Question 14 Marks
If $15(2x^2 – y^2) = 7xy,$ find $x: y;$ if $x$ and $y$ both are positive.
Answer$15\left(2 x^2-y^2\right)=7 x y$
$\frac{2 x^2-y^2}{x y}=\frac{7}{15}$
$\frac{2 x}{y}-\frac{y}{x}=\frac{7}{15}$
$\text { Let } \frac{x}{y}=a$
$\therefore 2 a-\frac{1}{a}=\frac{7}{15}$
$\frac{2 a^2-1}{a}=715$
$30 a^2-15=7 a$
$30 a^2-7 a-15=0$
$30 a^2-25 a+18 a-15=0$
$5 a(6 a-5)+3(6 a-5)=0$
$(6 a-5)(5 a+3)=0$
$a=\frac{5}{6},-\frac{3}{5}$
But a cannot be negative
$\therefore a = 5/6$
$\Rightarrow \frac{x}{y}=\frac{5}{6}$
$\Rightarrow x: y=5: 6$
View full question & answer→Question 24 Marks
Find x, if $16\left(\frac{a-x}{a+x}\right)^3=\frac{a+x}{a-x}$
Answer$\left(\frac{a-x}{a+x}\right)^3=\frac{a+x}{a-x}$
$\Rightarrow 16=\left(\frac{a-x}{a+x}\right)^4 $
$ \Rightarrow(2)^4=\left(\frac{a-x}{a+x}\right)^4$
$\Rightarrow \frac{a-x}{a+x}= \pm 2$
$ \Rightarrow \frac{a-x}{a+x}=\frac{2}{1} \text { or } \frac{a-x}{a+x}=\frac{-2}{1}$
Applying componendo and Dividendo, we get
$\Rightarrow \frac{ a + x + a - x }{ a + x - a + x }=\frac{3}{1} \text { or } \frac{ a + x + a - x }{ a + x - a + x }=\frac{-1}{-3} $
$ \Rightarrow \frac{2 a }{2 x }=3 \text { or } \frac{2 a }{2 x }=\frac{1}{3}$
$ \Rightarrow x =\frac{ a }{3} \text { or } x =3 a $
View full question & answer→Question 34 Marks
If $\frac{7 m+2 n}{7 m-2 n}=\frac{5}{3}$ use properties of proportion to find:
(i) $m : n$
(ii) $\frac{ m ^2+ n ^2}{ m ^2- n ^2}$
Answer(i) $\frac{7 m+2 n}{7 m-2 n}=\frac{5}{3}$
Applying Componendo and dividendo, we get
$\frac{7 m +2 n +7 m -2 n }{7 m -2 n -7 m +2 n }=\frac{5+3}{5-3}$
$ \Rightarrow \frac{14 m }{4 n }=\frac{8}{2} $
$\Rightarrow \frac{ m }{ n }=\frac{8 \times 4}{2 \times 14} $
$ \Rightarrow \frac{ m }{ n }=\frac{8}{7} $
$ \Rightarrow m : n =8: 7$
(ii) From i,
$\frac{ m }{ n }=\frac{8}{7} $
$ \Rightarrow \frac{ m ^2}{ n ^2}=\frac{64}{49}$
Applying Componendo and Dividendo,we get
$\frac{m^2+n^2}{m^2-n^2}=\frac{64+49}{64-49}$
$\Rightarrow \frac{ m ^2+ n ^2}{ m ^2- n ^2}=\frac{64+49}{64-49}$
$\Rightarrow \frac{ m ^2+ n ^2}{ m ^2- n ^2}=\frac{113}{15}=7 \frac{8}{15}$
View full question & answer→Question 44 Marks
If $b$ is the mean proportion between $a$ and $c,$ show that:
$\frac{a^4+a^2 b^2+b^4}{b^4+b^2 c^2+c^4}=\frac{a^2}{c^2}$
AnswerGiven that $b$ is the mean proportion between $a$ and $c.$
$\frac{ a }{ b }=\frac{ b }{ c }= k$
$\Rightarrow b = ck ; a = bk =( ck ) k = ck ^2$
$L.H.S =\frac{a^4+a^2 b^2+b^4}{b^4+b^2 c^2+c^4}$
$=\frac{\left( ck ^2\right)^4+\left( ck ^2\right)^2( ck )^2+( ck )^4}{( ck )^4+(c k)^2 c ^2+ c ^4}$
$=\frac{ c ^4 k ^8+ c ^4 k ^6+ c ^4 k ^4}{ c ^4 k ^4+ c ^4 k ^2+ c ^4}$
$=\frac{ c ^4 k ^4\left( k ^4+ k ^2+1\right)}{ c ^4\left( k ^4+ k ^2+1\right)}$
$= k ^4 \ldots \ldots \ldots . \text { (i) }$
$R.H.S =\frac{ a ^2}{ c ^2}=\frac{\left( ck ^2\right)^2}{ c ^2}=\frac{ c ^2 k ^4}{ c ^2}= k ^4 \ldots \ldots. (ii)$
From $(i)$ and $(ii)$ we get
$L.H.S = R.H.S$
$\Rightarrow \frac{ a ^4+ a ^2 b ^2+ b ^4}{ b ^4+ b ^2 c ^2+ c ^4}=\frac{ a ^2}{ c ^2}$
Hence proved.
