Class 10: Ratio and Proportion – Miscellaneous Problems

$\displaystyle \textbf{Question 1: } \text{(i) If }2x+3y:3x+5y=18:29,\text{ find }x:y.$
$\displaystyle \text{(ii) If }x:y=2:3,\text{ find the value of }(3x+2y):(2x+5y).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, }2x+3y:3x+5y=18:29$
$\displaystyle \Rightarrow \frac{2x+3y}{3x+5y}=\frac{18}{29}$
$\displaystyle \Rightarrow 29(2x+3y)=18(3x+5y)$
$\displaystyle \Rightarrow 58x+87y=54x+90y$
$\displaystyle \Rightarrow 4x=3y$
$\displaystyle \Rightarrow \frac{x}{y}=\frac{3}{4}$
$\displaystyle \therefore x:y=3:4$
$\displaystyle \text{(ii) Given, }x:y=2:3\Rightarrow \frac{x}{y}=\frac{2}{3}$
$\displaystyle \frac{3x+2y}{2x+5y}=\frac{3\left(\frac{x}{y}\right)+2}{2\left(\frac{x}{y}\right)+5}$
$\displaystyle =\frac{3\left(\frac{2}{3}\right)+2}{2\left(\frac{2}{3}\right)+5}$
$\displaystyle =\frac{4}{\frac{19}{3}}=\frac{12}{19}$
$\displaystyle \therefore (3x+2y):(2x+5y)=12:19$
$\\$

$\displaystyle \textbf{Question 2: } \text{If }a:b=5:3,\text{ find }(5a+8b):(6a-7b).\hfill \text{[ICSE 2002]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a:b=5:3\Rightarrow a=5x,\ b=3x$
$\displaystyle \frac{5a+8b}{6a-7b}=\frac{5(5x)+8(3x)}{6(5x)-7(3x)}$
$\displaystyle =\frac{25x+24x}{30x-21x}$
$\displaystyle =\frac{49x}{9x}$
$\displaystyle \therefore (5a+8b):(6a-7b)=49:9$
$\\$

$\displaystyle \textbf{Question 3: } \text{Two numbers are in the ratio }3:5.\text{ If }8\text{ is added to each,}$
$\displaystyle \text{the ratio becomes }2:3.\text{ Find the numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numbers be }3x\text{ and }5x.$
$\displaystyle \text{Given, }\frac{3x+8}{5x+8}=\frac{2}{3}$
$\displaystyle \Rightarrow 3(3x+8)=2(5x+8)$
$\displaystyle \Rightarrow 9x+24=10x+16$
$\displaystyle \Rightarrow x=8$
$\displaystyle \therefore \text{The required numbers are }3x=24\text{ and }5x=40.$
$\\$

$\displaystyle \textbf{Question 4: } \text{(i) What quantity must be added to each term of the ratio }8:15$
$\displaystyle \text{so that it becomes equal to }3:5\text{?}$
$\displaystyle \text{(ii) What quantity must be subtracted from each term of the ratio }a:b$
$\displaystyle \text{so that it becomes }c:d\text{?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Let }x\text{ be added to each term of the ratio }8:15.$
$\displaystyle \frac{8+x}{15+x}=\frac{3}{5}$
$\displaystyle \Rightarrow 5(8+x)=3(15+x)$
$\displaystyle \Rightarrow 40+5x=45+3x$
$\displaystyle \Rightarrow 2x=5$
$\displaystyle \Rightarrow x=\frac{5}{2}=2\frac{1}{2}$
$\displaystyle \text{(ii) Let }x\text{ be subtracted from each term.}$
$\displaystyle \frac{a-x}{b-x}=\frac{c}{d}$
$\displaystyle \Rightarrow ad-dx=bc-cx$
$\displaystyle \Rightarrow cx-dx=bc-ad$
$\displaystyle \Rightarrow x(c-d)=bc-ad$
$\displaystyle \therefore x=\frac{bc-ad}{c-d}$
$\\$

