MCQ Questions for Class 8 Maths Rational Numbers Quiz 5

State true or false.

$\dfrac {-3}{5}$ lies to the left of

$0$ on the number line.

0%
True
0%
False

Explanation

Since

$\dfrac {-3}{5}$ is a negative number, all the negative numbers are

$< 0$ and so all the negative numbers lie to the left of

$0$ on the number line.

$\therefore \dfrac {-3}{5}$ lies to the left of

$0$ . So the above statement is True.

How many rational numbers are there between two rational numbers?

0%
$1$
0%
$0$
0%
unlimited
0%
$100$

Explanation

Consider

$a\,\&\,b$ as rational numbers

Then a rational number

$n_1$ can be find in between

$a\,\&\,b$ such that

$a<n_1<b$

using the formula

$n_1=\dfrac{a+b}{2}$

similarly, again we can find a rational number between

$a\,\&\,n_1$ and

$n_1\,\&\,b$

and so on...

Hence, between any two distinct rational numbers there are infinitely many rational numbers.

State true or false.

The rational number $\dfrac{-3}{4}$ lies to the right of zero on the number line.

0%
True
0%
False

$\dfrac{-3}{8} + \dfrac{1}{7} = \dfrac{1}{7} + \left(\dfrac{-3}{8}\right)$ is an example to show that

0%
Addition of rational numbers is commutative.
0%
Rational numbers are closed under addition.
0%
Addition of rational number is associative.
0%
Rational numbers are distributive under addition.

Explanation

Commutative Property states that

$a+b=b+a$

Here,

$\dfrac{-3}8+\dfrac17=\dfrac{-13}{56}$

and

$\dfrac17+\left(\dfrac{-3}8\right)=\dfrac{-13}{56}$

$\Rightarrow$ Addition of rational numbers is commutative.

$\therefore$ Option A is the correct answer.

Which of the following is an example of distribute property of multiplication over addition for rational numbers ?

0%
$\dfrac{-1}{4} \times \left \{ \dfrac{2}{3} + \left ( \dfrac{-4}{7} \right ) \right \} = \left [ -\dfrac{1}{4} \times \dfrac{2}{3} \right ] + \left [ -\dfrac{1}{4} \times \left ( \dfrac{-4}{7} \right ) \right ]$
0%
$\dfrac{-1}{4} \times \left \{ \dfrac{2}{3} + \left ( \dfrac{-4}{7} \right ) \right \} = \left [ \dfrac{1}{4} \times \dfrac{2}{3} \right ] - \left ( \dfrac{-4}{7} \right )$
0%
$\dfrac{-1}{4} \times \left \{ \dfrac{2}{3} + \left ( \dfrac{-4}{7} \right ) \right \} = \dfrac{2}{3} + \left ( -\dfrac{1}{4} \right ) \times \left ( \dfrac{-4}{7} \right )$
0%
$\dfrac{-1}{4} \times \left \{ \dfrac{2}{3} + \left ( \dfrac{-4}{7} \right ) \right \} = \left \{ \dfrac{2}{3} + \left ( -\dfrac{4}{7} \times \right ) \right \} - \dfrac{1}{4}$

Explanation

Distributive property states that

$a \times (b + c) = (a \times b) + (a \times c)$

Let

$a=\dfrac{-1}{4} , b= \dfrac{2}{3}, c= \dfrac{-4}{7}$

Then

$\dfrac{-1}{4} \times \left \{ \dfrac{2}{3} + \left ( \dfrac{-4}{7} \right ) \right \} = \left [ -\dfrac{1}{4} \times \dfrac{2}{3} \right ] + \left [ -\dfrac{1}{4} \times \left ( \dfrac{-4}{7} \right ) \right ]$

Hence option A is the correct answer.

State whether the statement is true (T) or false (F).
For all rational numbers x and y,

$x - y = y - x$

0%
True
0%
False

Explanation

The given statement is false since subtraction of two rational numbers is not commutative.

in fact,

$(x - y) = -(y - x)$

Which of the following expressions shows that rational numbers are associative under multiplication ?