View full question & answer→Question 54 Marks
If $\frac{ x }{ a }=\frac{ y }{ b }=\frac{ z }{ c }$ show that:
$\frac{x^3}{a^3}+\frac{y^3}{b^3}+\frac{z^3}{c^3}=\frac{3 x y z}{a b c}$
AnswerLet $\frac{ x }{ a }=\frac{ y }{ b }=\frac{ z }{ c }$ = k
$\Rightarrow x=a k, y=b k, z=c k$
$ \text { L.H.S } =\frac{x^3}{a^3}+\frac{y^3}{b^3}+\frac{z^3}{c^3} $
$ =\frac{(a k)^3}{a^3}+\frac{(b k)^3}{b^3}+\frac{(c k)^3}{c^3} $
$=\frac{a^3 k^3}{a^3}+\frac{b^3 k^3}{b^3}+\frac{c^3 k^3}{c^3} $
$=k^3+k^3+k^3$
$ =3 k^3$
$\text { R.H.S }=\frac{3 x y z}{a b c} $
$ \quad=\frac{3( ak )( bk )( ck )}{ abc } $
$ \quad=3 k ^3$
$\Rightarrow \text { L.H.S }=\text { R.H.S }$
$ \text { i.e } \frac{ x ^3}{ a ^3}+\frac{ y ^3}{ b ^3}+\frac{ z ^3}{ c ^3}=\frac{3 xyz }{ abc }$
View full question & answer→Question 64 Marks
If $x=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}+\sqrt{a-1}}$ using properties of proportion show that $x^2-2 a x+1=0$
AnswerGiven that $x =\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}$
By applying Componendo-Dividendo
$\Rightarrow \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})} $
$ \Rightarrow \frac{x+1}{x-1}=\frac{2 \sqrt{a+1}}{2 \sqrt{a-1}} $
$\Rightarrow \frac{x+1}{x-1}=\frac{\sqrt{a+1}}{\sqrt{a-1}}$
Squaring both the sides of the equation we have
$\Rightarrow\left(\frac{x+1}{x-1}\right)^2=\frac{a+1}{a-1} $
$ \Rightarrow(x+1)^2(a-1)=(x-1)^2(a+1) $
$ \Rightarrow\left(x^2+2 x+1\right)-\left(x^2+2 x+1\right)=a\left(x^2-2 x+1\right)+\left(x^2-2 x+1\right) $
$\Rightarrow 4 a x=2 x^2+2$
$\Rightarrow 2 a x=x^2+1 $
$ \Rightarrow x^2-2 a x+1=0$
View full question & answer→Question 74 Marks
Using componendo and dividendo find the value of $x$
$\frac{\sqrt{3 x+4}+\sqrt{3 x-5}}{\sqrt{3 x+5}-\sqrt{3 x-5}}=9$
Answer$\frac{\sqrt{3 x+4}+\sqrt{3 x-5}}{\sqrt{3 x+5}-\sqrt{3 x-5}}=\frac{9}{1}$
Applying componendo and dividendo we have
$\frac{\sqrt{3 x+1}+\sqrt{3 x-5}+\sqrt{3 x+4}-\sqrt{3 x-5}}{\sqrt{3 x+4}+\sqrt{3 x-5}-\sqrt{3 x+4}+\sqrt{3 x-5}}=\frac{9+1}{9-1} $
$ \Rightarrow \frac{2 \sqrt{3 x+4}}{2 \sqrt{3 x-5}}=\frac{5}{4}$
Squaring both sides we have
$\frac{3 x+4}{3 x-5}=\frac{25}{16}$
$\Rightarrow 16(3 x+4)=25(3 x-5)$
$ \Rightarrow 48 x-64=75 x-125 $
$\Rightarrow 75 x-48 x=64+125 $
$ \Rightarrow27 x=189 $
$ \Rightarrow x=189 / 27$
$ \Rightarrow x=7$
View full question & answer→Question 84 Marks
