Properties of proportion - class-XII

If $a : b :: c : d$ then $b : a :: d : c \Rightarrow $ invertendo property

$\because \dfrac {x}{y}=\dfrac {3}{4}\Rightarrow \dfrac {y-x}{y+x}=\dfrac{4-3}{4+3}=\dfrac {1}{7}$
$\because \dfrac {x}{2z}=\dfrac {3}{2}\Rightarrow 2x=6z$ or, $x=3z$
$\therefore \dfrac {2x+z}{x-2z}+\left (\dfrac {6}{7}+\dfrac {y-x}{y+x}\right )=\dfrac {6z+z}{3z-2z}+\left (\dfrac {6}{7}+\dfrac {1}{7}\right )$
$=\dfrac {7z}{z}+\dfrac {7}{7}=7+1=8$.

$\because a=2+\sqrt 3$
$\therefore \frac {1}{a}=\frac {1}{2+\sqrt 3}\times \frac {2-\sqrt 3}{2-\sqrt 3}=\frac {2-\sqrt 3}{4-3}=2-\sqrt 3$
$a+\frac {1}{a}=2+\sqrt 3+2-\sqrt 3=4$

If $a : b :: c : d$ then $b : a :: d : c$ is invertendo property of ratios

Given,
$x=7-4\sqrt 3$
$\therefore \displaystyle\frac{1}{x}=\displaystyle\frac{1}{(7-4\sqrt 3)}\times \displaystyle\frac{7+4\sqrt 3}{7+4\sqrt 3}$
$=\displaystyle\frac{7+4\sqrt 3}{(49-48)}$
$=7+4\sqrt 3$
Now,
$x+\displaystyle\frac{1}{x}=(7-4\sqrt 3)+(7+4\sqrt 3)$
$=14$
$=> \left( x+\displaystyle\frac{1}{x}\right)^2=(14)^2$
Using $(a+b)^2=a^2+b^2+2ab$ we get,
$=> x^{ 2 }+\frac { 1 }{ x^{ 2 } } +2(x)(\frac { 1 }{ x } )=196$
$=> x^2+\displaystyle\frac{1}{x^2}=196-2$
$\therefore x^2+\displaystyle\frac{1}{x^2}=194$

$x=3+\sqrt 8\Rightarrow \frac {1}{x}=\frac {1}{3+\sqrt 8}\times \frac {3-\sqrt 8}{3-\sqrt 8}=\frac {3-\sqrt 8}{9-8}$
$=3-\sqrt 8$
$x^2+\frac {1}{x^2}=\left (x+\frac {1}{x}\right )^2-2=36-2=34$

If $a : b :: c : d$ then $b : a :: d : c\Rightarrow $invertendo property.

If $a : b :: c : d$ then $b : a :: d : c$ is invertendo property of ratios

If $a : b :: c : d$ then $b : a :: d : c$ is invertendo property of ratios

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

If $a : b :: c : d$ then $b : a :: d : c$ is invertendo property of ratios

If $a : b :: c : d$ then $b : a :: d : c$ is invertendo property of ratios

If $a : b :: c : d$ then $b : a :: d : c$ is invertendo property of ratios

Given a/b = b/c = c/d = k, then c = dk, b = ck = dk^2, and a = bk = dk^3. Substituting these into the expression: (b^3 + c^3 + d^3) / (a^3 + b^3 + c^3) = (d^3k^6 + d^3k^3 + d^3) / (d^3k^9 + d^3k^6 + d^3k^3) = d^3(k^6 + k^3 + 1) / d^3k^3(k^6 + k^3 + 1) = 1/k^3. Since a/b = k, then b/a = 1/k, so (b/a)^3 = 1/k^3. Alternatively, d/a = d/(dk^3) = 1/k^3.

