Question 1: If \displaystyle a:b=3:5 , find: \displaystyle (10a+3b):(5a+2b)  

Answer:

\displaystyle \text{Given } a:b=3:5  

\displaystyle \Rightarrow a = \frac{3}{5} b  

\displaystyle \frac{(10a+3b)}{(5a+2b)} = \frac{(10 .\frac{3}{5} b+3b)}{(5 .\frac{3}{5} b+2b)} = \frac{9}{5}  

\displaystyle \\

Question 2: If \displaystyle 5x+6y:8x+5y=8:9 , find: \displaystyle x:y  

Answer:

\displaystyle \text{Given } 5x+6y:8x+5y=8:9  

\displaystyle 9(5x+6y)=8(8x+5y)  

\displaystyle 45x +54y=64x+40y  

\displaystyle 14y = 19x  

\displaystyle \Rightarrow x:y = 14:19  

\displaystyle \\

Question 3: If \displaystyle (3x-4y):(2x-3y)=(5x-6y):(4x-5y) , find: \displaystyle x:y  

Answer:

\displaystyle (3x-4y):(2x-3y)=(5x-6y):(4x-5y)  

\displaystyle (3x-4y)(4x-5y) = (5x-6y)(2x-3y)  

\displaystyle 12x^2-16xy-15xy+20y^2 = 10x^2-12xy-15xy+18y^2  

\displaystyle 12x^2-31xy+20y^2 = 10x^2-27xy+18y^2  

\displaystyle 2x^2-4xy+2y^2 = 0  

\displaystyle x^2-2xy+y^2 = 0  

\displaystyle (x-y)^2 = 0  

\displaystyle \text{or } x-y = 0  

\displaystyle \text{or } x:y = 1:1  

Note: You could have also done this by applying componendo and dividendo.

\displaystyle \\

Question 4: Find the:

\displaystyle \text{i) Duplicate ratio of } 2\sqrt{2}:3\sqrt{5}  

\displaystyle \text{ii) Triplicate ratio of } 2a:3b  

\displaystyle \text{iii) Sub-Duplicate ratio of } 9x^2 a^4:25y^6 b^2  

\displaystyle \text{iv) Sub-triplicate ratio of } 216:343  

\displaystyle \text{v) Reciprocal ratio of } 3:5  

vi) Ratio compounded of the \displaystyle \text{Duplicate ratio of } 5:6 , the \displaystyle \text{Reciprocal ratio of } 25:42 and the \displaystyle \text{Sub-Duplicate ratio of } 36:49 .

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^3:81b^3  

\displaystyle \text{iii) Sub-Duplicate ratio of } 9x^2 a^4:25y^6 b^2 = \sqrt{9x^2 a^4}:\sqrt{25y^6 b^2} = 3xa^2:5y^3b  

\displaystyle \text{iv) Sub-triplicate ratio of } 216:343 = \sqrt[3]{216}:\sqrt[3]{343}  

\displaystyle \text{v) Reciprocal ratio of } 3:5 = 5:3  

vi) Ratio compounded of the \displaystyle \text{Duplicate ratio of } 5:6 , the \displaystyle \text{Reciprocal ratio of } 25:42 and the \displaystyle \text{Sub-Duplicate ratio of } 36:49 .

\displaystyle \text{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 } \frac{25}{36} \times \frac{42}{25} \times \frac{6}{7} = \frac{1}{1} \ or \ 1:1  

\displaystyle \\

Question 5: Find the value of \displaystyle x , 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 .

Answer:

\displaystyle \text{i) } (2x+3):(5x-38)  \text{ is the  Duplicate ratio of } \sqrt{5}:\sqrt{6}  

\displaystyle \frac{2x+3}{5x-38} = \frac{5}{6}  

\displaystyle 12x+18 = 25x-190  

\displaystyle 13x = 208 \Rightarrow x = 16  

\displaystyle \text{ii) } (2x+1):(3x+13)  \text{is the  Sub-Duplicate ratio of } 9:25 .

