Class 10: Ratio and Proportion – ICSE Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \textbf{Question 1: } \text{If } x, y \text{ and } z \text{ are in continued proportion, prove that: } \\ \\ \frac{(x+y)^2}{(y+z)^2} = \frac{x}{y} \text{. [ICSE 2010]}$
$\displaystyle \text{Answer:}$
If $\displaystyle x, y \text{ and } z$ are in continued proportion, then
$\displaystyle \frac{x}{y} = \frac{y}{z} \Rightarrow x = \frac{y^2}{z}$
Applying componendo and dividendo
$\displaystyle \frac{x+y}{x-y} = \frac{y+z}{y-z}$
$\displaystyle \Rightarrow \frac{x+y}{y+z} = \frac{x-y}{y-z}$
Squaring both sides
$\displaystyle \Rightarrow \frac{(x+y)^2}{(y+z)^2} = (\frac{x-y}{y-z})^2$
$\displaystyle \Rightarrow \frac{(x+y)^2}{(y+z)^2} = (\frac{x-y}{y-z})^2$
Substituting
$\displaystyle \Rightarrow \frac{(x+y)^2}{(y+z)^2} = (\frac{\frac{y^2}{z}-y}{y-z})^2$
$\displaystyle \Rightarrow \frac{(x+y)^2}{(y+z)^2} = (\frac{y^2-yz}{z(y-z)})^2 = \frac{y^2}{z^2} = \frac{zx}{z^2} = \frac{x}{z}$

$\displaystyle \textbf{Question 2: } \text{Given } x= \frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 }} \\ \\ \text{ Use componendo and dividendo to prove that: } x^2= \frac{2a^2 x}{x^2+1} \text{ [ICSE 2010 ] }$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given } x= \frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 }}$
Applying componendo and dividendo
$\displaystyle \frac{x+1}{x-1} = \frac{(\sqrt{a^2+b^2}+\sqrt{a^2-b^2})+(\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 })}{(\sqrt{a^2+b^2}+\sqrt{a^2-b^2})-(\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 })}$
Simplifying
$\displaystyle \frac{x+1}{x-1} = \frac{\sqrt{a^2+b^2}}{\sqrt{a^2-b^2 }}$
Square both sides
$\displaystyle \frac{x^2+1+2x}{x^2-2x+1} = \frac{a^2+b^2}{a^2-b^2}$
Applying componendo and dividendo
$\displaystyle \frac{x^2+1+2x+x^2-2x+1}{x^2+1+2x-x^2+2x-1} = \frac{a^2+b^2+a^2-b^2}{a^2+b^2-a^2+b^2}$
$\displaystyle \frac{2(x^2+1)}{4x} = \frac{2a^2}{2b^2}$
$\displaystyle \frac{x^2+1}{2x} = \frac{a^2}{b^2}$
Simplifying
$\displaystyle b^2 = \frac{2a^2x}{x^2+1}$

$\displaystyle \textbf{Question 3: } \text{If } \frac{x^2+y^2}{x^2-y^2} =2 \frac{1}{8} \text{ , find: }$
$\displaystyle \text{i) } \frac{x}{y}$ $\displaystyle \text{ii) } \frac{x^3+y^3}{x^3-y^3 } \text{ [ICSE 2010] }$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) } \text{Given } \frac{x^2+y^2}{x^2-y^2} =2 \frac{1}{8} = \frac{17}{8}$
Applying componendo and dividendo
$\displaystyle \frac{x^2+y^2+x^2-y^2}{x^2+y^2-x^2+y^2} = \frac{17+8}{17-8}$
$\displaystyle \frac{2x^2}{2y^2} = \frac{25}{9}$
Simplifying, we get
$\displaystyle \frac{x}{y} = \frac{5}{3}$
$\displaystyle \text{ii) } \frac{x^3+y^3}{x^3-y^3 }$
Applying componendo and dividendo
$\displaystyle \frac{x^3+y^3+x^3-y^3 }{x^3+y^3-x^3+y^3 }$
$\displaystyle \frac{2x^3}{2y^3} = \Big( \frac{x}{y} \Big)^3 = \Big( \frac{5}{3} \Big)^3 = \frac{125}{9}$

