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Questions Related to maths

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

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


$7+4=11.$

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

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

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

The ratio of $\left(\displaystyle\dfrac{1}{3}\text{ of Rs. }9.30\right)$ to $(0.6\text{ of Rs. }1.55)$ is ___________.

  1. $1:3$
  2. $10:3$
  3. $3:10$
  4. $3:1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle\frac{1}{3}$ of Rs. $9.30=Rs. \left(\displaystyle\frac{1}{3}\times 9.30\right)=Rs. 3.10$
$0.6$ of Rs. $1.55=$Rs. $(0.6\times 1.55)=$ Rs. $0.93$
$\therefore$ Required ratio$=3.10:0.93=10:3$.

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

Cost of a candy is $50$ paise and cost of an ice cream is Rs. $20$, then the ratio of the cost of a candy to the cost of an ice-cream is __________.

  1. $40:1$
  2. $1:40$
  3. $1:20$
  4. $1:15$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Cost of $1$ candy $=50$ paise
Cost of $1$ ice-cream $=Rs. 20=20\times 100$ paise
$=2000$ paise
$\therefore$ Required ratio $=50:2000=1:40$.

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

Out of $30$ students in a class, $6$ like football, $12$ like cricket and remaining like tennis. The ratio of number of students who like tennis to the total number of students, is:

  1. $2:3$
  2. $5:1$
  3. $2:5$
  4. $5:2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Total number of students$=30$
Number of students who like football$=6$
Number of students who like cricket$=12$
$\therefore$ Number of students who like tennis $=30-6-12=12$
$\therefore$ Required ratio $=12:30=2:5$.
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Compare the following pairs of surds. $\sqrt[8]{80}, \sqrt[4]{40}$    

  1. $\sqrt[8]{80} < \sqrt[4]{40}$
  2. $\sqrt[8]{80} \neq \sqrt[4]{40}$
  3. $\sqrt[8]{80} = \sqrt[4]{40}$
  4. $\sqrt[8]{80} > \sqrt[4]{40}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

   $\sqrt[8]{80}, \sqrt[4]{40}$
$={80}^{\frac{1}{8}}, {40}^{\frac{1}{4}}$
$={80}^{\frac{1}{8}}, {40}^{\frac{2}{8}}$
$={80}^{\frac{1}{8}}, {1600}^{\frac{1}{8}}$
Now,
   $80<1600$
$=>{80}^{\frac{1}{8}}<{1600}^{\frac{1}{8}}$
$=>\sqrt[8]{80}< \sqrt[4]{40}$

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Which among the following numbers is the greatest?
$\displaystyle 0.07+\sqrt{0.16},\sqrt{1.44},1.2\times 0.83,1.02-\frac{0.6}{24}$

  1. $\displaystyle \sqrt {1.44}$
  2. $\displaystyle 0.07+\sqrt{0.16}$
  3. $1.2\times 0.83$
  4. $1.02-\dfrac{0.6}{24}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\Rightarrow 0.07+\sqrt{0.16}=0.07+0.4=0.47$
$\Rightarrow \sqrt{1.44}=1.2$
$\Rightarrow 1.2 \times 0.83 = 0.996$
$\Rightarrow 1.02-\cfrac{0.6}{24}=1.02-\cfrac {6}{240}=1.02-0.025=0.995$
$ \therefore$ The greatest number is $1.2$ i.e. $\sqrt {1.44}$
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

State whether the following equality is true or false:

$\displaystyle \frac{2\sqrt{3}}{\sqrt{5}} = $$\displaystyle \frac{2\sqrt{15}}{\sqrt{5}}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Rationalizing factor of  $\displaystyle \frac{2\sqrt{3}}{\sqrt{5}}$ is $\sqrt{5}$


$\therefore \displaystyle \frac{2\sqrt{3}}{\sqrt{5}}$  $=\displaystyle \frac{2\sqrt{3}\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}$

$= \dfrac{2\sqrt{15}}{5}$