MCQ Questions for Class 8 Maths Rational Numbers Quiz 6

Rational numbers between

$\displaystyle \frac{3}{8}$ and

$\displaystyle \frac{7}{12}$ are

0%
$\displaystyle \frac{3}{8}, \frac{41}{96}, \frac{23}{48}, \frac{7}{12}$
0%
$\displaystyle \frac{3}{8}, \frac{41}{196}, \frac{23}{48}, \frac{7}{12}$
0%
$\displaystyle \frac{3}{8}, \frac{41}{96}, \frac{23}{148}, \frac{7}{12}$
0%
none of the above

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

$\dfrac {3}{8} ,\dfrac {41}{96}, \dfrac {23}{48}, \dfrac {7}{12}$

Addition of rational numbers does not satisfy which of the following property?

0%
Commutative
0%
Associative
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Closure
0%
None

________ are rational numbers between

$\displaystyle -\dfrac{3}{4}$ and

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

0%
$\dfrac{-7}{16}, \dfrac{-1}{8}, \dfrac{9}{16}$
0%
$\dfrac{-15}{16}, \dfrac{-1}{8}, \dfrac{3}{16}$
0%
$\dfrac{-7}{16}, \dfrac{-1}{8}, \dfrac{3}{16}$
0%
none of the above

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

$- \dfrac {3}{4}, - \dfrac {7}{16}, - \dfrac {1}{8}, \dfrac {3}{16}, \dfrac {1}{2}$

State TRUE or FALSE

The three rational number between

$\dfrac{1}{3}$ and

$\dfrac{1}{2}$ are

$\displaystyle\frac{9}{24},\frac{10}{24},\frac{11}{24}$ .

0%
True
0%
False

Explanation

$x=\frac { 1 }{ 3 } =\frac { 8 }{ 24 } \quad and\quad y=\frac { 1 }{ 2 } =\frac { 12 }{ 24 } ,\\$

$\frac { 9 }{ 24 } ,\frac { 10 }{ 24 } \ and\ \frac { 11 }{ 24 } \quad are\quad three\quad rational\quad numbers\quad lying\quad between\quad x\quad and\quad y.\quad \quad \\$

State True or False.

The five rational numbers between

$\dfrac{3}{5}$ and

$\dfrac{4}{5}$ are

$\displaystyle \frac{19}{30},\frac{20}{30},\frac{21}{30},\frac{22}{30},\frac{23}{30}$ .

0%
True
0%
False

Explanation

Since we want five numbers we write

$\dfrac{3}{5}$ and

$\dfrac{4}{5}$
So multiply in numerator and denominator by

$5+1 = 6$ we get

$\Rightarrow \dfrac{3}{5}=\dfrac{3\times 6}{5\times 6}=\dfrac{18}{30}$

$\Rightarrow \dfrac{4}{5}=\dfrac{4\times 6}{5\times 6}=\dfrac{24}{30}$

We know that

$18<19<20<21<22<23<24$

$\Rightarrow \displaystyle \frac{18}{30}<\frac{19}{30}<\frac{20}{30}<\frac{21}{30}<\frac{22}{30}<\frac{23}{30}<\frac{24}{30}$

Hence 5 rational number between

$\dfrac{3}{5} and \dfrac{4}{5}$ are

$\displaystyle \frac{19}{30},\frac{20}{30},\frac{21}{30},\frac{22}{30},\frac{23}{30},$

________ are rational numbers between

$\displaystyle \frac{1}{3}$ and

$\displaystyle \frac{1}{4}$

0%
$\displaystyle \frac{1}{3}, \frac{7}{64}, \frac{13}{48}, \frac{1}{4}$
0%
$\displaystyle \frac{1}{3}, \frac{7}{24}, \frac{13}{48}, \frac{1}{4}$
0%
$\displaystyle \frac{1}{3}, \frac{7}{24}, \frac{13}{68}, \frac{1}{4}$
0%
none of the above

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

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

Write the multiplicative inverse of each of the following rational numbers:

$7$ ;

$-11$ ;

$\displaystyle\frac{2}{5}$ ;

