}{3}$
$\displaystyle \therefore \frac{a}{b}=\frac{4}{3}$
$\displaystyle \therefore a:b=4:3$
$\\$

$\displaystyle \textbf{Question 29: If }\frac{x}{a}=\frac{y}{b}=\frac{z}{c},\text{ then show that }\frac{x^3}{a^3}+\frac{y^3}{b^3}+\frac{z^3}{c^3}=\frac{3xyz}{abc}.\hfill \text{[ICSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k$
$\displaystyle \therefore x=ak,\ y=bk,\ z=ck$
$\displaystyle \text{L.H.S.}=\frac{x^3}{a^3}+\frac{y^3}{b^3}+\frac{z^3}{c^3}$
$\displaystyle =\frac{(ak)^3}{a^3}+\frac{(bk)^3}{b^3}+\frac{(ck)^3}{c^3}$
$\displaystyle =k^3+k^3+k^3$
$\displaystyle =3k^3$
$\displaystyle \text{R.H.S.}=\frac{3xyz}{abc}$
$\displaystyle =\frac{3(ak)(bk)(ck)}{abc}$
$\displaystyle =\frac{3abck^3}{abc}$
$\displaystyle =3k^3$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\\$

$\displaystyle \textbf{Question 30: If }b\text{ is the mean proportional between }a\text{ and }c,\text{ show that}$
$\displaystyle \frac{a^4+a^2b^2+b^4}{b^4+b^2c^2+c^4}=\frac{a^2}{c^2}.\hfill \text{[ICSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }b\text{ is the mean proportional between }a\text{ and }c$
$\displaystyle \frac{a}{b}=\frac{b}{c}$
$\displaystyle b^2=ac$
$\displaystyle \text{L.H.S.}=\frac{a^4+a^2b^2+b^4}{b^4+b^2c^2+c^4}$
$\displaystyle =\frac{a^4+a^2(ac)+(ac)^2}{(ac)^2+(ac)c^2+c^4}$
$\displaystyle =\frac{a^4+a^3c+a^2c^2}{a^2c^2+ac^3+c^4}$
$\displaystyle =\frac{a^2(a^2+ac+c^2)}{c^2(a^2+ac+c^2)}$
$\displaystyle =\frac{a^2}{c^2}$
$\displaystyle =\text{R.H.S.}$
$\\$

$\displaystyle \textbf{Question 31: Find }x\text{ from the following equation using properties of proportion:}$
$\displaystyle \frac{x^2-x+1}{x^2+x+1}=\frac{14(x-1)}{13(x+1)}.\hfill \text{[ICSE Specimen 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{x^2-x+1}{x^2+x+1}=\frac{14(x-1)}{13(x+1)}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{(x^2-x+1)+(x^2+x+1)}{(x^2-x+1)-(x^2+x+1)}=\frac{14(x-1)+13(x+1)}{14(x-1)-13(x+1)}$
$\displaystyle \frac{2x^2+2}{-2x}=\frac{14x-14+13x+13}{14x-14-13x-13}$
$\displaystyle \frac{2x^2+2}{-2x}=\frac{27x-1}{x-27}$
$\displaystyle (2x^2+2)(x-27)=-2x(27x-1)$
$\displaystyle 2x^3-54x^2+2x-54=-54x^2+2x$
$\displaystyle 2x^3-54=0$
$\displaystyle x^3=27$
$\displaystyle x=3$
$\\$

$\displaystyle \textbf{Question 32: If }\frac{7m+2n}{7m-2n}=\frac{5}{3},\text{ use properties of proportion to find:}$
$\displaystyle \text{(i) }m:n\qquad\text{(ii) }\frac{m^2+n^2}{m^2-n^2}.\hfill \text{[ICSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, }\frac{7m+2n}{7m-2n}=\frac{5}{3}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{(7m+2n)+(7m-2n)}{(7m+2n)-(7m-2n)}=\frac{5+3}{5-3}$
$\displaystyle \frac{14m}{4n}=\frac{8}{2}$
$\displaystyle \frac{7m}{2n}=4$
$\displaystyle 7m=8n$
$\displaystyle \frac{m}{n}=\frac{8}{7}$
$\displaystyle \therefore m:n=8:7$
$\displaystyle \text{(ii) }\frac{m^2+n^2}{m^2-n^2}=\frac{\frac{m^2}{n^2}+1}{\frac{m^2}{n^2}-1}$
$\displaystyle =\frac{\left(\frac{m}{n}\right)^2+1}{\left(\frac{m}{n}\right)^2-1}$
$\displaystyle =\frac{\left(\frac{8}{7}\right)^2+1}{\left(\frac{8}{7}\right)^2-1}$
$\displaystyle =\frac{\frac{64}{49}+1}{\frac{64}{49}-1}$
$\displaystyle =\frac{\frac{113}{49}}{\frac{15}{49}}$
$\displaystyle =\frac{113}{15}$
$\\$

