Tag: rational numbers on the number line

Questions Related to rational numbers on the number line

 Rational numbers between $\displaystyle \frac{3}{8}$ and $\displaystyle \frac{7}{12}$ are

maths fractions, decimals and rational numbers representation of rational numbers on number line rational numbers on the number line rational numbers between two rational numbers

  1. $\displaystyle \frac{3}{8}, \frac{41}{96}, \frac{23}{48}, \frac{7}{12}$

  2. $\displaystyle \frac{3}{8}, \frac{41}{196}, \frac{23}{48}, \frac{7}{12}$

  3. $\displaystyle \frac{3}{8}, \frac{41}{96}, \frac{23}{148}, \frac{7}{12}$

  4. none of the above

Show answer
Correct Option: A
Explanation:

A rational number between two numbers $ a $ and $ b = \dfrac {(a +

b)}{2} $
So, a rational number between $\dfrac {3}{8} $ and $ \dfrac {7}{12}$

$ = \dfrac {\dfrac {3}{8} + \dfrac {7}{12}}{2} = \dfrac {23}{48} $

Now, another rational number between $ \dfrac {3}{8} $ and $ \dfrac {23}{48} $

$= \dfrac {\dfrac {3}{8} + \dfrac {23}{48}}{2} = \dfrac {41}{96} $ 

Hence, required two rational numbers between $\dfrac {3}{8} $ and $ \dfrac {7}{12} $ are $\dfrac {3}{8} ,\dfrac {41}{96}, \dfrac {23}{48}, \dfrac {7}{12}$

__________ are rational numbers between between 5 and -2.

maths fractions, decimals and rational numbers representation of rational numbers on number line rational numbers on the number line rational numbers between two rational numbers

  1. $\displaystyle 5, \frac{33}{4},\ \frac{3}{2}, -\frac{1}{4}, -2$

  2. $\displaystyle  \frac{13}{4},\ \frac{3}{2}, -\frac{1}{4} $

  3. $\displaystyle 5, \frac{13}{4},\ \frac{13}{2}, -\frac{1}{4}, -2$

  4. none of the above

Show answer
Correct Option: B
Explanation:

A rational number between two numbers $ a $ and $ b = \dfrac {(a + b)}{2} $

So, a rational number between $ 5 $ and $ - 2 = \dfrac

{( 5 - 2 )}{2} = \dfrac {3}{2} $


Now, another rational number between $ 5 $ and $ \dfrac {3}{2} =

\dfrac {( 5 + \dfrac {3}{2})}{2} = \dfrac {13}{4} $

Another rational number between $ \dfrac {3}{2} $ and $ -2 =

\dfrac {( \dfrac {3}{2}) - 2}{2} = -\dfrac {1}{4} $

 $ \therefore \dfrac {13}{4},  \dfrac {3}{2}, - \dfrac {1}{4} $  are the rational numbers between $ 5 $ and $ -2 $

________ are rational numbers between $\displaystyle \frac{1}{3}$ and $\displaystyle \frac{1}{4}$

maths fractions, decimals and rational numbers representation of rational numbers on number line rational numbers on the number line rational numbers between two rational numbers

  1. $\displaystyle \frac{1}{3}, \frac{7}{64}, \frac{13}{48}, \frac{1}{4}$

  2. $\displaystyle \frac{1}{3}, \frac{7}{24}, \frac{13}{48}, \frac{1}{4}$

  3. $\displaystyle \frac{1}{3}, \frac{7}{24}, \frac{13}{68}, \frac{1}{4}$

  4. none of the above

Show answer
Correct Option: B
Explanation:

A

rational number between two numbers $ a $ and $ b = \frac {(a +

b)}{2} $

So, a

rational number between $\frac {1}{3} $ and $ \frac {1}{4} = \frac {(\frac {1}{3} + \frac {1}{4})}{2} = \frac {7}{24} $

Now, another rational number between $ \frac {7}{24} $ and $ \frac {1}{4} = \frac {(\frac {7}{24} + \frac {1}{4})}{2} = \frac {13}{48} $



Hence, required two rational numbers between $\frac {1}{3} $ and $ \frac {1}{4} $ are $\frac {1}{3} ,\frac {7}{24}, \frac {13}{48}, \frac {1}{4}$

 ________ are rational numbers between $\displaystyle -\dfrac{3}{4}$ and $\displaystyle \dfrac{1}{2}.$

maths fractions, decimals and rational numbers representation of rational numbers on number line rational numbers on the number line rational numbers between two rational numbers

  1. $\dfrac{-7}{16}, \dfrac{-1}{8}, \dfrac{9}{16}$

  2. $\dfrac{-15}{16}, \dfrac{-1}{8}, \dfrac{3}{16}$

  3. $\dfrac{-7}{16}, \dfrac{-1}{8}, \dfrac{3}{16}$

  4. none of the above

Show answer
Correct Option: C
Explanation:
A rational number between two numbers $ a $ and $ b = \dfrac {(a + b)}{2} $ 

So,
a rational number between $ - \dfrac {3}{4} $ and $ \dfrac {1}{2} $
$= \dfrac {-\dfrac {3}{4} + \dfrac {1}{2}}{2} = - \dfrac {1}{8} $

Now,
another rational number between $ - \dfrac {3}{4} $ and $ - \dfrac {1}{8} $
$=\dfrac {- \dfrac {3}{4} - \dfrac {1}{8}}{2} = - \dfrac {7}{16} $ 

Another rational number between $ - \dfrac {1}{8} $ and $ \dfrac {1}{2} =$

$\dfrac { - \dfrac {1}{8} + \dfrac {1}{2}} {2} =  \dfrac {3}{16} $ 

Hence, required three rational numbers between $ - \dfrac {3}{4} $ and $  \dfrac {1}{2} $ are $ - \dfrac {3}{4}, - \dfrac {7}{16},  - \dfrac {1}{8}, \dfrac {3}{16}, \dfrac {1}{2} $