Question 4: Using componendo and dividendo, find the value of $\displaystyle x$ : $\displaystyle \frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}} =9 \text{ [ICSE 2010] }$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given } \frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}} =9$
Applying componendo and dividendo
$\displaystyle \frac{(\sqrt{3x+4}+\sqrt{3x-5})+(\sqrt{3x+4}-\sqrt{3x-5})}{(\sqrt{3x+4}+\sqrt{3x-5})-(\sqrt{3x+4}-\sqrt{3x-5})} = \frac{9+1}{9-1}$
$\displaystyle \frac{2\sqrt{3x+4}}{2\sqrt{3x-5}} = \frac{10}{8}$
Simplifying
$\displaystyle \frac{\sqrt{3x+4}}{\sqrt{3x-5}} = \frac{5}{4}$
Square both sides
$\displaystyle \frac{3x+4}{3x-5} = \frac{25}{14}$
$\displaystyle 42x+56 = 75x-125$
Simplifying we get $\displaystyle x = 7$

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

$\displaystyle \textbf{Question 6: } \text{Given, } \frac{a}{b} = \frac{c}{d} \text{, prove that: } \frac{3a-5b}{3a+5b} = \frac{3c-5d}{3c+5d} \text{ [ICSE 2000] }$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given } \frac{a}{b} = \frac{c}{d}$
$\displaystyle \Rightarrow \frac{3a}{5b} = \frac{3c}{5d}$
By componendo and dividendo
$\displaystyle \frac{3a+5b}{3a-5b} = \frac{3c+5d}{3c-5d}$
By Alternendo
$\displaystyle \frac{3a-5b}{3a+5b} = \frac{3c-5d}{3c+5d}$

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

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

Question 9: What least number must be added to each of the numbers $\displaystyle 6, 15, 20, \text{ and } 43$ to make them proportional. [ICSE 2005, 2013]
$\displaystyle \text{Answer:}$
Let the number added be $\displaystyle x$
$\displaystyle \text{Therefore } (6+x): (15+x) = (20+x): (43+x)$
$\displaystyle \Rightarrow (6+x) \times (43+x) = (20+x) \times (15+x)$
$\displaystyle \Rightarrow x^2+49x+258 = x^2+ 35x +300$
$\displaystyle \Rightarrow x = 3$

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. [ICSE 2012]
$\displaystyle \text{Answer:}$
Let monthly pocket of Rave and Sanjeev by $\displaystyle x \text{ and } y$ respectively.
$\displaystyle \frac{x}{y} = \frac{5}{7} \Rightarrow x = \frac{5}{7} y$
$\displaystyle \frac{x-80}{y-80} = \frac{3}{5}$
Substituting
$\displaystyle \frac{ \frac{5}{7} y-80}{y-80} = \frac{3}{5}$
$\displaystyle \frac{25}{7} x-400=3x-240 \Rightarrow x=280$
Substituting
$\displaystyle y = \frac{5}{7} \times 280 = 200$

Question 11: If $\displaystyle (x-9):(3x+6)$ is the triplicate ratio of $\displaystyle 4:9$ , find $\displaystyle x$ . [ICSE 2014]
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x-9}{3x+6} = \frac{4^2}{9^2} = \frac{16}{81}$
$\displaystyle 81x-729=48x+96$
$\displaystyle x=25$

Question 12: If $\displaystyle a:b=5:3$ , find $\displaystyle (5a+8b):(6a-7b)$ . [ICSE 2002]
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given } a:b=5:3$
$\displaystyle \text{or } \frac{a}{b} = \frac{5}{3} \Rightarrow a = b \frac{5}{3}$
Now substituting
$\displaystyle \frac{5a+8b}{6a-7b} = \frac{5 \times b\frac{5}{3}+8b}{6 \times b\frac{5}{3}-7b} = \frac{25+24}{30-21} = \frac{49}{9}$
Hence $\displaystyle (5a+8b):(6a-7b) = \frac{49}{9}$

