Rational Numbers Test

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Rational Numbers Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

Multiplicative inverse of $$\dfrac{2}{-3}$$ is

A
$$\dfrac{2}{3}$$

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

C
$$\dfrac{3}{2}$$

D
None of these

Solution

$${\textbf{Step -1: Write the number whose inverse is to be done}}{\textbf{.}}$$
$$\dfrac{2}{{ - 3}}$$

$${\textbf{Step -2: Get it's reciprocal to get the multiplicative inverse}}{\textbf{.}}$$
$$\text{Multiplicative inverse }=\dfrac{1}{\dfrac{2}{-3}}=\dfrac{{ - 3}}{2}$$

$${\textbf{Hence, multiplicative inverse of}}$$ $${\mathbf{\dfrac{2}{{ - 3}}}}$$ $${\textbf{is}}$$ $${\mathbf{\dfrac{{ - 3}}{2.}}}$$
Question 2
1 / -0

The additive inverse of $$\dfrac{-7}{19}$$ is

A
$$\dfrac{-7}{19}$$

B
$$\dfrac{7}{19}$$

C
$$\dfrac{19}{7}$$

D
$$\dfrac{-19}{7}$$

Solution

Sum of any number $$a$$ and its additive inverse $$b$$ should be zero.
That is, $$a + b = 0$$

Let $$a = \dfrac{-7}{19}$$ and $$b$$ be the additive inverse of $$a$$

Then,
$$b - \dfrac{7}{19} = 0$$

$$\Rightarrow b = \dfrac{7}{19}$$

Therefore, the additive inverse of $$\dfrac{-7}{19}$$ is $$\dfrac{7}{19}$$.

Hence, option B is the correct.
Question 3
1 / -0

Which of the following rational numbers is equal to its reciprocal?

A
$$1$$

B
$$2$$

C
$$\dfrac{1}{2}$$

D
$$0$$

Solution

(a) Reciprocal of $$1=\dfrac{1}{1}=1$$

(b) Reciprocal of $$2=\dfrac{1}{2}$$

(c) Reciprocal of $$\dfrac{1}{2}=\dfrac{1}{\dfrac{1}{2}}=2$$

(d) Reciprocal of $$0=\dfrac{1}{0}$$

Therefore, only $$1$$ is the only number that is equal to its reciprocal.
Question 4
1 / -0

If $$x$$ be any rational number, then $$x + 0$$ is equal to

A
$$x$$

B
$$0$$

C
$$-x$$

D
Not defined

Solution

Given $$x$$ is a rational number.
To find the value of $$(x+0)$$.

We know that, the sum of any rational number and '$$0$$' is the rational number itself.

$$\therefore x+0=x$$

Hence, $$Op-A$$ is correct.
Question 5
1 / -0

Which of the following statements is always true ?

A
$$\dfrac{x - y}{2}$$ is a rational number between x and y

B
$$\dfrac{x + y}{2}$$ is a rational number between x and y

C
$$\dfrac{x \times y}{2}$$ is a rational number between x and y

D
$$\dfrac{x \div y}{2}$$ is a rational number between x and y

Solution
Question 6
1 / -0

The multiplicative inverse of $$-1 \dfrac{1}{7}$$ is

A
$$\dfrac{8}{7}$$

B
$$\dfrac{-8}{7}$$

C
$$\dfrac{7}{8}$$

D
$$\dfrac{7}{-8}$$

Solution

Given number is $$-1\dfrac{1}{7}$$
$$-1\dfrac{1}{7}=-\dfrac{7+1}{7}$$
$$=-\dfrac{8}{7}$$
Since we know that the multiplicative inverse of any non-zero number $$a$$ is $$\dfrac{1}{a}$$
$$\therefore$$ the multiplicative inverse of $$-\dfrac{8}{7}$$ is $$\dfrac{1}{-\dfrac{8}{7}}$$
$$=-\dfrac{7}{8}$$
Question 7
1 / -0