$\displaystyle \textbf{Question 33: Using properties of proportion, solve for }x,\text{ given that }x\text{ is positive:}$
$\displaystyle \frac{2x+\sqrt{4x^2-1}}{2x-\sqrt{4x^2-1}}=4.\hfill \text{[ICSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{2x+\sqrt{4x^2-1}}{2x-\sqrt{4x^2-1}}=\frac{4}{1}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{(2x+\sqrt{4x^2-1})+(2x-\sqrt{4x^2-1})}{(2x+\sqrt{4x^2-1})-(2x-\sqrt{4x^2-1})}=\frac{4+1}{4-1}$
$\displaystyle \frac{4x}{2\sqrt{4x^2-1}}=\frac{5}{3}$
$\displaystyle \frac{2x}{\sqrt{4x^2-1}}=\frac{5}{3}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle \frac{4x^2}{4x^2-1}=\frac{25}{9}$
$\displaystyle 36x^2=100x^2-25$
$\displaystyle 64x^2=25$
$\displaystyle x^2=\frac{25}{64}$
$\displaystyle x=\pm\frac{5}{8}$
$\displaystyle \text{Since }x\text{ is positive, }x=\frac{5}{8}$
$\\$

$\displaystyle \textbf{Question 34: Using properties of proportion, solve for }x,\text{ given}$
$\displaystyle \frac{\sqrt{5x}+\sqrt{2x-6}}{\sqrt{5x}-\sqrt{2x-6}}=4.\hfill \text{[ICSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{\sqrt{5x}+\sqrt{2x-6}}{\sqrt{5x}-\sqrt{2x-6}}=\frac{4}{1}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{(\sqrt{5x}+\sqrt{2x-6})+(\sqrt{5x}-\sqrt{2x-6})}{(\sqrt{5x}+\sqrt{2x-6})-(\sqrt{5x}-\sqrt{2x-6})}=\frac{4+1}{4-1}$
$\displaystyle \frac{2\sqrt{5x}}{2\sqrt{2x-6}}=\frac{5}{3}$
$\displaystyle \frac{\sqrt{5x}}{\sqrt{2x-6}}=\frac{5}{3}$
$\displaystyle 3\sqrt{5x}=5\sqrt{2x-6}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle 9(5x)=25(2x-6)$
$\displaystyle 45x=50x-150$
$\displaystyle 5x=150$
$\displaystyle x=30$
$\\$

$\displaystyle \textbf{Question 35: The numbers }K+3,\ K+2,\ 3K-7\text{ and }2K-3\text{ are in proportion.}$
$\displaystyle \text{Find the value of }K.\hfill \text{[ICSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }K+3,\ K+2,\ 3K-7\text{ and }2K-3\text{ are in proportion}$
$\displaystyle \frac{K+3}{K+2}=\frac{3K-7}{2K-3}$
$\displaystyle (K+3)(2K-3)=(K+2)(3K-7)$
$\displaystyle 2K^2+3K-9=3K^2-K-14$
$\displaystyle K^2-4K-5=0$
$\displaystyle K^2-5K+K-5=0$
$\displaystyle K(K-5)+1(K-5)=0$
$\displaystyle (K-5)(K+1)=0$
$\displaystyle K=5\text{ or }K=-1$
$\\$

$\displaystyle \textbf{Question 36: Using properties of proportion, find }x:y\text{ if}$
$\displaystyle \frac{x^3+12x}{6x^2+8}=\frac{y^3+27y}{9y^2+27}.\hfill \text{[ICSE Specimen 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{x^3+12x}{6x^2+8}=\frac{y^3+27y}{9y^2+27}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{x^3+12x+6x^2+8}{x^3+12x-6x^2-8}=\frac{y^3+27y+9y^2+27}{y^3+27y-9y^2-27}$
$\displaystyle \frac{(x+2)^3}{(x-2)^3}=\frac{(y+3)^3}{(y-3)^3}$
$\displaystyle \frac{x+2}{x-2}=\frac{y+3}{y-3}$
$\displaystyle \text{Again applying componendo and dividendo,}$
$\displaystyle \frac{(x+2)+(x-2)}{(x+2)-(x-2)}=\frac{(y+3)+(y-3)}{(y+3)-(y-3)}$
$\displaystyle \frac{2x}{4}=\frac{2y}{6}$
$\displaystyle \frac{x}{2}=\frac{y}{3}$
$\displaystyle \frac{x}{y}=\frac{2}{3}$
$\displaystyle \therefore x:y=2:3$
$\\$