$\displaystyle\frac{-7}{15}$

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$\displaystyle\frac{-1}{7};\;\displaystyle\frac{1}{-11};\;\displaystyle\frac{5}{2};\;\displaystyle\frac{15}{-7}$
0%
$\displaystyle\frac{1}{7};\;\displaystyle\frac{1}{-11};\;\displaystyle\frac{5}{2};\;\displaystyle\frac{15}{-7}$
0%
$\displaystyle\frac{1}{7};\;\displaystyle\frac{1}{11};\;\displaystyle\frac{5}{2};\;\displaystyle\frac{15}{-7}$
0%
$\displaystyle\frac{1}{7};\;\displaystyle\frac{1}{-11};\;\displaystyle\frac{-5}{2};\;\displaystyle\frac{15}{-7}$

Explanation

Multiplicative inverse of

$7$ is

$\displaystyle\frac{1}{7}$ i.e.

$7^{-1}=\displaystyle\frac{1}{7}$

Multiplicative inverse of

$-11$ is

$\displaystyle\frac{1}{-11}$ i.e.

$(-11)^{-1}=\displaystyle\frac{1}{-11}$

Multiplicative inverse of

$\displaystyle\frac{2}{5}$ is

$\displaystyle\frac{5}{2}$ i.e.

$\begin{pmatrix}\displaystyle\frac{2}{5}\end{pmatrix}^{-1}=\displaystyle\frac{5}{2}$

Multiplicative inverse of

$\displaystyle\frac{-7}{15}$ is

$\displaystyle\frac{15}{-7}$ i.e

$\begin{pmatrix}\displaystyle\frac{-7}{15}\end{pmatrix}^{-1}=\displaystyle\frac{15}{-7}$

If A : Rational numbers are always closed under division and
R : Division by Zero is not defined, then which of the following statement is correct?

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A is True and R is the correct explanation of A
0%
A is False and R is the correct explanation of A
0%
A is True and R is False
0%
None of these

Explanation

Let us take two rational numbers

$\displaystyle \frac{3}{5}$ and

$\displaystyle \frac{0}{3}$
Then the division

$\dfrac{3}{5}$

$\div$

$\dfrac{0}{3}$ is not a rational number.
So division is not closure because division by zero is not defined.

$\therefore$ option

$B$ is correct

Find five rational numbers between

$\displaystyle\frac{-3}{2}$ and

$\displaystyle\frac{5}{3}$ .

0%
$\displaystyle\frac{-8}{6},\,\displaystyle\frac{-13}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$ and $\displaystyle\frac{2}{6}$
0%
$\displaystyle\frac{-8}{6},\,\displaystyle\frac{-7}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$ and $\displaystyle\frac{13}{6}$
0%
$\displaystyle\frac{-8}{6},\,\displaystyle\frac{-7}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$ and $\displaystyle\frac{11}{6}$
0%
$\displaystyle\frac{-8}{6},\,\displaystyle\frac{-7}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$ and $\displaystyle\frac{2}{6}$

Explanation

Converting the given rational numbers with the same denominators

$\cfrac{-3}{2}=\cfrac{-3\times3}{2\times3}=\cfrac{-9}{6}$ and

$\cfrac{5}{3}=\cfrac{5\times2}{3\times2}=\cfrac{10}{6}$

We know that

$-9\,<\,-8\,<\,-7\,<\,-6\,<\,\dots\,<\,0\,<\,1\,<\,2\,<\,8\,<\,9\,<\,10$

$\Rightarrow\cfrac{-9}{6}\,<\,\cfrac{-8}{6}\,<\,\cfrac{-7}{6}\,<\,\cfrac{-6}{6}\,<\,\dots\,<\,\cfrac{0}{6}\,<\,\cfrac{1}{6}\,<\,\cfrac{2}{6}\,<\,\dots\,<\,\cfrac{8}{6}\,<\,\cfrac{9}{6}\,<\,\cfrac{10}{6}$ .

Thus, we have the following five rational numbers between

$\cfrac{-3}{2}$ and

$\cfrac{5}{3}$

$\Rightarrow \cfrac{-8}{6},\,\cfrac{-7}{6},\,\cfrac{0}{6},\,\cfrac{1}{6}and\cfrac{2}{6}$

Write any

$10$ rational numbers between

$0\;and\;2$ .

