Rational Numbers Test

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Rational Numbers Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following represents a rational number between $$-6$$ and $$-7$$?

A
$$\dfrac {-6 - 7}{2}$$

B
$$\dfrac {-6 + 7}{2}$$

C
$$\dfrac {6 + 7}{2}$$

D
$$-6 - 7$$

Solution

Take the average average of the $$-6$$ and $$-7$$:
Mean $$= \dfrac {(-6) + (-7)}{2} = \dfrac {-6 -7}{2}$$.
Mean of two numbers always lies between the two numbers.
So, answer is option $$A.$$
Question 2
1 / -0

Which of the following rational numbers lies between $$\dfrac {3}{4}$$ and $$\dfrac {13}{8}$$?

A
$$\dfrac {11}{16}$$

B
$$\dfrac {12}{16}$$

C
$$\dfrac {19}{16}$$

D
$$\dfrac {26}{16}$$

Solution

$$Mean = \dfrac{\left (\dfrac {3}{4} + \dfrac {13}{8}\right )}{2} = \dfrac {6 + 13}{8} \times \dfrac {1}{2} = \dfrac {19}{16}$$.
Mean of two numbers always lies between the two numbers.
So, answer is option $$C.$$
Question 3
1 / -0

Which of the following rational number lies between $$\dfrac {4}{9}$$ and $$\dfrac {4}{5}$$?

A
$$-1$$

B
$$\dfrac {28}{45}$$

C
$$0$$

D
$$1$$

Solution

$$Mean = \dfrac{\left (\dfrac {4}{9} + \dfrac {4}{5}\right )}{2} = \left (\dfrac {20 + 36}{45}\right ) \times \dfrac {1}{2} = \dfrac {56}{45}\times \dfrac {1}{2}$$
$$= \dfrac {28}{45}$$
Mean of two numbers always lies between the two numbers.
So, answer is option $$B.$$
Question 4
1 / -0

Which number is represented by $$A$$, in the following number line?

A
$$13$$

B
$$\dfrac {6}{13}$$

C
$$6$$

D
$$\dfrac {0}{13}$$

Solution

In the given number line, every number increases by $$\dfrac {1}{13}$$ as we move right.
$$\therefore \dfrac {5}{13} + \dfrac {1}{13} = \dfrac {6}{13} \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \dfrac {A}{13}$$
$$\therefore A = 6$$.
Question 5
1 / -0

Which number is represented by $$x$$, in the given number line?

A
$$-1$$

B
$$1$$

C
$$0$$

D
$$3$$

Solution

The numbers represented on the number line are ranging. $$\dfrac {-3}{3}, \dfrac {-2}{3}, \dfrac {-1}{3}, 0, \dfrac {1}{3}, \dfrac {2}{3}, .....$$ i.e. $$\dfrac {1}{3}$$ is added to every right element. Thus, $$x = \dfrac {-1}{3} + \dfrac {1}{3} = 0$$.
Question 6
1 / -0

The values of $$A$$ and $$B$$ represented on the number line are

A
$$1, 1$$

B
$$-1, 1$$

C
$$1, -1$$

D
$$-1, -1$$

Solution

The denominators of all the numbers represented on the number line here , are $$1$$. Hence, $$A = 1$$ and $$B = 1$$.
Question 7
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What rational number lies exactly halfway between $$\dfrac{2}{3}$$ and $$\dfrac{3}{4}$$?

A
$$\dfrac{3}{5}$$

B
$$\dfrac{5}{6}$$

C
$$\dfrac{7}{12}$$

D
$$\dfrac{9}{16}$$

E
$$\dfrac{17}{24}$$

Solution

Consider $$3 \times 4 = 12$$,

So, $$\dfrac {2}{3} = \dfrac{8}{12}$$

$$\dfrac 34 = \dfrac{9}{12}$$

Multiplying the numerator and denominator by $$2$$:

$$\dfrac{16}{24}$$ and $$\dfrac{18}{24}$$.

