Tag: other quantities

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Questions Related to other quantities

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.

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

In a box, the ratio of the number of red marbles to that of blue marbles is $4 : 7$. which of the following could be the total number of in the box?

  1. $14$
  2. $21$
  3. $22$
  4. $28$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$With\quad ratios,add\quad all\quad the\quad parts\quad together\quad to\quad get\quad a\quad total.In\quad this\quad case,it\quad is\quad 4red\quad and\quad 7blue.$


$7+4=11.$

$Therefore,the\quad lowest\quad total\quad number\quad of\quad marbles\quad is11.The\quad answer\quad will\quad be\quad any\quad number\quad divisible\quad by\quad 11.$

$Hence\quad it\quad is\quad 22.$