Rational Numbers Test

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Rational Numbers Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

Choose the rational number which does not lie between rational numbers $$ \displaystyle \frac{3}{5} $$ and $$ \displaystyle \frac{2}{3} $$ :

A
$$ \displaystyle \frac{46}{75} $$

B
$$ \displaystyle \frac{47}{75} $$

C
$$ \displaystyle \frac{49}{75} $$

D
$$ \displaystyle \frac{50}{75} $$

Solution

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.
Question 2
1 / -0

Choose the rational number which does not lie between rational numbers $$\displaystyle\frac{3}{5}$$ and $$\displaystyle\frac{2}{3}$$

A
$$\displaystyle\frac{46}{75}$$

B
$$\displaystyle\frac{47}{75}$$

C
$$\displaystyle\frac{49}{75}$$

D
$$\displaystyle\frac{50}{75}$$

Solution

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.

.
Question 3
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 4
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 5
1 / -0

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 6
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 7
1 / -0

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 8
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If A : The quotient of two integers is always a rational number, and
R : $$\displaystyle \frac{1}{0}$$ is not rational, then which of the following statements is true ?

A
A is True and R is the correct explanation of A

B
A is False and R is the correct explanation of A

C
A is True and R is False

D
Both A and R are False

Solution

Since $$\displaystyle \frac{1}{0}$$ is not rational, the quotient of two integers is not rational.
Question 9
1 / -0

Find the rational number which is not equal to $$ \displaystyle \frac{ 2}{3} $$

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

B
$$ \displaystyle \frac{ -4}{+6} $$

C
$$ \displaystyle \frac{ 8}{12} $$

D
$$ \displaystyle \frac{ 6}{9} $$

Solution

We are given the rational number $$\dfrac{2}{3}$$. We want to find a rational number, out of the given options, that is not equal to $$\dfrac{2}{3}$$.

i) $$\dfrac{-2}{-3}$$ = $$\dfrac{2}{3}$$ , after cancelling the $$-$$ signs

ii) $$\dfrac{-4}{6} = \dfrac{-2}{3}\neq \dfrac{2}{3}$$ , after dividing the numerator and denominator by $$2$$

iii) $$\dfrac{8}{12} = \dfrac{2}{3}$$ , after dividing the numerator and denominator by $$4$$

iv) $$\dfrac{6}{9} = \dfrac{2}{3}$$ , after dividing the numerator and denominator by $$3$$.

Thus, the correct answer is Option $$B$$
Question 10
1 / -0

A rational number between $$\dfrac {-9}{10}$$ and $$\dfrac {4}{5}$$ is:

A
$$\left (\dfrac {-9}{10} + \dfrac {4}{5}\right ) \times \dfrac {1}{2}$$

B
$$\left (\dfrac {-9}{10} - \dfrac {4}{5}\right ) + \dfrac {1}{2}$$

C
$$\left (\dfrac {-9}{10} + \dfrac {4}{5}\right ) \times 2$$

D
All above are correct

Solution

Mean of two numbers always lies between the two numbers.

Mean

$$= \dfrac{\left (\dfrac {-9}{10} + \dfrac {4}{5}\right )}{2}$$

$$= \left (\dfrac {-9}{10} + \dfrac {4}{5}\right )\times \dfrac {1}{2}$$.

So, answer is option $$A.$$

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Re-Attempt Weekly Quiz Competition

Rational Numbers Test - 17

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Rational Numbers Test - 17

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Question 1

$$ \displaystyle \frac{46}{75} $$

$$ \displaystyle \frac{47}{75} $$

$$ \displaystyle \frac{49}{75} $$

$$ \displaystyle \frac{50}{75} $$

Solution

Question 2

$$\displaystyle\frac{46}{75}$$

$$\displaystyle\frac{47}{75}$$

$$\displaystyle\frac{49}{75}$$

$$\displaystyle\frac{50}{75}$$

Solution

Question 3

$$\displaystyle\frac{-8}{6},\,\displaystyle\frac{-13}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$$ 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 4

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

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

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

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

Solution

Question 5

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

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

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

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

Solution

Question 6

$$\displaystyle\frac{3}{2},\,\displaystyle\frac{1}{4}$$ 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 7

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

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

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

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

Solution

Question 8

A is True and R is the correct explanation of A

A is False and R is the correct explanation of A

A is True and R is False

Both A and R are False

Solution

Question 9

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

$$ \displaystyle \frac{ -4}{+6} $$

$$ \displaystyle \frac{ 8}{12} $$

$$ \displaystyle \frac{ 6}{9} $$

Solution

Question 10

$$\left (\dfrac {-9}{10} + \dfrac {4}{5}\right ) \times \dfrac {1}{2}$$

$$\left (\dfrac {-9}{10} - \dfrac {4}{5}\right ) + \dfrac {1}{2}$$

$$\left (\dfrac {-9}{10} + \dfrac {4}{5}\right ) \times 2$$

All above are correct

Solution

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