ML Aggarwal Solution Class 10 Chapter 7 Ratio and Proportion Exercise 7.3

 Exercise 7.3

Question 1

If a : b : : c : d, prove that

(i) $\frac{2 a+5 b}{2 a-5 b}=\frac{2 c+5 d}{2 c-5 d}$

(ii) $\frac{5 a+11 b}{5 c+11 d}=\frac{5 a-11 b}{5 c-11 d}$

(iii) (2a + 3b)(2c – 3d) = (2a – 3b)(2c + 3d)

(iv) (la + mb) : (lc + mb) :: (la – mb) : (lc – mb)
Sol :
(i) a : b : : c : d
then $\frac{a}{b}=\frac{c}{d}$
$\Rightarrow \frac{2 a}{5 b}=\frac{2 c}{5 d}$ (multiply by $\left.\frac{2}{5}\right)$

Applying componendo and dividendo,

$\frac{2 a+5 b}{2 a-5 b}=\frac{2 c+5 d}{2 c-5 d}$


(ii) ∵ a: b:: c: d

$\therefore \frac{a}{b}=\frac{c}{d} \Rightarrow \frac{5 a}{11 b}=\frac{5 c}{11 d}\left(\right.$ Multiply by $\left.\frac{5}{11}\right)$

Applying componendo and dividendo,

$\frac{5 a+11 b}{5 a-11 b}=\frac{5 c+11 d}{5 c-11 d}$

$\Rightarrow \frac{5 a+11 b}{5 c+11 d}=\frac{5 a-11 b}{5 c-11 d}$ (Applying alternendo)


(iii) ∵ a: b:: c: d

$\therefore \frac{a}{b}=\frac{c}{d} \Rightarrow \frac{2 a}{3 b}=\frac{2 c}{3 d}\left(\right.$ Multiply by $\left.\frac{2}{3}\right)$

Applying componendo and dividendo,

$\frac{2 a+3 b}{2 a-3 b}=\frac{2 c+3 d}{2 c-3 d}$

⇒(2 a+3 b)(2 c-3 d)

=(2 a-3 b)(2 c+3 d) (By cross multiplication)


(iv) ∵ a: b:: c: d

$\therefore \frac{a}{b}=\frac{c}{d}$

$\Rightarrow \quad \frac{l a}{m b}=\frac{l c}{m d}\left(\right.$ Multiply by $\left.\frac{l}{m}\right)$

Applying componendo and dividendo,

$\frac{l a+m b}{l a-m b}=\frac{l c+m d}{l c-m d}$

$\Rightarrow \quad \frac{l a+m b}{l c+m d}=\frac{l a-m b}{l c-m d} \quad$ (By alternendo)

⇒(l a+m b):(l c+m d)::(l a-m b) :(l c-m d)


Question 2

(i) If $\frac{5 x+7 y}{5 u+7 y}=\frac{5 x-7 y}{5 u-7 y},$ Show that $\frac{x}{y}=\frac{u}{v}$

(ii) $\frac{8 a-5 b}{8 c-5 d}=\frac{8 a+5 b}{8 c+5 d},$ prove that $\frac{a}{b}=\frac{c}{d}$

Sol :

(i) $\frac{5 x+7 y}{5 x+7 y}=\frac{5 x-7 y}{5 x-7 y}$

Applying alternendo $\frac{5 x+7 y}{52+7 y}=\frac{5 x-7 y}{54-7}$

Applying componendo and dividendo

$\frac{5 x+7 y+5 x-7 y}{5 x+7 y-5 x+7 y}=\frac{5 u+7 v+5 u-7 v}{5 u+7 u-5 u+7 v}$

$\Rightarrow \frac{10 x}{14 y}=\frac{10 u}{14 v} \Rightarrow \frac{x}{y}=\frac{u}{v}$

Hence proved. ( Dividing by $\left.=\frac{10}{14}\right)$


(ii) $\frac{8 a-5 b}{8 c-5 d}=\frac{8 a+5 b}{8 c+5 d}$

$\Rightarrow \frac{8 a+5 b}{8 a-5 b}=\frac{8 c+5 d}{8 c-5 d} \quad$ (using alternendo)

Applying compounde and dividendo,

$\frac{8 a+5 b+8 a-5 b}{8 a+5 b-8 a+5 b}=\frac{8 c+5 d+8 c-5 c}{8 c+5 d-8 c+5 d}$

$\therefore \frac{16 a}{10 b}=\frac{16 c}{10 d}$

$\Rightarrow \frac{a}{b}=\frac{c}{d}$

(Dividing by $\left.\frac{16}{10}\right)$

Hence proved.