Question 13: The work done by $\displaystyle (x-3)$ men in $\displaystyle (2x+1)$ days and the work done by $\displaystyle (2x+1)$ men in $\displaystyle (x+4)$ days are in the ratio $\displaystyle 3:10$ . Find the value of $\displaystyle x$ . [ICSE 2003]
$\displaystyle \text{Answer:}$
Amount of work done by $\displaystyle (x-3)$ men in $\displaystyle (2x+1)$ days $\displaystyle = (x-3)(2x+1)$
Similarly, amount of work done by $\displaystyle (2x+1)$ men in $\displaystyle (x+4)$ days $\displaystyle = (2x+1)(x+4)$
$\displaystyle \text{Given } \frac{(x-3)(2x+1)}{(2x+1)(x+4)} = \frac{3}{10}$
$\displaystyle 10(2x^2+x-6x-3)=3(2x^2+8x+x+4)$
Simplifying
$\displaystyle 2x^2-11x-6=0$
$\displaystyle (x-6)(2x+1) = 0 \Rightarrow x = 6 or x=- \frac{1}{2} (not possible)$
$\displaystyle \text{Therefore } x = 6$
$\displaystyle \\$

Question 14: What number should be subtracted from each of the numbers $\displaystyle 23, 30, 57 \text{ and } 78$ ; so that the ratios are in proportion. [ICSE 2004]
$\displaystyle \text{Answer:}$
Let the number subtracted $\displaystyle = x$
$\displaystyle \text{Therefore } (23-x):(30-x)=(57-x):(78-x)$
$\displaystyle \frac{23-x}{30-x} = \frac{57-x}{78-x}$
Simplifying
$\displaystyle x^2-101x+1794 = x^2-87x+1710 \Rightarrow x =6$

Question 15: $\displaystyle 6$ is the mean proportion between two numbers $\displaystyle x \text{ and } y \text{ and } 48$ is the third proportion to $\displaystyle x \text{ and } y$ . Find the numbers. [ICSE 2011]
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given } 6$ is the mean proportion between two numbers $\displaystyle x \text{ and } y$
$\displaystyle \text{Therefore } \frac{x}{6} ={6}{y} \Rightarrow xy=36 \Rightarrow x = \frac{36}{y}$ … … … … … … i)
Also $\displaystyle \text{Given } 48$ is the third proportion to $\displaystyle x \text{ and } y$
$\displaystyle \text{Therefore } \frac{x}{y} = \frac{y}{48} \Rightarrow y^2=48x$ … … … … … … ii)
Solving i) and ii)
$\displaystyle y^2 = 48 \frac{36}{y}$
$\displaystyle y^3 = 2^3 \times 6^3 \Rightarrow y = 12$
$\displaystyle \text{Hence } x = \frac{36}{12} = 3$
Hence the numbers are $\displaystyle 3 \text{ and } 12$ .

$\displaystyle \textbf{Question 16: } \text{If } \frac{8a-5b}{8c-5d} = \frac{8a+5b}{8c+5d} \text{, prove that } \frac{a}{b} = \frac{c}{d} \text{ [ICSE 2008] }$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given } \frac{8a-5b}{8c-5d} = \frac{8a+5b}{8c+5d}$
$\displaystyle \text{or } \frac{8c+5d}{8c-5d} = \frac{8a+5b}{8a-5b}$
Applying Componendo and Dividendo
$\displaystyle \frac{8c+5d+8c-5d}{8c+5d-8c+5d} = \frac{8a+5b+8a-5b}{8a+5b-8a+5b}$
$\displaystyle \frac{16c}{10d} = \frac{16a}{10b}$
$\displaystyle \frac{c}{d} = \frac{a}{b}$
$\displaystyle \text{or } \frac{a}{b} = \frac{c}{d} \text{ Hence proved.}$

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

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

$\displaystyle \textbf{Question 19: }\text{If }x,\ y\text{ and }z\text{ are in continued proportion, then prove that } \\ \frac{(x+y)^2}{(y+z)^2}=\frac{x}{z}.\ \text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }x,y\text{ and }z\text{ are in continued proportion,}$
$\displaystyle \frac{x}{y}=\frac{y}{z}\Rightarrow y^2=xz$
$\displaystyle \text{To prove, }\frac{(x+y)^2}{(y+z)^2}=\frac{x}{z}$
$\displaystyle \text{LHS}=\frac{(x+y)^2}{(y+z)^2}=\frac{x^2+y^2+2xy}{y^2+z^2+2yz}$
$\displaystyle =\frac{x^2+xz+2xy}{xz+z^2+2yz}\ \ \text{[since }y^2=xz\text{]}$
$\displaystyle =\frac{x(x+z+2y)}{z(x+z+2y)}=\frac{x}{z}=\text{RHS}\ \text{Hence proved.}$