Given that

$\dfrac{a+3d}{a+9d}=\dfrac{a+d}{a+5d}$

$\dfrac{2-7x}{1-5x}=\dfrac{3+7x}{4+5x}$

Suppose that,

The length of larger part be$=x$

And the smaller part be $=y$

So, their ratio is $x$ is to

$ \dfrac{x}{y}=\dfrac{kx+ky}{kx} $

$ \Rightarrow \dfrac{x}{y}=\dfrac{x+y}{x} $

$ \Rightarrow {{x}^{2}}=xy+{{y}^{2}} $

$ \Rightarrow {{x}^{2}}-{{y}^{2}}=xy $

Let $y=1$

$ {{x}^{2}}-{{1}^{2}}=x $

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

$ \Rightarrow x=\dfrac{1\pm \sqrt{5}}{2}\,\,\,\,\left( \text{On}\,\text{solving}\,\text{by}\,\text{second}\,\text{degree}\,\text{equation}\,\text{rule} \right) $

Negative value cannot be considered

So,

$x=\dfrac{1+\sqrt{5}}{2}$ so, the ratio

$ \dfrac{x}{y}=\dfrac{\dfrac{1+\sqrt{5}}{2}}{1} $

$ \Rightarrow x:y=\left( 1+\sqrt{5} \right):2 $

Therefore,

$ \dfrac{y}{x+y}=\dfrac{2}{\left( 1+\sqrt{5}+2 \right)} $

$ \Rightarrow \dfrac{y}{x+y}=\dfrac{2}{\left( 3+\sqrt{5} \right)} $

Hence, this is the answer.

All of these.

The ratio of $A,B,C =12:15:25$
Then, total of the ratio is $12+15+25=52$
Then, $ A=\displaystyle \frac{12}{52}\times 312=72$

Given, $x:y=1:1$

If $\dfrac{a}{b}=\dfrac{c}{d},$ then by dividendo property

If $a : b :: c : d$ then $b : a :: d : c$ is invertendo property of ratios

$\displaystyle \frac{x}{y}=\frac{6}{5}$

Given, $x = 4ab/(a+b)$

$\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{e}{f}=k$

Using the property of componendo and dividendo on the given equation (a+be^y)/(a-be^y) = (b+ce^y)/(b-ce^y), we get a/(be^y) = b/(ce^y). This simplifies to a/b = b/c, which implies b^2 = ac, meaning a, b, and c are in G.P. Extending this to the third term confirms a, b, c, and d are in G.P.

$\dfrac{{{a^3} + 3a{b^2}}}{{3{a^2}b + {b^3}}} = \dfrac{{{x^3} + 3x{y^2}}}{{3{x^2}y + {y^3}}}$

$x=\cfrac { 2\sqrt { 5 }  }{ \sqrt { 3 } +\sqrt { 5 }  } \Rightarrow \cfrac { x }{ \sqrt { 3 }  } =\cfrac { 2\sqrt { 5 }  }{ \sqrt { 3 } +\sqrt { 5 }  } $
and $\cfrac { x }{ \sqrt { 5 }  } =\cfrac { 2\sqrt { 3 }  }{ \sqrt { 3 } +\sqrt { 5 }  } $
Applying components and dividendo, we get
$\cfrac { x+\sqrt { 5 }  }{ x-\sqrt { 5 }  } =-\left( 7+2\sqrt { 15 }  \right) $
and
$\cfrac { x+\sqrt { 3 }  }{ x-\sqrt { 3 }  } =9+2\sqrt { 15 } $
$\Rightarrow \cfrac { x+\sqrt { 5 }  }{ x-\sqrt { 5 }  } +\cfrac { x+\sqrt { 3 }  }{ x-\sqrt { 3 }  } =2$

A.  $18 : 8 = \dfrac{18}{8}$

Let $a = 3x$ and $b = 5x$.
$\therefore \displaystyle \frac{a-b}{a+b}=\frac{3x-5x}{3x+5x}=\frac{-2x}{8x}=-\frac{1}{4}$

$\dfrac { 1 }{ x } :\dfrac { 1 }{ y } :\dfrac { 1 }{ z } =\quad 2:3:5\ \dfrac { yz:xz:xy }{ xyz } =\quad 2:3:5\ yz:xz:xy\quad =\quad 2xyz:3xyz:5xyz\ 1:1:1=\quad 2x:3y:5z\ x:y:z=\quad \dfrac { 1 }{ 2 } :\dfrac { 1 }{ 3 } :\dfrac { 1 }{ 5 } =\dfrac { 15:10:6 }{ 30 } \ So,\quad x:y:z\quad is\quad 15:10:6$