\displaystyle \frac{2x+1}{3x+13} = \frac{\sqrt{9}}{\sqrt{25}}  

\displaystyle \frac{2x+1}{3x+13} = \frac{3}{5}  

\displaystyle 10x+5 = 9x+39  

\displaystyle x = 34  

\displaystyle \text{iii) } (3x-7):(4x+3)  \text{is the  Sub-triplicate ratio of } 8:27  

\displaystyle (3x-7):(4x+3) = \frac{\sqrt[3]{8}}{\sqrt[3]{27}} = \frac{2}{3}  

\displaystyle 9x-21 = 8x+6  

\displaystyle \text{or } x = 27  

\displaystyle \\

Question 6: What quantity must be added to each term of the ratio \displaystyle x:y so that it may become equal to \displaystyle c:d ?

Answer:

Let us add quantity \displaystyle a  

Therefore

\displaystyle \frac{x+a}{y+a} = \frac{c}{d}  

\displaystyle dx+da = cy+ca  

\displaystyle (d-c)a = cy-dx  

\displaystyle \text{or } a = \frac{cy-dx}{d-c}  

\displaystyle \\

Question 7: Two numbers are in the ratio \displaystyle 5:7 . If \displaystyle 3 is subtracted from each of them, the ratio between them becomes \displaystyle 2:3 find the numbers.

Answer:

Let the two numbers be \displaystyle x \ and \ y  

Given

\displaystyle \frac{x}{y} = \frac{5}{7}  

\displaystyle \Rightarrow x = \frac{5}{7} y  

\displaystyle \text{Therefore if } 3 is subtracted from each of them, then

\displaystyle \frac{x-3}{y-3} = \frac{2}{3}  

\displaystyle 3x-9 = 2y-6  

Substituting

\displaystyle 3 \frac{5}{7} y - 9 = 2y-6  

\displaystyle 15y-63 = 14y -42  

\displaystyle y = 21  

\displaystyle \text{and } x = \frac{5}{7} y = 15  

\displaystyle \text{Hence } x = 15 \ and \ y = 21  

\displaystyle \\

Question 8: If \displaystyle 15(2x^2-y^2 )=7xy , find \displaystyle x:y ; if \displaystyle x \ and \ y both are positive.

Answer:

\displaystyle \text{Given } 15(2x^2-y^2 )=7xy  

\displaystyle 30x^2-15y^2=7xy  

Dividing both sides by \displaystyle xy , we get

\displaystyle 30 \frac{x}{y} - 15 \frac{y}{x} = 7  

Let \displaystyle \frac{x}{y} = a  

\displaystyle 30a - \frac{15}{a} =7  

\displaystyle 30a^2-7a-15=0  

\displaystyle (6a-5)(5a+3)=0  

\displaystyle \Rightarrow a= \frac{5}{6} \ or \ \frac{-3}{5} ( not possible as both \displaystyle x \ and \ y are positive)

\displaystyle \text{Hence } \frac{x}{y} = \frac{5}{6} \ or\ x:y=5:6  

\displaystyle \\

Question 9: Find the:

i) Fourth Proportional to \displaystyle 2xy, x^2 \ and \ y^2 .

ii) Third proportional to \displaystyle a^2-b^2 \ and \ a+b  

iii) Mean proportional to \displaystyle (x-y) \ and \ (x^3-x^2 y)  

Answer:

i) Fourth Proportional to \displaystyle 2xy, x^2 \ and \ y^2 .

Let the fourth proportional be \displaystyle a  

Therefore \displaystyle 2xy: x^2 = y^2 : a  

 \displaystyle \Rightarrow a = \frac{y^2. x^2}{2xy} = \frac{xy}{2}  

ii) Third proportional to \displaystyle a^2-b^2 \ and \ a+b  

Let the third proportional be \displaystyle x  

Therefore \displaystyle (a^2-b^2): (a+b) = (a+b) : x  

 \displaystyle \Rightarrow \frac{a^2-b^2}{a+b} = \frac{a+b}{x}  

Simplifying \displaystyle x = \frac{a+b}{a-b}  

iii) Mean proportional to \displaystyle (x-y) \ and \ (x^3-x^2 y)  

Let \displaystyle a be the mean proportion.