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

$\displaystyle \textbf{Question 21: }\text{Given, }\frac{a}{b}=\frac{c}{d},\text{ prove that }\frac{3a-5b}{3a+5b}=\frac{3c-5d}{3c+5d}.\ \text{[ICSE 2000]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{To prove, }\frac{3a-5b}{3a+5b}=\frac{3c-5d}{3c+5d}$
$\displaystyle \text{Given, }\frac{a}{b}=\frac{c}{d}$
$\displaystyle \Rightarrow \frac{3a}{5b}=\frac{3c}{5d}\ \ \text{[multiplying both sides by }\frac{3}{5}\text{]}$
$\displaystyle \text{Applying componendo and dividendo, we get}$
$\displaystyle \frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}$
$\displaystyle \text{Applying invertendo, we get}$
$\displaystyle \frac{3a-5b}{3a+5b}=\frac{3c-5d}{3c+5d}\ \ \text{Hence proved.}$

$\displaystyle \textbf{Question 22: }\text{Two numbers are in the ratio of }3:5.\text{ If }8\text{ is added to each number,} \\ \text{then the ratio becomes }2:3.\text{ Find the numbers.}\ \text{[ICSE 2001]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }(3a+2b):(5a+3b)=18:29$
$\displaystyle \frac{3a+2b}{5a+3b}=\frac{18}{29}$
$\displaystyle \frac{\frac{3a}{b}+2}{\frac{5a}{b}+3}=\frac{18}{29}$
$\displaystyle 29\left(\frac{3a}{b}+2\right)=18\left(\frac{5a}{b}+3\right)$
$\displaystyle 87\frac{a}{b}+58=90\frac{a}{b}+54$
$\displaystyle 87\frac{a}{b}-90\frac{a}{b}=54-58$
$\displaystyle -3\frac{a}{b}=-4$
$\displaystyle \frac{a}{b}=\frac{4}{3}$
$\displaystyle \therefore a:b=4:3$

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

$\displaystyle \textbf{Question 24: }\text{If }x=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}},\text{ then using the properties of proportion,} \\ \text{show that }x^2-2ax+1=0.\ \text{[ICSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{To prove, }x^2-2ax+1=0$
$\displaystyle \text{Given, }\frac{x}{1}=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}$
$\displaystyle \text{Applying componendo and dividendo rule, we get}$
$\displaystyle \frac{x+1}{x-1}=\frac{\sqrt{a+1}+\sqrt{a-1}+\sqrt{a+1}-\sqrt{a-1}}{\sqrt{a+1}+\sqrt{a-1}-(\sqrt{a+1}-\sqrt{a-1})}$
$\displaystyle \Rightarrow \frac{x+1}{x-1}=\frac{2\sqrt{a+1}}{2\sqrt{a-1}}=\frac{\sqrt{a+1}}{\sqrt{a-1}}$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle \frac{(x+1)^2}{(x-1)^2}=\frac{(\sqrt{a+1})^2}{(\sqrt{a-1})^2}$
$\displaystyle \Rightarrow \frac{x^2+2x+1}{x^2-2x+1}=\frac{a+1}{a-1}$
$\displaystyle \text{Again, applying componendo and dividendo, we get}$
$\displaystyle \frac{x^2+2x+1+x^2-2x+1}{x^2+2x+1-(x^2-2x+1)}=\frac{(a+1)+(a-1)}{(a+1)-(a-1)}$
$\displaystyle \Rightarrow \frac{2(x^2+1)}{4x}=\frac{2a}{2}$
$\displaystyle \Rightarrow \frac{x^2+1}{2x}=a$
$\displaystyle x^2+1=2ax$
$\displaystyle x^2-2ax+1=0\ \ \text{Hence proved.}$