$\displaystyle \textbf{Question 5: } \text{The work done by }(x-3)\text{ men in }(2x+1)\text{ days and}$
$\displaystyle \text{the work done by }(2x+1)\text{ men in }(x+4)\text{ days are in the ratio }3:10.$
$\displaystyle \text{Find the value of }x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Work done by }(x-3)\text{ men in }(2x+1)\text{ days}=(x-3)(2x+1)$
$\displaystyle \text{Work done by }(2x+1)\text{ men in }(x+4)\text{ days}=(2x+1)(x+4)$
$\displaystyle \text{According to the given statement,}$
$\displaystyle \frac{(x-3)(2x+1)}{(2x+1)(x+4)}=\frac{3}{10}$
$\displaystyle \Rightarrow \frac{x-3}{x+4}=\frac{3}{10}$
$\displaystyle \Rightarrow 10(x-3)=3(x+4)$
$\displaystyle \Rightarrow 10x-30=3x+12$
$\displaystyle \Rightarrow 7x=42$
$\displaystyle \therefore x=6$
$\\$

$\displaystyle \textbf{Question 6: } \text{When the fare of a certain journey by an airliner was increased}$
$\displaystyle \text{in the ratio }5:7,\text{ the cost of the ticket became Rs. }1421.\text{ Find the increase.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Original fare : Increased fare}=5:7$
$\displaystyle \Rightarrow \frac{\text{Original fare}}{1421}=\frac{5}{7}$
$\displaystyle \Rightarrow \text{Original fare}=\frac{5\times1421}{7}=\text{Rs. }1015$
$\displaystyle \therefore \text{Increase in fare}=1421-1015=\text{Rs. }406$
$\\$

$\displaystyle \textbf{Question 7: } \text{In a regiment, the ratio of number of officers to soldiers was }3:31$
$\displaystyle \text{before a battle. In the battle }6\text{ officers and }22\text{ soldiers were killed.}$
$\displaystyle \text{The ratio now is }1:13.\text{ Find the numbers before the battle.}\hfill \text{[ICSE 1992]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of officers before the battle be }3x.$
$\displaystyle \therefore \text{Number of soldiers before the battle}=31x$
$\displaystyle \text{After the battle, number of officers}=3x-6$
$\displaystyle \text{After the battle, number of soldiers}=31x-22$
$\displaystyle \text{Given, }\frac{3x-6}{31x-22}=\frac{1}{13}$
$\displaystyle \Rightarrow 13(3x-6)=31x-22$
$\displaystyle \Rightarrow 39x-78=31x-22$
$\displaystyle \Rightarrow 8x=56$
$\displaystyle \Rightarrow x=7$
$\displaystyle \therefore \text{Number of officers before the battle}=3x=3\times7=21$
$\displaystyle \text{Number of soldiers before the battle}=31x=31\times7=217$
$\\$

$\displaystyle \textbf{Question 8: } \text{If }\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\text{ and }a+b+c=0,$
$\displaystyle \text{show that each given ratio is equal to }-1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }a+b+c=0$
$\displaystyle \Rightarrow b+c=-a,\quad c+a=-b,\quad a+b=-c$
$\displaystyle \therefore \frac{a}{b+c}=\frac{a}{-a}=-1$
$\displaystyle \frac{b}{c+a}=\frac{b}{-b}=-1$
$\displaystyle \frac{c}{a+b}=\frac{c}{-c}=-1$
$\displaystyle \therefore \text{Each given ratio is equal to }-1.$
$\\$

$\displaystyle \textbf{Question 9: } \text{If }\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\text{ and }a+b+c\neq0,$
$\displaystyle \text{show that each given ratio is equal to }\frac{1}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}$
$\displaystyle =\frac{a+b+c}{(b+c)+(c+a)+(a+b)}$
$\displaystyle =\frac{a+b+c}{2(a+b+c)}$
$\displaystyle =\frac{1}{2}$
$\displaystyle \therefore \text{Each given ratio is equal to }\frac{1}{2}.$
$\\$

$\displaystyle \textbf{Question 10: } \text{Find the compound ratio of:}$
$\displaystyle \text{(i) }3a:2b,\ 2m:n\text{ and }4x:3y$
$\displaystyle \text{(ii) }a-b:a+b,\ (a+b)^2:a^2+b^2\text{ and }a^4-b^4:(a^2-b^2)^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Required compound ratio}$
$\displaystyle =(3a\times2m\times4x):(2b\times n\times3y)$
$\displaystyle =24amx:6bny$
$\displaystyle =4amx:bny$
$\displaystyle \text{(ii) Required compound ratio}$
$\displaystyle =[(a-b)(a+b)^2(a^4-b^4)]:[(a+b)(a^2+b^2)(a^2-b^2)^2]$
$\displaystyle =[(a-b)(a+b)^2(a^2-b^2)(a^2+b^2)]$
$\displaystyle \quad :[(a+b)(a^2+b^2)(a^2-b^2)^2]$
$\displaystyle =[(a-b)(a+b)^2(a^2+b^2)(a-b)(a+b)]$
$\displaystyle \quad :[(a+b)(a^2+b^2)(a-b)^2(a+b)^2]$
$\displaystyle =1:1$
$\\$

