Rational Numbers Test

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Rational Numbers Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

Write the additive inverse of each of the following rational numbers:
$$\displaystyle\frac{4}{9}$$; $$\displaystyle\frac{-13}{7}$$; $$\displaystyle\frac{5}{-11}$$; $$\displaystyle\frac{-11}{-14}$$

A
$$\displaystyle\frac{-4}{9};\;\displaystyle\frac{13}{7};\;\displaystyle\frac{5}{11};\;\displaystyle\frac{-13}{14}$$

B
$$\displaystyle\frac{-4}{9};\;\displaystyle\frac{1}{7};\;\displaystyle\frac{5}{11};\;\displaystyle\frac{-11}{14}$$

C
$$\displaystyle\frac{-4}{9};\;\displaystyle\frac{13}{7};\;\displaystyle\frac{5}{11};\;\displaystyle\frac{-11}{14}$$

D
$$\displaystyle\frac{-5}{9};\;\displaystyle\frac{13}{7};\;\displaystyle\frac{5}{11};\;\displaystyle\frac{-11}{14}$$

Solution

The additive inverse of $$\displaystyle\frac{4}{9}$$ is $$-\displaystyle\frac{4}{9}=\displaystyle\frac{-4}{9}$$

The additive inverse of $$\displaystyle\frac{-13}{7}$$ is $$-\begin{pmatrix}\displaystyle\frac{-13}{7}\end{pmatrix}=-\begin{pmatrix}-\displaystyle\frac{13}{7}\end{pmatrix}=\displaystyle\frac{13}{7}$$

We have, $$\displaystyle\frac{5}{-11}=\displaystyle\frac{-5}{11}$$
The additive inverse of $$\displaystyle\frac{-5}{11}$$ is $$-\begin{pmatrix}\displaystyle\frac{-5}{11}\end{pmatrix}$$$$=-\begin{pmatrix}-\displaystyle\frac{5}{11}\end{pmatrix}=\displaystyle\frac{5}{11}$$

We have, $$\displaystyle\frac{-11}{-14}=\displaystyle\frac{11}{14}$$
The additive inverse of $$\displaystyle\frac{11}{14}$$ is $$-\begin{pmatrix}\displaystyle\frac{11}{14}\end{pmatrix}$$ $$=\displaystyle\frac{-11}{14}$$
Question 2
1 / -0

The statement ''Rational numbers are closed under division'' is

A
True

B
False

C
Cannot be determined

D
None
Question 3
1 / -0

$$x$$ is the multiplicative identity. The value of $$x$$ is

A
1

B
2

C
0

D
None

Solution

Multiplying a number by 1 leaves it unchanged

$$a\times 1 =1\times a=a$$

$$\therefore$$ $$1$$ is the multiplicative identity.
Question 4
1 / -0

Choose the correct option for following statement.
Subtraction of rational numbers is not commutative.

A
True

B
False

C
Cannot be determined

D
None of these

Solution

Subtraction of two rational numbers is not commutative.
If $$\dfrac{a}{b}$$ and $$\dfrac{c}{d}$$ are any two rational numbers,
Then, $$\dfrac{a}{b}-\dfrac{c}{d}\ne \dfrac{c}{d}-\dfrac{a}{b}$$
For example:
$$\dfrac{5}{4}-\dfrac{3}{4}=\dfrac{2}{4}=\dfrac{1}{2}$$

$$\dfrac{3}{4}-\dfrac{5}{4}=\dfrac{-2}{4}=\dfrac{-1}{2}$$

$$\therefore$$ $$\dfrac{5}{4}-\dfrac{3}{4}\ne$$ $$\dfrac{3}{4}-\dfrac{5}{4}$$

$$\therefore$$ The given statement is true.
Question 5
1 / -0

$$x$$ and $$\displaystyle\frac{8}{9}$$ are the additive inverse of each other. Then the value of $$ x$$ is

A
$$\displaystyle\frac{-8}{9}$$

B
$$\displaystyle\frac{-4}{9}$$

C
$$\displaystyle\frac{-5}{9}$$

D
$$\displaystyle\frac{-7}{9}$$

Solution

$$\displaystyle\frac{-8}{9}+\displaystyle\frac{8}{9}=0$$
Thus $$\displaystyle\frac{-8}{9}$$ and $$\displaystyle\frac{8}{9}$$are additive inverse of each other.
$$\therefore$$ the value of $$x$$ is$$\displaystyle\frac{-8}{9}$$
Question 6
1 / -0