Question 3

If (4a + 5b) (4c – 5d) = (4a – 5d) (4c + 5d), prove that a, b, c, d are in proporton.

Sol :

(4a + 5b) (4c – 5d) = (4a – 5d) (4c + 5d)

$\Rightarrow \frac{4 a+5 b}{4 a-5 b}=\frac{4 c+5 d}{4 c-5 d}$

Applying componendo and dividendo

$\frac{4 a+5 b+4 a-5 b}{4 a+5 b-4 a+5 b}=\frac{4 c+5 d+4 c-5 d}{4 c+5 d-4 c+5 d}$
$\Rightarrow \frac{8 a}{10 b}=\frac{8 c}{10 d} \Rightarrow \frac{a}{b}=\frac{c}{d}$

Hence, a,b,c,d are in proportion


Question 4

If (pa + qb) : (pc + qd) :: (pa – qb) : (pc – qd) prove that a : b : : c : d

Sol :

(pa + qb) : (pc + qd) :: (pa – qb) : (pc – qd)

$\Rightarrow \frac{p \alpha+q b}{p c+q d}=\frac{p q-q b}{p c-q d}$
$\Rightarrow \frac{p a+q b}{p c-q d}=\frac{p q+q b}{p c-q d}$

Applying componendo and dividendo

$\Rightarrow \frac{p a+q b+p \alpha-q b}{p a+q b-p \alpha+q b}=\frac{p c+q d+p c-q d}{p c-q d-p c+q d}$

$\Rightarrow \frac{2 p a}{2 q b}=\frac{2 p c}{2 q d}$

$\Rightarrow \frac{a}{b}=\frac{c}{d}$ $\left(\right.$ Dividing by $\left.\frac{2 p}{2 q}\right)$

hence a : b:: c : d


Question 5

If (ma + nb): b :: (mc + nd) : d, prove that a, b, c, d are in proportion.

Sol :

(ma + nb): b :: (mc + nd) : d

$\Rightarrow \frac{m a+n b}{b}=\frac{m c+n d}{d}$

⇒ mad + nbd = mbc + nbd

⇒ mad = mbc

⇒ ad = bc

$\Rightarrow \frac{a}{b}=\frac{c}{d}$

Hence a : b :: c : d.


Question 6

If (11a² + 13b²) (11c² – 13d²) = (11a² – 13b²)(11c² + 13d²), prove that a : b :: c : d.

Sol :

(11a² + 13b²) (11c² – 13d²) = (11a² – 13b²)(11c² + 13d²)

$\Rightarrow \frac{11 a+13 b^{2}}{11 a^{2}-13 b^{2}}=\frac{11 c^{2}+13 d^{2}}{11 c^{2}-13 d^{2}}$

Applying componendo and dividendo

$\frac{11 a^{2}+13 b^{2}+11 a^{2}-13 b^{2}}{11 a^{2}+13 b^{2}-11 a^{2}+13 b^{2}}=\frac{11 c^{2}+13 d^{2}+11 c^{2}-13 d^{2}}{11 c^{2}+13 d^{2}-11 c^{2}+13 d^{2}}$

$\Rightarrow \frac{22 a^{2}}{26 b^{2}}=\frac{22 c^{2}}{26 d^{2}}$

$\Rightarrow \frac{a^{2}}{b^{2}}=\frac{c^{2}}{d^{2}}$ $\left(\right.$ Dividing by $\left.\frac{22}{26}\right)$

$\Rightarrow \frac{a}{b}=\frac{c}{d}$

hence a:b::c:d


Question 7

If (a + 3b + 2c + 6d) (a – 3b – 2c + 6d) = (a + 3b – 2c – 6d) (a – 3b + 2c – 6d), prove that a : b :: c : d.