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$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{5}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{88}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$
0%
$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{21}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$
0%
$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{35}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$
0%
$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{5}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$

Explanation

Let us write

$0$ as

$\dfrac{0}{10}$ and

$2$ as

$\dfrac{20}{10}.$

So, ten rational numbers between these will be,

$\dfrac{1}{10},\;\dfrac{2}{10},\;\dfrac{3}{10},\;\dfrac{4}{10},\;\dfrac{5}{10},\;\dfrac{6}{10},\;\dfrac{7}{10},\ \dfrac{8}{10},\ \dfrac{9}{10},\; 1$

Hence, option

$D$ is correct.

Which of the following alternatives is wrong? Given that
(i) difference of two rational numbers is a rational number
(ii) subtraction is commutative on rational numbers
(iii) addition is not commutative on rational numbers.

0%
(ii) and (iii)
0%
(i) only
0%
(i) and (iii)
0%
All the above

Which of the following statements is true?

0%
The reciprocals $1$ and $-1$ are themselves
0%
Zero has no reciprocal
0%
The product of two rational numbers is a rational number
0%
All the above

Explanation

A. Reciprocal of a number is the number divided by

$1$ . So

$1$ and

$-1$ when divided by

$1$ give the same result.

Hence, True.

B. When

$1$ is divided by

$0$ , the result is not defined. Hence, no reciprocal of zero.

C. When a rational number is multiplied, add or subtracted by any rational number, result is a rational number. For division, denominator must not be zero.

Hence, all the above statements are correct.

Name the property of multiplication illustrated by

$\displaystyle \frac{-4}{3} \times \left(\displaystyle \frac{6}{5} + \frac{8}{7} \right) = \left(\displaystyle \frac{-4}{3} \times \frac{6}{5} \right) + \left(\displaystyle \frac{-4}{3} \times \frac{8}{7} \right)$

0%
Associative property
0%
Commutative property
0%
Distributive property
0%
None of these

Explanation

We know that for

$a , b , c$ in

$R$ ;the distributive property is given by,

$a ( b + c ) = a b + a c$ .

$\therefore - \dfrac{4}{3} \times \left ( \dfrac{6}{5} + \dfrac{8}{7} \right) = \left ( \dfrac{-4}{3} \times \dfrac{6}{5} \right ) + \left ( -\dfrac{4}{3} \times \dfrac{8}{7} \right)$

can be illustrated by distributive property considering

$a = - \dfrac{4}{3} ; \, b = \dfrac{6}{5} ; \, c = \dfrac{8}{7}$

Hence, option C distributive property is correct answer.

Find the multiplicative inverse of the following:

$-13$ ,

$\displaystyle-\frac{13}{19}$ and

$-\displaystyle\frac{5}{8}\times\displaystyle-\frac{3}{7}$

0%
$-\displaystyle\frac{1}{13}\;;\;\displaystyle-\frac{19}{13}\;;\;\displaystyle\frac{56}{15}$
0%
$-\displaystyle\frac{1}{13}\;;\;\displaystyle\frac{19}{13}\;;\;\displaystyle\frac{56}{15}$
0%
$\displaystyle\frac{1}{13}\;;\;-\displaystyle\frac{19}{13}\;;\;\displaystyle\frac{56}{15}$
0%
$\displaystyle\frac{1}{13}\;;\;\displaystyle\frac{19}{13}\;;\;\displaystyle\frac{56}{15}$

Explanation

The multiplicative inverse of

$-13$ is

$(-13)^{-1}=\displaystyle\frac{1}{-13}$

The multiplicative inverse of

$\;\;\;\;\displaystyle\frac{-13}{19}$

is

$\begin{pmatrix}\displaystyle\frac{-13}{19}\end{pmatrix}^{-1}=\displaystyle\frac{19}{-13}$ .