The multiplication inverse of $$10^{-100}$$ is

A
$$10$$

B
$$100$$

C
$$10^{100}$$

D
$$10^{-100}$$

Solution

For multiplicative inverse, let $$a$$ be the multiplicative of $$10^{-100}$$
So, $$a\times b=1$$
Therefore $$a\times\ 10^{-100}=1$$
$$\Rightarrow \ a=\dfrac {1}{10^{-100}} =\dfrac {1}{\dfrac {1}{10^{100}}}=10^{100}$$
Question 8
1 / -0

The Multiplicative inverse of $$\dfrac{-4}{9}$$ is

A
$$\dfrac{4}{9}$$

B
$$\dfrac{-9}{4}$$

C
$$\dfrac{9}{4}$$

D
none of these

Solution

Multipiicative
inverse of $$\dfrac{-4}{9}$$

$$=\dfrac{9}{-4}=\dfrac{9(-1)}{-4(-1)}=\dfrac{-9}{4}$$
Question 9
1 / -0

The multiplicative inverse of $$\left(-\dfrac {5}{9} \right)^{99}$$ is

A
$$\left(-\dfrac {5}{9} \right)$$

B
$$\left(-\dfrac {5}{9} \right)^{99}$$

C
$$\left(-\dfrac {9}{-5} \right)^{99}$$

D
$$\left(-\dfrac {9}{5} \right)^{99}$$

Solution

For multiplicative inverse $$a$$ is called multiplicative inverse $$b$$, if $$a\times b=1$$
Put $$\Rightarrow a\times \left(\dfrac {-5}{9}\right)^{99}=1\quad \left[As, a^{-m}=\dfrac {1}{a^m}\right]$$
$$\Rightarrow a=\dfrac {1}{\left(\dfrac {-5}{9}\right)^{99}}\Rightarrow a=\left(-\dfrac {5}{9}\right)^{-99}=\left(-\dfrac {9}{5}\right)^{99}$$
Question 10
1 / -0

The product of rational number $$\dfrac{-2}{5}$$ and its multiplicative inverse is

A
1

B
0

C
$$\dfrac{4}{25}$$

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

Solution

Product of
rational number $$\dfrac{-2}{5}$$ and its

multiplicative
inverse is $$\dfrac{-2}{5} \times \dfrac{-5}{2}=1$$

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Rational Numbers Test - 33

Result

Rational Numbers Test - 33

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SHARING IS CARING

Question 1

$$\dfrac{2}{3}$$

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

$$\dfrac{3}{2}$$

None of these

Solution

Question 2

$$\dfrac{-7}{19}$$

$$\dfrac{7}{19}$$

$$\dfrac{19}{7}$$

$$\dfrac{-19}{7}$$

Solution

Question 3

$$1$$

$$2$$

$$\dfrac{1}{2}$$

$$0$$

Solution

Question 4

$$x$$

$$0$$

$$-x$$

Not defined

Solution

Question 5

$$\dfrac{x - y}{2}$$ is a rational number between x and y

$$\dfrac{x + y}{2}$$ is a rational number between x and y

$$\dfrac{x \times y}{2}$$ is a rational number between x and y

$$\dfrac{x \div y}{2}$$ is a rational number between x and y

Solution

Question 6

$$\dfrac{8}{7}$$

$$\dfrac{-8}{7}$$

$$\dfrac{7}{8}$$

$$\dfrac{7}{-8}$$

Solution

Question 7

$$10$$

$$100$$

$$10^{100}$$

$$10^{-100}$$

Solution

Question 8

$$\dfrac{4}{9}$$

$$\dfrac{-9}{4}$$

$$\dfrac{9}{4}$$

none of these

Solution

Question 9

$$\left(-\dfrac {5}{9} \right)$$

$$\left(-\dfrac {5}{9} \right)^{99}$$

$$\left(-\dfrac {9}{-5} \right)^{99}$$

$$\left(-\dfrac {9}{5} \right)^{99}$$

Solution

Question 10

1

0

$$\dfrac{4}{25}$$

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

Solution

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