Sol :

$\frac{a+3 b+2 c+6 d}{a-3 b+2 c-6 d}=\frac{a+3 b-2 c-6 d}{a-3 b-2 c+6 d}$

$\Rightarrow \frac{a+3 b+2 c+6 d}{a+3 b-2 c-6 d}=\frac{a-3 b+2 c-6 d}{a-3 b-2 c+6 d}$ (by altenendo)

Applying componendo and dividendo

$\frac{a+3 b+2 c+6 d+a+3 b-2 c-6 d}{a+3 b+2 c+6 d-a-3 b+2 c+6 d}$

$=\frac{a-3 b+2 c-6 d+a-3 b-2 c+6 d}{a-3 b+2 c-6 d-a+3 b+2 c-6 d}$

$\Rightarrow \frac{2(a+3 b)}{2(2 c+6 d)}=\frac{2(a-3 b)}{2(2 c-6 d)}$ (Dividing by 2)

$\Rightarrow \frac{a+3 b}{2 c+6 d}=\frac{a-3 b}{2 c-6 d}$ (by alternendo)

Again applying componendo and dividendo)

$\frac{a+3 b+a-3 b}{a+3 b-a+3 b}=\frac{2 c+6 d+2 c-6 d}{2 c+6 d-2 c+6 d}$

$\Rightarrow \frac{2 a}{6 b}=\frac{4 c}{12 d}=\frac{2 c}{6 d}$

$\Rightarrow \frac{a}{b}=\frac{c}{d}$  $\left[\right.$ Dividing by $\left.\frac{2}{6}\right]$


Question 8

If $x=\frac{2 a b}{a+b}$ find the value of $\frac{x+a}{x-a}+\frac{x+b}{x-b}$

Sol :

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

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

Applying componendo and dividendo,

$\frac{x+a}{x-a}=\frac{2 b+a+b}{2 b-a-b}=\frac{3 b+a}{b-a}$...(i)

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

Applying componendo and dividendo,

$\frac{x+b}{x-b}=\frac{2 a+a+b}{2 a-a-b}=\frac{3 a+b}{a-b}$..(ii)

Adding (i) and (ii)

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

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

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

$=\frac{2 a-2 b}{a-b}=\frac{2(a-b)}{a-b}=2$


Question 9

If $x=\frac{8 a b}{a+b}$ find the value of $\frac{x+4 a}{x-4 a}+\frac{x+4 b}{x-4 b}$

Sol :

$x=\frac{8 a b}{a+b}$

$\Rightarrow \frac{x}{4 a}=\frac{2 b}{a+b}$

Applying componendo and dividendo,

$\frac{x+4 a}{x-4 a}=\frac{2 b+a+b}{2 b-a-b}=\frac{3 b+a}{b-a} \quad \ldots(i)$

$\operatorname{Again} \frac{x}{4 b}=\frac{2 a}{a+b}$

Applying compondndo and dividendo,

$\frac{x+4 b}{x-4 b}=\frac{2 a+a+b}{2 a-a-b}=\frac{3 a+b}{a-b}$..(ii)

Adding (i) and (ii)

$\frac{x+4 a}{x-4 a}+\frac{x+4 b}{x-4 b}=\frac{3 b+a}{b-a}+\frac{3 a+b}{a-b}$

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

$=\frac{-a-3 b+3 a+b}{a-b}=\frac{2 a-2 b}{a-b}$

$=\frac{2(a-b)}{a-b}=2$


Question 10

If $x=\frac{4 \sqrt{6}}{\sqrt{2}+\sqrt{3}}$ find the value of $\frac{x+2 \sqrt{2}}{x-2 \sqrt{2}}+\frac{x+2 \sqrt{3}}{x-2 \sqrt{3}}$

Sol :

$x=\frac{4 \sqrt{6}}{\sqrt{2}+\sqrt{3}}$

$\Rightarrow \frac{4 \sqrt{2} \times \sqrt{3}}{\sqrt{2}+\sqrt{3}}$

$\frac{x}{2 \sqrt{2}}=\frac{2 \sqrt{3}}{\sqrt{2}+\sqrt{3}}$

Applying componendo and dividendo,

$\frac{x+2 \sqrt{2}}{x-2 \sqrt{2}}=\frac{2 \sqrt{3}+\sqrt{2}+\sqrt{3}}{2 \sqrt{3}-\sqrt{2}-\sqrt{3}}$

