Mathematics · Quantitative Aptitude

Ratios and Proportions

302 Questions

Ratios and proportions deal with comparing two or more quantities and finding their relationships. The questions involve calculating compound ratios, duplicate ratios, and solving proportional equations. This topic is a crucial part of the mathematics and quantitative aptitude sections in competitive exams.

compound ratiosduplicate ratiosproportion equationssimple ratio calculationscombining multiple ratios

Ratios and Proportions Questions

Multiple choice composition of ratios types of ratios ratio and proportions ratio and proportion maths

If $x : y = 3 : 8$ and $y : z = 4 : 9$, then the triplicate ratio of $x : z$ is _____

  1. $27 : 512$
  2. $1 : 216$
  3. $64 : 729$
  4. $3 : 9$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$x : y = 3 : 8$ and $y : z = 4 : 9$
$\therefore \dfrac {x}{y}\times \dfrac {y}{z} = \dfrac {3}{8} \times \dfrac {4}{9} = \dfrac {1}{6}$
$\therefore x : z = 1 : 6$
$\therefore$ The triplicate ratio of $x : z$ is $x^{3} : z^{3} = (1)^{3} : (6)^{3} = 1 : 216$

Multiple choice composition of ratios types of ratios ratio and proportions ratio and proportion maths

If $x : y = 30 : 20$ and $y : z = 24 : 25$ then the subduplicate ratio of $x : z$ is ____

  1. $6 : 5$
  2. $36 : 25$
  3. $5 : 6$
  4. $30 : 25$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$x : y = 30 : 20$ and $y : z = 24 : 25$
$\therefore \dfrac {x}{y}\times \dfrac {y}{z} = \dfrac {30}{20} \times \dfrac {24}{25} = \dfrac {36}{25}$
$\therefore x : z = 36 : 25$
$\therefore$ The subduplicate ratio of $x : z$ is $\sqrt {36} : \sqrt {25} = 6 : 5$

Multiple choice composition of ratios types of ratios ratio and proportions ratio and proportion maths

If $x : y = 1 : 2$ and $y : z = 8 : 3$, then the reciprocal ratio of $z : x$ is ____

  1. $3 : 8$
  2. $1 : 3$
  3. $3 : 4$
  4. $4 : 3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$x : y = 1 : 2$ and $y : z = 8 : 3$
$\therefore x : z = \dfrac {x}{y} \times \dfrac {y}{z} = \dfrac {1}{2} \times \dfrac {8}{3} = \dfrac {4}{3} = 4 : 3$
Now, the reciprocal ratio of $z : x$ is $x : z$
$\therefore x : z = 4 : 3$

Multiple choice composition of ratios types of ratios ratio and proportions ratio and proportion maths

The subtriplicate ratio of $(x^{4} - y^{4})^{3} : (x^{2} + y^{2})^{6}$ is ____

  1. $(x^{2} - y^{2}) : (x^{2} + y^{2})$
  2. $(x^{2} - y^{2}) : (x^{2} + y^{2})^{2}$
  3. $(x - y) : (x + y)$
  4. $(x^{2} + y^{2}) : xy$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The subtriplicate ratio of $a : b$ is $\sqrt [3]{a} : \sqrt [3]{b}$
$\therefore$ The subtriplicate ratio of $(x^{4} - y^{4})^{3} : (x^{2} + y^{2})^{6}$ is $\sqrt [3]{(x^{4} - y^{4})^{3}} : \sqrt [3]{(x^{2} + y^{2})^{6}} = (x^{4} - y^{4}) : (x^{2} + y^{2})^{2}$
$= \dfrac {x^{4} - y^{4}}{(x^{2} + y^{2})^{2}} = \dfrac {(x^{2} - y^{2})(x^{2} + y^{2})}{(x^{2} + y^{2})^{2}}$
$= \dfrac {x^{2} - y^{2}}{x^{2} + y^{2}} = x^{2} - y^{2} : x^{2} + y^{2}$

Multiple choice composition of ratios types of ratios ratio and proportions ratio and proportion maths

If $a : b = 2 : 3$ and $b : c = 4 : 7$ then the reciprocal ratio of $a : c$ is ____

  1. $8 : 21$
  2. $21 : 8$
  3. $7 : 4$
  4. $3 : 2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$a : b = 2 : 3$ and $b : c = 4 : 7$
$\therefore a : c = \dfrac {a}{b} \times \dfrac {b}{c} = \dfrac {2}{3} \times \dfrac {4}{7}$
$= \dfrac {8}{21}$
$\therefore a : c = 8 : 21$
$\therefore$ The reciprocal ratio of $a : c$ is $21 : 8$.

Multiple choice composition of ratios types of ratios ratio and proportions ratio and proportion maths

_____ is the subtriplicate ratio of $(a + b)^{3} : (a^{2} - b^{2})^{3}$

  1. $(a + b) : 1$
  2. $1 : (a + b)$
  3. $1 : (a - b)$
  4. $(a - b) : 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The subtriplicate ratio of $a : b$ is $\sqrt [3]{a} : \sqrt [3]{b}$
$\therefore$ The subtriplicate ratio of $(a + b)^{3} : (a^{2} - b^{2})^{3}$ is $\sqrt [3]{(a + b)^{3}} : \sqrt [3]{(a^{2} - b^{2})^{3}} = (a + b) : (a^{2} - b^{2})$
$= \dfrac {a + b}{a^{2} - b^{2}} = \dfrac {a + b}{(a - b)(a + b)} = \dfrac {1}{a - b} = 1 : (a - b)$.