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 $b^2 = (2a^2x)/(x^2+ 1)$
Answer$x=\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}$
by 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}} $
$ \frac{x+1}{x-1}=\frac{2 \sqrt{a^2+b^2}}{2 \sqrt{a^2-b^2}}$
Squaring both sides
$\frac{x^2+2 x+1}{x^2-2 x+1}=\frac{a^2+b^2}{a^2-b^2}$
By componendo and dividendo
$\frac{\left(x^2+2 x+1\right)+\left(x^2-2 x+1\right)}{\left(x^2+2 x+1\right)-\left(x^2-2 x+1\right)}=\frac{\left(a^2+b^2\right)+\left(a^2-b^2\right)}{\left(a^2+b^2\right)-\left(a^2-b^2\right)} $
$ \Rightarrow \frac{2\left(x^2+1\right)}{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}$
View full question & answer→Question 94 Marks
If $\frac{a-2 b-3 c+4 d}{a+2 b-3 c-4 d}=\frac{a-2 b+3 c-4 d}{a+2 b+3 c+4 d}$ show that 2ad = 3bc
Answer$\frac{a-2 b-3 c+4 d}{a+2 b-3 c-4 d}=\frac{a-2 b+3 c-4 d}{a+2 b+3 c+4 d}$
Applying componendo and dividendo
$\Rightarrow \frac{a-2 b-3 c+4 d+a+2 b-3 c-4 d}{a-2 b-3 c+4 d-a-2 b+3 c+4 d}=\frac{a-2 b+3 c-4 d+a+2 b+3 c+4 d}{a-2 b+3 c-4 d-a-2 b-3 c-4 d}$
$\begin{aligned} & \Rightarrow \frac{2 a-6 c}{-4 b+8 d}=\frac{2 a+6 c}{-4 b-8 d} \\ & \Rightarrow \frac{2(a-3 c)}{-4(b-2 d)}=\frac{2(a+3 c)}{4(b+2 d)} \\ & \Rightarrow \frac{a-3 c}{a+3 c}=\frac{b-2 d}{b+2 d}\end{aligned}$
Applying componendo and dividendo
$\begin{aligned} & \Rightarrow \frac{a-3 c+a+3 c}{a-3 c-a-3 c}=\frac{b-2 d+b+2 d}{b-2 d-b-2 d} \\ & \Rightarrow \frac{2 a}{-6 c}=\frac{2 b}{-4 d} \\ & \Rightarrow-4 \mathrm{da}=-6 \mathrm{cb} \\ & \Rightarrow 2 \mathrm{ad}=3 \mathrm{bc}\end{aligned}$
View full question & answer→Question 104 Marks
If $=\frac{4 \sqrt{6}}{\sqrt{2}+\sqrt{3}}$ find the value of $\frac{a+2 \sqrt{2}}{a-2 \sqrt{2}}+\frac{a+2 \sqrt{3}}{a-2 \sqrt{3}}$
Answer$a=\frac{4 \sqrt{6}}{\sqrt{2}+\sqrt{3}}$
$\frac{a}{2 \sqrt{2}}=\frac{2 \sqrt{3}}{\sqrt{2}+\sqrt{3}}$
Applying componendo and divendo
$\frac{a+2 \sqrt{2}}{a-2 \sqrt{2}}=\frac{2 \sqrt{3}+\sqrt{2}+\sqrt{3}}{2 \sqrt{3}-\sqrt{2}-\sqrt{3}}$
$\frac{a+2 \sqrt{2}}{a-2 \sqrt{2}}=\frac{3 \sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}...(1)$
$\frac{a}{2 \sqrt{3}}=\frac{2 \sqrt{2}}{\sqrt{2}+\sqrt{3}}$
Applying componendo and dividendo
$\frac{a+2 \sqrt{3}}{a-2 \sqrt{3}}=\frac{2 \sqrt{2}+\sqrt{2}+\sqrt{3}}{2 \sqrt{2}-\sqrt{2}-\sqrt{3}}$
$\frac{a+2 \sqrt{3}}{a-2 \sqrt{3}}=\frac{3 \sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}...(2)$
