\displaystyle \textbf{Question 1: } \text{If }a:b=3:5,\text{ find }(10a+3b):(5a+2b).
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a:b=3:5
\displaystyle \Rightarrow \frac{a}{b}=\frac{3}{5}
\displaystyle \frac{10a+3b}{5a+2b}=\frac{10\left(\frac{a}{b}\right)+3}{5\left(\frac{a}{b}\right)+2}
\displaystyle =\frac{10\left(\frac{3}{5}\right)+3}{5\left(\frac{3}{5}\right)+2}=\frac{6+3}{3+2}=\frac{9}{5}
\displaystyle \therefore (10a+3b):(5a+2b)=9:5
\\

\displaystyle \textbf{Question 2: } \text{If }5x+6y:8x+5y=8:9,\text{ find }x:y.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }5x+6y:8x+5y=8:9
\displaystyle \Rightarrow 9(5x+6y)=8(8x+5y)
\displaystyle \Rightarrow 45x+54y=64x+40y
\displaystyle \Rightarrow 14y=19x
\displaystyle \Rightarrow \frac{x}{y}=\frac{14}{19}
\displaystyle \therefore x:y=14:19
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\displaystyle \textbf{Question 3: } \text{If }(3x-4y):(2x-3y)=(5x-6y):(4x-5y),\text{ find }x:y.
\displaystyle \text{Answer:}
\displaystyle (3x-4y):(2x-3y)=(5x-6y):(4x-5y)
\displaystyle \Rightarrow (3x-4y)(4x-5y)=(5x-6y)(2x-3y)
\displaystyle \Rightarrow 12x^2-16xy-15xy+20y^2=10x^2-12xy-15xy+18y^2
\displaystyle \Rightarrow 12x^2-31xy+20y^2=10x^2-27xy+18y^2
\displaystyle \Rightarrow 2x^2-4xy+2y^2=0
\displaystyle \Rightarrow x^2-2xy+y^2=0
\displaystyle \Rightarrow (x-y)^2=0
\displaystyle \Rightarrow x=y
\displaystyle \therefore x:y=1:1
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\displaystyle \textbf{Question 4: } \text{Find the:}
\displaystyle \text{(i) Duplicate ratio of }2\sqrt{2}:3\sqrt{5}\qquad \text{(ii) Triplicate ratio of }2a:3b
\displaystyle \text{(iii) Sub-duplicate ratio of }9x^2a^4:25y^6b^2\qquad \text{(iv) Sub-triplicate ratio of }216:343
\displaystyle \text{(v) Reciprocal ratio of }3:5
\displaystyle \text{(vi) Ratio compounded of duplicate ratio of }5:6,\text{ reciprocal ratio of }25:42\text{ and sub-duplicate ratio of }36:49.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Duplicate ratio of }2\sqrt{2}:3\sqrt{5}=(2\sqrt{2})^2:(3\sqrt{5})^2=8:45
\displaystyle \text{(ii) Triplicate ratio of }2a:3b=(2a)^3:(3b)^3=8a^3:27b^3
\displaystyle \text{(iii) Sub-duplicate ratio of }9x^2a^4:25y^6b^2
\displaystyle =\sqrt{9x^2a^4}:\sqrt{25y^6b^2}=3xa^2:5y^3b
\displaystyle \text{(iv) Sub-triplicate ratio of }216:343=\sqrt[3]{216}:\sqrt[3]{343}=6:7
\displaystyle \text{(v) Reciprocal ratio of }3:5=5:3
\displaystyle \text{(vi) Duplicate ratio of }5:6=25:36
\displaystyle \text{Reciprocal ratio of }25:42=42:25
\displaystyle \text{Sub-duplicate ratio of }36:49=6:7
\displaystyle \text{Compound ratio }=(25\times42\times6):(36\times25\times7)=1:1
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\displaystyle \textbf{Question 5: } \text{Find the value of }x,\text{ if:}
\displaystyle \text{(i) }(2x+3):(5x-38)\text{ is the duplicate ratio of }\sqrt{5}:\sqrt{6}.
\displaystyle \text{(ii) }(2x+1):(3x+13)\text{ is the sub-duplicate ratio of }9:25.
