Tag: properties of irrational numbers

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Questions Related to properties of irrational numbers

Multiple choice maths number systems existence of irrational numbers irrational numbers properties of irrational numbers

$A,B,C$ and $D$ are all different digits between $0$ and $9$. If $AB+DC=7B\ (AB,DC$ and $7B$ are two digit numbers), then the value of $C$ is

  1. $0$
  2. $1$
  3. $2$
  4. $3$
  5. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

AB + DC = 7B. (10A + B) + (10D + C) = 70 + B. 10A + 10D + C = 70. A + D + C/10 = 7. Since A, D, C are digits, C must be 0 for the equation to hold with integer digits A and D. If C=0, A+D=7.

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

If $\sqrt{a}$ is an irrational number, what is a? 

  1. Rational

  2. Irrational
  3. $0$

  4. Real
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider the given irrational number$\sqrt{a}$ ,

Definition  of rational number- which number can be write in the form of $\dfrac{p}{q}$ but $q\ne 0$ is called rational number.

Hence, $a=\dfrac{a}{1}$

That why  $a$ is rational number

 

Hence, this is the answer.

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

Which of the following is irrational

  1. $\sqrt {\dfrac{4}{9}} $
  2. $\dfrac{4}{5}$
  3. $\sqrt 7 $
  4. $\sqrt {81} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
A $=\sqrt{\dfrac{4}{9}}=\dfrac{2}{3}$         Rational

B $=\dfrac{4}{5}$                       Rational

C $=\sqrt7$                     Irrational

D $=\sqrt{81}=9$          Rational
Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

Which of the following rational number represents a terminating decimal expansion?

  1. $

    \dfrac { 77 } { 210 }

    $
  2. $

    \dfrac { 13 } { 125 }

    $
  3. $

    \dfrac { 2 } { 15 }

    $
  4. $

    \dfrac { 17 } { 18 }

    $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Any rational number its denominator is in the form of $2^m\times 5^n$, where $m,n$ are positive integer s are terminating decimals.

Solution is $B$ as $A$ is non terminating decimals.
$A =\dfrac{77}{210}= 0.366......$

$B =\dfrac{13}{125}= 0.104$

$C =\dfrac{2}{15}= 0.133.....$

$D =\dfrac{17}{18}=  0. 9444....$
Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

Read out each of the following numbers carefully and specify the natural numbers in it.
$87, 54, 0, -13, -4.7, \sqrt{7}, 2{1}{7}, \sqrt{15}, -{8}{7}, 3\sqrt{7}, 4.807, 0.002, \sqrt{16}$ and $2+\sqrt{3}.$

  1. $0,87,54,\sqrt{16}$
  2. $87, 54,$ $\sqrt{16}$, $217$
  3. $0, -13, -4,7, 217, 54, 87$
  4. $\sqrt{7}$, $\sqrt{15}$, $3 \sqrt{7}$, $\sqrt{16}$, $2 + \sqrt{3}$,
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Natural numbers from the given list are 87, 54,  $\sqrt { 16 } =4$ and 217

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

There can be a pair of irrational numbers whose sum is irrational 

Such as: $\displaystyle \sqrt{3}+2$ and $\displaystyle 5+\sqrt{2}$

  1. True

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

To get the sum as irrational, the numbers need to have an irrational part as well which are different from each other.

Example, the pair of numbers $ \sqrt{3} + 2 $ and $ 5 + \sqrt {2} $ have the sum $ \sqrt{3} + 2 + 5 + \sqrt {2} = 7 + \sqrt {2} + \sqrt {3} $ which is an irrational number too.

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

Simplify : 

$\displaystyle \sqrt{2}\times \sqrt[3]{3} \times \sqrt[4]{4}$.

  1. $\sqrt[3]{12}$
  2. $\sqrt[3]{24}$
  3. $\sqrt[3]{20}$
  4. $\sqrt[3]{25}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ \sqrt{2} \times \sqrt[3] {3} \times \sqrt[4]{4}$
$=2^{ \frac { 1 }{ 2 }  } \times 3^{ \frac { 1 }{ 3 }  }\times 2^{ \frac { 2 }{ 4 }  }$
$=2^{ \frac { 1 }{ 2 }  } \times 2^{ \frac { 1 }{ 2 }  }\times 3^{ \frac { 1 }{ 3 }  }$
$=2  \times3^{ \frac { 1 }{ 3 }  }$
$=2^{ \frac { 3 }{ 3 }  }\times3^{ \frac { 1 }{ 3 }  }  $
$=\sqrt [ 3 ]{ 2^{ 3 } }\times\sqrt[3]{3}$
$=\sqrt[3]{8\times3}$
$=\sqrt[3]{24}$