Tag: ratio, proportion and unitary method

Questions Related to ratio, proportion and unitary method


select the missing number based on the given
$27: 65 :: 54: $____

  1. $130$

  2. $127$

  3. $126$

  4. $125$


Correct Option: A
Explanation:

$let\> the\> missing\> number\> be \>x \\ then\> (\frac{27}{65})=(\frac{54}{x})\\  \>x=(\frac{54\times 65}{27})\\= 130$

A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains $25$%. The percentage of water in the mixture is

  1. $4$%

  2. $6\frac { 1 }{ 4 } $%

  3. $20$%

  4. $25$%


Correct Option: C
Explanation:

According to the question,

Let's assume milkman has $100$ liter of milk, If he added $x$ liter of water, the percentage of water in the mixture$=\dfrac {x}{(100+x)}\times 100$

The milk gain $25$%, he must have added $25$ liter water in $100$ liter of milk. Then the percentage of water in the mixture$=\dfrac {25}{(100+25)}\times 100=20\%$


In the new mixture, If the milk is $80$% then $80$% of total mixture should be $100$ liter
$(100+x)\times \dfrac{80}{100}=100$
$8x=200$
$x=25$

Then, the percentage of water in the mixture $=\dfrac {25}{(100+25)}\times 100=20\%$

Convert the following into percentage.
$\dfrac{2}{3}$.

  1. $66\%$

  2. $65\%$

  3. $33\%$

  4. $66.67\%$


Correct Option: D
Explanation:
$\dfrac{2}{3}$ into percentage
$=\dfrac{2}{3}\times 100\%= \dfrac{200}{3}\% = 66.67\%$

Convert the following into percentage.
$\dfrac{5}{8}$.

  1. $62.5 \%$

  2. $65.5\%$

  3. $66.6\%$

  4. $64\%$


Correct Option: A
Explanation:
$\dfrac{5}{8}$
$=\dfrac{5}{8}\times 100\%=\dfrac{500}{8}\%=62.5\%$

If $\dfrac{1}{x} + y = 3$ and $x + \dfrac{1}{y} = 2$ then $x:y$ is 

  1. $3:2$

  2. $2:3$

  3. $1:2$

  4. $2:1$


Correct Option: B
Explanation:
The question states '...then x: is' 
It should state '..then x:y is'

Given
$\dfrac { 1 }{ x } +y=3$ --- Eqn (1)
$x+\dfrac { 1 }{ y } =2$ ---Eqn (2)

Multiplying Eqn (1) by x and Eqn (2) by y, we get:

$x\left( \dfrac { 1 }{ x } +y \right) =3x\quad \Rightarrow 1+yx=3x$ --- Eqn (3)
$y\left( x+\dfrac { 1 }{ y }  \right) =2y\quad \Rightarrow xy+1=2y$ --- Eqn (4)

Subtracting Eqn (4) and Eqn (5), we get:

$1+yx-1-yx=3x-2y$
$\Rightarrow 0=3x-2y$
$\Rightarrow 3x=2y$
$\Rightarrow \dfrac { x }{ y } =\dfrac { 2 }{ 3 } $
$\therefore x:y=2:3$

Hence the answer is B

Find the ratio of: $36$ to $64$

  1. $9:11$

  2. $9:12$

  3. $9:16$

  4. $9:17$


Correct Option: C
Explanation:

$\dfrac{36}{64}=\dfrac{36\div 4}{64\div 4} =\dfrac{9}{16}$

The ratio of $40\ min$ to $2.5$ hours is

  1. $4:17$

  2. $4:18$

  3. $4:13$

  4. $4:15$


Correct Option: D
Explanation:
$\dfrac {40\ min}{2.5\ hr}$
$=\dfrac {10\ min}{2hr +30\ min}=\dfrac {40}{2\times 60+30}$
$=\dfrac {40}{150}\ \Rightarrow \boxed {4:15}$
Option $D$ is correct

If ${x+y}{ax+by}=\dfrac{y+z}{ay+bz}=\dfrac{z+x}{az+bx}$, then "each of these ratio is equal to $\dfrac{2}{a+b}$, unless $x+y+z=0$." this statement is ____

  1. True

  2. False


Correct Option: A

$30$ cricket players and $20$ kho-kho players are training on a field. What is the ratio cricket players to the total number of players?

  1. $\dfrac {3}{2}$

  2. $\dfrac {2}{5}$

  3. $\dfrac {3}{5}$

  4. $\dfrac {1}{5}$


Correct Option: C
Explanation:

$\dfrac{\text{number of cricket players}}{\text{total number of players}}=\dfrac{30}{30+20}=\dfrac{30}{50} =\dfrac{3}{5}$

Snehal has a red ribbon that is $80cm$ long and a blue ribbon, $220m$ long. What is the ratio of the length of the red ribbon to that of the blue ribbon?

  1. $\dfrac {4}{21}$

  2. $\dfrac {4}{5}$

  3. $\dfrac {5}{11}$

  4. $\dfrac {4}{11}$


Correct Option: D
Explanation:

$\dfrac{\text{length of red ribbon}}{\text{length of blue ribbon}}= \dfrac{80cm}{220cm} = \dfrac{8}{22} = \dfrac{4}{11}$