$=\frac{3 \sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$..(i)

Again $\frac{x}{2 \sqrt{3}}=\frac{2 \sqrt{2}}{\sqrt{2}+\sqrt{3}}$

Applying componendo and dividendo,

$\frac{x+2 \sqrt{3}}{x-2 \sqrt{3}}=\frac{2 \sqrt{2}+\sqrt{2}+\sqrt{3}}{2 \sqrt{2}-\sqrt{2}-\sqrt{3}}$

$=\frac{3 \sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}$..(ii)

Adding (i) and (ii)

$\frac{x+2 \sqrt{2}}{x-2 \sqrt{2}}+\frac{x+2 \sqrt{3}}{x-2 \sqrt{3}}$

$=\frac{3 \sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{3 \sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}$

$=\frac{3 \sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{3 \sqrt{2}+\sqrt{3}}{\sqrt{3}-\sqrt{2}}$

$=\frac{3 \sqrt{3}+\sqrt{2}-3 \sqrt{2}-\sqrt{3}}{\sqrt{3}-\sqrt{2}}$

$=\frac{2 \sqrt{3}-2 \sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{2(\sqrt{3}-\sqrt{2})}{\sqrt{3}-\sqrt{2}}=2$


Question 11

Solve $x: \frac{\sqrt{36 x+1}+6 \sqrt{x}}{\sqrt{36 x+1}-6 \sqrt{x}}=9$

Sol :

$\frac{\sqrt{36 x+1}+6 \sqrt{x}}{\sqrt{36 x+1}-6 \sqrt{x}}=\frac{9}{1}$

Applying componendo and dividendo,

$\frac{\sqrt{36 x+1}+6 \sqrt{x}+\sqrt{36 x+1}-6 \sqrt{x}}{\sqrt{36 x+1}+6 \sqrt{x}-\sqrt{36 x-1}+6 \sqrt{x}}$
$=\frac{9+1}{9-1}$
$ \Rightarrow \frac{2 \sqrt{36 x+1}}{12 \sqrt{x}}=\frac{10}{8}$
$\Rightarrow \frac{\sqrt{36 x+1}}{6 \sqrt{x}}=\frac{5}{4}$ (Squaring both sides)
$\frac{36 x+1}{36 x}=\frac{25}{16}$
$\Rightarrow  36 x \times 25=16(36 x+1)$

$\Rightarrow 900 x=576 x+16$

$ \Rightarrow 900 x-576 x=16$

$\Rightarrow 324 x=16$

$\therefore x=\frac{16}{324}=\frac{4}{81}$


Question 12

Find x from the following equations :

(i) $\frac{\sqrt{2-x}+\sqrt{2+x}}{\sqrt{2-x}-\sqrt{2+x}}=3$
(ii) $\frac{\sqrt{x+4}+\sqrt{x-10}}{\sqrt{x+4}-\sqrt{x-10}}=\frac{5}{2}$
(iii) $\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{a}{b}$
(iv) $\frac{\sqrt{12 x+1}+\sqrt{2 x-3}}{\sqrt{12 x+1}-\sqrt{2 x-3}}=\frac{3}{2}$
(v) $\frac{3 x+\sqrt{9 x^{2}-5}}{3 x-\sqrt{9 x^{2}-5}}=5$
(vi) $\frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}=\frac{c}{d}$
Sol :
(i) $\frac{\sqrt{2-x}+\sqrt{2+x}}{\sqrt{2-x}-\sqrt{2+x}}=3$

Applying componendo and dividendo,

$\frac{\sqrt{2-x}+\sqrt{2+x}+\sqrt{2-x}-\sqrt{2+x}}{\sqrt{2-x}+\sqrt{2+x}-\sqrt{2-x}+\sqrt{2+x}}=\frac{3+1}{3-1}$

$\Rightarrow \frac{2 \sqrt{2-x}}{2 \sqrt{2+x}}=\frac{4}{2} \Rightarrow \frac{\sqrt{2-x}}{\sqrt{2+x}}=\frac{2}{1}$

Squaring both sides

$\frac{2-x}{2+x}=\frac{4}{1} \Rightarrow 8+4 x=2-x$

$4 x+x=2-8 \Rightarrow 5 x=-6$

$\therefore x=\frac{-6}{5}$ Ans.