0%
$\dfrac{2}{3} \times (\dfrac{-6}{7} \times \dfrac{3}{5}) = (\dfrac{2}{3} \times \dfrac{-6}{7}) \times \dfrac{3}{5}$
0%
$\dfrac{2}{3} \times (\dfrac{-6}{7} \times \dfrac{3}{5}) = \dfrac{2}{3} \times (\dfrac{3}{5} \times \dfrac{-6}{7})$
0%
$\dfrac{2}{3} \times (\dfrac{-6}{7} \times \dfrac{3}{5}) = (\dfrac{3}{5} \times \dfrac{2}{3}) \times \dfrac{-6}{7}$
0%
$(\dfrac{2}{3} \times \dfrac{-6}{7}) \times \dfrac{3}{5} = (\dfrac{-6}{7} \times \dfrac{3}{5}) \times \dfrac{2}{3}$

Explanation

For associative multiplication property-

$a \times (b \times c) = (a \times b) \times c$

let

$a = \dfrac{2}{3}, b = \dfrac{-6}{7}, c = \dfrac{3}{5}$

Then,

$\dfrac{2}{3} \times (\dfrac{-6}{7} \times \dfrac{3}{5}) = (\dfrac{2}{3} \times \dfrac{-6}{7}) \times \dfrac{3}{5}$

Hence option A is the correct answer.

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

0%
Between x and y
0%
Less than x and y both.
0%
Greater than x and y both.
0%
Less than x but greater than y.

Explanation

$\dfrac{x+ y}{2}$ lies in between x and y.

Hence, option A is the correct answer.

The additive inverse of

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

0%
$\dfrac{-7}{19}$
0%
$\dfrac{7}{19}$
0%
$\dfrac{19}{7}$
0%
$\dfrac{-19}{7}$

Explanation

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.

The numerical expression

$\dfrac{3}{8} + \dfrac{(-5)}{7} = \dfrac{-19}{56}$ shows that

0%
Rational numbers are not closed under addition.
0%
Rational numbers are closed under addition.
0%
Rational numbers are closed under multiplication.
0%
Addition of rational numbers is not commutative.

Explanation

The sum of any two rational numbers is always a rational number.

Here,

$\dfrac{3}{8}$ and

$\dfrac{-5}{7}$ are rational numbers and their sum

$\dfrac{-19}{56}$ is also a rational number.

$\therefore$ Option B is the correct answer.

State whether the statement is true (T) or false (F).
The rational number

$\dfrac{-8}{-3}$ lies neither to the right nor to the left of zero on the number line.

0%
True
0%
False

Explanation

The given number is

$\dfrac{-8}{-3}=\dfrac{-8}{-3}\times \dfrac{-1}{-1}\quad$ [Multiplying numerator and denominator both by

$-1$ ]

$=\dfrac83\quad$

which is a positive rational number that lies on the right side of zero.

Hence the given statement is false

State whether the following statement is true of false:
Number of rational numbers between

$15$ and

$18$ is finite.

0%
True
0%
False

Explanation

The given statement is false.

because, there lies infinitely many rational numbers between any two rational number. Hence, number of rational numbers between

$15$ and

$18$ are infinite.

State whether the statements are true (T) or false (F).

The rational numbers

$\dfrac{-5}{-7}$ and

$\dfrac{7}{-9}$ lie on opposite sides of zero on the number line.

0%
True
0%
False

Explanation

On a number line,

$\dfrac{-5}{-7}=\dfrac{5}{7}$ and $\dfrac{7}{-9}=\dfrac{-7}{9}$ lie on opposite sides of zero. ( $\because$ opposite sign)

State whether the following statement is true/ false?

$-5/7$ is the additive inverse of

$5/7$ .

0%
True
0%
False

Explanation

The additive inverse of

$-5/7=-(-5/7)=5/7$

The given statement is true.

State whether the following statement is true/ false?
Commutative property holds for subtraction of rational numbers.

0%
True
0%
False

Explanation

Let us take two numbers

$5/4$ and

$3/4$ ,

Now

$5/4 - 3/4 = 2/4$
And we know that,

$3/4 - 5/4 = -2/4$

Since,

$2/4\neq -2/4$
Therefore, the given statement is false.

State whether the following statement is true/ false?

$0$ is the additive inverse of its own.

0%
True
0%
False

Explanation

Let the additive inverse of

$0$ be

$x$

Then

$0+x=0$

$\therefore x=0$

Therefore, the additive inverse of 0 is 0 itself.

State whether the following statement is true/ false?
The associative property does not hold for subtraction of rational numbers.

0%
True
0%
False

Explanation

Let us take three numbers

$5/4,3/4$ and

$1/4$ ,

Now

$5/4 - (3/4-1/4)$

$= 5/4-2/4=3/4$

And we know that,

$(5/4 - 3/4)-1/4$

$= 2/4-1/4=1/4$

Since,

$3/4\neq -1/4$
Therefore, the given statement is true.