Therefore \displaystyle (x-y) : a = a: (x^3-x^2 y)  

\displaystyle \Rightarrow \frac{x-y}{a} = \frac{a}{x^3-x^2 y}  

Simplifying

\displaystyle a = x(x-y)  

\displaystyle \\

Question 10: Find two numbers such that the mean proportional between them is \displaystyle 14 and third proportional to them is \displaystyle 112 .

Answer:

Let the two numbers be \displaystyle a \ and \ b .

Given, mean proportional between them is \displaystyle 14  

Therefore \displaystyle a:14=14:b  

\displaystyle \text{or } ab = 196 … … … … … … … … i)

Also given that third proportional to them is \displaystyle 112  

Therefore \displaystyle a:b = b: 112  

\displaystyle \text{or } b^2 = 112a … … … … … … … … ii)

Solving i) and ii)

\displaystyle b^2 = 112 \Big( \frac{196}{b} \Big)  

\displaystyle b^3 = 112. 196 \ or \ b = 28  

Substituting back in i),

\displaystyle a = \frac{196}{28} = 7  

Hence \displaystyle a = 7 \ and \ b = 28  

\displaystyle \\

Question 11: If \displaystyle x \ and \ y be unequal \displaystyle \text{and } x:y is the \displaystyle \text{Duplicate ratio of } x+z \ and \ y+z , prove that \displaystyle z is mean proportional between \displaystyle x \ and \ y .

Answer:

Given

\displaystyle \frac{x}{y} = \frac{(x+z)^2}{(y+z)^2}  

\displaystyle x(y^2+2yz+z^2) = y(x^2+2xz+z^2)  

\displaystyle xy^2+2xyz+xz^2 = yx^2+2xyz+yz^2  

\displaystyle xy(y-x)=z^2(y-x)  

\displaystyle xy=z^2  

\displaystyle \text{or } x:z=z:y  

\displaystyle \\

Question 12: If \displaystyle q is the proportional between \displaystyle p \ and \ r prove that:

\displaystyle \frac{p^3+q^3+r^3}{p^2 q^2 r^2 } = \frac{1}{p^3} + \frac{1}{q^3} + \frac{1}{r^3}  

Answer:

\displaystyle \text{Given } q is the proportional between \displaystyle p \ and \ r  

\displaystyle \text{or } q^2 = pr  

\displaystyle \text{LHS } = \frac{p^3+q^3+r^3}{p^2 q^2 r^2 }  

Substituting \displaystyle q^2 = pr  

\displaystyle = \frac{p^3+pqr+r^3}{p^2 q^2 r^2 }  

\displaystyle = \frac{1}{r^3}+ \frac{q}{p^2r^2} + \frac{1}{p^3}  

\displaystyle = \frac{1}{p^3} + \frac{1}{q^3} + \frac{1}{r^3} = RHS

Hence proved.

\displaystyle \\

Question 13: If \displaystyle a, b \ and \ c are in continued proportion. Prove that: \displaystyle a:c=(a^2+b^2 ):(b^2+c^2)  

Answer:

\displaystyle \text{Given } a:b=b:c  

\displaystyle \text{or } \frac{a}{b} = \frac{b}{c} = k  

Therefore \displaystyle a = bk \ and \ b= ck  

\displaystyle \text{Now, LHS } = \frac{a}{c}= \frac{ck^2}{c} =k^2  

\displaystyle \text{RHS } = \frac{a^2+b^2}{b^2+c^2}  

 \displaystyle = \frac{(bk)^2+b^2}{(ck)^2+c^2}  

 \displaystyle = \frac{(ckk)^2+(ck)^2}{(ck)^2+c^2}  

 \displaystyle = \frac{c^2k^2(k^2+1)}{c^2(k^2+1)}  

 \displaystyle =k^2  

LHS = RHS

\displaystyle \\

\displaystyle \text{Question 14: If } x= \frac{2ab}{a+b} \text{ , find the value of: } \frac{x+a}{x-a} + \frac{x+b}{x-b} .

