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

Given $a:b=3:5$

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

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

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

Given $5x+6y:8x+5y=8:9$

$9(5x+6y)=8(8x+5y)$

$45x +54y=64x+40y$

$14y = 19x$

$\Rightarrow x:y = 14:19$

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

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

$(3x-4y)(4x-5y) = (5x-6y)(2x-3y)$

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

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

$2x^2-4xy+2y^2 = 0$

$x^2-2xy+y^2 = 0$

$(x-y)^2 = 0$

or $x-y = 0$

or $x:y = 1:1$

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

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Question 4: Find the:

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

ii) Triplicate ratio of $2a:3b$

iii) Sub-duplicate ratio of $9x^2 a^4:25y^6 b^2$

iv) Sub-triplicate ratio of $216:343$

v) Reciprocal ratio of $3:5$

vi) Ratio compounded of the duplicate ratio of $5:6$ , the reciprocal ratio of $25:42$ and the sub-duplicate ratio of $36:49$ .

i) Duplicate ratio of $2\sqrt{2}:3\sqrt{5} = (2\sqrt{2})^2:(3\sqrt{5})^2 = 8:45$

ii) Triplicate ratio of $2a:3b = (2a)^3:(3b)^3 = 8a^3:81b^3$

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

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

v) Reciprocal ratio of $3:5 = 5:3$

vi) Ratio compounded of the duplicate ratio of $5:6$ , the reciprocal ratio of $25:42$ and the sub-duplicate ratio of $36:49$.

Duplicate ratio of $5:6 = 25:36$

Reciprocal ratio of $25:42 = 42:25$

Sub-duplicate ratio of $36:49 = 6:7$

Compound ratio $\frac{25}{36} \times \frac{42}{25} \times \frac{6}{7} = \frac{1}{1} or 1:1$

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Question 5: Find the value of $x$, if:

i) $(2x+3):(5x-38)$ is the duplicate ratio of $\sqrt{5}:\sqrt{6}$ .

ii) $(2x+1):(3x+13)$ is the sub-duplicate ratio of $9:25$.

iii) $(3x-7):(4x+3)$ is the sub-triplicate ratio of $8:27$.

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

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

$12x+18 = 25x-190$

$13x = 208 \Rightarrow x = 16$

ii) $(2x+1):(3x+13)$ is the sub-duplicate ratio of $9:25$.

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

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

$10x+5 = 9x+39$

$x = 34$

iii) $(3x-7):(4x+3)$ is the sub-triplicate ratio of $8:27$

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

$9x-21 = 8x+6$

or $x = 27$

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

Let us add quantity $a$

Therefore

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

$dx+da = cy+ca$

$(d-c)a = cy-dx$

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

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

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

Given

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

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

Therefore if  $3$ is subtracted from each of them, then

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

$3x-9 = 2y-6$

Substituting

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

$15y-63 = 14y -42$

$y = 21$

and $x = \frac{5}{7} y = 15$

Hence $x = 15 and y = 21$

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

Given $15(2x^2-y^2 )=7xy$

$30x^2-15y^2=7xy$

Dividing both sides by  $xy$, we get

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

Let  $\frac{x}{y} = a$

$30a -\frac{15}{a}=7$

$30a^2-7a-15=0$

$(6a-5)(5a+3)=0$

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

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

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Question 9: Find the:

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

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

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

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

Let the fourth proportional be $a$

Therefore $2xy: x^2 = y^2 : a$

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

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

Let the third proportional be $x$

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

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

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

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

Let $a$ be the mean proportion.

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

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

Simplifying

$a = x(x-y)$

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

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

Given, mean proportional between them is $14$

Therefore $a:14=14:b$

or $ab = 196$ … … … … … … … … i)

Also given that third proportional to them is $112$

Therefore $a:b = b: 112$

or $b^2 = 112a$  … … … … … … … … ii)

Solving i) and ii)

$b^2 = 112. (\frac{196}{b})$

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

Substituting back in i),

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

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

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Question 11: If $x \ and \ y$ be unequal and $x:y$ is the duplicate ratio of $x+z \ and \ y+z$, prove that $z$ is mean proportional between $x \ and \ y$.

