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

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 $\displaystyle x = 7$

$\displaystyle \\$

$\displaystyle \text{Question 26: If } x=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}} \text{, using properties of proportion show that: } \\ \\ x^2-2ax+1 [2012]$