Tag: ratio, proportion and unitary method

Quizzes (5) Questions (125)

Questions Related to ratio, proportion and unitary method

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

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-

  1. $21:1$
  2. $1:14$
  3. $7:30$
  4. $30:7$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $

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

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. 

  1. $80 : 50$
  2. $3 : 5$
  3. $5 : 3$

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $5:3$.

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

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

  1. $4 : 15 : 6$
  2. $4 : 5 : 6$
  3. $20 : 15 : 12$
  4. $20 : 15 : 18$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $5 : 6$.

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

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

  1. $2 : 5 : 7$
  2. $2 : 5 : 6$
  3. $12 : 30 : 35$
  4. $12 : 10 : 42$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $12 : 30 : 35$

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

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?

  1. $20$ years
  2. $22$ years
  3. $24$ years
  4. $25$ years
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. 149

  2. 150

  3. 151

  4. 152
Reveal answer Fill a bubble to check yourself
B Correct answer
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.