$\displaystyle \textbf{Question 37: If }x=\frac{\sqrt{2a+1}+\sqrt{2a-1}}{\sqrt{2a+1}-\sqrt{2a-1}},\text{ prove that}$
$\displaystyle x^2-4ax+1=0.\hfill \text{[ICSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{x}{1}=\frac{\sqrt{2a+1}+\sqrt{2a-1}}{\sqrt{2a+1}-\sqrt{2a-1}}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{x+1}{x-1}=\frac{(\sqrt{2a+1}+\sqrt{2a-1})+(\sqrt{2a+1}-\sqrt{2a-1})}{(\sqrt{2a+1}+\sqrt{2a-1})-(\sqrt{2a+1}-\sqrt{2a-1})}$
$\displaystyle \frac{x+1}{x-1}=\frac{2\sqrt{2a+1}}{2\sqrt{2a-1}}$
$\displaystyle \frac{x+1}{x-1}=\frac{\sqrt{2a+1}}{\sqrt{2a-1}}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle \frac{(x+1)^2}{(x-1)^2}=\frac{2a+1}{2a-1}$
$\displaystyle \frac{x^2+2x+1}{x^2-2x+1}=\frac{2a+1}{2a-1}$
$\displaystyle \text{Again applying componendo and dividendo,}$
$\displaystyle \frac{(x^2+2x+1)+(x^2-2x+1)}{(x^2+2x+1)-(x^2-2x+1)}=\frac{(2a+1)+(2a-1)}{(2a+1)-(2a-1)}$
$\displaystyle \frac{2x^2+2}{4x}=\frac{4a}{2}$
$\displaystyle \frac{x^2+1}{2x}=2a$
$\displaystyle x^2+1=4ax$
$\displaystyle x^2-4ax+1=0$
$\displaystyle \therefore x^2-4ax+1=0$
$\\$

$\displaystyle \textbf{Question 38: Using properties of proportion, find }x:y,\text{ given}$
$\displaystyle \frac{x^2+2x}{2x+4}=\frac{y^2+3y}{3y+9}.\hfill \text{[ICSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{x^2+2x}{2x+4}=\frac{y^2+3y}{3y+9}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{x^2+2x+2x+4}{x^2+2x-2x-4}=\frac{y^2+3y+3y+9}{y^2+3y-3y-9}$
$\displaystyle \frac{x^2+4x+4}{x^2-4}=\frac{y^2+6y+9}{y^2-9}$
$\displaystyle \frac{(x+2)^2}{(x+2)(x-2)}=\frac{(y+3)^2}{(y+3)(y-3)}$
$\displaystyle \frac{x+2}{x-2}=\frac{y+3}{y-3}$
$\displaystyle \text{Again applying componendo and dividendo,}$
$\displaystyle \frac{(x+2)+(x-2)}{(x+2)-(x-2)}=\frac{(y+3)+(y-3)}{(y+3)-(y-3)}$
$\displaystyle \frac{2x}{4}=\frac{2y}{6}$
$\displaystyle \frac{x}{2}=\frac{y}{3}$
$\displaystyle \frac{x}{y}=\frac{2}{3}$
$\displaystyle \therefore x:y=2:3$
$\\$

$\displaystyle \textbf{Question 39: If }x,\ y,\ z\text{ are in continued proportion, then }(y^2+z^2):(x^2+y^2)$
$\displaystyle \text{is equal to: (a) }z:x\qquad\text{(b) }x:z\qquad\text{(c) }zx\qquad\text{(d) }(y+z):(x+y)\hfill \text{[ICSE Semester I 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }x,\ y,\ z\text{ are in continued proportion}$
$\displaystyle \frac{x}{y}=\frac{y}{z}$
$\displaystyle y^2=xz$
$\displaystyle \frac{y^2+z^2}{x^2+y^2}=\frac{xz+z^2}{x^2+xz}$
$\displaystyle =\frac{z(x+z)}{x(x+z)}$
$\displaystyle =\frac{z}{x}$
$\displaystyle \therefore (y^2+z^2):(x^2+y^2)=z:x$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\\$