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

$\displaystyle \textbf{Question 26: }\text{If }\frac{x^2+y^2}{x^2-y^2}=\frac{17}{8},\text{ then find the value of}$
$\displaystyle \text{(i) }x:y \qquad \text{(ii) }\frac{x^3+y^3}{x^3-y^3}.\ \text{[ICSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{x^2+y^2}{x^2-y^2}=\frac{17}{8}$
$\displaystyle (i)\ \text{Applying componendo and dividendo rule, we get}$
$\displaystyle \frac{(x^2+y^2)+(x^2-y^2)}{(x^2+y^2)-(x^2-y^2)}=\frac{17+8}{17-8}$
$\displaystyle \Rightarrow \frac{2x^2}{2y^2}=\frac{25}{9}\Rightarrow \frac{x^2}{y^2}=\left(\frac{5}{3}\right)^2$
$\displaystyle \text{On taking square root both sides, we get}$
$\displaystyle \frac{x}{y}=\frac{5}{3}\ \ \text{(i)}$
$\displaystyle \therefore x:y=5:3$
$\displaystyle (ii)\ \text{From Eq.\ (i), we have }\frac{x}{y}=\frac{5}{3}$
$\displaystyle \text{On cubing both sides, we get}$
$\displaystyle \frac{x^3}{y^3}=\left(\frac{5}{3}\right)^3$
$\displaystyle \Rightarrow \frac{x^3}{y^3}=\frac{125}{27}$
$\displaystyle \text{Applying componendo and dividendo rule, we get}$
$\displaystyle \frac{x^3+y^3}{x^3-y^3}=\frac{125+27}{125-27}$
$\displaystyle \Rightarrow \frac{x^3+y^3}{x^3-y^3}=\frac{152}{98}=\frac{76}{49}$

$\displaystyle \textbf{Question 27: }\text{If }(x-9):(3x+6)\text{ is the duplicate ratio of }4:9,\text{ then find the} \\ \text{value of }x.\ \text{[ICSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, two numbers are in the ratio of }3:5$
$\displaystyle \text{Let the two numbers be }3x\text{ and }5x$
$\displaystyle \text{According to the question, }\frac{3x+8}{5x+8}=\frac{2}{3}$
$\displaystyle 3(3x+8)=2(5x+8)$
$\displaystyle 9x+24=10x+16$
$\displaystyle 10x-9x=24-16$
$\displaystyle x=8$
$\displaystyle \text{First number }=3x=24\text{ and second number }=5x=40$

$\displaystyle \textbf{Question 28: }\text{If }(3a+2b):(5a+3b)=18:29,\text{ then find }a:b.\ \text{[ICSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k$
$\displaystyle \text{Then, }x=ak,\ y=bk,\ z=ck$
$\displaystyle \text{Now, LHS}=\frac{x^3}{a^3}+\frac{y^3}{b^3}+\frac{z^3}{c^3}$
$\displaystyle =\frac{a^3k^3}{a^3}+\frac{b^3k^3}{b^3}+\frac{c^3k^3}{c^3}$
$\displaystyle =k^3+k^3+k^3=3k^3$
$\displaystyle \text{and RHS}=\frac{3xyz}{abc}=\frac{3(ak)(bk)(ck)}{abc}$
$\displaystyle =\frac{3abck^3}{abc}=3k^3$
$\displaystyle \therefore \text{LHS}=\text{RHS}\ \text{Hence proved.}$

$\displaystyle \textbf{Question 29: }\text{If }\frac{x}{a}=\frac{y}{b}=\frac{z}{c},\text{ then show that }\frac{x^3}{a^3}+\frac{y^3}{b^3}+\frac{z^3}{c^3}=\frac{3xyz}{abc}.\ \text{[ICSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Using the definition, the duplicate ratio of }a:b\text{ is }a^2:b^2$
$\displaystyle \text{Given, }(x-9):(3x+6)\text{ is the duplicate ratio of }4:9$
$\displaystyle \therefore \frac{x-9}{3x+6}=\left(\frac{4}{9}\right)^2$
$\displaystyle \frac{x-9}{3x+6}=\frac{16}{81}$
$\displaystyle 81(x-9)=16(3x+6)$
$\displaystyle 81x-729=48x+96$
$\displaystyle 81x-48x=729+96$
$\displaystyle 33x=825$
$\displaystyle x=25$