$\displaystyle \textbf{Question 11: } \text{Find the ratio compounded of the duplicate ratio of }5:6,$
$\displaystyle \text{the reciprocal ratio of }25:42\text{ and the sub-triplicate ratio of }216:343.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Duplicate ratio of }5:6=5^2:6^2=25:36$
$\displaystyle \text{Reciprocal ratio of }25:42=\frac{1}{25}:\frac{1}{42}=42:25$
$\displaystyle \text{Sub-triplicate ratio of }216:343=\sqrt[3]{216}:\sqrt[3]{343}=6:7$
$\displaystyle \therefore \text{Required compounded ratio}$
$\displaystyle =(25\times42\times6):(36\times25\times7)$
$\displaystyle =6300:6300$
$\displaystyle =1:1$
$\\$

$\displaystyle \textbf{Question 12: } \text{Find:}$
$\displaystyle \text{(i) the fourth proportional to }3,\ 6\text{ and }4.5$
$\displaystyle \text{(ii) the mean proportional between }6.25\text{ and }0.16$
$\displaystyle \text{(iii) the third proportional to }1.2\text{ and }1.8$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Let the fourth proportional be }x.$
$\displaystyle 3:6=4.5:x$
$\displaystyle \Rightarrow 3x=6\times4.5$
$\displaystyle \Rightarrow x=9$
$\displaystyle \text{(ii) Let the mean proportional be }x.$
$\displaystyle 6.25:x=x:0.16$
$\displaystyle \Rightarrow x^2=6.25\times0.16$
$\displaystyle \Rightarrow x^2=1$
$\displaystyle \Rightarrow x=1$
$\displaystyle \text{(iii) Let the third proportional be }x.$
$\displaystyle 1.2:1.8=1.8:x$
$\displaystyle \Rightarrow 1.2x=1.8\times1.8$
$\displaystyle \Rightarrow x=\frac{3.24}{1.2}=2.7$
$\\$

$\displaystyle \textbf{Question 13: } \text{Quantities }a,\ 2,\ 10\text{ and }b\text{ are in continued proportion.}$
$\displaystyle \text{Find the values of }a\text{ and }b.$
$\displaystyle \text{Answer:}$
$\displaystyle a,\ 2,\ 10\text{ and }b\text{ are in continued proportion.}$
$\displaystyle \Rightarrow \frac{a}{2}=\frac{2}{10}=\frac{10}{b}$
$\displaystyle \Rightarrow \frac{a}{2}=\frac{2}{10}\text{ and }\frac{2}{10}=\frac{10}{b}$
$\displaystyle \Rightarrow a=\frac{4}{10}=0.4$
$\displaystyle \Rightarrow b=\frac{100}{2}=50$
$\\$

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

$\displaystyle \textbf{Question 15: } \text{What should be added to each of the numbers }13,\ 17\text{ and }22$
$\displaystyle \text{so that the resulting numbers are in continued proportion?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required number be }x.$
$\displaystyle \therefore 13+x,\ 17+x\text{ and }22+x\text{ are in continued proportion.}$
$\displaystyle \Rightarrow \frac{13+x}{17+x}=\frac{17+x}{22+x}$
$\displaystyle \Rightarrow (13+x)(22+x)=(17+x)^2$
$\displaystyle \Rightarrow 286+35x+x^2=289+34x+x^2$
$\displaystyle \Rightarrow x=3$
$\\$