We

have,

$\displaystyle\frac{-5}{8}\times\displaystyle\frac{-3}{7}=\displaystyle\frac{-5\times-3}{8\times7}=\displaystyle\frac{15}{56}$

$\;\;\;\;$ The multiplicative inverse of

$\displaystyle\frac{15}{56}$ is

$\;\;\;\;\begin{pmatrix}\displaystyle\frac{15}{56}\end{pmatrix}^{-1}=\displaystyle\frac{56}{15}$

The multiplicative inverse of

$\displaystyle \left ( \frac{1}{3} \right )^{-2}$ is

0%
$9$
0%
$\displaystyle \frac{1}{9}$
0%
$\displaystyle \frac{1}{4}$
0%
None of these

Explanation

Given rational number is

$\left( \cfrac { 1 }{ 3 }\right ) ^{ -2}$

To find out,

The multiplicative inverse of the given rational number.

Let the multiplicate inverse be

$x$ .

We know that, the multiplicative inverse of a number is another number which when multiplied by the original number yields

$1$ as a product.

We also know that,

$\left( \cfrac { a }{ b }\right ) ^{ -m}=\left( \cfrac { b}{ a }\right ) ^{ m}$

$\therefore\ \left( \cfrac { 1 }{ 3 }\right ) ^{ -2} =3^{ 2 }$

$= 9$

Now,

$9\times x=1$

$\therefore \ x=\dfrac19$

Hence, the multiplicative inverse of

$\left( \cfrac { 1 }{ 3 }\right ) ^{ -2}$ is

$\cfrac 19$ .

Choose the rational number which does not lie between rational numbers

$\displaystyle \frac{3}{5}$ and

$\displaystyle \frac{2}{3}$ :

0%
$\displaystyle \frac{46}{75}$
0%
$\displaystyle \frac{47}{75}$
0%
$\displaystyle \frac{49}{75}$
0%
$\displaystyle \frac{50}{75}$

Explanation

For

$\dfrac{3}{5}$ multiply numerator and denominator by

$15$ to make denominator

$75$ that comes into

$\dfrac{45}{75}.$
Similarly doing for second then we have

$\dfrac{50}{75}.$
Now question is asking about rational lying between them.

So, we need to check the numerator only that lies in between

$45$ and

$50$ or not.
Clearly

$D$ is correct.

Choose the rational number which does not lie between rational numbers

$\displaystyle\frac{3}{5}$ and

$\displaystyle\frac{2}{3}$

0%
$\displaystyle\frac{46}{75}$
0%
$\displaystyle\frac{47}{75}$
0%
$\displaystyle\frac{49}{75}$
0%
$\displaystyle\frac{50}{75}$

Explanation

Consider the given rational numbers,

$\dfrac{3}{5}$ and $\dfrac{2}{3}$

Now,

$\dfrac{3\times 15}{5\times 15}$ and $\dfrac{2\times 25}{3\times 25}$

$\dfrac{45}{75}$ and $\dfrac{50}{75}$

Now, rational number which does not lie between thes rational numbers is,

$\dfrac{50}{75}$

Hence, this is the answer.

Find

$3$ rational numbers between

$0$ and

$1$ .

0%
$\displaystyle\frac{3}{2},\,\displaystyle\frac{1}{4}$ and $\displaystyle\frac{3}{4}$
0%
$\displaystyle\frac{1}{2},\,\displaystyle\frac{1}{4}$ and $\displaystyle\frac{3}{4}$
0%
$\displaystyle\frac{1}{2},\,\displaystyle\frac{5}{4}$ and $\displaystyle\frac{3}{4}$
0%
$\displaystyle\frac{1}{2},\,\displaystyle\frac{1}{4}$ and $\displaystyle\frac{7}{4}$

Explanation

The mean of

$0$ and

$1$ is

$\cfrac{0+1}{2}=\cfrac{1}{2}$ .

The mean of

$0$ and

$\cfrac{1}{2}$ is

$\begin{pmatrix}0+\cfrac{1}{2}\end{pmatrix}\div2=\cfrac{1}{2}\div2=\cfrac{1}{2}\times\cfrac{1}{2}=\cfrac{1}{4}$

The mean of

$\cfrac{1}{2}$ and

$1$ is

$\begin{pmatrix}\cfrac{1}{2}+1\end{pmatrix}\div2$

$=\begin{pmatrix}\cfrac{1+2}{2}\end{pmatrix}\div2=\cfrac{3}{2}\div2=\cfrac{3}{2}\times\cfrac{1}{2}=\cfrac{3}{4}$ .