$\displaystyle \textbf{Question 40: If }a,\ b,\ c\text{ and }d\text{ are proportional, then }\frac{a+b}{a-b}\text{ is equal to:}$
$\displaystyle \text{(a) }\frac{c}{d}\qquad\text{(b) }\frac{c-d}{c+d}\qquad\text{(c) }\frac{d}{c}\qquad\text{(d) }\frac{c+d}{c-d}\hfill \text{[ICSE Semester I 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) Given, }a,\ b,\ c\text{ and }d\text{ are proportional}$
$\displaystyle \frac{a}{b}=\frac{c}{d}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{a+b}{a-b}=\frac{c+d}{c-d}$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\\$

$\displaystyle \textbf{Question 41: If }\frac{5a}{7b}=\frac{4c}{3d},\text{ then by componendo and dividendo:}$
$\displaystyle \text{(a) }\frac{5a+7b}{5a-7b}=\frac{4c-3d}{4c+3d}\qquad\text{(b) }\frac{5a-7b}{5a+7b}=\frac{4c+3d}{4c-3d}$
$\displaystyle \text{(c) }\frac{5a+7b}{5a-7b}=\frac{4c+3d}{4c-3d}\qquad\text{(d) }\frac{5a+7b}{5a+7b}=\frac{4c-3d}{4c-3d}\hfill \text{[ICSE Semester I 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) Given, }\frac{5a}{7b}=\frac{4c}{3d}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{5a+7b}{5a-7b}=\frac{4c+3d}{4c-3d}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\\$

$\displaystyle \textbf{Question 42: If }x,\ 5.4,\ 5,\ 9\text{ are in proportion, then }x\text{ is:}$
$\displaystyle \text{(a) }3\qquad\text{(b) }9.72\qquad\text{(c) }25\qquad\text{(d) }\frac{25}{3}\hfill \text{[ICSE Semester I 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) Given, }x,\ 5.4,\ 5,\ 9\text{ are in proportion}$
$\displaystyle \frac{x}{5.4}=\frac{5}{9}$
$\displaystyle x=\frac{5\times5.4}{9}$
$\displaystyle =\frac{5\times54}{9\times10}$
$\displaystyle =\frac{5\times6}{10}$
$\displaystyle =3$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\\$

$\displaystyle \textbf{Question 43: If }1.5,\ 3,\ x\text{ and }8\text{ are in proportion, then }x\text{ is equal to:}$
$\displaystyle \text{(a) }6\qquad\text{(b) }4\qquad\text{(c) }4.5\qquad\text{(d) }16\hfill \text{[ICSE Semester I 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Given, }1.5,\ 3,\ x\text{ and }8\text{ are in proportion}$
$\displaystyle \frac{1.5}{3}=\frac{x}{8}$
$\displaystyle 3x=1.5\times8$
$\displaystyle 3x=12$
$\displaystyle x=4$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\\$

$\displaystyle \textbf{Question 44: The mean proportional between }4\text{ and }9\text{ is:}\hfill \text{[ICSE 2023]}$
$\displaystyle \text{(a) }4\qquad\text{(b) }6\qquad\text{(c) }9\qquad\text{(d) }36$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Mean proportional between two numbers }x\text{ and }y\text{ is }\sqrt{xy}$
$\displaystyle \text{Mean proportional between }4\text{ and }9=\sqrt{4\times9}$
$\displaystyle =\sqrt{36}=6$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\\$

$\displaystyle \textbf{Question 45: Using componendo and dividendo, solve for }x:$
$\displaystyle \frac{\sqrt{2x+2}+\sqrt{2x-1}}{\sqrt{2x+2}-\sqrt{2x-1}}=3.\hfill \text{[ICSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{\sqrt{2x+2}+\sqrt{2x-1}}{\sqrt{2x+2}-\sqrt{2x-1}}=3$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{(\sqrt{2x+2}+\sqrt{2x-1})+(\sqrt{2x+2}-\sqrt{2x-1})}{(\sqrt{2x+2}+\sqrt{2x-1})-(\sqrt{2x+2}-\sqrt{2x-1})}=\frac{3+1}{3-1}$
$\displaystyle \frac{2\sqrt{2x+2}}{2\sqrt{2x-1}}=\frac{4}{2}$
$\displaystyle \frac{\sqrt{2x+2}}{\sqrt{2x-1}}=2$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle \frac{2x+2}{2x-1}=4$
$\displaystyle 2x+2=4(2x-1)$
$\displaystyle 2x+2=8x-4$
$\displaystyle 6x=6$
$\displaystyle x=1$
$\\$