$\displaystyle \textbf{Question 16: } \text{If }(a^2+c^2),\ (ab+cd)\text{ and }(b^2+d^2)\text{ are in}$
$\displaystyle \text{continued proportion, prove that }a,\ b,\ c\text{ and }d\text{ are in proportion.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }(a^2+c^2),\ (ab+cd)\text{ and }(b^2+d^2)\text{ are in continued proportion.}$
$\displaystyle \therefore \frac{a^2+c^2}{ab+cd}=\frac{ab+cd}{b^2+d^2}$
$\displaystyle \Rightarrow (a^2+c^2)(b^2+d^2)=(ab+cd)^2$
$\displaystyle \Rightarrow a^2b^2+a^2d^2+b^2c^2+c^2d^2=a^2b^2+2abcd+c^2d^2$
$\displaystyle \Rightarrow a^2d^2+b^2c^2-2abcd=0$
$\displaystyle \Rightarrow (ad-bc)^2=0$
$\displaystyle \Rightarrow ad-bc=0$
$\displaystyle \Rightarrow ad=bc$
$\displaystyle \Rightarrow \frac{a}{b}=\frac{c}{d}$
$\displaystyle \therefore a,\ b,\ c\text{ and }d\text{ are in proportion.}$
$\\$

$\displaystyle \textbf{Question 17: } \text{If }p:q::q:r,\text{ prove that }p:r=p^2:q^2.$
$\displaystyle \text{Answer:}$
$\displaystyle p:q::q:r$
$\displaystyle \Rightarrow q^2=pr$
$\displaystyle \therefore p^2:q^2=\frac{p^2}{q^2}=\frac{p^2}{pr}=\frac{p}{r}=p:r$
$\displaystyle \therefore p:r=p^2:q^2$
$\\$

$\displaystyle \textbf{Question 18: } \text{If }a\neq b\text{ and }a:b\text{ is the duplicate ratio of }a+c$
$\displaystyle \text{and }b+c,\text{ prove that }c\text{ is the mean proportional between }a\text{ and }b.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{a}{b}=\frac{(a+c)^2}{(b+c)^2}$
$\displaystyle \Rightarrow a(b+c)^2=b(a+c)^2$
$\displaystyle \Rightarrow a(b^2+c^2+2bc)=b(a^2+c^2+2ac)$
$\displaystyle \Rightarrow ab^2+ac^2+2abc=a^2b+bc^2+2abc$
$\displaystyle \Rightarrow ac^2-bc^2=a^2b-ab^2$
$\displaystyle \Rightarrow c^2(a-b)=ab(a-b)$
$\displaystyle \Rightarrow c^2=ab\qquad [\because a\neq b]$
$\displaystyle \therefore c\text{ is the mean proportional between }a\text{ and }b.$
$\\$

$\displaystyle \textbf{Question 19: } \text{If }a+c=mb\text{ and }\frac{1}{b}+\frac{1}{d}=\frac{m}{c},\text{ prove that }$
$\displaystyle a,\ b,\ c\text{ and }d\text{ are in proportion.}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{1}{b}+\frac{1}{d}=\frac{m}{c}$
$\displaystyle \Rightarrow \frac{d+b}{bd}=\frac{m}{c}$
$\displaystyle \Rightarrow cd+bc=mbd$
$\displaystyle \Rightarrow cd+bc=(a+c)d$
$\displaystyle \Rightarrow cd+bc=ad+cd$
$\displaystyle \Rightarrow bc=ad$
$\displaystyle \Rightarrow \frac{a}{b}=\frac{c}{d}$
$\displaystyle \therefore a,\ b,\ c\text{ and }d\text{ are in proportion.}$
$\\$

$\displaystyle \textbf{Question 20: } \text{If }q\text{ is the mean proportional between }p\text{ and }r,\text{ prove that:}$
$\displaystyle p^2-q^2+r^2=q^4\left(\frac{1}{p^2}-\frac{1}{q^2}+\frac{1}{r^2}\right)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }q\text{ is the mean proportional between }p\text{ and }r,$
$\displaystyle q^2=pr$
$\displaystyle \text{RHS}=q^4\left(\frac{1}{p^2}-\frac{1}{q^2}+\frac{1}{r^2}\right)$
$\displaystyle =\frac{q^4}{p^2}-\frac{q^4}{q^2}+\frac{q^4}{r^2}$
$\displaystyle =\frac{q^4}{p^2}-q^2+\frac{q^4}{r^2}$
$\displaystyle =\frac{p^2r^2}{p^2}-q^2+\frac{p^2r^2}{r^2}$
$\displaystyle =r^2-q^2+p^2$
$\displaystyle =p^2-q^2+r^2=\text{LHS}$
$\\$