From (1) and (2)
$\frac{a+2 \sqrt{2}}{a-2 \sqrt{2}}+\frac{a+2 \sqrt{3}}{a-2 \sqrt{3}}=\frac{3 \sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{3 \sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}} $
$ \frac{a+2 \sqrt{2}}{a-2 \sqrt{2}}+\frac{a+2 \sqrt{3}}{a-2 \sqrt{3}}=\frac{3 \sqrt{2}+\sqrt{3}-3 \sqrt{3}-\sqrt{2}}{\sqrt{2}-\sqrt{3}}$
$ \frac{a+2 \sqrt{2}}{a-2 \sqrt{2}}+\frac{a+2 \sqrt{3}}{a-2 \sqrt{3}}=\frac{2 \sqrt{2}-2 \sqrt{3}}{\sqrt{2}-\sqrt{3}}=2$
View full question & answer→Question 114 Marks
if x $=\frac{6 a b}{a+b}$ find the value of $\frac{x+3 a}{x-3 a}=\frac{x+3 b}{x-3 b}$
Answer$x=\frac{6 a b}{a+b}$
$\Rightarrow \frac{x}{3 a}=\frac{2 b}{a+b}$
Applying compinendo and dividendo
$\frac{x+3 a}{x-3 a}=\frac{2 b+a+b}{2 b-a-b}$
$\frac{x+3 a}{a-3 a}=\frac{3 b+a}{b-a}...(1)$
Again $x =\frac{6 a b}{a+b}$
$\Rightarrow \frac{x}{3 b}=\frac{2 a}{a+b}$
Applying componendo and dividendo
$\frac{x+3 b}{x-3 b}=\frac{2 a+a+b}{2 a-a-b}$
$\frac{x+3 b}{x-3 b}=\frac{3 a+b}{a-b}...(2)$
From (1) and (2)
$\frac{x+3 a}{x-3 a}+\frac{x+3 b}{x-3 b}=\frac{3 b+a}{b-a}+\frac{3 a+b}{a-b}$
$\frac{x+3 a}{x-3 a}+\frac{x+3 b}{x-3 b}=\frac{-3 b-a+3 a+b}{a-b}$
$\frac{x+3 a}{x-3 a}+\frac{x+3 b}{x-3 b}=\frac{2 a-2 b}{a-b}=2$
View full question & answer→Question 124 Marks
If $x=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}$ express n in terms of x and m
Answer$x=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}$
Applying componendo and dividendo
$\frac{x+1}{x-1}=\frac{\sqrt{m+n}+\sqrt{m-n}+\sqrt{m+n}-\sqrt{m-n}}{\sqrt{m+n}+\sqrt{m-n}-\sqrt{m+n}+\sqrt{m-n}} $
$\frac{x+1}{x-1}=\frac{2 \sqrt{m+n}}{2 \sqrt{m-n}}$
Squaring both sides,
$\frac{x^2+2 x+1}{x^2-2 x+1}=\frac{m+n}{m-n}$
Applying componendo and dividendo
$\frac{x^2+2 x+1+x^2-2 x+1}{x^2+2 x+1-x^2+2 x-1}=\frac{m+n+m-n}{m+n-m+n}$
$ \frac{2 x^2+2}{4 x}=\frac{2 m}{2 n} $
$ \frac{x^2+1}{2 x}=\frac{m}{n} $
$\frac{x^2+1}{2 m x}=\frac{1}{n}$
$n=\frac{2 m x}{x^2+1}$
View full question & answer→Question 134 Marks
Using the properties of proportion solve for x given $\frac{x^4+1}{2 x^2}=\frac{17}{8}$
Answer$\frac{x^4+1}{2 x^2}=\frac{17}{8}$
Applying componendo and dividendo we get
$\frac{x^4+1+2 x^2}{x^4+1-2 x^2}=\frac{17+8}{17-8}$
$ \Rightarrow \frac{\left(x^2\right)^2+(1)^2+2 \times x^2+1}{\left(x^2\right)^2+(1)^2-2 \times x^2 \times 1}=\frac{25}{9}$
