Tag: maths

Questions Related to maths

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

maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

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

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

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

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

Show answer
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?

maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

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

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

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

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

Show answer
Correct Option: D
Explanation:

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

If a:b=$3$:$5$, find 
$\left( {10a + 3b} \right):\left( {5a + 2b} \right)$

maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

  1. $9:8$

  2. $3:5$

  3. $9:5$

  4. $7:5$

Show answer
Correct Option: C
Explanation:

Given $a:b=3:5$, Let $a=3k$ and $b=5k$


Then, $(10a+3b):(5a+2b)$

$=\dfrac{10\times3k+3\times5k}{5\times3k+2\times5k}$

$=\dfrac{30k+15k}{15k+10k}$

$=\dfrac{45k}{25k}$

$=\dfrac{9}{5}$

$(10a+3b):(5a+2b)=9:5$

The total population of a village is  $3540,$  out of which  $2065$  are males. Find the ratio of males to females.

maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

  1. $\dfrac 57$

  2. $\dfrac 35$

  3. $\dfrac 75$

  4. $\dfrac 65$

Show answer
Correct Option: C

Sides of two similar triangles are in the ratio $4:9$.Area of these triangles are in the ratio

maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

  1. $2:3$

  2. $4:9$

  3. $81:16$

  4. $16:81$

Show answer
Correct Option: D
Explanation:
Given:Ratio of sides of similar triangles$=\dfrac{4}{9}$

We know that if two triangles are similar, 

ratio of areas is equal to the ratio of squares of corresponding sides.

So, $\dfrac{area \,\, of\,\, triangle\,\, 1}{area \,\, of\,\, triangle\,\, 2}=\dfrac{{\left(side\,\, of\,\, triangle\,\, 1\right)}^{2}}{{\left(side\,\, of\,\, triangle\,\, 2\right)}^{2}}$

$\dfrac{area \,\, of\,\, triangle\,\, 1}{area \,\, of\,\, triangle\,\, 2}={\left(\dfrac{4}{9}\right)}^{2}=\dfrac{16}{81}$

$\dfrac{area \,\, of\,\, triangle\,\, 1}{area \,\, of\,\, triangle\,\, 2}=\dfrac{16}{81}$

The ratio of the present ages of two brothers is $1:2$ and $5$ years back the ratio was $1:3$. What will be the ratio of their ages after $5$ years?

maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

  1. $1:4$

  2. $2:3$

  3. $3:5$

  4. $5:6$

Show answer
Correct Option: C
Explanation:

Let the age of the two brothers be $x$ and $y$ respectively


Given


At present 

$\dfrac { x }{ y } =\dfrac { 1 }{ 2 } \Rightarrow y=2x$

Five years ago

$\dfrac { x-5 }{ y-5 } =\dfrac { 1 }{ 3 }$

substitute $y=2x$ 

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

$\Rightarrow 3x-15=2x-5\Rightarrow x=10$

$\Rightarrow y=2x=2(10)=20$

$\therefore\ x=10$ and $y=20$

Required ratio:

$\displaystyle \frac { x+5 }{ y+5 } =\frac { 10+5 }{ 20+5 } =\frac { 15 }{ 25 } =\frac { 3 }{ 5 }$

Hence option (C) is the correct option.

A jar contains black and white marbles. If there are 25 marbles in the jar, then which of the following could not be the ratio of black to white marbles?

maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

  1. $\dfrac{12}{13}$

  2. $\dfrac{11}{14}$

  3. $\dfrac{1}{10}$

  4. $\dfrac{8}{17}$

Show answer
Correct Option: C
Explanation:

In this Question the sum of ratio of marbles must be 25.
A. $12+13=25$
B. $11+14=25$
C. $1+10=11$
D. $8+17=25$

If $A\colon\,B=2\colon3,\,B\colon\,C=4\colon5\,$ and $\,C\colon\,D=6\colon7$, then $A\colon\,D=?$

maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

  1. $\;2\colon7$

  2. $\;7\colon8$

  3. $\;16\colon35$

  4. $\;4\colon13$

Show answer
Correct Option: C
Explanation:

Given, $A:B=2:3$, $B:C=4:5$ and $C:D=6:7$


Then $\dfrac{A}{B}\times \dfrac{B}{C}=\dfrac{2}{3}\times \dfrac{4}{5}\Rightarrow \dfrac{A}{C}=\dfrac{8}{15}$

And $\dfrac{A}{C}\times \dfrac{C}{D}=\dfrac{8}{15}\times \dfrac{6}{7}$

$\Rightarrow \dfrac{A}{D}=\dfrac{16}{35}$

The condition for two ratios to be equal is

maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

  1. Product of means is equal to antecedents

  2. Product of extremes is equal to consequents

  3. Antecedents are equal to consequents

  4. Product of means is equal to product of extremes

Show answer
Correct Option: D
Explanation:

For the ratio to be equal the product of mean =product of extremes
for ex   $\dfrac{a}{b}=\dfrac{c}{d}$
product of $a\times d=b\times c$
where
$a\times d$ $=$product of extreme.
and$b\times c$ $=$ product of means.

If $a:b = 3:4$ and $b:c = 8:9$ , then $a:c =$ ?

maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

  1. $1:2$

  2. $3:2$

  3. $1:3$

  4. $2:3$

Show answer
Correct Option: D
Explanation:

Given  $a:b = 3:4$ and $b:c = 8:9$
Then $\displaystyle \frac{a}{b}\times \frac{b}{c}=\frac{3}{4}\times \frac{8}{9}$

$\therefore \displaystyle \frac{a}{c}=\frac{2}{3}$

$\therefore a:c=2:3$