Tag: representation of rational numbers on number line

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Multiple choice 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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\dfrac {3}{8} ,\dfrac {41}{96}, \dfrac {23}{48}, \dfrac {7}{12}$

Multiple choice 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

__________ are rational numbers between between 5 and -2.

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $

Multiple choice 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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\frac {1}{3} ,\frac {7}{24}, \frac {13}{48}, \frac {1}{4}$

Multiple choice 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

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

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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} $
Multiple choice 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

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

  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. $\displaystyle \frac{117}{300}$,$\displaystyle \frac{287}{400}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice 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

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?

  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. $\frac {a^2+b^2+c^2}{a+c-b}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For (a*sqrt(2)+b)/(b*sqrt(2)+c) to be rational, the irrational parts must cancel. This implies a/b = b/c, so b^2 = ac. Testing option D: (a^2+b^2+c^2)/(a+c-b) is a standard algebraic identity related to this condition.

Multiple choice 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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $x/y=\displaystyle\frac{1}{\sqrt{2}}\;\in\;Q^c$

Multiple choice 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

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.

  1. $I$ and $II$
  2. $II$ and $III$
  3. $III$ and $I$

  4. all three
Reveal answer Fill a bubble to check yourself
C Correct answer
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

Multiple choice 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

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

  1. $\dfrac {7}{2}$
  2. $1$
  3. $\dfrac {7}{24}$
  4. $0$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\dfrac {5}{24}.$

Multiple choice 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

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

  1. $\dfrac {-1}{2}$
  2. $\dfrac {-3}{2}$
  3. $\dfrac {1}{2}$
  4. $\dfrac {3}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $B.$