$\Rightarrow \frac{\left(x^2+1\right)^2}{\left(x^2-1\right)^2}=\frac{5^2}{3^2} $
$ \Rightarrow\left(\frac{x^2+1}{x^2-1}\right)^2=\left(\frac{5}{3}\right)^2$
$\Rightarrow \frac{x^2+1}{x^2-1}=\frac{5}{3}$
Applying componeddo and diividendo we get
$\frac{x^2+1+x^2-1}{x^2+1-x^2+1}=\frac{5+3}{5-3} $
$ \Rightarrow \frac{2 x^2}{2}=\frac{8}{2}$
$ \Rightarrow x^2=4 $
$ \Rightarrow x= \pm 2$
View full question & answer→Question 144 Marks
Using properties of proportion solve for x:
$\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\frac{4 x-1}{2}$
Answer$\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\frac{4 x-1}{2}$
Applying componendo and dividendo
$\frac{\sqrt{x+1}+\sqrt{x-1}+\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}-\sqrt{x+1}+\sqrt{x-1}}=\frac{4 x-1+2}{4 x-1-2}$
$\frac{2 \sqrt{x+1}}{2 \sqrt{x-1}}=\frac{4 x+1}{4 x-3}$
Squaring both sides
$\frac{x+1}{x-1}=\frac{16 x^2+1+8 x}{16 x^2+9-24 x}$
Applying componedo and dividendo
$\frac{x+1+x-1}{x+1-x+1}=\frac{16 x^2+1+8 x+16 x^2+9-24 x}{16 x^2+1+8 x-16 x^2-9+24 x} $
$ \frac{2 x}{2}=\frac{32 x^2+10-16 x}{32 x-8} $
$ x=\frac{16 x^2+5-8 x}{16 x-4} $
$16 x^2-4 x=16 x^2+5-8 x$
$4 x=5$
$x=5 / 4$
View full question & answer→Question 154 Marks
Using properties of proportion solve for x
$\frac{\sqrt{x+5}+\sqrt{x-16}}{\sqrt{x+5}-\sqrt{x-16}}=\frac{7}{3}$
Answer$\frac{\sqrt{x+5}+\sqrt{x-16}}{\sqrt{x+5}-\sqrt{x-16}}=\frac{7}{3}$
Applying componendo and dividendo
$\frac{\sqrt{x+5}+\sqrt{x-16}+\sqrt{x+5}-\sqrt{x-16}}{\sqrt{x+5}+\sqrt{x-16}-\sqrt{x+5}+\sqrt{x-16}}=\frac{7+3}{7-3} $
$ \frac{2 \sqrt{x+5}}{2 \sqrt{x-16}}=\frac{10}{4} $
$ \frac{\sqrt{x+5}}{\sqrt{x-16}}=\frac{5}{2}$
Squaring both sides
$\frac{x+5}{x-16}=\frac{25}{4} $
$ 4 x+20=25 x-400 $
$21 x=420$
$ x=\frac{420}{21}=20$
View full question & answer→Question 164 Marks
If $y$ is the mean proportional between $x$ and $z$; show that $xy + yz$ is the mean proportional between $x^2+y^2$ and $y^2+ z^2$.
AnswerSince $y$ is the mean proportion between $x$ and $z$
Therefore, $y^2= xz$
Now, we have to prove that $xy+yz$ is the mean proportional between $x^2+y^2$ and $y^2+z^2,$
i.e.,$(x y+y z)^2=\left(x^2+y^2\right)\left(y^2+z^2\right)$
$LHS = (xy + yz)^2$
$=[y(x+z)]^2$
$=y^2(x+z)^2$
$=x z(x+z)^2$
$\text { RHS }=\left(x^2+y^2\right)\left(y^2+z^2\right)$
$=\left(x^2+x z\right)\left(x z+z^2\right)$
$=x(x+z) z(x+z)$
$=x z(x+z)^2$
$LHS = RHS$
Hence, proved.