$\displaystyle \textbf{Question 21: } \text{If }\frac{a}{b}=\frac{c}{d},\text{ prove that each given ratio}$
$\displaystyle \left(\frac{a}{b}\text{ and }\frac{c}{d}\right)\text{ is equal to:}$
$\displaystyle \text{(i) }\frac{3a-5c}{3b-5d}\qquad \text{(ii) }\sqrt{\frac{2a^2+9c^2}{2b^2+9d^2}}$
$\displaystyle \text{(iii) }\left(\frac{5a^3-13c^3}{5b^3-13d^3}\right)^{\frac{1}{3}}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\frac{a}{b}=\frac{c}{d}=k.$
$\displaystyle \Rightarrow a=bk\text{ and }c=dk$
$\displaystyle \text{(i) }\frac{3a-5c}{3b-5d}=\frac{3bk-5dk}{3b-5d}$
$\displaystyle =\frac{k(3b-5d)}{3b-5d}=k$
$\displaystyle \text{(ii) }\sqrt{\frac{2a^2+9c^2}{2b^2+9d^2}}=\sqrt{\frac{2b^2k^2+9d^2k^2}{2b^2+9d^2}}$
$\displaystyle =\sqrt{\frac{k^2(2b^2+9d^2)}{2b^2+9d^2}}$
$\displaystyle =k$
$\displaystyle \text{(iii) }\left(\frac{5a^3-13c^3}{5b^3-13d^3}\right)^{\frac{1}{3}}$
$\displaystyle =\left(\frac{5b^3k^3-13d^3k^3}{5b^3-13d^3}\right)^{\frac{1}{3}}$
$\displaystyle =\left(\frac{k^3(5b^3-13d^3)}{5b^3-13d^3}\right)^{\frac{1}{3}}$
$\displaystyle =(k^3)^{\frac{1}{3}}=k$
$\displaystyle \therefore \text{Each expression is equal to the given ratios.}$
$\\$

$\displaystyle \textbf{Question 22: } \text{If }a,\ b,\ c\text{ and }d\text{ are in proportion, prove that:}$
$\displaystyle \text{(i) }\frac{a-b}{c-d}=\sqrt{\frac{3a^2+8b^2}{3c^2+8d^2}}\qquad \text{(ii) }\frac{5a^2+12c^2}{5b^2+12d^2}=\sqrt{\frac{3a^4-7c^4}{3b^4-7d^4}}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\frac{a}{b}=\frac{c}{d}=k.$
$\displaystyle \Rightarrow a=bk\text{ and }c=dk$
$\displaystyle \text{(i) LHS}=\frac{a-b}{c-d}=\frac{bk-b}{dk-d}$
$\displaystyle =\frac{b(k-1)}{d(k-1)}=\frac{b}{d}$
$\displaystyle \text{RHS}=\sqrt{\frac{3a^2+8b^2}{3c^2+8d^2}}$
$\displaystyle =\sqrt{\frac{3b^2k^2+8b^2}{3d^2k^2+8d^2}}$
$\displaystyle =\sqrt{\frac{b^2(3k^2+8)}{d^2(3k^2+8)}}$
$\displaystyle =\sqrt{\frac{b^2}{d^2}}=\frac{b}{d}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \text{(ii) LHS}=\frac{5a^2+12c^2}{5b^2+12d^2}$
$\displaystyle =\frac{5b^2k^2+12d^2k^2}{5b^2+12d^2}$
$\displaystyle =\frac{k^2(5b^2+12d^2)}{5b^2+12d^2}=k^2$
$\displaystyle \text{RHS}=\sqrt{\frac{3a^4-7c^4}{3b^4-7d^4}}$
$\displaystyle =\sqrt{\frac{3b^4k^4-7d^4k^4}{3b^4-7d^4}}$
$\displaystyle =\sqrt{\frac{k^4(3b^4-7d^4)}{3b^4-7d^4}}$
$\displaystyle =\sqrt{k^4}=k^2$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\\$