$\displaystyle \textbf{Question 46: Solve for }x,\text{ using the properties of proportion:}$
$\displaystyle \frac{\sqrt{2+x}+\sqrt{3-x}}{\sqrt{2+x}-\sqrt{3-x}}=3.\hfill \text{[ICSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{\sqrt{2+x}+\sqrt{3-x}}{\sqrt{2+x}-\sqrt{3-x}}=3$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{(\sqrt{2+x}+\sqrt{3-x})+(\sqrt{2+x}-\sqrt{3-x})}{(\sqrt{2+x}+\sqrt{3-x})-(\sqrt{2+x}-\sqrt{3-x})}=\frac{3+1}{3-1}$
$\displaystyle \frac{2\sqrt{2+x}}{2\sqrt{3-x}}=\frac{4}{2}$
$\displaystyle \frac{\sqrt{2+x}}{\sqrt{3-x}}=2$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle \frac{2+x}{3-x}=4$
$\displaystyle 2+x=4(3-x)$
$\displaystyle 2+x=12-4x$
$\displaystyle 5x=10$
$\displaystyle x=2$
$\\$

$\displaystyle \textbf{Question 47: }3,\ 9,\ m,\ 81\text{ and }n\text{ are in continued proportion. Find}$
$\displaystyle \text{the values of }m\text{ and }n.\hfill \text{[ICSE Specimen 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }3,\ 9,\ m,\ 81\text{ and }n\text{ are in continued proportion}$
$\displaystyle \therefore \frac{3}{9}=\frac{9}{m}=\frac{m}{81}=\frac{81}{n}$
$\displaystyle \frac{3}{9}=\frac{9}{m}$
$\displaystyle 3m=9\times9$
$\displaystyle m=27$
$\displaystyle \frac{m}{81}=\frac{81}{n}$
$\displaystyle mn=81^2$
$\displaystyle n=\frac{81^2}{27}$
$\displaystyle n=\frac{81\times81}{27}$
$\displaystyle n=243$
$\displaystyle \therefore \text{The required values are }m=27\text{ and }n=243$
$\\$

$\displaystyle \textbf{Question 48: If }\frac{(a+b)^3}{(a-b)^3}=\frac{64}{27}.$
$\displaystyle \text{(i) Find }\frac{a+b}{a-b}.\qquad\text{(ii) Hence, using properties of proportion, find }a:b.\hfill \text{[ICSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, }\frac{(a+b)^3}{(a-b)^3}=\frac{64}{27}$
$\displaystyle \text{Taking cube root of both sides,}$
$\displaystyle \frac{a+b}{a-b}=\frac{4}{3}$
$\displaystyle \text{(ii) Applying componendo and dividendo,}$
$\displaystyle \frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{4+3}{4-3}$
$\displaystyle \frac{2a}{2b}=\frac{7}{1}$
$\displaystyle \frac{a}{b}=7$
$\displaystyle \therefore a:b=7:1$
$\\$

$\displaystyle \textbf{Question 49: The given table shows the distance covered and the time taken by a train}$
$\displaystyle \text{moving at a uniform speed along a straight track.}\hfill \text{[ICSE 2024]}$
$\displaystyle \begin{array}{c|ccc} \text{Distance (in m)} & 60 & 90 & y\\ \hline \text{Time (in sec)} & 2 & x & 5 \end{array}$
$\displaystyle \text{The values of }x\text{ and }y\text{ are:}$
$\displaystyle \text{(a) }x=4,\ y=150\qquad\text{(b) }x=3,\ y=100\qquad\text{(c) }x=4,\ y=100\qquad\text{(d) }x=3,\ y=150$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) The speed of the train is uniform.}$
$\displaystyle \text{Speed}=\frac{\text{Distance}}{\text{Time}}=\frac{60}{2}=30\text{ m/sec}$
$\displaystyle \text{For a distance of }90\text{ m, }\frac{90}{x}=30$
$\displaystyle 30x=90$
$\displaystyle x=3$
$\displaystyle \text{Also, }\frac{y}{5}=30$
$\displaystyle y=30\times5=150$
$\displaystyle \therefore x=3\text{ and }y=150$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\\$