View full question & answer→Question 174 Marks
If $\frac{ a }{ b }=\frac{ c }{ d }$, show that: $\frac{a^3 c+a c^3}{b^3 d+b d^3}=\frac{( a + c )^4}{( b + d )^4}$
Answer$\begin{aligned} & \text { Let } \frac{a}{b}=\frac{c}{d}=k \\ & \Rightarrow a=b k \text { and } c=d k \\ & \text { L.H.S }=\frac{a^3 c+a c^3}{b^3 d+b d^3} \\ & =\frac{(b k)^3 \times(d k)+(b k)\left(d k^3\right)}{b^3 d+b d^3} \\ & =\frac{b^3 k^3 d k+b k d^3 k^3}{b^3 d+b d^3} \\ & =\frac{b^3 d k^4+b d^3 k^4}{b^3 d+b d^3} \\ & =\frac{k^4\left(b^3 d+b d^3\right)}{b^3 d+b d^3} \\ & =k^4\end{aligned}$
$\begin{aligned} & \text { R.H.S }=\frac{(a+c)^4}{(b+d)^4} \\ & =\frac{(b k+d k)^4}{(b+d)^4} \\ & =\frac{k^4(b+d)^4}{(b+d)^4} \\ & =k^4 \\ & \therefore \text { L.H.S = R.H.S }\end{aligned}$
View full question & answer→Question 184 Marks
If $a, b, c$ are in continued proportion and $a(b -c) = 2b$, prove that: $a-c=\frac{2(a+b)}{a}$
AnswerSince a, b, c are in continued proportion,
$\frac{a}{b}=\frac{b}{c}$
$\Rightarrow b^2 = ac ...(i)$
a(b - c) = 2b
$\Rightarrow \frac{a(b-c)}{b}=2 \ldots$ (ii)
Now,
$\begin{aligned} & \text { R.H.S. }=\frac{2(a+b)}{a} \\ & \Rightarrow \frac{a(b-c)}{b} \times \frac{a+b}{a} \quad \ldots \text { Using equation (ii) } \\ & \Rightarrow \frac{\not a(b-c)}{b} \times \frac{a+b}{\not a} \\ & \Rightarrow \frac{b-c}{b} \times a+b \\ & \Rightarrow \frac{b a+b^2-a c-b c}{b} \\ & \Rightarrow \frac{b a+a c-a c-b c}{b} \quad \ldots .(\text { Using equation (I)) } \\ & \Rightarrow \frac{b a-b c}{b} \\ & \Rightarrow \frac{\not b(a-c)}{\not b} \\ & \Rightarrow \mathrm{a}-\mathrm{c}\end{aligned}$
View full question & answer→Question 194 Marks
The monthly pocket money of Ravi and Sanjeev are in the ratio 5:7. Their expenditures are in the ratio 3:5. If each saves Rs. 80 every month, find their monthly pocket money.
AnswerLet the monthly pocket money of Ravi and Sanjeev be 5x and 7x respectively.
Let their expenditures be 3y and 5y respectively.
Ravi’s savings = 5x – 3y
Sanjeev’s savings = 7x – 5y
By the given information,
5x – 3y = 80 … (1)
7x – 5y = 80 … (2)
From (1) and (2), we have
5x – 3y = 7x – 5y
⇒ x = y
From equation (1),
5x – 3x = 80
⇒ 2x = 80
⇒ x = 40
Hence,
Monthly pocket money of Ravi = 5 × 40 = Rs. 200
Monthly pocket money of Sanjeev = 7 × 40 = Rs. 280
View full question & answer→Question 204 Marks
If $x$ and $y$ both are positive and $(2x^2- 5y^2): xy = 1: 3,$ find $x: y.$
Answer$\left(2 x^2-5 y^2\right): x y=1: 3$
$\Rightarrow \frac{2 x^2-5 y^2}{x y}=\frac{1}{3}$
$\Rightarrow \frac{2 x}{y}-\frac{5 y}{x}=\frac{1}{3}$
Put $\frac{ x }{ v }= a$ we get
$\Rightarrow 2 a-5 \frac{1}{a}=\frac{1}{3}$
$\Rightarrow 3\left(2 a^2-5\right)=a$
$\Rightarrow 6 a 2-a-15=0$
$\Rightarrow 6 a^2+9 a-10 a-15=0$
$\Rightarrow 3 a(2 a+3)-5(2 a+3)=0$
$\Rightarrow(2 a+3)(3 a-5)=0$
$\Rightarrow(2 a+3)=0 \text { or }(3 a-5)=0$
$\Rightarrow a =-\frac{3}{2}$ or $a =\frac{5}{3}$
$a=-\frac{3}{2}$ is not acceptable, as $x$ and $y$ both are positive.