\displaystyle \text{(iii) }(3x-7):(4x+3)\text{ is the sub-triplicate ratio of }8:27.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Duplicate ratio of }\sqrt{5}:\sqrt{6}=5:6
\displaystyle \therefore \frac{2x+3}{5x-38}=\frac{5}{6}
\displaystyle \Rightarrow 12x+18=25x-190
\displaystyle \Rightarrow 13x=208
\displaystyle \Rightarrow x=16
\displaystyle \text{(ii) Sub-duplicate ratio of }9:25=\sqrt{9}:\sqrt{25}=3:5
\displaystyle \therefore \frac{2x+1}{3x+13}=\frac{3}{5}
\displaystyle \Rightarrow 10x+5=9x+39
\displaystyle \Rightarrow x=34
\displaystyle \text{(iii) Sub-triplicate ratio of }8:27=\sqrt[3]{8}:\sqrt[3]{27}=2:3
\displaystyle \therefore \frac{3x-7}{4x+3}=\frac{2}{3}
\displaystyle \Rightarrow 9x-21=8x+6
\displaystyle \Rightarrow x=27
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\displaystyle \textbf{Question 6: } \text{What quantity must be added to each term of the ratio }x:y\text{ so that it becomes }c:d?
\displaystyle \text{Answer:}
\displaystyle \text{Let the quantity to be added be }a.
\displaystyle \therefore \frac{x+a}{y+a}=\frac{c}{d}
\displaystyle \Rightarrow d(x+a)=c(y+a)
\displaystyle \Rightarrow dx+da=cy+ca
\displaystyle \Rightarrow (d-c)a=cy-dx
\displaystyle \therefore a=\frac{cy-dx}{d-c}
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\displaystyle \textbf{Question 7: } \text{Two numbers are in the ratio }5:7.\text{ If }3\text{ is subtracted from each,}
\displaystyle \text{the ratio becomes }2:3.\text{ Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two numbers be }x\text{ and }y.
\displaystyle \text{Given, }\frac{x}{y}=\frac{5}{7}
\displaystyle \Rightarrow x=\frac{5}{7}y
\displaystyle \text{If }3\text{ is subtracted from each, then}
\displaystyle \frac{x-3}{y-3}=\frac{2}{3}
\displaystyle \Rightarrow 3x-9=2y-6
\displaystyle \text{Substituting }x=\frac{5}{7}y,
\displaystyle 3\left(\frac{5}{7}y\right)-9=2y-6
\displaystyle \Rightarrow 15y-63=14y-42
\displaystyle \Rightarrow y=21
\displaystyle \Rightarrow x=\frac{5}{7}\times21=15
\displaystyle \therefore \text{The required numbers are }15\text{ and }21.
\\

\displaystyle \textbf{Question 8: } \text{If }15(2x^2-y^2)=7xy,\text{ find }x:y,\text{ if }x\text{ and }y\text{ are positive.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }15(2x^2-y^2)=7xy
\displaystyle \Rightarrow 30x^2-15y^2=7xy
\displaystyle \text{Dividing by }xy,
\displaystyle 30\frac{x}{y}-15\frac{y}{x}=7
\displaystyle \text{Let }\frac{x}{y}=a.
\displaystyle \Rightarrow 30a-\frac{15}{a}=7
\displaystyle \Rightarrow 30a^2-7a-15=0
\displaystyle \Rightarrow (6a-5)(5a+3)=0
\displaystyle \Rightarrow a=\frac{5}{6}\text{ or }a=-\frac{3}{5}
\displaystyle \text{Since }x\text{ and }y\text{ are positive, }a\neq-\frac{3}{5}.
\displaystyle \therefore \frac{x}{y}=\frac{5}{6}
\displaystyle \therefore x:y=5:6
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\displaystyle \textbf{Question 9: } \text{Find the:}
\displaystyle \text{(i) Fourth proportional to }2xy,\;x^2\text{ and }y^2.
\displaystyle \text{(ii) Third proportional to }a^2-b^2\text{ and }a+b.
\displaystyle \text{(iii) Mean proportional to }(x-y)\text{ and }(x^3-x^2y).
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let the fourth proportional be }a.
\displaystyle \therefore 2xy:x^2=y^2:a
\displaystyle \Rightarrow a=\frac{x^2\cdot y^2}{2xy}=\frac{xy}{2}
\displaystyle \therefore \text{The fourth proportional is }\frac{xy}{2}.
\displaystyle \\
\displaystyle \text{(ii) Let the third proportional be }x.
\displaystyle \therefore (a^2-b^2):(a+b)=(a+b):x
\displaystyle \Rightarrow x=\frac{(a+b)^2}{a^2-b^2}
\displaystyle =\frac{(a+b)^2}{(a+b)(a-b)}=\frac{a+b}{a-b}
\displaystyle \therefore \text{The third proportional is }\frac{a+b}{a-b}.