Given

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

$x(y^2+2yz+z^2) = y(x^2+2xz+z^2)$

$xy^2+2xyz+xz^2 = yx^2+2xyz+yz^2$

$xy(y-x)=z^2(y-x)$

$xy=z^2$

or $x:z=z:y$

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Question 12: If $q$ is the proportional between $p \ and \ r$ prove that: $\frac{p^3+q^3+r^3}{p^2 q^2 r^2 }=\frac{1}{p^3} +\frac{1}{q^3} +\frac{1}{r^3}$

Given $q$ is the proportional between $p \ and \ r$

or $q^2 = pr$

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

Substituting  $q^2 = pr$

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

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

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

Hence proved.

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

Given $a:b=b:c$

or $\frac{a}{b}=\frac{b}{c}= k$

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

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

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

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

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

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

$=k^2$

LHS = RHS

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

Given $x=\frac{2ab}{a+b}$

Therefore

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

Applying componendo and dividendo

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

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

Similarly Applying componendo and dividendo

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

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

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

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

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

Given $(4a+9b)(4c-9d)=(4a-9b)(4c+9d)$

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

Applying componendo and dividendo

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

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

or $a:b=c:d$

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Question 16: If $\frac{a}{b}=\frac{c}{d}$, show that $(a+b):(c+d)=\sqrt{a^2+b^2 }:\sqrt{c^2+d^2}$

Given $\frac{a}{b}=\frac{c}{d} = k$

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

LHS $= \frac{a+b}{c+d}$

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

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

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

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

Hence, LHS = RHS

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

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

Therefore

$x = ak \ and \ y=bk \ and \ z=ck$

LHS  $= \frac{ax-by}{(a+b)(x-y)}+\frac{by-cz}{(b+c)(y-z)}+\frac{cz-ax}{(c+a)(z-x)}$

$= \frac{a(ak)-b(bk)}{(a+b)(ak-bk)}+\frac{b(bk)-c(ck)}{(b+c)(bk-ck)}+\frac{c(ck)-a(ak)}{(c+a)(ck-ak)}$

$= \frac{k(a^2-b^2)}{k(a^2-b^2)}+\frac{k(b^2-c^2)}{k(b^2-c^2)}+\frac{k(c^2-a^2)}{k(c^2-a^2)} = 3$

Hence LHS = RHS

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

Let the number of girls $= x$

The the number of boys $= 3x$

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

Therefore Boys $= 27$  and Girls $= 9$

Therefore $\frac{27}{9+n} = \frac{9}{5}$

or $135 = 81 +9n$

or $n = 6$

Hence $6$ more girls should be added.

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

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

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

LHS $= ax+by+cz$

$= ak(b-c)+bk(c-a) + ck(a-b)$

$=k(ab-ac+bc-ab+ca-bc) = 0$

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

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

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

Given $7x-15y=4x+y$

$3x = 16y$

or $\frac{x}{y}=\frac{16}{3}$

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

Divide Numerator and Denominator by $y$, we get

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

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

$= \frac{53}{43}$

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

Divide Numerator and Denominator by $y^2$, we get

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

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

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

i) $m:n$

ii) $\frac{2m^2-11n^2}{2m^2+11n^2 }$.