$a=\frac{5}{3} \Rightarrow \frac{x}{y}=\frac{5}{3}$
$\Rightarrow x: y=5: 3$
View full question & answer→Question 214 Marks
Given $\frac{x^3+12 x}{6 x^2+8}=\frac{y^3+27 y}{9 y^2+27}$ using componendo and dividendo, find $x : y$
Answer$\frac{x^3+12 x}{6 x^2+8}=\frac{y^3+27 y}{9 y^2+27}$
Applying componendo and dividendo, we get
$\frac{x^3+12 x+6 x^2+8}{x^3+12 x-6 x^2-8}=\frac{y^3+27+9 y^2+27}{y^3+27 y-9 y^2-27}$
$\Rightarrow\ \frac{ x ^3+3(1)(4) x +3(1)(2) x ^2+2^3}{ x ^3+3(1)(4) x +3(1)(2) x ^2+2^3}=\frac{ y ^3+3(1)(9) y +3(1)(3) y ^2+3^3}{ y ^3+3(1)(9) y +3(1)(3) y ^2+3^3}$
$\Rightarrow \frac{ x ^3+3(1)(4) x +3(1)(2) x ^2+2^3}{ x ^3+3(1)(2) x +3(1)(4) x ^2+2^3}=\frac{ y ^3+3(1)(9) y +3(1)(3) y ^2+3^3}{ y ^3+3(1)(3) y +3(1)(9) y ^2+3^3}$
$\Rightarrow \frac{( x +2)^3}{( x -2)^3}=\frac{( y +3)^3}{( y -3)^3}$
Again applying componendo and 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}{6}$
$ \Rightarrow \frac{x}{2}=\frac{y}{3} $
Applying alternendo, we get
$ \frac{x}{y}=\frac{2}{3}$
$ \Rightarrow x: y=2: 3 $
View full question & answer→Question 224 Marks
If $x=\frac{2 a b}{a+b}$ find the value of $\frac{x+a}{x-a}+\frac{x+b}{x-b}$
Answer$ \begin{aligned} & x=\frac{2 a b}{a+b} \\ & \frac{x}{a}=\frac{2 b}{a+b} \end{aligned} $
Applying componendo and dividendo
$ \begin{aligned} & \frac{x+a}{x-a}=\frac{2 b+a+b}{2 b-a-b} \\ & \frac{x=a}{x-a}=\frac{3 b+a}{b-a} \ldots(1) \end{aligned} $
Also $\mathrm{x}=\frac{2 a b}{a+b}$
$ \frac{x}{b}=\frac{2 a}{a+b} $
Apllying componendo and dividendo
$ \begin{aligned} & \frac{x+b}{x-b}=\frac{2 a+a+b}{2 a-a-b} \\ & \frac{x+a}{x-a}=\frac{3 b+a}{b-a} \ldots(1) \end{aligned} $
Also $\mathrm{x}=\frac{2 a b}{a+b}$
$ \frac{x}{b}=\frac{2 a}{a+b} $
Applying componendo and dividendo
$ \begin{aligned} & \frac{x+b}{x-b}=\frac{2 a+a+b}{2 a-a-b} \\ & \frac{x+b}{x-b}=\frac{3 a+b}{a-b} ....... (2)\end{aligned} $
From (1) and (2)
$ \begin{aligned} & \frac{x+a}{x-a}=\frac{2 a+a+b}{2 a-a-b} \\ & \frac{x+b}{x-b}=\frac{3 a+b}{a-b} \ldots(2) \end{aligned} $
From (1) and (2)
$ \begin{aligned} & \frac{x+a}{x-a}+\frac{x+a}{x-b}=\frac{3 b+a}{b-a}+\frac{3 a+b}{a-b} \\ & \frac{x+a}{x-a}+\frac{x+b}{x-b}=\frac{-3 b-a+3 a+b}{a-b} \\ & \frac{x+a}{x-a}+\frac{x+b}{x-b}=\frac{2 a-2 b}{a-b} \\ & \frac{x+a}{x-a}+\frac{x+b}{x-b}=\frac{2 a-2 b}{a-b}=2 \end{aligned} $
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