\displaystyle \\
\displaystyle \text{(iii) Let }a\text{ be the mean proportional.}
\displaystyle \therefore (x-y):a=a:(x^3-x^2y)
\displaystyle \Rightarrow a^2=(x-y)(x^3-x^2y)
\displaystyle =(x-y)\cdot x^2(x-y)=x^2(x-y)^2
\displaystyle \Rightarrow a=x(x-y)
\displaystyle \therefore \text{The mean proportional is }x(x-y).
\\

\displaystyle \textbf{Question 10: } \text{Find two numbers such that the mean proportional between them is }14
\displaystyle \text{and the third proportional to them is }112.
\displaystyle \text{Answer:}
\displaystyle \text{Let the two numbers be }a\text{ and }b.
\displaystyle \text{Given, mean proportional between them is }14.
\displaystyle \therefore a:14=14:b
\displaystyle \Rightarrow ab=196 \qquad \cdots(1)
\displaystyle \text{Also, third proportional to them is }112.
\displaystyle \therefore a:b=b:112
\displaystyle \Rightarrow b^2=112a \qquad \cdots(2)
\displaystyle \text{From (1), }a=\frac{196}{b}
\displaystyle \text{Substituting in (2),}
\displaystyle b^2=112\left(\frac{196}{b}\right)
\displaystyle \Rightarrow b^3=112\times196
\displaystyle \Rightarrow b^3=21952
\displaystyle \Rightarrow b=28
\displaystyle \Rightarrow a=\frac{196}{28}=7
\displaystyle \therefore \text{The two numbers are }7\text{ and }28.
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\displaystyle \textbf{Question 11: } \text{If }x\text{ and }y\text{ are unequal and }x:y\text{ is the duplicate ratio of }(x+z):(y+z),
\displaystyle \text{prove that }z\text{ is the mean proportional between }x\text{ and }y.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\frac{x}{y}=\frac{(x+z)^2}{(y+z)^2}
\displaystyle \Rightarrow x(y+z)^2=y(x+z)^2
\displaystyle \Rightarrow x(y^2+2yz+z^2)=y(x^2+2xz+z^2)
\displaystyle \Rightarrow xy^2+2xyz+xz^2=x^2y+2xyz+yz^2
\displaystyle \Rightarrow xy(y-x)=z^2(y-x)
\displaystyle \Rightarrow xy=z^2 \qquad (\because x\neq y)
\displaystyle \Rightarrow x:z=z:y
\displaystyle \therefore z\text{ is the mean proportional between }x\text{ and }y.
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\displaystyle \textbf{Question 12: } \text{If }q\text{ is the mean proportional between }p\text{ and }r,\text{ prove that}
\displaystyle \frac{p^3+q^3+r^3}{pqr}=\frac{1}{p^3}+\frac{1}{q^3}+\frac{1}{r^3}.
\displaystyle \text{Answer:}
\displaystyle \text{This statement is incorrect as written.}
\displaystyle \text{For example, if }p=1,\ q=2,\ r=4,\text{ then }q^2=pr.
\displaystyle \text{But }\frac{p^3+q^3+r^3}{p^2q^2r^2}=\frac{73}{64}
\displaystyle \text{and }\frac{1}{p^3}+\frac{1}{q^3}+\frac{1}{r^3}=1+\frac{1}{8}+\frac{1}{64}=\frac{73}{64}.
\displaystyle \text{So the original statement is actually correct.}
\displaystyle \text{Given, }q^2=pr
\displaystyle \therefore p^2q^2r^2=(pqr)^2=(q^3)^2=q^6
\displaystyle \text{Also, }pr=q^2
\displaystyle \frac{p^3+q^3+r^3}{p^2q^2r^2}
\displaystyle =\frac{p^3}{p^2q^2r^2}+\frac{q^3}{p^2q^2r^2}+\frac{r^3}{p^2q^2r^2}
\displaystyle =\frac{p}{q^2r^2}+\frac{q}{p^2r^2}+\frac{r}{p^2q^2}
\displaystyle =\frac{1}{r^3}+\frac{1}{q^3}+\frac{1}{p^3}
\displaystyle =\frac{1}{p^3}+\frac{1}{q^3}+\frac{1}{r^3}
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\displaystyle \textbf{Question 13: } \text{If }a,\ b\text{ and }c\text{ are in continued proportion, prove that}
\displaystyle a:c=(a^2+b^2):(b^2+c^2).
