Rational Numbers Test

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Rational Numbers Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

Find $$3$$ rational numbers between $$0$$ and $$1$$.

A
$$\displaystyle\frac{3}{2},\,\displaystyle\frac{1}{4}$$ and $$\displaystyle\frac{3}{4}$$

B
$$\displaystyle\frac{1}{2},\,\displaystyle\frac{1}{4}$$ and $$\displaystyle\frac{3}{4}$$

C
$$\displaystyle\frac{1}{2},\,\displaystyle\frac{5}{4}$$ and $$\displaystyle\frac{3}{4}$$

D
$$\displaystyle\frac{1}{2},\,\displaystyle\frac{1}{4}$$ and $$\displaystyle\frac{7}{4}$$

Solution

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 $$ \cfrac{3}{4}$$.
Question 2
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Choose the rational number which does not lie between rational numbers $$-\cfrac {2}{5}$$ and $$-\cfrac {1}{5}$$

A
$$-\dfrac {1}{4}$$

B
$$-\dfrac {3}{10}$$

C
$$\dfrac {3}{10}$$

D
$$-\dfrac {7}{20}$$

Solution

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 $$-\dfrac {1}{5}$$
Question 3
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If $$D$$ be subset of the set of all rational numbers, then $$D$$ is closed under the binary operations of ..............

A
addition, subtraction and division.

B

addition, multiplication and division.

C
addition, subtraction and multiplication.

D
subtraction, multiplication and division.

Solution

Rational numbers are closed under all these operations except for division. Add two rational numbers, get a rational number; multiply two rationals, get a rational, etc. Problem with division is that if we divide any number by '0' we cannot tell the output. Some people says it is infinity but that's not true. Division by '0' is not defined.
So correct answer will be option C
Question 4
1 / -0

Find five rational numbers between $$\displaystyle\frac{-3}{2}$$ and $$\displaystyle\frac{5}{3}$$.

A
$$\displaystyle\frac{-8}{6},\,\displaystyle\frac{-13}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$$ and $$\displaystyle\frac{2}{6}$$

B
$$\displaystyle\frac{-8}{6},\,\displaystyle\frac{-7}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$$ and $$\displaystyle\frac{13}{6}$$

C
$$\displaystyle\frac{-8}{6},\,\displaystyle\frac{-7}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$$ and $$\displaystyle\frac{11}{6}$$

D
$$\displaystyle\frac{-8}{6},\,\displaystyle\frac{-7}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$$ and $$\displaystyle\frac{2}{6}$$

Solution

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 $$\cfrac{5}{3}$$
$$\Rightarrow \cfrac{-8}{6},\,\cfrac{-7}{6},\,\cfrac{0}{6},\,\cfrac{1}{6}and\cfrac{2}{6}$$
Question 5
1 / -0

Find $$9$$ rational numbers between $$-\displaystyle\frac{1}{9}\;and\;\displaystyle\frac{1}{5}$$.

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

B
$$\displaystyle\frac{-4}{45},\,\displaystyle\frac{-3}{45},\,\displaystyle\frac{-2}{45},\,\displaystyle\frac{-1}{45},\,\displaystyle\frac{10}{45},\,\displaystyle\frac{3}{45},\,\displaystyle\frac{4}{45},\,\displaystyle\frac{5}{45},\,\displaystyle\frac{8}{45}$$

C
$$\displaystyle\frac{-4}{45},\,\displaystyle\frac{-3}{45},\,\displaystyle\frac{-2}{45},\,\displaystyle\frac{-1}{45},\,\displaystyle\frac{2}{45},\,\displaystyle\frac{9}{45},\,\displaystyle\frac{4}{45},\,\displaystyle\frac{5}{45},\,\displaystyle\frac{8}{45}$$

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

Solution

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

$$\displaystyle\frac{-4}{45},\,\displaystyle\frac{-3}{45},\,\displaystyle\frac{-2}{45},\,\displaystyle\frac{-1}{45},\,\displaystyle\frac{2}{45},\,\displaystyle\frac{3}{45},\,\displaystyle\frac{4}{45},\,\displaystyle\frac{5}{45},\,\displaystyle\frac{8}{45}$$
Question 6
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Write any $$10$$ rational numbers between $$0\;and\;2$$.

A
$$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{5}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{88}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$$

B
$$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{21}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$$

C
$$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{35}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$$

D
$$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{5}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$$

Solution

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.
Question 7
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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.

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

B
$$b\neq 0, q\neq 0$$

C
$$b\neq 1, q\neq 1$$

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

Solution

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.
Question 8
1 / -0

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

A
$$\dfrac {28}{15}$$

B
$$\dfrac {2}{5}$$

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

D
$$\dfrac {-3}{15}$$

Solution

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

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

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

$$= \dfrac {-2}{5}$$
Question 9
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Simplify using associative property :
$$\dfrac {-8}{9}\times \left (\dfrac {27}{32} \times \dfrac {-8}{21}\right )$$

A
$$\dfrac {2}{7}$$

B
$$\dfrac {-2}{7}$$

C
$$\dfrac {-8}{21}$$

D
$$\dfrac {-3}{4}$$

Solution

$$\dfrac {-8}{9}\times \left (\dfrac {27}{32} \times \dfrac {-8}{21}\right ) = \left (\dfrac {-8}{9} \times \dfrac {27}{32}\right ) \times \dfrac {-8}{21}$$ ($$\because$$ associative property)
$$= \left (\dfrac {-3}{4}\right ) \times \dfrac {-8}{21}$$
$$= \dfrac {+2}{7}$$
Question 10
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Choose the rational number which does not lie between rational numbers $$\displaystyle-\frac{2}{5}$$ and $$\displaystyle-\frac{1}{5}$$

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

B
$$\displaystyle-\frac{3}{10}$$

C
$$\displaystyle\frac{3}{10}$$

D
$$\displaystyle-\frac{7}{20}$$

Solution

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.

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Rational Numbers Test - 28

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Rational Numbers Test - 28

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Question 1

$$\displaystyle\frac{3}{2},\,\displaystyle\frac{1}{4}$$ and $$\displaystyle\frac{3}{4}$$

$$\displaystyle\frac{1}{2},\,\displaystyle\frac{1}{4}$$ and $$\displaystyle\frac{3}{4}$$

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

$$\displaystyle\frac{1}{2},\,\displaystyle\frac{1}{4}$$ and $$\displaystyle\frac{7}{4}$$

Solution

Question 2

$$-\dfrac {1}{4}$$

$$-\dfrac {3}{10}$$

$$\dfrac {3}{10}$$

$$-\dfrac {7}{20}$$

Solution

Question 3

addition, subtraction and division.

addition, multiplication and division.

addition, subtraction and multiplication.

subtraction, multiplication and division.

Solution

Question 4

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

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

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

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

Solution

Question 5

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

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

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

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

Solution

Question 6

$$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{5}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{88}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$$

$$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{21}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$$

$$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{35}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$$

$$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{5}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$$

Solution

Question 7

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

$$b\neq 0, q\neq 0$$

$$b\neq 1, q\neq 1$$

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

Solution

Question 8

$$\dfrac {28}{15}$$

$$\dfrac {2}{5}$$

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

$$\dfrac {-3}{15}$$

Solution

Question 9

$$\dfrac {2}{7}$$

$$\dfrac {-2}{7}$$

$$\dfrac {-8}{21}$$

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

Solution

Question 10

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

$$\displaystyle-\frac{3}{10}$$

$$\displaystyle\frac{3}{10}$$

$$\displaystyle-\frac{7}{20}$$

Solution

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