$\displaystyle \textbf{Question 30: }\text{If }b\text{ is the mean proportional between }a\text{ and }c,\text{ show that } \\ \frac{a^4+a^2b^2+b^4}{b^4+b^2c^2+c^4}=\frac{a^2}{c^2}.\ \text{[ICSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }b\text{ is the mean proportional between }a\text{ and }c$
$\displaystyle \therefore \frac{a}{b}=\frac{b}{c}\Rightarrow b^2=ac$
$\displaystyle \text{Now, LHS}=\frac{a^4+a^2b^2+b^4}{b^4+b^2c^2+c^4}$
$\displaystyle =\frac{a^4+a^2(ac)+(ac)^2}{(ac)^2+(ac)c^2+c^4}$
$\displaystyle =\frac{a^4+a^3c+a^2c^2}{a^2c^2+ac^3+c^4}$
$\displaystyle =\frac{a^2(a^2+ac+c^2)}{c^2(a^2+ac+c^2)}$
$\displaystyle =\frac{a^2}{c^2}=\text{RHS}\ \text{Hence proved.}$

$\displaystyle \textbf{Question 31: }\text{Find }x\text{ from the following equation using properties of proportion}$
$\displaystyle \frac{x^2-x+1}{x^2+x+1}=\frac{14(x-1)}{13(x+1)}.\ \text{[ICSE Specimen 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{x^2-x+1}{x^2+x+1}=\frac{14(x-1)}{13(x+1)}$
$\displaystyle \text{On applying componendo and dividendo property, we get}$
$\displaystyle \frac{(x^2-x+1)+(x^2+x+1)}{(x^2-x+1)-(x^2+x+1)}=\frac{(14x-14)+(13x+13)}{(14x-14)-(13x+13)}$
$\displaystyle \Rightarrow \frac{2x^2+2}{-2x}=\frac{27x-1}{x-27}$
$\displaystyle \Rightarrow (2x^2+2)(x-27)=-2x(27x-1)$
$\displaystyle 2x^3-54x^2+2x-54=-54x^2+2x$
$\displaystyle 2x^3-54=0$
$\displaystyle 2x^3=54$
$\displaystyle x^3=27=3^3$
$\displaystyle \text{On taking cube root on both sides, we get }x=3$

$\displaystyle \textbf{Question 32: }\text{If }\frac{7m+2n}{7m-2n}=\frac{5}{3},\text{ use properties of proportion to find}$
$\displaystyle \text{(i) }m:n \qquad \text{(ii) }\frac{m^2+n^2}{m^2-n^2}.\ \text{[ICSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle (i)\ \text{Given, }\frac{7m+2n}{7m-2n}=\frac{5}{3}$
$\displaystyle \text{On applying componendo and dividendo rule, we get}$
$\displaystyle \frac{(7m+2n)+(7m-2n)}{(7m+2n)-(7m-2n)}=\frac{5+3}{5-3}$
$\displaystyle \Rightarrow \frac{14m}{4n}=\frac{8}{2}$
$\displaystyle \Rightarrow \frac{7m}{2n}=4$
$\displaystyle \Rightarrow \frac{m}{n}=\frac{8}{7}\ \ \text{(i)}$
$\displaystyle \therefore m:n=8:7$
$\displaystyle (ii)\ \frac{m^2+n^2}{m^2-n^2}=\frac{\frac{m^2}{n^2}+1}{\frac{m^2}{n^2}-1}=\frac{\left(\frac{m}{n}\right)^2+1}{\left(\frac{m}{n}\right)^2-1}\ \ \text{[by dividing numerator and denominator by }n^2\text{]}$
$\displaystyle \text{On putting }\frac{m}{n}=\frac{8}{7}\text{ from Eq.\ (i), we get}$
$\displaystyle \frac{m^2+n^2}{m^2-n^2}=\frac{\left(\frac{8}{7}\right)^2+1}{\left(\frac{8}{7}\right)^2-1}=\frac{\frac{64}{49}+1}{\frac{64}{49}-1}=\frac{\frac{64+49}{49}}{\frac{64-49}{49}}=\frac{113}{15}$