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a:b=b:c
\displaystyle \Rightarrow \frac{a}{b}=\frac{b}{c}=k
\displaystyle \Rightarrow b=ck\text{ and }a=bk=ck^2
\displaystyle \text{Now, }\frac{a}{c}=\frac{ck^2}{c}=k^2
\displaystyle \text{Also, }\frac{a^2+b^2}{b^2+c^2}
\displaystyle =\frac{(ck^2)^2+(ck)^2}{(ck)^2+c^2}
\displaystyle =\frac{c^2k^4+c^2k^2}{c^2k^2+c^2}
\displaystyle =\frac{c^2k^2(k^2+1)}{c^2(k^2+1)}=k^2
\displaystyle \therefore \frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}
\displaystyle \therefore a:c=(a^2+b^2):(b^2+c^2)
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\displaystyle \textbf{Question 14: If }x=\frac{2ab}{a+b},\text{ find the value of }\frac{x+a}{x-a}+\frac{x+b}{x-b}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }x=\frac{2ab}{a+b}
\displaystyle \therefore \frac{x}{a}=\frac{2b}{a+b}\qquad\text{and}\qquad\frac{x}{b}=\frac{2a}{a+b}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{x+a}{x-a}=\frac{2b+a+b}{2b-a-b}=\frac{a+3b}{b-a}
\displaystyle \text{Similarly, applying componendo and dividendo,}
\displaystyle \frac{x+b}{x-b}=\frac{2a+a+b}{2a-a-b}=\frac{3a+b}{a-b}
\displaystyle \therefore \frac{x+a}{x-a}+\frac{x+b}{x-b}=\frac{a+3b}{b-a}+\frac{3a+b}{a-b}
\displaystyle =\frac{-(a+3b)+(3a+b)}{a-b}=\frac{2a-2b}{a-b}=2
\\

\displaystyle \textbf{Question 15: If }(4a+9b)(4c-9d)=(4a-9b)(4c+9d),\text{ prove that }a:b=c:d.
\displaystyle \text{Answer:}
\displaystyle \text{Given }(4a+9b)(4c-9d)=(4a-9b)(4c+9d)
\displaystyle \Rightarrow \frac{4a+9b}{4a-9b}=\frac{4c+9d}{4c-9d}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{(4a+9b)+(4a-9b)}{(4a+9b)-(4a-9b)}=\frac{(4c+9d)+(4c-9d)}{(4c+9d)-(4c-9d)}
\displaystyle \Rightarrow \frac{8a}{18b}=\frac{8c}{18d}
\displaystyle \Rightarrow \frac{a}{b}=\frac{c}{d}
\displaystyle \therefore a:b=c:d
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\displaystyle \textbf{Question 16: If }\frac{a}{b}=\frac{c}{d},\text{ show that }(a+b):(c+d)=\sqrt{a^2+b^2}:\sqrt{c^2+d^2}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\frac{a}{b}=\frac{c}{d}=k
\displaystyle \Rightarrow a=bk\text{ and }c=dk
\displaystyle \text{LHS}=\frac{a+b}{c+d}
\displaystyle =\frac{bk+b}{dk+d}=\frac{b(k+1)}{d(k+1)}=\frac{b}{d}
\displaystyle \text{RHS}=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}
\displaystyle =\frac{\sqrt{b^2k^2+b^2}}{\sqrt{d^2k^2+d^2}}
\displaystyle =\frac{\sqrt{b^2(k^2+1)}}{\sqrt{d^2(k^2+1)}}=\frac{b}{d}
\displaystyle \therefore (a+b):(c+d)=\sqrt{a^2+b^2}:\sqrt{c^2+d^2}
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\displaystyle \textbf{Question 17: If }\frac{x}{a}=\frac{y}{b}=\frac{z}{c},\text{ prove that}
\displaystyle \frac{ax-by}{(a+b)(x-y)}+\frac{by-cz}{(b+c)(y-z)}+\frac{cz-ax}{(c+a)(z-x)}=3.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k
\displaystyle \Rightarrow x=ak,\ y=bk,\ z=ck
\displaystyle \text{LHS}=\frac{ax-by}{(a+b)(x-y)}+\frac{by-cz}{(b+c)(y-z)}+\frac{cz-ax}{(c+a)(z-x)}
\displaystyle =\frac{a(ak)-b(bk)}{(a+b)(ak-bk)}+\frac{b(bk)-c(ck)}{(b+c)(bk-ck)}+\frac{c(ck)-a(ak)}{(c+a)(ck-ak)}
\displaystyle =\frac{k(a^2-b^2)}{k(a+b)(a-b)}+\frac{k(b^2-c^2)}{k(b+c)(b-c)}+\frac{k(c^2-a^2)}{k(c+a)(c-a)}
\displaystyle =1+1+1=3
\\

\displaystyle \textbf{Question 18: } \text{There are }36\text{ members in a student's council and the ratio of boys to girls is }3:1.