Multiple choice

What is the golden ratio?

  1. 16:9

  2. 3:1

  3. 4:3

  4. 1:1.618
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The golden ratio is a mathematical ratio approximately equal to 1:1.618, often used in art and design to create aesthetically pleasing compositions.

Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

If m is the variance of P.D., then the ratio of sum of the terms in odd places to the sum of the terms in even places is

  1. $e^{-m}\cosh m$
  2. $e^{-m}\sinh m$
  3. $\coth m$
  4. $\tanh m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If m is the variance of P. D, then  P$(x; \mu) = \frac { { e }^{-\mu}{\mu}^{x}}{x!} $
Sum of the terms in odd places = [ P$(0; \mu) + P(2; \mu)  + P(4; \mu) + .........$ ]


                                       =  [$\dfrac { { e }^{-\mu}{\mu}^{0}}{0!} + \dfrac { { e }^{-\mu}{\mu}^{2}}{2!} + \dfrac { { e }^{-\mu}{\mu}^{4}}{4!} + ....... $]

                                      =  $  { e }^{-\mu} [ 1 + \dfrac {{\mu}^{2}}{2!} + \dfrac {{\mu}^{4}}{4!} + .......]$       --------------------- (1)

  Since   ${ e }^{ x }=1+\dfrac { x }{ 1! } +\dfrac { { x }^{ 2 } }{ 2! } +\dfrac { { x }^{ 3 } }{ 3! } +\dfrac { { x }^{4} }{ 4! } +....$

    ${ e }^{ -x }=1+\dfrac {( -x )}{ 1! } +\dfrac { { (-x) }^{ 2 } }{ 2! } +\dfrac { {( -x) }^{ 3 } }{ 3! } +\dfrac { { (-x) }^{4} }{ 4! } +....$ 

on adding , we get  
 ${e}^{x} + { e }^{-x} =  2[ 1 + \dfrac {{x}^{2}}{2!} + \dfrac {{x}^{4}}{4!} + .....] $
$ \dfrac {{e}^{x} + { e }^{-x}}{2} =  [ 1 +\dfrac { { x }^{ 2} }{ 2! } + \dfrac {{x}^{4}}{4!} + .....] $
So, 
$ \dfrac {{e}^{\mu} + { e }^{-\mu}}{2} =  [ 1+\dfrac { { \mu }^{ 2 } }{ 2! } + \dfrac {{\mu}^{4}}{4!} + .....] $

put this value in equation (1),
Sum of the terms in odd places = $  { e }^{-\mu} [ 1 + \dfrac {{\mu}^{2}}{2!} + \dfrac {{\mu}^{4}}{4!} + .......]$
                    =  $  { e }^{-\mu} (\dfrac {{e}^{\mu} + { e }^{-\mu}}{2}) $
                    = $ { e }^{-\mu} \cosh { \mu  } $
                    = $ { e }^{-m} \cosh {m} $       [ Variance (m) is equal to mean ($\mu$) in Poisson distribution ]
Similarly Sum of the terms in even places =  $ { e }^{-m} \sinh {m} $ 
The ratio of sum of the terms in odd places to the sum of the terms in even places is = $ \dfrac { { e }^{ -m }\cosh { m }  }{ { e }^{ -m }\sinh { m }  } =\coth { m }   $ 

Multiple choice maths average arithmetic mean of ap introduction to averages means

If $n\ AM's$ are inserted between $1$ and $31$ and ratio of ${7}^{th}$ and $(n-1)^{th}$ $A.M.$ is $5:9$ then $n$ equals ?

  1. $12$
  2. $13$
  3. $14$
  4. $None$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the formula for the k-th arithmetic mean between a and b, A_k = a + k(b-a)/(n+1). Setting up the ratio A_7 / A_(n-1) = 5/9 and solving for n yields n = 14.

Multiple choice maths average arithmetic mean of ap introduction to averages means

The ratio of sum of n arithmetic means between two given numbers to that of single arithmetic mean between them id

  1. n : 1
  2. n$^2$ : 1

  3. 1 : 1
  4. $\sqrt{n}$ : 1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of n arithmetic means between a and b is n * (a+b)/2. The single arithmetic mean is (a+b)/2. The ratio is therefore n : 1.

Multiple choice maths fraction lowest form of a fraction simplest ratio lowest form of fractions

Three number $ A, B$ and $C$ are in the ratio of $12 : 15 : 25 .$ If the some of these numbers be $364$ find the ratio between the difference of $B$ and $A$ and the difference of $C$and $B ?$

  1. $3 : 2$
  2. $3 : 10$
  3. $3 : 5$
  4. $4 : 2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $A=12k,B=15k, C=25k$

Now $A+B+C=364$
$12k+15k+25k=364$
$52k=364$
$k=\dfrac{364}{52}=7$
$\dfrac{B-A}{C-B}=\dfrac{15k-12k}{25k-15k}=\dfrac{3k}{10k}=\dfrac{3}{10}$
Hence the correct option is (B).