The rational number lying between the numbers $\displaystyle \frac{1}{3}$ and $\displaystyle \frac{3}{4}$ are

maths fractions, decimals and rational numbers representation of rational numbers on number line rational numbers on the number line rational numbers between two rational numbers

  1. $\displaystyle \frac{97}{300}$,$\displaystyle \frac{299}{500}$

  2. $\displaystyle \frac{99}{300}$,$\displaystyle \frac{301}{400}$

  3. $\displaystyle \frac{95}{300}$,$\displaystyle \frac{301}{400}$

  4. $\displaystyle \frac{117}{300}$,$\displaystyle \frac{287}{400}$

Show answer
Correct Option: D
Explanation:

To insert rational numbers between $2$ numbers, we will arrange the options and check if they are in ascending order.
$\dfrac { 1 }{ 3 } { ? }\dfrac { 117 }{ 300 } \ \Longrightarrow 300<351\ \qquad \dfrac { 3 }{ 4 } { ? }\dfrac { 287 }{ 400 } \ \Longrightarrow 1200>1148$
They are in ascending order, i.e., $\dfrac { 1 }{ 3 } ,\dfrac { 117 }{ 300 } ,\dfrac { 287 }{ 400 } ,\dfrac { 3 }{ 4 } $
From the given options only option $D$ satisfies this condition. Hence, $D$ is the answer.

Let a, b, c be positive integers such that $\frac {a\sqrt 2+b}{b\sqrt 2+c}$ is a rational number, then which of the following is always an integers?

maths fractions, decimals and rational numbers representation of rational numbers on number line rational numbers on the number line rational numbers between two rational numbers

  1. $\frac {2a^2+b^2}{2b^2+c^2}$

  2. $\frac {a^2+b^2-c^2}{a+b-c}$

  3. $\frac {a^2 _2b^2}{b^2+2c^2}$

  4. $\frac {a^2+b^2+c^2}{a+c-b}$

Show answer
Correct Option: D

Let $x\;\in\;Q,\;y\;\in\;Q^c$, which of the following statement is always WRONG ?

maths fractions, decimals and rational numbers representation of rational numbers on number line rational numbers on the number line rational numbers between two rational numbers

  1. $xy\;\in\;Q^c$

  2. $y/x\;\in\;Q$, whenever defined

  3. $\sqrt{2}x+y\;\in\;Q$

  4. $x/y\;\in\;Q^c$, whenever defined

Show answer
Correct Option: B
Explanation:

Let $x=1,\;y=\sqrt{2}$
Then $xy=\sqrt{2}\;\in\;Q^c$
Obvious
$x=-1,\;y=\sqrt{2}$ then $\sqrt{2}x+y=0\;\in\;Q$
$x=1,\;y=\sqrt{2}$ then $x/y=\displaystyle\frac{1}{\sqrt{2}}\;\in\;Q^c$

Which of these is true?
$(I)$ $5\sqrt {3}$ is not a rational number
$(II)$ $1$ is not the cube of a rational number
$(III)$ If a is rational and $n$ is an integer greater than $1$, then ${a}^{n}$ is rational.

maths fractions, decimals and rational numbers representation of rational numbers on number line rational numbers on the number line rational numbers between two rational numbers

  1. $I$ and $II$

  2. $II$ and $III$

  3. $III$ and $I$

  4. all three

Show answer
Correct Option: C
Explanation:

(I) In $5\sqrt{3}$

5 ia s rational number and $\sqrt{3}$ is an Irrational number
As we know, The product of a rational and irrational number is an irrational number.
So, $5\sqrt{3}$ is not a rational number.
Hence, the option (I) is true
(II) 1 is a rational number 
and cube of 1 is 1 only, which is a rational number
Hence the option (II) is False
(III) We know that product of two rational number is always a rational number
Hence if a is a rational number and n is greater than one 
Then,
a2 = a x a is a rational number.

a3 = a2 x a is a rational number,

a4 = a3x a is a rational number,

......

......

∴ an = an-1 x a is a rational number.

So, the option (III) is true

Which of the following numbers lies between $\dfrac {5}{24}$ and $\dfrac {3}{8}$?

maths fractions, decimals and rational numbers representation of rational numbers on number line rational numbers on the number line rational numbers between two rational numbers

  1. $\dfrac {7}{2}$

  2. $1$

  3. $\dfrac {7}{24}$

  4. $0$

Show answer
Correct Option: C
Explanation:

Mean $= \dfrac {\dfrac {3}{8} + \dfrac {5}{24}}{2} = \dfrac {\dfrac {9 + 5}{24}}{2} = \dfrac{\left (\dfrac {14}{24}\right )}{2}$


$= \dfrac {7}{12}\times \dfrac {1}{2}$

$= \dfrac {7}{24}$

Mean of two numbers lies between the two numbers.  
So, $ \dfrac {7}{24}$ lies between $\dfrac {3}{8}$ and $\dfrac {5}{24}.$

Which of the following numbers lies between $-1$ and $-2$?

maths fractions, decimals and rational numbers representation of rational numbers on number line rational numbers on the number line rational numbers between two rational numbers

  1. $\dfrac {-1}{2}$

  2. $\dfrac {-3}{2}$

  3. $\dfrac {1}{2}$

  4. $\dfrac {3}{2}$

Show answer
Correct Option: B
Explanation:

Mean $= \dfrac {(-1) + (-2)}{2} = \dfrac {-1 -2}{2} = \dfrac {-3}{2}$.

Mean of two numbers always lies between the two numbers.
So, answer is option $B.$