\displaystyle \text{How many more girls should be added so that the ratio of boys to girls becomes }9:5?
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of girls be }x.
\displaystyle \therefore \text{Number of boys }=3x
\displaystyle 3x+x=36
\displaystyle \Rightarrow 4x=36
\displaystyle \Rightarrow x=9
\displaystyle \therefore \text{Girls }=9\text{ and boys }=27
\displaystyle \text{Let }n\text{ girls be added.}
\displaystyle \therefore \frac{27}{9+n}=\frac{9}{5}
\displaystyle \Rightarrow 135=81+9n
\displaystyle \Rightarrow 9n=54
\displaystyle \Rightarrow n=6
\displaystyle \therefore 6\text{ more girls should be added.}
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\displaystyle \textbf{Question 19: If }\frac{x}{b-c}=\frac{y}{c-a}=\frac{z}{a-b},\text{ prove that }ax+by+cz=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\frac{x}{b-c}=\frac{y}{c-a}=\frac{z}{a-b}=k
\displaystyle \Rightarrow x=k(b-c),\ y=k(c-a),\ z=k(a-b)
\displaystyle \text{LHS}=ax+by+cz
\displaystyle =ak(b-c)+bk(c-a)+ck(a-b)
\displaystyle =k(ab-ac+bc-ab+ca-bc)=0
\displaystyle \therefore ax+by+cz=0
\\

\displaystyle \textbf{Question 20: If }7x-15y=4x+y,\text{ find }x:y.\text{ Hence, find:}
\displaystyle \text{(i) }\frac{9x+5y}{9x-5y}\qquad \text{(ii) }\frac{3x^2+2y^2}{3x^2-2y^2}
\displaystyle \text{Answer:}
\displaystyle \text{Given }7x-15y=4x+y
\displaystyle \Rightarrow 3x=16y
\displaystyle \Rightarrow \frac{x}{y}=\frac{16}{3}
\displaystyle \therefore x:y=16:3
\displaystyle \text{(i) }\frac{9x+5y}{9x-5y}=\frac{9\left(\frac{x}{y}\right)+5}{9\left(\frac{x}{y}\right)-5}
\displaystyle =\frac{9\left(\frac{16}{3}\right)+5}{9\left(\frac{16}{3}\right)-5}
\displaystyle =\frac{48+5}{48-5}=\frac{53}{43}
\displaystyle \text{(ii) }\frac{3x^2+2y^2}{3x^2-2y^2}=\frac{3\left(\frac{x}{y}\right)^2+2}{3\left(\frac{x}{y}\right)^2-2}
\displaystyle =\frac{3\left(\frac{16}{3}\right)^2+2}{3\left(\frac{16}{3}\right)^2-2}
\displaystyle =\frac{\frac{256}{3}+2}{\frac{256}{3}-2}=\frac{262}{250}=\frac{131}{125}
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\displaystyle \textbf{Question 21: If } \frac{4m+3n}{4m-3n}=\frac{7}{4}\text{, use properties of proportion to find:}
\displaystyle \text{(i) }m:n \qquad \text{(ii) }\frac{2m^2-11n^2}{2m^2+11n^2}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given }\frac{4m+3n}{4m-3n}=\frac{7}{4}
\displaystyle \text{Applying componendo and dividendo}
\displaystyle \frac{(4m+3n)+(4m-3n)}{(4m+3n)-(4m-3n)}=\frac{7+4}{7-4}
\displaystyle \frac{8m}{6n}=\frac{11}{3}
\displaystyle \frac{4m}{3n}=\frac{11}{3}
\displaystyle \Rightarrow \frac{m}{n}=\frac{11}{4}
\displaystyle \therefore m:n=11:4
\displaystyle \text{(ii) }\frac{2m^2-11n^2}{2m^2+11n^2}
\displaystyle =\frac{2\left(\frac{m}{n}\right)^2-11}{2\left(\frac{m}{n}\right)^2+11}
\displaystyle =\frac{2\left(\frac{11}{4}\right)^2-11}{2\left(\frac{11}{4}\right)^2+11}
\displaystyle =\frac{\frac{242}{16}-11}{\frac{242}{16}+11}
\displaystyle =\frac{242-176}{242+176}
\displaystyle =\frac{66}{418}=\frac{33}{209}
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\displaystyle \textbf{Question 22: If }x,\ y\text{ and }z\text{ are in continued proportion, prove that:}\\ \\ \frac{(x+y)^2}{(y+z)^2}=\frac{x}{y}.\text{ [2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }x,\ y\text{ and }z\text{ are in continued proportion,}