(ii) $\frac{\sqrt{x+4}+\sqrt{x-10}}{\sqrt{x+4}-\sqrt{x-10}}=\frac{5}{2}$

Applying componendo and dividendo,

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

$\Rightarrow \quad \frac{2 \sqrt{x+4}}{2 \sqrt{x-10}}=\frac{7}{3} \Rightarrow \frac{\sqrt{x+4}}{\sqrt{x-10}}=\frac{7}{3}$

Squaring both sides,

$\frac{x+4}{x-10}=\frac{49}{9}$

$ \Rightarrow 49 x-490=9 x+36$

$\Rightarrow 49 x-9 x=36+490 \Rightarrow \quad 40 x=526$

$\therefore x=\frac{526}{40}=\frac{263}{20}$


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

Applying componendo and dividendo,

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

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

Squaring both sides, $\frac{1+x}{1-x}=\frac{(a+b)^{2}}{(a-b)^{2}}$

Again applying componendo and dividendo,

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

$\Rightarrow \frac{2}{2 x}=\frac{2\left(a^{2}+b^{2}\right)}{4 a b}$

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

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


(iv) $\frac{\sqrt{12 x+1}+\sqrt{2 x-3}}{\sqrt{12 x+1}-\sqrt{2 x-3}}=\frac{3}{2}$

Applying componendo and dividendo,

$\frac{\sqrt{12 x+1}+\sqrt{2 x-3}+\sqrt{12 x+1}-\sqrt{2 x-3}}{\sqrt{12 x+1}+\sqrt{2 x-3}-\sqrt{12 x+1}+\sqrt{2 x-3}}$

$=\frac{3+2}{3-2}$

$\Rightarrow \frac{2 \sqrt{12 x+1}}{2 \sqrt{2 x-3}}=\frac{5}{1}$

$\Rightarrow \frac{\sqrt{12 x+1}}{\sqrt{2 x-3}}=\frac{5}{1}$

Squaring both sides, $\frac{12 x+1}{2 x-3}=\frac{25}{1}$

$\Rightarrow \quad 50 x-75=12 x+1$

$\Rightarrow \quad 50 x-12 x=1+75$

$\Rightarrow \quad 38 x=76 \Rightarrow x=\frac{76}{38}=2$

∴ x=2


(v) $\frac{3 x+\sqrt{9 x^{2}-5}}{3 x-\sqrt{9 x^{2}-5}}=\frac{5}{1}$

Applying componendo and dividendo.

$\frac{3 x+\sqrt{9 x^{2}-5}+3 x-\sqrt{9 x^{2}-5}}{3 x+\sqrt{9 x^{2}-5}-3 x+\sqrt{9 x^{2}-5}}=\frac{5+1}{5-1}$

$\frac{6 x}{2 \sqrt{9 x^{2}-5}}=\frac{6}{4} \Rightarrow \frac{3 x}{\sqrt{9 x^{2}-5}}=\frac{3}{2}$

Squaring both sides

$\frac{9 x^{2}}{9 x^{2}-5}=\frac{9}{4} \Rightarrow 81 x^{2}-45=36 x^{2}$

$\Rightarrow \quad 81 x^{2}-36 x^{2}=45 \Rightarrow 45 x^{2}=45$

$\Rightarrow \quad x^{2}=1 \Rightarrow x=\pm 1$

$\therefore x=1,-1$

CHECK 

(i) When x=1, then in the given equation

$\frac{3 \times 1+\sqrt{9 \times 1-5}}{3 \times 1-\sqrt{9 \times 1-5}}=\frac{3+\sqrt{4}}{3-\sqrt{4}}=\frac{3+2}{3-2}=\frac{5}{1}$

which is given

∴x=1

(ii) When x=-1, then

$3(-1)+\sqrt{9(-1)^{2}-5}$

$3(-1)-\sqrt{9(-1)^{2}-5}$

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

$=\frac{-3+2}{-3-2}=\frac{-1}{-5}=\frac{1}{5} \neq \frac{5}{1}$

$\therefore x=-1$ is not its solution.

