Tag: ratio and proportion

Questions Related to ratio and proportion

If two ratios are equal then their inverse ratios are equal. This property is known as

  1. Dividendo property

  2. Invertendo property

  3. Componendo property

  4. Alternendo property


Correct Option: B
Explanation:

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

If $\dfrac {x}{y}=\dfrac {3}{4}$ and $\dfrac {x}{2z}=\dfrac {3}{2}$, then $\dfrac {2x+z}{x-2z}+\left (\dfrac {6}{7}+\dfrac {y-x}{y+x}\right )$ will be equivalent to

  1. 8

  2. 9

  3. 11

  4. 12


Correct Option: A
Explanation:

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

If $a=2+\sqrt 3$ then the value of $a+\frac {1}{a}$ is

  1. 1

  2. 2

  3. 3

  4. 4


Correct Option: D
Explanation:

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

After applying invertendo to $3:7::2:9$ we get

  1. $3:7::2:9$

  2. $3:2::7:9$

  3. $9:7::2:3$

  4. $7:3::9:2$


Correct Option: D
Explanation:

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

We have $3:7::2:9$
After applying invertendo we get
$7:3::9:2$
Option D is correct

If $x=7-4\sqrt 3$, the value of $x^2+\displaystyle\frac{1}{x^2}$ will be

  1. $146$

  2. $148$

  3. $194$

  4. $196$


Correct Option: C
Explanation:

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,
$=> x^{ 2 }+\frac { 1 }{ x^{ 2 } } +2(x)(\frac { 1 }{ x } )=196$
$=> x^2+\displaystyle\frac{1}{x^2}=196-2$
$\therefore x^2+\displaystyle\frac{1}{x^2}=194$

If $x=3+\sqrt 8$ then the value of $x^2+\frac {1}{x^2}$ is

  1. 30

  2. 32

  3. 34

  4. 36


Correct Option: C
Explanation:

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

If $a : b :: c : d$ then $b : a :: d : c$. This property is known as :

  1. Invertendo property

  2. Alternendo property

  3. Componendo property

  4. Dividendo property


Correct Option: A
Explanation:

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

proof
$\dfrac{a}{b}=\dfrac{c}{d}$

inverting both sides, we get
$\Rightarrow \dfrac{b}{a}=\dfrac{d}{c}$

After applying invertendo to $31:15::18:19$  we get $15:a::b:18$ 
What is the value of $a+b$

  1. $40$

  2. $50$

  3. $30$

  4. $60$


Correct Option: B
Explanation:

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

We have $31:15::18:19$
After applying invertendo we get
$15:31::19:18\equiv  15:a::b:18\ a+b=31+19=50$
Option B is correct

Find the value of $a$ and $b$ respectively
After applying invertendo to $2:5::3:9$  we get $a:2::9:b$ 

  1. $9$ and $2$

  2. $2$ and $9$

  3. $3$ and $5$

  4. $5$ and $3$


Correct Option: D
Explanation:

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

We have $2:5::3:9$
After applying invertendo we get
$5:2::9:3\equiv a:2::9:b$
$\Rightarrow a=5,b=3$
Option D is correct

Here 'x' in the following is : $\dfrac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}=b$

  1. $\dfrac{2ab}{(b^2+1)}$

  2. $\dfrac{2ab}{a+b}$

  3. $\dfrac{a+b}{2ab}$

  4. $\dfrac{b^2+1}{2ab}$


Correct Option: A
Explanation:

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

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

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

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

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

$\Rightarrow x=\dfrac{2ab}{b^2+1}$