So, the

$3$ rational numbers between

$0$ and

$1$ are

$\cfrac{1}{2},\,\cfrac{1}{4}$ and

$\cfrac{3}{4}$ .

Choose the rational number which does not lie between rational numbers

$-\cfrac {2}{5}$ and

$-\cfrac {1}{5}$

0%
$-\dfrac {1}{4}$
0%
$-\dfrac {3}{10}$
0%
$\dfrac {3}{10}$
0%
$-\dfrac {7}{20}$

Explanation

Since the given rational numbers

$-\dfrac {2}{5}$ and

$-\dfrac {1}{5}$ are negative rational numbers, therefore, none of the positive rational number can lie between them.

Hence, the rational number

$\dfrac {3}{10}$ does not lie between the rational numbers

$-\dfrac {2}{5}$ and

$-\dfrac {1}{5}$

$D$ be subset of the set of all rational numbers, then

$D$ is closed under the binary operations of ..............

0%
addition, subtraction and division.
0%

addition, multiplication and division.
0%
addition, subtraction and multiplication.
0%
subtraction, multiplication and division.

Find

$9$ rational numbers between

$-\displaystyle\frac{1}{9}\;and\;\displaystyle\frac{1}{5}$ .

0%
$\displaystyle\frac{-7}{45},\,\displaystyle\frac{-3}{45},\,\displaystyle\frac{-2}{45},\,\displaystyle\frac{-1}{45},\,\displaystyle\frac{2}{45},\,\displaystyle\frac{3}{45},\,\displaystyle\frac{4}{45},\,\displaystyle\frac{5}{45},\,\displaystyle\frac{8}{45}$
0%
$\displaystyle\frac{-4}{45},\,\displaystyle\frac{-3}{45},\,\displaystyle\frac{-2}{45},\,\displaystyle\frac{-1}{45},\,\displaystyle\frac{10}{45},\,\displaystyle\frac{3}{45},\,\displaystyle\frac{4}{45},\,\displaystyle\frac{5}{45},\,\displaystyle\frac{8}{45}$
0%
$\displaystyle\frac{-4}{45},\,\displaystyle\frac{-3}{45},\,\displaystyle\frac{-2}{45},\,\displaystyle\frac{-1}{45},\,\displaystyle\frac{2}{45},\,\displaystyle\frac{9}{45},\,\displaystyle\frac{4}{45},\,\displaystyle\frac{5}{45},\,\displaystyle\frac{8}{45}$
0%
$\displaystyle\frac{-4}{45},\,\displaystyle\frac{-3}{45},\,\displaystyle\frac{-2}{45},\,\displaystyle\frac{-1}{45},\,\displaystyle\frac{2}{45},\,\displaystyle\frac{3}{45},\,\displaystyle\frac{4}{45},\,\displaystyle\frac{5}{45},\,\displaystyle\frac{8}{45}$

Explanation

Convert the rational numbers into equivalent rational numbers with the same denominator.

LCM of

$9$ and

$5$ is

$45$ .

$-\displaystyle\frac{1}{9}=\displaystyle\frac{-1\times5}{9\times5}=\displaystyle\frac{-5}{45}$ and

$\displaystyle\frac{1}{5}=\displaystyle\frac{1\times9}{5\times9}=\displaystyle\frac{9}{45}$

The integers between

$-5$ and

$9$ are

$-4,\,-3,\,-2,\,-1,\,0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8$ .

The corresponding rational numbers are

$\displaystyle\frac{-4}{45},\,\displaystyle\frac{-3}{45},\,\displaystyle\frac{-2}{45},\,\displaystyle\frac{-1}{45},\,\displaystyle\frac{0}{45},\,\displaystyle\frac{1}{45},\,\displaystyle\frac{2}{45},\,\displaystyle\frac{3}{45},\,\displaystyle\frac{4}{45},\,\displaystyle\frac{5}{45},\,\displaystyle\frac{6}{45},\,\displaystyle\frac{7}{45},\,\displaystyle\frac{8}{45}$