Hence x=1


(vi) $\frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}=\frac{c}{d}$

Applying componendo and dividendo,

$\frac{\sqrt{a+x}+\sqrt{a-x}+\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}-\sqrt{a+x}+\sqrt{a-x}}=\frac{c+d}{c-d}$

$\Rightarrow \frac{2 \sqrt{a+x}}{2 \sqrt{a-x}}=\frac{c+d}{c-d} \Rightarrow \frac{\sqrt{a+x}}{\sqrt{a-x}}=\frac{c+d}{c-d}$

Squaring both sides $\frac{a+x}{a-x}=\frac{(c+d)^{2}}{(c-d)^{2}}$

Again applying componendo and dividendo

$\frac{a+x+a-x}{a+x-a+x}=\frac{(c+d)^{2}+(c-d)^{2}}{(c+d)^{2}-(c-d)^{2}}$

$\Rightarrow \frac{2 a}{2 x}=\frac{2\left(c^{2}+d^{2}\right)}{4 c d} \Rightarrow \frac{a}{x}=\frac{c^{2}+d^{2}}{2 c d}$

$\Rightarrow  x\left(c^{2}+d^{2}\right)=2 a c d$

$\Rightarrow  x=\frac{2 a c d}{c^{2}+d^{2}}$


Question 13

Solve $\frac{1+x+x^{2}}{1-x+x^{2}}=\frac{62(1+x)}{63(1-x)}$

Sol :

$\frac{1+x+x^{2}}{1-x+x^{2}}=\frac{62(1+x)}{63(1-x)}$

$\Rightarrow \frac{(1-x)\left(1+x+x^{2}\right)}{(1+x)\left(1-x+x^{2}\right)}=\frac{62}{63}$

$\Rightarrow \frac{(1+x)\left(1-x+x^{2}\right)}{(1-x)\left(1+x+x^{2}\right)}=\frac{63}{62}$

$\Rightarrow \quad \frac{1+x^{3}}{1-x^{3}}=\frac{63}{62}$

Applying componendo and dividendo,

$\frac{1+x^{3}+1-x^{3}}{1+x^{3}-1+x^{3}}=\frac{63+62}{63-62}$

$\Rightarrow \frac{2}{2 x^{3}}=\frac{125}{1}$

$ \Rightarrow  \frac{1}{x^{3}}=\frac{125}{1}$

$\Rightarrow x^{3}=\frac{1}{125}=\left(\frac{1}{5}\right)^{3}$

$\therefore x=\frac{1}{5}$ Ans.


Question 14

Solve for $x: 16\left(\frac{a-x}{a+x}\right)^{3}=\frac{a+x}{a-x}$

Sol :

$x : 16\left(\frac{a-x}{a+x}\right)^{3}=\frac{a+x}{a-x}$

$\Rightarrow\left(\frac{a+x}{a-x}\right) \times\left(\frac{a+x}{a-x}\right)^{3}=16$

$\Rightarrow\left(\frac{a+x}{a-x}\right)^{4}=16=(\pm 2)^{4}$

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

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

Applying componendo and dividendo,

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

$ \Rightarrow=\frac{2 a}{2 x}=\frac{-1}{-3}$

$\Rightarrow \frac{a}{x}=\frac{1}{3}$

$ \Rightarrow x=3 a$

Hence $x=\frac{a}{3}, 3 a$ 


Question 15

If $x=\frac{\sqrt{a+x}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}$ , using properties of proportion , show that x² – 2ax + 1 = 0

Sol :

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

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

(Applying componendo and dividendo)

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

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

(Again applying componendo and dividendo)

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

$\Rightarrow \frac{2 x^{2}+2}{4 x}=a \Rightarrow \frac{2\left(x^{2}+1\right)}{4 x}=a$

$\Rightarrow \frac{\left(x^{2}+1\right)}{2 x}=a \Rightarrow 2 a x=x^{2}+1$

$\Rightarrow x^{2}-2 a x+1=0 \quad$


Question 16

Given : $x=\frac{\sqrt{\alpha^{2}+b^{2}}+\sqrt{\alpha^{2}-b^{2}}}{\sqrt{\alpha^{2}+b^{2}}-\sqrt{\alpha^{2}-b^{2}}}$  Use componendo and dividendo to prove that 