Answer:

\displaystyle \text{Given } x= \frac{2ab}{a+b}  

Therefore

\displaystyle \frac{x}{a} = \frac{2b}{a+b} \ and \ \frac{x}{b} = \frac{2a}{a+b}  

Applying componendo and dividendo

\displaystyle \frac{x+a}{x-a} = \frac{2b+a+b}{2b-a-b}  

\displaystyle \Rightarrow \frac{x+a}{x-a} = \frac{3b+a}{b-a}  

Similarly Applying componendo and dividendo

\displaystyle \frac{x+b}{x-b} = \frac{2a+a+b}{2a-a-b}  

\displaystyle \Rightarrow \frac{x+b}{x-b} = \frac{3a+b}{a-b}  

\displaystyle \text{Adding } \frac{x+a}{x-a} + \frac{x+b}{x-b} = \frac{3b+a}{b-a} + \frac{3a+b}{a-b}  

\displaystyle \text{or } \frac{x+a}{x-a} + \frac{x+b}{x-b} = \frac{2a-2b}{a-b} =2  

\displaystyle \\

Question 15: If \displaystyle (4a+9b)(4c-9d)=(4a-9b)(4c+9d) , prove that: \displaystyle a:b=c:d .

Answer:

\displaystyle \text{Given } (4a+9b)(4c-9d)=(4a-9b)(4c+9d)  

\displaystyle \text{or } \frac{(4a+9b)}{(4a-9b)} = \frac{(4c+9d)}{(4c-9d)}  

Applying componendo and dividendo

\displaystyle \frac{(4a+9b)+(4a-9b)}{(4a+9b)-(4a-9b)} = \frac{(4c+9d)+(4c-9d)}{(4c+9d)-(4c-9d)}  

\displaystyle \frac{8a)}{18b} = \frac{8c}{18d}  

\displaystyle \text{or } a:b=c:d  

\displaystyle \\

\displaystyle \text{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}  

Answer:

\displaystyle \text{Given } \frac{a}{b} = \frac{c}{d} = k  

therefore \displaystyle a = bk \ and \ c = dk  

\displaystyle \text{LHS } = \frac{a+b}{c+d}  

\displaystyle = \frac{bk+b}{dk+d} = \frac{b}{d}  

\displaystyle \text{RHS } = \frac{\sqrt{a^2+b^2 }}{\sqrt{c^2+d^2}}  

\displaystyle = \frac{\sqrt{(bk)^2+b^2 }}{\sqrt{(dk)^2+d^2}}  

\displaystyle = \frac{\sqrt{b^2 }}{\sqrt{d^2}} = \frac{b}{d}  

Hence, LHS = RHS

\displaystyle \\

\displaystyle \text{Question 17: If } \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \text{ , prove that: } \frac{ax-by}{(a+b)(x-y)} + \frac{by-cz}{(b+c)(y-z)} + \frac{cz-ax}{(c+a)(z-x)}=3  

Answer:

\displaystyle \text{Given } \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k  

Therefore

\displaystyle x = ak \ and \ y=bk \ and \ 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^2-b^2)} + \frac{k(b^2-c^2)}{k(b^2-c^2)} + \frac{k(c^2-a^2)}{k(c^2-a^2)} = 3  

Hence LHS = RHS

\displaystyle \\

Question 18: There are \displaystyle 36 members in a student’s council in a school and the ratio of the number of boys to the numbers of girls is \displaystyle 3:1 . How many more girls should be added to the council so that the ratio of the number of boys to the number of girls may be \displaystyle 9:5 .

Answer:

Let the number of girls \displaystyle = x  

The the number of boys \displaystyle = 3x  

Therefore \displaystyle 3x+x = 36 \Rightarrow x = 9  

Therefore Boys \displaystyle = 27 and Girls \displaystyle = 9  

Let n girls be added

\displaystyle \text{Therefore } \frac{27}{9+n} = \frac{9}{5}  

\displaystyle \text{or } 135 = 81 +9n  

\displaystyle \text{or } n = 6  

Hence \displaystyle 6 more girls should be added.