\displaystyle \frac{x}{y}=\frac{y}{z}\Rightarrow y^2=xz
\displaystyle \text{LHS}=\frac{(x+y)^2}{(y+z)^2}
\displaystyle =\frac{\left(\frac{y^2}{z}+y\right)^2}{(y+z)^2}
\displaystyle =\frac{\left(\frac{y(y+z)}{z}\right)^2}{(y+z)^2}
\displaystyle =\frac{y^2}{z^2}
\displaystyle =\frac{xz}{z^2}
\displaystyle =\frac{x}{z}
\displaystyle =\frac{x^2}{xy}
\displaystyle =\frac{x}{y}\qquad\left(\because\ xz=y^2\right)
\displaystyle \therefore \frac{(x+y)^2}{(y+z)^2}=\frac{x}{y}. \text{ Hence proved.}
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\displaystyle \textbf{Question 23: 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}.\text{ [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}}
\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 \Rightarrow \frac{x+1}{x-1}=\frac{\sqrt{a^2+b^2}}{\sqrt{a^2-b^2}}
\displaystyle \text{Squaring both sides,}
\displaystyle \frac{x^2+2x+1}{x^2-2x+1}=\frac{a^2+b^2}{a^2-b^2}
\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^2+b^2+a^2-b^2}{a^2+b^2-a^2+b^2}
\displaystyle \Rightarrow \frac{2(x^2+1)}{4x}=\frac{2a^2}{2b^2}
\displaystyle \Rightarrow \frac{x^2+1}{2x}=\frac{a^2}{b^2}
\displaystyle \Rightarrow b^2=\frac{2a^2x}{x^2+1}
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\displaystyle \textbf{Question 24: If }\frac{x^2+y^2}{x^2-y^2}=2\frac{1}{8},\text{ find:}
\displaystyle \text{(i) }\frac{x}{y}\qquad \text{(ii) }\frac{x^3+y^3}{x^3-y^3}.\text{ [2014]}
\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 \Rightarrow \frac{2x^2}{2y^2}=\frac{25}{9}
\displaystyle \Rightarrow \frac{x^2}{y^2}=\frac{25}{9}
\displaystyle \Rightarrow \frac{x}{y}=\frac{5}{3}
\displaystyle \text{(ii) }\frac{x^3+y^3}{x^3-y^3}=\frac{\left(\frac{x}{y}\right)^3+1}{\left(\frac{x}{y}\right)^3-1}
\displaystyle =\frac{\left(\frac{5}{3}\right)^3+1}{\left(\frac{5}{3}\right)^3-1}
\displaystyle =\frac{\frac{125}{27}+1}{\frac{125}{27}-1}
\displaystyle =\frac{152}{98}=\frac{76}{49}
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\displaystyle \textbf{Question 25: } \text{Using componendo and dividendo, find }x:\frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}}=9.\text{ [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 \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{Squaring both sides,}
\displaystyle \frac{3x+4}{3x-5}=\frac{25}{16}
\displaystyle \Rightarrow 16(3x+4)=25(3x-5)
\displaystyle \Rightarrow 48x+64=75x-125
\displaystyle \Rightarrow 27x=189
\displaystyle \Rightarrow x=7
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\displaystyle \textbf{Question 26: } \text{If }x=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}},\text{ using properties of proportion show that}
\displaystyle x^2-2ax+1=0.\text{ [2012]}
\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 \Rightarrow \frac{x+1}{x-1}=\frac{\sqrt{a+1}}{\sqrt{a-1}}
\displaystyle \text{Squaring both sides,}
\displaystyle \frac{x^2+2x+1}{x^2-2x+1}=\frac{a+1}{a-1}
\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+1)+(a-1)}{(a+1)-(a-1)}
\displaystyle \Rightarrow \frac{2x^2+2}{4x}=\frac{2a}{2}
\displaystyle \Rightarrow x^2+1=2ax
\displaystyle \Rightarrow x^2-2ax+1=0
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