$\displaystyle \textbf{Question 23: }6\text{ is the mean proportional between two numbers }x\text{ and }y,$
$\displaystyle \text{and }48\text{ is the third proportional to }x\text{ and }y.\text{ Find the numbers.}\hfill \text{[ICSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }6\text{ is the mean proportional between }x\text{ and }y,$
$\displaystyle x:6=6:y$
$\displaystyle \Rightarrow xy=36\qquad \text{...(i)}$
$\displaystyle \text{Since }48\text{ is the third proportional to }x\text{ and }y,$
$\displaystyle x:y=y:48$
$\displaystyle \Rightarrow y^2=48x\qquad \text{...(ii)}$
$\displaystyle \text{From (i), }x=\frac{36}{y}$
$\displaystyle \text{Substituting }x=\frac{36}{y}\text{ in (ii), we get}$
$\displaystyle y^2=48\times\frac{36}{y}$
$\displaystyle \Rightarrow y^3=1728$
$\displaystyle \Rightarrow y=12$
$\displaystyle \therefore x=\frac{36}{12}=3$
$\displaystyle \text{Hence, the required numbers are }3\text{ and }12.$
$\\$

$\displaystyle \textbf{Question 24: } \text{If }\frac{8x+13y}{8x-13y}=\frac{9}{7},\text{ find }x:y.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{(8x+13y)+(8x-13y)}{(8x+13y)-(8x-13y)}=\frac{9+7}{9-7}$
$\displaystyle \Rightarrow \frac{16x}{26y}=\frac{16}{2}$
$\displaystyle \Rightarrow \frac{16x}{26y}=8$
$\displaystyle \Rightarrow 16x=208y$
$\displaystyle \Rightarrow x=13y$
$\displaystyle \therefore x:y=13:1$
$\\$

$\displaystyle \textbf{Question 25: } \text{If }a:b=c:d,\text{ show that }3a+2b:3a-2b=3c+2d:3c-2d.$
$\displaystyle \text{Answer:}$
$\displaystyle a:b=c:d$
$\displaystyle \Rightarrow \frac{a}{b}=\frac{c}{d}$
$\displaystyle \Rightarrow \frac{3a}{2b}=\frac{3c}{2d}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{3a+2b}{3a-2b}=\frac{3c+2d}{3c-2d}$
$\displaystyle \therefore 3a+2b:3a-2b=3c+2d:3c-2d$
$\\$

$\displaystyle \textbf{Question 26: } \text{If }\frac{8a-5b}{8c-5d}=\frac{8a+5b}{8c+5d},\text{ prove that }\frac{a}{b}=\frac{c}{d}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{8a-5b}{8c-5d}=\frac{8a+5b}{8c+5d}$
$\displaystyle \Rightarrow \frac{8a-5b}{8a+5b}=\frac{8c-5d}{8c+5d}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{(8a-5b)+(8a+5b)}{(8a-5b)-(8a+5b)}=\frac{(8c-5d)+(8c+5d)}{(8c-5d)-(8c+5d)}$
$\displaystyle \Rightarrow \frac{16a}{-10b}=\frac{16c}{-10d}$
$\displaystyle \Rightarrow \frac{a}{b}=\frac{c}{d}$
$\\$

$\displaystyle \textbf{Question 27: } \text{If }p=\frac{4xy}{x+y},\text{ find the value of }$
$\displaystyle \frac{p+2x}{p-2x}+\frac{p+2y}{p-2y}.$
$\displaystyle \text{Answer:}$
$\displaystyle p=\frac{4xy}{x+y}\Rightarrow \frac{p}{2x}=\frac{2y}{x+y}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{p+2x}{p-2x}=\frac{2y+x+y}{2y-x-y}$
$\displaystyle \Rightarrow \frac{p+2x}{p-2x}=\frac{x+3y}{y-x}$
$\displaystyle \text{Also, }p=\frac{4xy}{x+y}\Rightarrow \frac{p}{2y}=\frac{2x}{x+y}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{p+2y}{p-2y}=\frac{2x+x+y}{2x-x-y}$
$\displaystyle \Rightarrow \frac{p+2y}{p-2y}=\frac{3x+y}{x-y}$
$\displaystyle \therefore \frac{p+2x}{p-2x}+\frac{p+2y}{p-2y}=\frac{x+3y}{y-x}+\frac{3x+y}{x-y}$
$\displaystyle =\frac{x+3y}{y-x}-\frac{3x+y}{y-x}$
$\displaystyle =\frac{x+3y-3x-y}{y-x}=\frac{-2x+2y}{y-x}=2$
$\\$

