Tag: ratio, proportion and unitary method

Questions Related to ratio, proportion and unitary method

If A:B =$3 :4$ and B:C =$5:6$, then A:C is

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

  1. $5 : 9$

  2. $3 : 4$

  3. $1 : 2$

  4. $5 : 8$

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Correct Option: D
Explanation:

Given that,  If $A:B=3:4,B:C=5:6$,then$A:C$

Given$ a:b=3:4$  and $b:c= 5:6$  

$a:b::b:c =3:4::5:6$ 

$a:c= 3/4*5/6=5/8$ 


Hence, this is the answer. 

In an office the working hours are 9:30 AM to 4:30 PM and in between 20 minutes are spent on lunch Find the ratio of office hours to the time spent for lunch-

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

  1. $21:1$

  2. $1:14$

  3. $7:30$

  4. $30:7$

Show answer
Correct Option: A
Explanation:

Total time in office $ = 9:30  to  4:30  = 7  hours  $

Ratio of two quantities can be found when they are of same units.

So, $ 7  hours   = 7 \times 60 = 420  minutes   $
Hence, ratio $ = \dfrac {420  min }{20  min } = 21:1 $

In an office the working hours are 10.30 AM to 5.30 PM and in between 30 minutes are spent on lunch. Find the ratio of office hours to the time spent for lunch.

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

  1. 7:30`

  2. 1:14

  3. 14:1

  4. 30:7

Show answer
Correct Option: C
Explanation:

Office hours
$= 10.30$ AM to $5.30$ AM=
 $i.e., 7 hrs = 420 min$.
Lunch time $= 30$ min
$420:30$
14:1

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

Show answer
Correct Option: C
Explanation:

To find the ratio of two numbers, we have to consider their fraction. 


Here, it is given that there are $50$ boys and $30$ girls in the class, then the ratio of the number of boys to the number of girls is:

$\dfrac { 50 }{ 30 } =\dfrac { 5 }{ 3 } =5:3$

Hence, the ratio of the number of boys to the number of girls is $5:3$.

If $a:b=5:7$ and $b:c=6:11$, then $a:b:c=$

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

  1. $55:77:66$

  2. $30:42:77$

  3. $35:49:42$

  4. $50:48:49$

Show answer
Correct Option: B
Explanation:

$b:c=\begin{pmatrix}6\times\dfrac{7}{6}\end{pmatrix}:\begin{pmatrix}11\times\dfrac{7}{6}\end{pmatrix}=7:\dfrac{77}{6}$
$a:b:c=5:7:\dfrac{77}{6}=30:42:77$

The continued ratio of $4 : 3$ and $5 : 6$ is ____

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

  1. $4 : 15 : 6$

  2. $4 : 5 : 6$

  3. $20 : 15 : 12$

  4. $20 : 15 : 18$

Show answer
Correct Option: D
Explanation:

$Ratio 1 = 4 : 3$
$Ratio 2 = 5 : 6$
Multiplying ratio $1$ by antecedent of ratio $2$ and ratio $2$ by consequent of ratio $1$,
Ratio $1 = 20 : 15$
Ratio $2 = 15 : 18$
Thus, for two ratios $a : b$ and $b : c, a : b : c$ is called the continued ratio.
$\therefore 20 : 15 : 18$ is the continued ratio for $4 : 3$ and $5 : 6$.

The continued ratio of $2 : 5$ and $6 : 7$ is _____

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

  1. $2 : 5 : 7$

  2. $2 : 5 : 6$

  3. $12 : 30 : 35$

  4. $12 : 10 : 42$

Show answer
Correct Option: C
Explanation:

$R _{1}$ = $2 : 5$
$R _{2}$ =$ 6 : 7$
Multiplying the LCM of the consequent of ratio $1$ and antecedent of ratio $2$, to both the ratios.
$R _1$ = $12 : 30$
$R _2$ = $30 : 35$
Thus, the continued ration for $2 : 5$ and $6 : 7$ is $12 : 30 : 35$

The ratio between the ages of $A$ and $B$ is $2 : 5$. After $8$ years, their ages will be in the ratio $1 : 2$. What is the difference between their present ages?

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

  1. $20$ years

  2. $22$ years

  3. $24$ years

  4. $25$ years

Show answer
Correct Option: C
Explanation:

Let $A = 2x; B = 5x$ be the present ages of $A$ and $B$ respectively.
After $8$ years, their ages will be $(2x + 8)$ years and $(5x + 8)$ years respectively.
Therefore, $ (2x + 8) : (5x + 8) : : 1 : 2$
$\Rightarrow  (5x + 8)\times 1 = 2(2x + 8)$
$\Rightarrow x = 8$
Thus difference of their present ages is $5x - 2x = 3x$ i.e., $24$ years.

For $\dfrac { { 2 }^{ 2 }+{ 4 }^{ 2 }+{ 6 }^{ 2 }+....+{ \left( 2n \right)  }^{ 2 } }{ { 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+....+{ \left( 2n-1 \right)  }^{ 2 } }$ to exceed $1.01$, the maximum value of $n$ is

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

  1. 149

  2. 150

  3. 151

  4. 152

Show answer
Correct Option: B
Explanation:

Given


$\dfrac { { 2 }^{ 2 }+{ 4 }^{ 2 }+{ 6 }^{ 2 }....+{ (2n) }^{ 2 } }{ { 1 }^{ 2 }{ +3 }^{ 2 }{ +5 }^{ 2 }{ ....+(2n-1) }^{ 2 } } =\dfrac { \sum { { (2n) }^{ 2 } }  }{ \sum { { (2n-1) }^{ 2 } }  } $

$\sum { { (2n) }^{ 2 }=\sum { 4{ n }^{ 2 } } =4\times \sum { { n }^{ 2 } } =\dfrac { 4(n)(n+1)(2n+1) }{ 6 }  } $[since $\sum { { n }^{ 2 } } =\dfrac { (n)(n+1)(2n+1) }{ 6 } $]

$\sum { { (2n-1) }^{ 2 }=\sum { 4{ n }^{ 2 }+1-4n } =4\sum { { n }^{ 2 }+\sum { 1 }  }  } -4\sum { n } =\dfrac { 4(n)(n+1)(2n+1) }{ 6 } +n-\dfrac { 4(n)(n+1) }{ 2 } $[since $\sum { { n }^{ 2 }= } \dfrac { (n)(n+1) }{ 2 } $]

Now solving numerator and denominator we get

$\dfrac { { 4n }^{ 2 }+6n+2 }{ 4{ n }^{ 2 }-1 } $ to exceed $1.01$

 $n\Rightarrow$  $\in[0,150]$

Therefore maximim value of $n$ is 150.

The sum of two number is $50$. If the number are in the ration $2 : 3$. Find the number.

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

  1. $20,30$

  2. $10,40$

  3. $15,35$

  4. $25,25$

Show answer
Correct Option: A
Explanation:

Let the two numbers be $2x$ and $3x.$ Thus,


$2x+3x=50$

$5x=50$

$x=10$

Thus, the required numbers are $20$ and $30$.