Tag: maths

Questions Related to maths

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 $30$.

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?

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

  1. $14$

  2. $21$

  3. $22$

  4. $28$

Show answer
Correct Option: C
Explanation:

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


$7+4=11.$

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

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

If A : B = 7 : 5 and B : C = 9 : 11, then A : B : C is equal to

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

  1. 55:45:63

  2. 63:45:55

  3. 45:63:55

  4. None of these

Show answer
Correct Option: B
Explanation:

$A:B=7:5\B:C=9:11$

Let $A=7p, B=5p\B=9q,C=11q\\because 5p=99\ \therefore p=\cfrac{99}{5}\ \therefore A=7\times\cfrac{99}{5}=\cfrac{639}{5}\ \therefore A:B:C=\cfrac{639}{5}:9q:11q\A:B:C=63:45:55$
Option B

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

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

  1. $1:3$

  2. $10:3$

  3. $3:10$

  4. $3:1$

Show answer
Correct Option: B
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$=3.10:0.93=10:3$.

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 __________.

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

  1. $40:1$

  2. $1:40$

  3. $1:20$

  4. $1:15$

Show answer
Correct Option: B
Explanation:

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

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:

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

  1. $2:3$

  2. $5:1$

  3. $2:5$

  4. $5:2$

Show answer
Correct Option: C
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 $=12:30=2:5$.

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

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

  1. $\sqrt[8]{80} < \sqrt[4]{40}$

  2. $\sqrt[8]{80} \neq \sqrt[4]{40}$

  3. $\sqrt[8]{80} = \sqrt[4]{40}$

  4. $\sqrt[8]{80} > \sqrt[4]{40}$

Show answer
Correct Option: A
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,
   $80<1600$
$=>{80}^{\frac{1}{8}}<{1600}^{\frac{1}{8}}$
$=>\sqrt[8]{80}< \sqrt[4]{40}$

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

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

  1. $\displaystyle \sqrt {1.44}$

  2. $\displaystyle 0.07+\sqrt{0.16}$

  3. $1.2\times 0.83$

  4. $1.02-\dfrac{0.6}{24}$

Show answer
Correct Option: A
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. $\sqrt {1.44}$

State whether the following equality is true or false:

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

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

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

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


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

$= \dfrac{2\sqrt{15}}{5}$