Sol :

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

Applying componendo and dividendo both sides

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

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

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

Squaring, both sides we have

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

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

Applying componendo and dividendo both sides

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

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

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


Question 17

Given that $\frac{a^{3}+3 a b^{2}}{b^{3}+3 a^{2} b}=\frac{63}{62}$. Using componendo and dividendo find a: b. (2009)

Sol :
Given that $\frac{a^{3}+3 a b^{2}}{b^{3}+3 a^{2} b}=\frac{63}{62}$
By componendo and dividendo 
$\frac{a^{3}+3 a b^{2}+b^{3}+3 a^{2} b}{a^{3}+3 a b^{2}-b^{3}-3 a^{2} b}=\frac{63+62}{63-62}=\frac{125}{1}$

$\Rightarrow \frac{(a+b)^{3}}{(a-b)^{3}}=\left(\frac{5}{1}\right)^{3}$

$ \Rightarrow \frac{a+b}{a-b}=5 \Rightarrow a+b=5 a-5 b$

⇒5a-a-5b-b=0

⇒4a-6b=0

⇒4a=60

$\Rightarrow \frac{a}{b}=\frac{6}{4}$

$ \Rightarrow \frac{a}{b}=\frac{3}{2}$

a : b = 3 : 2


Question 18

Give $\frac{x^{3}+12 x}{6 x^{2}+8}=\frac{y^{3}+27 y}{9 y^{2}+27}$ Using componendo and dividendo find x : y.

Sol :
Given : 
$\frac{x^{3}+12 x}{6 x^{2}+8}=\frac{y^{3}+27 y}{9 y^{2}+27}$

Using componendo-dividendo, we have

$\frac{x^{3}+12 x+6 x^{2}+8}{x^{3}+12 x-6 x^{2}-8}=\frac{y^{3}+27 y+9 y^{2}+27}{y^{3}+27 y-9 y^{2}-27}$

$\Rightarrow \frac{(x+2)^{3}}{(x-2)^{3}}=\frac{(y+3)^{3}}{(9 y-3)^{3}}$

$\Rightarrow\left(\frac{x+2}{x-2}\right)^{3}=\left(\frac{y+3}{y-3}\right)^{3}$

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

Again using componendo-dividendo, we get

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

$\Rightarrow \frac{2 x}{4}=\frac{2 y}{3}$

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

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

Thus the required ratio is x: y=2: 3


Question 19

Using the properties of proportion, solve the following equation for x; given

$\frac{x^{3}+3 x}{3 x^{2}+1}=\frac{341}{91}$
Sol :
$\frac{x^{3}+3 x}{3 x^{2}+1}=\frac{341}{91}$

Applying componendo and dividendo

$\frac{x^{3}+3 x+3 x^{2}+1}{x^{3}+3 x-3 x^{2}-1}=\frac{341+91}{341-91}$

$\Rightarrow \frac{x^{3}+3 x^{2}+3 x+1}{x^{3}-3 x^{2}+3 x-1}=\frac{432}{250}=\frac{216}{125}$

$\Rightarrow \frac{(x+1)^{3}}{(x-1)^{3}}=\frac{216}{125}=\left(\frac{6}{5}\right)^{3}$

$\therefore \frac{x+1}{x-1}=\frac{6}{5}$

⇒6 x-6=5 x+5

x=11


Question 20

If $\frac{x+y}{a x+b y}=\frac{y+z}{a y+b z}=\frac{z+x}{a z+b x}$ , prove that each of these ratio is equal to $\frac{2}{a+b}$ unless x + y + z = 0

Sol :

$\frac{x+y}{a x+b y}=\frac{y+z}{a y+b z}=\frac{z+x}{a z+b x}$

$=\frac{x+y+y+z+z+x}{a x+b y+a y+b z+\alpha z+b x}$

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

$=\frac{2(x+y+z)}{(a+b)(x+y+z)}=\frac{2}{a+b}$ if $x+y+z \neq 0$

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