The number of rational numbers between two given rational numbers is

A
Infinite

B
Finite

C
Two

D
One

Solution

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

Like this, using this rational number $$ \frac {(a + b)}{2} $$ and $$ b $$, we can find another rational number.
Hence, if we continue this, we get infinite rational numbers between two given rational numbers
Question 7
1 / -0

The rational number between $$\cfrac{1}{2}$$ and $$\cfrac{6}{10}$$ is

A
$$\cfrac{1}{4}$$

B
$$\cfrac{3}{4}$$

C
$$\cfrac{21}{40}$$

D
$$\cfrac{33}{100}$$

Solution

$$\dfrac{1}{2}$$ =$$\dfrac{5}{10}$$
$$\dfrac{5}{10}$$ and $$\dfrac{6}{10}$$
Multiplying the denominator and numerator by $$4$$ we get,
$$\dfrac{5}{10}\times 4=\dfrac{20}{40}$$ and
$$\dfrac{6}{10}\times 4=\dfrac{24}{40}$$
Number between $$\dfrac{20}{40}$$ and $$\dfrac{24}{40}$$ is $$\dfrac{21}{40}$$
Question 8
1 / -0

Find the value of $$x$$ in $$\displaystyle\frac{13}{10}\times x=1$$

A
$$\dfrac{10}{13}$$

B
$$\dfrac{13}{10}$$

C
$$\dfrac{23}{10}$$

D
$$\dfrac{10}{23}$$

Solution

$$x$$ is a reciprocal of $$\dfrac{13}{10}$$. Hence, $$x = \dfrac{10}{13}$$
Question 9
1 / -0

Find the additive inverse $$\displaystyle\frac{24}{25}$$ is

A
$$\displaystyle\frac{-24}{-25}$$

B
$$\displaystyle\frac{-24}{25}$$

C
0

D
None

Solution

Let the additive inverse of $$\displaystyle\frac{24}{25}$$ be $$x$$
Thus, $$\displaystyle\frac{24}{25}+x=0$$
$$=>x=-\displaystyle\frac{24}{25}$$
$$\therefore$$ The additive inverse of $$\displaystyle\frac{24}{25}=-\displaystyle\frac{24}{25}$$
Question 10
1 / -0

How many rational numbers are there between $$-1$$ and $$0$$?

A
Infinite

B
1000

C
4990

D
None

Solution

There are infinite number of rational numbers between any two integers.

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Rational Numbers Test - 17

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Rational Numbers Test - 17

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Question 1

$$\displaystyle\frac{-4}{9};\;\displaystyle\frac{13}{7};\;\displaystyle\frac{5}{11};\;\displaystyle\frac{-13}{14}$$

$$\displaystyle\frac{-4}{9};\;\displaystyle\frac{1}{7};\;\displaystyle\frac{5}{11};\;\displaystyle\frac{-11}{14}$$

$$\displaystyle\frac{-4}{9};\;\displaystyle\frac{13}{7};\;\displaystyle\frac{5}{11};\;\displaystyle\frac{-11}{14}$$

$$\displaystyle\frac{-5}{9};\;\displaystyle\frac{13}{7};\;\displaystyle\frac{5}{11};\;\displaystyle\frac{-11}{14}$$

Solution

Question 2

True

False

Cannot be determined

None

Question 3

1

2

0

None

Solution

Question 4

True

False

Cannot be determined

None of these

Solution

Question 5

$$\displaystyle\frac{-8}{9}$$

$$\displaystyle\frac{-4}{9}$$

$$\displaystyle\frac{-5}{9}$$

$$\displaystyle\frac{-7}{9}$$

Solution

Question 6

Infinite

Finite

Two

One

Solution

Question 7

$$\cfrac{1}{4}$$

$$\cfrac{3}{4}$$

$$\cfrac{21}{40}$$

$$\cfrac{33}{100}$$

Solution

Question 8

$$\dfrac{10}{13}$$

$$\dfrac{13}{10}$$

$$\dfrac{23}{10}$$

$$\dfrac{10}{23}$$

Solution

Question 9

$$\displaystyle\frac{-24}{-25}$$

$$\displaystyle\frac{-24}{25}$$

0

None

Solution

Question 10

Infinite

1000

4990

None

Solution

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