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

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

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

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

$\displaystyle \textbf{Question 37: }\text{If }x=\frac{\sqrt{2a+1}+\sqrt{2a-1}}{\sqrt{2a+1}-\sqrt{2a-1}},\text{ prove that }x^2-4ax+1=0.\ \text{[ICSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{x}{1}=\frac{\sqrt{2a+1}+\sqrt{2a-1}}{\sqrt{2a+1}-\sqrt{2a-1}}$
$\displaystyle \text{On applying componendo and dividendo property, we get}$
$\displaystyle \frac{x+1}{x-1}=\frac{\sqrt{2a+1}+\sqrt{2a-1}+\sqrt{2a+1}-\sqrt{2a-1}}{\sqrt{2a+1}+\sqrt{2a-1}-\sqrt{2a+1}+\sqrt{2a-1}}$
$\displaystyle \Rightarrow \frac{x+1}{x-1}=\frac{\sqrt{2a+1}}{\sqrt{2a-1}}$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle \frac{x^2+2x+1}{x^2-2x+1}=\frac{2a+1}{2a-1}$
$\displaystyle \text{Again, applying componendo and dividendo property, we get}$
$\displaystyle \frac{x^2+2x+1+x^2-2x+1}{x^2+2x+1-x^2+2x-1}=\frac{2a+1+2a-1}{2a+1-2a+1}$
$\displaystyle \Rightarrow \frac{2x^2+2}{4x}=\frac{4a}{2}\Rightarrow \frac{x^2+1}{2x}=2a$
$\displaystyle x^2+1=4ax$
$\displaystyle x^2-4ax+1=0\ \ \text{Hence proved.}$

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

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

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

$\displaystyle \textbf{Question 41: }\text{If }\frac{5a}{7b}=\frac{4c}{3d},\text{ then by componendo and dividendo}\ \text{[ICSE Semester I 2022]}$
$\displaystyle (a)\ \frac{5a+7b}{5a-7b}=\frac{4c-3d}{4c+3d} \qquad (b)\ \frac{5a-7b}{5a+7b}=\frac{4c+3d}{4c-3d}$
$\displaystyle (c)\ \frac{5a+7b}{5a-7b}=\frac{4c+3d}{4c-3d} \qquad (d)\ \frac{5a+7b}{5a+7b}=\frac{4c-3d}{4c-3d}$
$\displaystyle \text{Answer:}$
$\displaystyle (c)\ \text{Given, }\frac{5a}{7b}=\frac{4c}{3d}$
$\displaystyle \text{Applying componendo and dividendo on both sides, we get}$
$\displaystyle \frac{5a+7b}{5a-7b}=\frac{4c+3d}{4c-3d}$

$\displaystyle \textbf{Question 42: }\text{If }x,\ 5.4,\ 5,\ 9\text{ are in proportion, then }x\text{ is}\ \text{[ICSE Semester I 2022]}$
$\displaystyle (a)\ 3 \qquad (b)\ 9.72 \qquad (c)\ 25 \qquad (d)\ \frac{25}{3}$
$\displaystyle \text{Answer:}$
$\displaystyle (a)\ \text{Given }x,5.4,5,9\text{ are in proportion, then}$
$\displaystyle \frac{x}{5.4}=\frac{5}{9}$
$\displaystyle x=\frac{5\times5.4}{9}=\frac{5\times54}{9\times10}=\frac{6}{2}=3$

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

$\displaystyle \textbf{Question 44: }\text{The mean proportional between }4\text{ and }9\text{ is}\ \text{[ICSE 2023]}$
$\displaystyle (a)\ 4 \qquad (b)\ 6 \qquad (c)\ 9 \qquad (d)\ 36$
$\displaystyle \text{Answer:}$
$\displaystyle (b)\ \text{We know, mean proportional between two numbers}$
$\displaystyle \text{is the square root of the product of the numbers}$
$\displaystyle \text{i.e. }M=\sqrt{x\times y}$
$\displaystyle \text{Let the given numbers be }x=4\text{ and }y=9$
$\displaystyle \text{Hence, the mean proportional between }4\text{ and }9\text{ is}$
$\displaystyle M=\sqrt{4\times9}$
$\displaystyle =\sqrt{36}=6$

