Tag: other quantities

Questions Related to other quantities

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. $16:81$

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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. $5:6$

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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. $\dfrac{8}{17}$

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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. $\;4\colon13$

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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

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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. $2:3$

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Correct Option: D
Explanation:

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

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

$\therefore a:c=2:3$

If $A\, :\, B\, =\, \displaystyle {\frac{1}{2}\, :\, \frac{3}{8},\, \, B\, :\, C\, =\, \frac{1}{3}\, :\, \frac{5}{9}\, , \, C\, :\, D\, =\, \frac{5}{6}\, :\, \frac{3}{4}}$, then the ratio $A : B : C : D$ is

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

  1. $4 : 6 : 8 : 10$

  2. $6 : 4 : 8 : 10$

  3. $6 : 8 : 9 : 10$

  4. $8 : 6 : 10 : 9$

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Correct Option: D
Explanation:

$A\, :\, B\, = \, \displaystyle {\frac{1}{2}\, :\, \frac{3}{8}\,  =\, 4\, :\, 3,}$


$B\, :\, C\, = \, \displaystyle {\frac{1}{3}\, :\, \frac{5}{9}\, =\, 3\, :\, 5,}$
$C\, :\, D\, = \, \displaystyle {\frac{5}{6}\, :\, \frac{3}{4}\, =\, 10\, :\, 9,}$ 
$\Rightarrow A : B = 4 : 3, B : C = 3 : 5 \, and \, C : D = 5 : \displaystyle \frac{9}{2}$
$\Rightarrow A : B : C : D = 4 : 3 : 5 : \displaystyle \frac{9}{2}$
$= 8 : 6 : 10 : 9$

If a : b = 3 : 4 and b : c = 5 : 9 , then a : b : c is

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

  1. 15 : 20 : 36

  2. 20 : 36 :15

  3. 3 : 20 : 9

  4. None of these

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Correct Option: A
Explanation:

To find a : b : c, b is made same in both the ratios. L.C.M of 4 and 5 is 20
$a\, :\, b\, =\, \displaystyle {\frac{3}{4}\, =\, \frac{3\, \times\, 5}{4\, \times\, 5}\, =\, \frac{15}{20}}$
$b\, :\, c\, =\, \displaystyle {\frac{5}{9}\, =\, \frac{5\, \times\, 4}{9\, \times\, 4}\, =\, \frac{20}{36}}$
$\therefore$   $ a : b : c = 15 : 20 : 36$

If a : b = 8 : 9, b : c = 18 : 40, then a : c is

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

  1. 1 : 5

  2. 5 : 2

  3. 2 : 5

  4. None of these

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Correct Option: C
Explanation:

Given a : b = 8 : 9, b : c = 18 : 40
$\displaystyle {\frac{a}{c}\, =\, \frac{a}{c}\, \times\, \frac{b}{b}\, =\, \frac{a}{b}\, \times\, \frac{b}{c}\, =\, \frac{8}{9}\, \times\, \frac{18}{40}\, =\, \frac{2}{5}}$
a : c = 2 : 5

In a class, there are $50$ boys and $30$ girls. The ratio of the number of boys to the number of girls in the class is:

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

  1. $80 : 50$

  2. $3 : 5$

  3. $5 : 3$

  4. none of the above

Show answer
Correct Option: C
Explanation:
Number of boys in class $=50$
Number of girls in class $=30$
Ratio of number of boys to the number of girls in the class $=50:30=5:3$