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

Applying componendo and dividendo

$= \frac{4m+3n+4m-3n}{4m+3n-4m+3n}=\frac{7+4}{7-4}$

$= \frac{8m}{6n}=\frac{11}{3}$

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

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

Applying componendo and dividendo

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

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

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

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

If $x, y \ and \ z$ are in continued proportion, then

$\frac{x}{y}=\frac{y}{z} \Rightarrow x = \frac{y^2}{z}$

Applying componendo and dividendo

$\frac{x+y}{x-y}=\frac{y+z}{y-z}$

$\Rightarrow \frac{x+y}{y+z}=\frac{x-y}{y-z}$

Squaring both sides

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

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

Substituting

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

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

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Question 23: Given $x=\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 }}$ . Use componendo and dividendo to prove that: $x^2=\frac{2a^2 x}{x^2+1}$.   [2010]

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

$\frac{x+1}{x-1}=\frac{(\sqrt{a^2+b^2}+\sqrt{a^2-b^2})+(\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 })}{(\sqrt{a^2+b^2}+\sqrt{a^2-b^2})-(\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 })}$

Simplifying

$\frac{x+1}{x-1}=\frac{\sqrt{a^2+b^2}}{\sqrt{a^2-b^2 }}$

Square both sides

$\frac{x^2+1+2x}{x^2-2x+1}=\frac{a^2+b^2}{a^2-b^2}$

Applying componendo and dividendo

$\frac{x^2+1+2x+x^2-2x+1}{x^2+1+2x-x^2+2x-1}=\frac{a^2+b^2+a^2-b^2}{a^2+b^2-a^2+b^2}$

$\frac{2(x^2+1)}{4x}=\frac{2a^2}{2b^2}$

$\frac{x^2+1}{2x}=\frac{a^2}{b^2}$

Simplifying

$b^2 = \frac{2a^2x}{x^2+1}$

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Question 24: If $\frac{x^2+y^2}{x^2-y^2}=2 \frac{1}{8}$, find:

i) $\frac{x}{y}$

ii) $\frac{x^3+y^3}{x^3-y^3 }$    [2014]

i) Given $\frac{x^2+y^2}{x^2-y^2}=2 \frac{1}{8} = \frac{17}{8}$

Applying componendo and dividendo

$\frac{x^2+y^2+x^2-y^2}{x^2+y^2-x^2+y^2} = \frac{17+8}{17-8}$

$\frac{2x^2}{2y^2} = \frac{25}{9}$

Simplifying, we get

$\frac{x}{y} = \frac{5}{3}$

ii) $\frac{x^3+y^3}{x^3-y^3 }$

Applying componendo and dividendo

$\frac{x^3+y^3+x^3-y^3 }{x^3+y^3-x^3+y^3 }$

$\frac{2x^3}{2y^3} = (\frac{x}{y})^3 = (\frac{5}{3})^3 = \frac{125}{9}$

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Question 25: Using componendo and dividendo, find the value of $x$$\frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}}=9$    [2011]

Given $\frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}}=9$

Applying componendo and dividendo

$\frac{(\sqrt{3x+4}+\sqrt{3x-5})+(\sqrt{3x+4}-\sqrt{3x-5})}{(\sqrt{3x+4}+\sqrt{3x-5})-(\sqrt{3x+4}-\sqrt{3x-5})}=\frac{9+1}{9-1}$

$\frac{2\sqrt{3x+4}}{2\sqrt{3x-5}}=\frac{10}{8}$

Simplifying

$\frac{\sqrt{3x+4}}{\sqrt{3x-5}}=\frac{5}{4}$

Square both sides

$\frac{3x+4}{3x-5} = \frac{25}{14}$

$42x+56 = 75x-125$

Simplifying we get $x = 7$

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

Given $x=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}$

Applying componendo and dividendo

$\frac{x+1}{x-1}=\frac{(\sqrt{a+1}+\sqrt{a-1})+(\sqrt{a+1}-\sqrt{a-1})}{(\sqrt{a+1}+\sqrt{a-1})-(\sqrt{a+1}-\sqrt{a-1})}$

Simplify

$\frac{x+1}{x-1}=\frac{\sqrt{a+1}}{\sqrt{a-1}}$

Now square both sides

$\frac{x^2+1+2x}{x^2-2x+1}=\frac{a+1}{a-1}$

Simplifying

$x^2+1 = 2ax$

or $x^2-2ax+1 = 0$

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