On selecting any

$9$ of them, we get

$9$ rational numbers between

$-\displaystyle\frac{1}{9}$ and

$\displaystyle\frac{1}{5}$ as

$\displaystyle\frac{-4}{45},\,\displaystyle\frac{-3}{45},\,\displaystyle\frac{-2}{45},\,\displaystyle\frac{-1}{45},\,\displaystyle\frac{2}{45},\,\displaystyle\frac{3}{45},\,\displaystyle\frac{4}{45},\,\displaystyle\frac{5}{45},\,\displaystyle\frac{8}{45}$

Choose the rational number which does not lie between rational numbers

$\displaystyle-\frac{2}{5}$ and

$\displaystyle-\frac{1}{5}$

0%
$\displaystyle-\frac{1}{4}$
0%
$\displaystyle-\frac{3}{10}$
0%
$\displaystyle\frac{3}{10}$
0%
$\displaystyle-\frac{7}{20}$

Explanation

Consider given the rational numbers $-\dfrac{2}{5}$ and $-\dfrac{1}{5}$

Now, given both rational numbers are negative numbers so the number which lies between them will be negative.

So, $\dfrac{3}{10}$ will not lie between them,

Hence, this is the answer.

Four rational number equivalent to

$\displaystyle \frac{5}{8}$ are:

0%
$\displaystyle \frac{10}{6}$ , $\displaystyle \frac{12}{24}$ , $\displaystyle \frac{20}{32}$ , $\displaystyle \frac{25}{40}$
0%
$\displaystyle \frac{10}{16}$ , $\displaystyle \frac{15}{24}$ , $\displaystyle \frac{20}{32}$ , $\displaystyle \frac{25}{40}$
0%
$\displaystyle \frac{10}{16}$ , $\displaystyle \frac{15}{24}$ , $\displaystyle \frac{8}{32}$ , $\displaystyle \frac{25}{40}$
0%
$\displaystyle \frac{8}{16}$ , $\displaystyle \frac{15}{24}$ , $\displaystyle \frac{20}{32}$ , $\displaystyle \frac{25}{40}$

Explanation

$\displaystyle \frac{5}{8}$ =

$\displaystyle \frac{5\times 2}{8\times 2}$ =

$\displaystyle \frac{5\times 3}{8\times 3}$ =

$\displaystyle \frac{5\times 4}{8\times 4}$ =

$\displaystyle \frac{5\times 5}{8\times 5}$

$\displaystyle \therefore \frac{5}{8}$ =

$\displaystyle \frac{10}{16}$ =

$\displaystyle \frac{15}{24}$ =

$\displaystyle \frac{20}{32}$ =

$\displaystyle \frac{25}{40}$

Thus four rational number equivalent to

$\displaystyle \frac{5}{8}$ are :

$\displaystyle \frac{10}{16}$ ,

$\displaystyle \frac{15}{24}$ ,

$\displaystyle \frac{20}{32}$ ,

$\displaystyle \frac{25}{40}$

The value of

$\dfrac {1}{2} \times \left (\dfrac {1}{3} + \dfrac {4}{9}\right )$ is

0%
$\dfrac {7}{18}$
0%
$\dfrac {1}{2}$
0%
$\dfrac {7}{12}$
0%
$\dfrac {7}{24}$

Explanation

$\dfrac {1}{2} \times \left (\dfrac {1}{3} + \dfrac {4}{9}\right ) = \dfrac {1}{2} \times \dfrac {1}{3} + \dfrac {1}{2} \times \dfrac {4}{9}$ (

$\because$ distributive property)

$= \dfrac {1}{6} + \dfrac {2}{9}$

$= \dfrac {3 + 4}{18}$

$= \dfrac {7}{18}$ .

For rational numbers

$\dfrac {a}{b}$ and

$\dfrac {p}{q}$ , where

$a, b, p, q\in Q$ and .............. condition exists, then they are closed under division.

0%
$b\neq 0, q\neq 1$
0%
$b\neq 0, q\neq 0$
0%
$b\neq 1, q\neq 1$
0%
$b\neq 1, q\neq 0$

Explanation

The numbers

$a\div 0$ and

$p\div 0$ are not rational numbers. So, for any rational numbers of the forms

$\dfrac {a}{b}$ and

$\dfrac {p}{q}$ where

$b\neq 0$ and

$q\neq 0$ and

$a, b, p, q$ are rational numbers, they are closed under division.