$\displaystyle \textbf{Question 45: }\text{Using Componendo and Dividendo solve for }x: \\ \frac{\sqrt{2x+2}+\sqrt{2x-1}}{\sqrt{2x+2}-\sqrt{2x-1}}=3\ \text{[ICSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression is }\frac{\sqrt{2x+2}+\sqrt{2x-1}}{\sqrt{2x+2}-\sqrt{2x-1}}=3$
$\displaystyle \text{Using componendo and dividendo rule, we get}$
$\displaystyle \frac{(\sqrt{2x+2}+\sqrt{2x-1})+(\sqrt{2x+2}-\sqrt{2x-1})}{(\sqrt{2x+2}+\sqrt{2x-1})-(\sqrt{2x+2}-\sqrt{2x-1})}=\frac{3+1}{3-1}$
$\displaystyle \frac{2\sqrt{2x+2}}{2\sqrt{2x-1}}=\frac{4}{2}$
$\displaystyle \frac{\sqrt{2x+2}}{\sqrt{2x-1}}=\frac{2}{1}$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle \frac{2x+2}{2x-1}=\frac{4}{1}$
$\displaystyle 2x+2=8x-4$
$\displaystyle 6x=6\Rightarrow x=1$

$\displaystyle \textbf{Question 46: }\text{Solve for }x,\text{ using the properties of proportion. } \\ \frac{\sqrt{2+x}+\sqrt{3-x}}{\sqrt{2+x}-\sqrt{3-x}}=3\ \text{[ICSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression is }\frac{\sqrt{2+x}+\sqrt{3-x}}{\sqrt{2+x}-\sqrt{3-x}}=3$
$\displaystyle \text{By componendo and dividendo rule, we get}$
$\displaystyle \frac{\sqrt{2+x}+\sqrt{3-x}+\sqrt{2+x}-\sqrt{3-x}}{\sqrt{2+x}+\sqrt{3-x}-(\sqrt{2+x}-\sqrt{3-x})}=\frac{3+1}{3-1}$
$\displaystyle \frac{2\sqrt{2+x}}{2\sqrt{3-x}}=\frac{4}{2}$
$\displaystyle \frac{\sqrt{2+x}}{\sqrt{3-x}}=2$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle \frac{2+x}{3-x}=4$
$\displaystyle 2+x=12-4x$
$\displaystyle 5x=10\Rightarrow x=2$

$\displaystyle \textbf{Question 47: }\text{ }3,\ 9,\ m,\ 81\text{ and }n\text{ are in continued proportion. Find} \\ \text{the values of }m\text{ and }n.\ \text{[ICSE Specimen 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }3,9,m,81\text{ and }n\text{ are in continued proportion.}$
$\displaystyle \text{We know that, if non-zero numbers }x,y,z,t,s\text{ are in continued proportion,}$
$\displaystyle \frac{x}{y}=\frac{y}{z}=\frac{z}{t}=\frac{t}{s}$
$\displaystyle \therefore \frac{3}{9}=\frac{9}{m}=\frac{m}{81}=\frac{81}{n}$
$\displaystyle \Rightarrow \frac{3}{9}=\frac{9}{m}\ \text{and }\frac{m}{81}=\frac{81}{n}$
$\displaystyle \Rightarrow m=\frac{9\times9}{3}\ \text{and }mn=81\times81$
$\displaystyle \Rightarrow m=27\text{ and }n=\frac{81\times81}{m}$
$\displaystyle \therefore n=\frac{81\times81}{27}$
$\displaystyle \Rightarrow n=81\times3\Rightarrow n=243$
$\displaystyle \text{Hence, the required values of }m\text{ and }n\text{ are }27\text{ and }243\text{ respectively.}$

$\displaystyle \textbf{Question 48: }\text{If }\frac{(a+b)^3}{(a-b)^3}=\frac{64}{27}.$
$\displaystyle \text{(i) Find }\frac{a+b}{a-b}.$
$\displaystyle \text{(ii) Hence, using properties of proportion, find }a:b.\ \text{[ICSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle (i)\ \text{Given, }\frac{(a+b)^3}{(a-b)^3}=\frac{64}{27}$
$\displaystyle \text{On taking cube root of both sides, we get}$
$\displaystyle \frac{a+b}{a-b}=\frac{4}{3}$
$\displaystyle (ii)\ \text{Now, we have }\frac{a+b}{a-b}=\frac{4}{3}$
$\displaystyle \text{By componendo and dividendo rule, we get}$
$\displaystyle \frac{a+b+a-b}{a+b-a+b}=\frac{4+3}{4-3}$
$\displaystyle \Rightarrow \frac{2a}{2b}=\frac{7}{1}$
$\displaystyle \text{or }a:b=7:1$

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