$\displaystyle \textbf{Question 28: } \text{If }a:b=c:d,\text{ prove that:}$
$\displaystyle (a^2+ac+c^2):(a^2-ac+c^2)=(b^2+bd+d^2):(b^2-bd+d^2).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\frac{a}{b}=\frac{c}{d}=k.$
$\displaystyle \Rightarrow a=bk\text{ and }c=dk$
$\displaystyle \frac{a^2+ac+c^2}{a^2-ac+c^2}=\frac{(bk)^2+(bk)(dk)+(dk)^2}{(bk)^2-(bk)(dk)+(dk)^2}$
$\displaystyle =\frac{k^2(b^2+bd+d^2)}{k^2(b^2-bd+d^2)}$
$\displaystyle =\frac{b^2+bd+d^2}{b^2-bd+d^2}$
$\displaystyle \therefore (a^2+ac+c^2):(a^2-ac+c^2)=(b^2+bd+d^2):(b^2-bd+d^2).$
$\\$

$\displaystyle \textbf{Question 29: } \text{If }x,\ y\text{ and }z\text{ are in continued proportion, prove that}$
$\displaystyle x^2-y^2:x^2+y^2=x-z:x+z.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }x,\ y\text{ and }z\text{ are in continued proportion,}$
$\displaystyle \frac{x}{y}=\frac{y}{z}=k$
$\displaystyle \Rightarrow x=yk,\ y=zk\text{ and }x=zk^2$
$\displaystyle \frac{x^2-y^2}{x^2+y^2}=\frac{y^2k^2-y^2}{y^2k^2+y^2}$
$\displaystyle =\frac{y^2(k^2-1)}{y^2(k^2+1)}=\frac{k^2-1}{k^2+1}$
$\displaystyle \text{Also, }\frac{x-z}{x+z}=\frac{zk^2-z}{zk^2+z}$
$\displaystyle =\frac{z(k^2-1)}{z(k^2+1)}=\frac{k^2-1}{k^2+1}$
$\displaystyle \therefore x^2-y^2:x^2+y^2=x-z:x+z$
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$\displaystyle \textbf{Question 30: } \text{Using the properties of proportion, solve the following equation for }x:$
$\displaystyle \frac{x^3+3x}{3x^2+1}=\frac{341}{91}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{x^3+3x+3x^2+1}{x^3+3x-3x^2-1}=\frac{341+91}{341-91}$
$\displaystyle \Rightarrow \frac{x^3+3x^2+3x+1}{x^3-3x^2+3x-1}=\frac{432}{250}$
$\displaystyle \Rightarrow \frac{(x+1)^3}{(x-1)^3}=\frac{216}{125}=\left(\frac{6}{5}\right)^3$
$\displaystyle \Rightarrow \frac{x+1}{x-1}=\frac{6}{5}$
$\displaystyle \text{Again, applying componendo and dividendo,}$
$\displaystyle \frac{x+1+x-1}{x+1-x+1}=\frac{6+5}{6-5}$
$\displaystyle \Rightarrow \frac{2x}{2}=11$
$\displaystyle \therefore x=11$
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$\displaystyle \textbf{Question 31: } \text{If }x=\frac{\sqrt{3a+2b}+\sqrt{3a-2b}}{\sqrt{3a+2b}-\sqrt{3a-2b}},\text{ prove that}$
$\displaystyle bx^2-3ax+b=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x}{1}=\frac{\sqrt{3a+2b}+\sqrt{3a-2b}}{\sqrt{3a+2b}-\sqrt{3a-2b}}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{x+1}{x-1}=\frac{2\sqrt{3a+2b}}{2\sqrt{3a-2b}}$
$\displaystyle \Rightarrow \frac{x+1}{x-1}=\frac{\sqrt{3a+2b}}{\sqrt{3a-2b}}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle \frac{(x+1)^2}{(x-1)^2}=\frac{3a+2b}{3a-2b}$
$\displaystyle \Rightarrow \frac{x^2+2x+1}{x^2-2x+1}=\frac{3a+2b}{3a-2b}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{(x^2+2x+1)+(x^2-2x+1)}{(x^2+2x+1)-(x^2-2x+1)}$
$\displaystyle =\frac{(3a+2b)+(3a-2b)}{(3a+2b)-(3a-2b)}$
$\displaystyle \Rightarrow \frac{2x^2+2}{4x}=\frac{6a}{4b}$
$\displaystyle \Rightarrow \frac{x^2+1}{2x}=\frac{3a}{2b}$
$\displaystyle \Rightarrow 2b(x^2+1)=6ax$
$\displaystyle \Rightarrow bx^2+b=3ax$
$\displaystyle \therefore bx^2-3ax+b=0$
$\\$