Find the value of

$x$ in

$\dfrac {4}{3} \times \left [x + \dfrac {1}{13} \right ] = \dfrac {4}{3}\times \dfrac {8}{11} + \dfrac {4}{3}\times \dfrac {1}{13}.$

0%
$\dfrac {8}{11}$
0%
$\dfrac {11}{8}$
0%
$\dfrac {4}{3}$
0%
$\dfrac {1}{13}$

Explanation

$\dfrac {4}{3} \times \left (x + \dfrac {1}{13} \right ) = \dfrac {4}{3}\times x + \dfrac {4}{3} \times \dfrac {1}{13}$

The distributive property over addition for rational numbers states that for any three rational numbers

$a, b$ and

$c :$

$a (b + c) = (a\times b) + (a\times c)$

$\dfrac {4}{3}\times \dfrac {8}{11} + \dfrac {4}{3} \times \dfrac {1}{13} = \dfrac {4}{3} \times \left (\dfrac {8}{11} \times \dfrac {1}{13}\right )$

Simplify using associative property :

$\dfrac {7}{16} \times \left (\dfrac {-24}{49} \times \dfrac {28}{15}\right )$

0%
$\dfrac {28}{15}$
0%
$\dfrac {2}{5}$
0%
$\dfrac {-2}{5}$
0%
$\dfrac {-3}{15}$

Explanation

$\dfrac {7}{16} \times \left (\dfrac {-24}{49} \times \dfrac {28}{15}\right ) = \left (\dfrac {7}{16}\times \dfrac {-24}{49}\right ) \times \dfrac {28}{15}$ (associative property)

$= \left (\dfrac {1}{2} \times \dfrac {-3}{7}\right )\times \dfrac {28}{15}$

$= \dfrac {-3}{14} \times \dfrac {28}{15}$

$= \dfrac {-2}{5}$

Simplify using associative property :

$\dfrac {-8}{9}\times \left (\dfrac {27}{32} \times \dfrac {-8}{21}\right )$

0%
$\dfrac {2}{7}$
0%
$\dfrac {-2}{7}$
0%
$\dfrac {-8}{21}$
0%
$\dfrac {-3}{4}$

Explanation

$\dfrac {-8}{9}\times \left (\dfrac {27}{32} \times \dfrac {-8}{21}\right ) = \left (\dfrac {-8}{9} \times \dfrac {27}{32}\right ) \times \dfrac {-8}{21}$ (

$\because$ associative property)

$= \left (\dfrac {-3}{4}\right ) \times \dfrac {-8}{21}$

$= \dfrac {+2}{7}$

The value of

$\dfrac {6}{11} \times \left [\left (\dfrac {-7}{6}\right ) - \left (\dfrac {11}{7}\right )\right ]$ is

0%
$\dfrac {17}{77}$
0%
$\dfrac {-17}{77}$
0%
$\dfrac {-115}{77}$
0%
$\dfrac {115}{77}$

Explanation

$\dfrac {6}{11} \times \left [\left (\dfrac {-7}{6}\right ) - \left (\dfrac {11}{7}\right )\right ]$

$= \dfrac {6}{11} \times \dfrac {-7}{6} - \dfrac {6}{11}\times \dfrac {11}{7}$ (

$\because$ distributive property)

$= \dfrac {-7}{11} - \dfrac {6}{7}$

$= \dfrac {-49 - 66}{77}$

$= \dfrac {-115}{77}$ .

$\displaystyle \frac{5}{13}+\_\_=\frac{5}{13}$

0%
$0$
0%
$1$
0%
$\displaystyle \frac{5}{13}$
0%
$\displaystyle \frac{2}{13}$

Explanation

If we add

$0$ to any number, we get the same number

$\therefore \dfrac{5}{13}+0=\dfrac{5}{13}$

Hence, the answer is

$0$ .

CBSE Questions for Class 8 Maths Rational Numbers Quiz 6 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Rational Numbers Quiz 6 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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