The mid point is $$\dfrac{17}{24}$$

Hence option $$E$$ is correct.
Question 8
1 / -0

An illustration of the associative law for multiplication is given by

A
$$\left (\dfrac{1}{3}\times 5\right)\times 8=\dfrac{1}{3}\times (5\times 8)$$

B
$$\dfrac{1}{3}\times 5\times 8=\dfrac{1}{3}\times 8\times 5$$

C
$$\dfrac{1}{3}\times 5+\dfrac{1}{3}\times 8=\dfrac{1}{3}\times 13$$

D
$$\dfrac{1}{3}\times 5\times 8=\left (\dfrac{1}{3}\times 5\right)\times \left (\dfrac{1}{3}\times 8\right)$$

E
none of these

Solution

Associative law for multiplication means $$(a\times b)\times c = a \times(b\times c)$$
Therefore, $$\left (\dfrac{1}{3} \times 5\right) \times 8 = \dfrac{1}{3} \times (5 \times 8)$$ implies associative law for multiplication.
Question 9
1 / -0

Subtract the additive inverse of $$\dfrac {5}{6}$$ from the multiplicative inverse of $$\dfrac {-5}{7}\times \dfrac {14}{15}$$.

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

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

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

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

Solution

$$Additive\quad inverse\quad of\quad \frac { 5 }{ 6 } =-\frac { 5 }{ 6 } \\ Now\quad \frac { -5 }{ 7 } \times \frac { 14 }{ 15 } =-\frac { 2 }{ 3 } \\ Multiplicative\quad inverse\quad of\quad -\frac { 2 }{ 3 } =-\frac { 3 }{ 2 } \\ now\quad subtracting\quad addtive\quad inverse\quad from\quad multiplicative\quad inverse\\ -\frac { 3 }{ 2 } -(-\frac { 5 }{ 6 } )=-\frac { 2 }{ 3 } $$
So correct answer will be option B
Question 10
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Identity the rational number that does not lie between $$ \cfrac{3}{5}$$ and $$ \cfrac{2}{3}$$.

A
$$ \cfrac{46}{75}$$

B
$$ \cfrac{47}{75}$$

C
$$ \cfrac{49}{75}$$

D
$$ \cfrac{50}{75}$$

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Re-Attempt Weekly Quiz Competition

Rational Numbers Test - 31

Result

Rational Numbers Test - 31

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Rank

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SHARING IS CARING

Question 1

$$\dfrac {-6 - 7}{2}$$

$$\dfrac {-6 + 7}{2}$$

$$\dfrac {6 + 7}{2}$$

$$-6 - 7$$

Solution

Question 2

$$\dfrac {11}{16}$$

$$\dfrac {12}{16}$$

$$\dfrac {19}{16}$$

$$\dfrac {26}{16}$$

Solution

Question 3

$$-1$$

$$\dfrac {28}{45}$$

$$0$$

$$1$$

Solution

Question 4

$$13$$

$$\dfrac {6}{13}$$

$$6$$

$$\dfrac {0}{13}$$

Solution

Question 5

$$-1$$

$$1$$

$$0$$

$$3$$

Solution

Question 6

$$1, 1$$

$$-1, 1$$

$$1, -1$$

$$-1, -1$$

Solution

Question 7

$$\dfrac{3}{5}$$

$$\dfrac{5}{6}$$

$$\dfrac{7}{12}$$

$$\dfrac{9}{16}$$

$$\dfrac{17}{24}$$

Solution

Question 8

$$\left (\dfrac{1}{3}\times 5\right)\times 8=\dfrac{1}{3}\times (5\times 8)$$

$$\dfrac{1}{3}\times 5\times 8=\dfrac{1}{3}\times 8\times 5$$

$$\dfrac{1}{3}\times 5+\dfrac{1}{3}\times 8=\dfrac{1}{3}\times 13$$

$$\dfrac{1}{3}\times 5\times 8=\left (\dfrac{1}{3}\times 5\right)\times \left (\dfrac{1}{3}\times 8\right)$$

none of these

Solution

Question 9

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

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

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

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

Solution

Question 10

$$ \cfrac{46}{75}$$

$$ \cfrac{47}{75}$$

$$ \cfrac{49}{75}$$

$$ \cfrac{50}{75}$$

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