\displaystyle \\

\displaystyle \text{Question 19: If } \frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b} \text{ , prove that: } ax+by+cz=0  

Answer:

\displaystyle \text{Given } \frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b} = k  

\displaystyle x = k(b-c) \ and \ y = k(c-a) \ and \ 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 \\

Question 20: If \displaystyle 7x-15y=4x+y , find the value of \displaystyle x:y . hence, use componendo and dividendo to find the value of:

\displaystyle \text{i) } \frac{9x+5y}{9x-5y}       \displaystyle \text{ii) } \frac{3x^2+2y^2}{3x^2-2y^2}  

Answer:

\displaystyle \text{Given } 7x-15y=4x+y  

\displaystyle 3x = 16y  

\displaystyle \text{or } \frac{x}{y} = \frac{16}{3}  

\displaystyle \text{i) } \frac{9x+5y}{9x-5y}  

Divide Numerator and Denominator by \displaystyle y , we get

\displaystyle = \frac{9 \times \frac{x}{y}+5}{9 \times \frac{x}{y}-5}  

\displaystyle = \frac{9 \times \frac{16}{3}+5}{9 \times \frac{16}{3}-5}  

\displaystyle = \frac{53}{43}  

\displaystyle \text{ii) } \frac{3x^2+2y^2}{3x^2-2y^2}  

Divide Numerator and Denominator by \displaystyle y^2 , we get

\displaystyle = \frac{3 \times (\frac{x}{y})^2+2}{3 \times (\frac{x}{y})^2-2}  

\displaystyle = \frac{262}{250}=\frac{131}{125}  

\displaystyle \\

\displaystyle \text{Question 21: If } \frac{4m+3n}{4m-3n} = \frac{7}{4} \text{ , use properties of proportion to find: }

i) \displaystyle m:n            \displaystyle \text{ii) } \frac{2m^2-11n^2}{2m^2+11n^2 } .

Answer:

\displaystyle \text{i)  Given } \frac{4m+3n}{4m-3n} = \frac{7}{4}  

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{m}{n} = \frac{11 \times 6}{3 \times 8} = \frac{11}{4}  

\displaystyle \text{ii)  } \frac{2m^2-11n^2}{2m^2+11n^2 }  

Applying componendo and dividendo

\displaystyle = \frac{2m^2-11n^2+2m^2+11n^2}{2m^2-11n^2-2m^2-11n^2 }  

\displaystyle = \frac{4m^2}{-22n^2 }  

\displaystyle = \frac{4}{-22} \Big( \frac{m}{n} \Big)^2 = \frac{4}{-22} \Big( \frac{11}{4} \Big)^2 = - \frac{11}{8}  

\displaystyle \\

\displaystyle \text{Question 22: If } x, y \ and \ z \text{ are in continued proportion, prove that: } \\ \\ \frac{(x+y)^2}{(y+z)^2} = \frac{x}{y} \text{ . [2010] }

Answer:

If \displaystyle x, y \ 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} =\Big( \frac{x-y}{y-z} \Big)^2  

\displaystyle \Rightarrow \frac{(x+y)^2}{(y+z)^2} = \Big( \frac{x-y}{y-z} \Big)^2  

Substituting

\displaystyle \Rightarrow \frac{(x+y)^2}{(y+z)^2} = \Big( \frac{\frac{y^2}{z}-y}{y-z} \Big)^2  

\displaystyle \Rightarrow \frac{(x+y)^2}{(y+z)^2} = \Big( \frac{y^2-yz}{z(y-z)} \Big)^2= \frac{y^2}{z^2} = \frac{zx}{z^2} = \frac{x}{z}  

\displaystyle \\

\displaystyle \text{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 }} \\ \\ \text{ Use componendo and dividendo to prove that: } x^2= \frac{2a^2 x}{x^2+1} \text{. [2010]}  

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 \\

\displaystyle \text{Question 24: 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 } [2014]

Answer:

\displaystyle \text{i)  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}  

\displaystyle \\

Question 25: 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{ [2011] }  

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 \\

\displaystyle \text{Question 26: 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 [2012]

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