Two rational numbers between

$\dfrac{1}{5}$ and

$\dfrac{4}{5}$ are :

0%
1 and $\dfrac{3}{5}$
0%
$\dfrac{2}{5}$ and $\dfrac{3}{5}$
0%
$\dfrac{1}{2}$ and $\dfrac{2}{1}$
0%
$\dfrac{3}{5}$ and $\dfrac{6}{5}$

Explanation

Since the denominator of both rational numbers are same. So, for getting the rational numbers between the given rational numbers, we only have to consider the numerators of the rational numbers.

Two numbers between 1 & 4 are 2 and 3.

So, two rational numbers between the given rational numbers will be

$\dfrac { 2 }{ 5 }$ and

$\dfrac { 3 }{ 5 }$

So, correct answer is option B.

Find the nine rational numbers between

$0$ and

$1$ .

0%
$0.1,0.2,0.3, ... ,0.9$
0%
$1.1,0.2 ,10.3, ... ,0.9$
0%
$0.1,0.2,0.3, ... ,20.9$
0%
$0.1 , 0.2 , 10.3 , ... , 0.9$

Explanation

$0 < (0+0.1)=0.1 <(0.1+0.1)= 0.2 < (0.2+0.1)=0.3 < ... < (0.8+1)=0.9 <(0.9+0.1)= 1$

$0 < 0.1 < 0.2 < 0.3 < ... < 0.9 < 1$

$\therefore$ The nine rational numbers between

$0$ and

$1$ are

$0.1,0.2,0.3, ... ,0.9$

Choose the rational number, which does not lie, between the rational numbers,

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

$-\dfrac{1}{5}$

0%
$-\dfrac{3}{10}$
0%
$\dfrac{3}{10}$
0%
$-\dfrac{1}{4}$
0%
$-\dfrac{7}{20}$

Explanation

The given rational numbers

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

$-\dfrac { 1 }{ 5 }$ are negative rational numbers because the numerator and denominator of both the rational numbers are of opposite signs that is the numerator of both the integers is negative while the denominators are positive.

Therefore, none of the positive rational number can lie between the given negative rational numbers

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

$-\dfrac { 1 }{ 5 }$ .

Hence,

$\dfrac { 3 }{ 10 }$ does not lie between the rational numbers

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

$-\dfrac { 1 }{ 5 }$ .

Choose the rational number which does not lie between rational numbers

$\dfrac{3}{5}$ and

$\dfrac{2}{3}$ .

0%
$\dfrac{46}{75}$
0%
$\dfrac{47}{75}$
0%
$\dfrac{49}{75}$
0%
$\dfrac{50}{75}$

Explanation

All the options have a denominator of

$75$ . Hence, let us convert

$\dfrac35$ and

$\dfrac23$ into equivalent fractions having denominator

$75$ .

$\dfrac{3}{5}$

$=\dfrac{3\times 15}{5\times 15}$

$=\dfrac{45}{75}$

$\dfrac{2}{3}$

$=\dfrac{2\times 25}{3\times 25}$

$=\dfrac{50}{75}$

Hence,

$\dfrac{50}{75}$ does not lie between the given rational numbers.

State true or false.

Five rational numbers between.

$\dfrac{2}{3}$ and

$\dfrac{4}{5}$ are

$\dfrac{41}{60},\dfrac{42}{60},\dfrac{43}{60},\dfrac{44}{60},\dfrac{45}{60}$

0%
True
0%
False

Explanation

To get the rational numbers between

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

$\displaystyle\frac{4}{5}$

Take an LCM of these two numbers:

$\displaystyle\frac{10}{15}$ and

$\displaystyle\frac{12}{15}$

Multiply numerator and denominator by 4:

$\displaystyle\frac{40}{60}$ and

$\displaystyle\frac{48}{60}$

All the numbers between

$\displaystyle\frac{40}{60}$ and

$\displaystyle\frac{48}{60}$ form the answer

Some of these numbers are

$\displaystyle\frac{41}{60}$ ,

$\displaystyle\frac{42}{60}$ ,

$\displaystyle\frac{43}{60}$ ,

$\displaystyle\frac{44}{60}$ ,

$\displaystyle\frac{45}{60}$

Hence the statement is true

State whether the given statement is true or false:

Five rational numbers between

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

$\dfrac{5}{3}$ are

$\dfrac{-8}{6},\dfrac{-7}{6},0,\dfrac{1}{6},\dfrac{2}{6}$ .

0%
True
0%
False

Explanation

Given rational numbers are

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

$\displaystyle\frac{5}{3}$

LCM of the denominators is

$6$

Hence

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

$\displaystyle\frac{5}{3}$ can be written as

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

$\displaystyle\frac{10}{6}$

The numbers between

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

$\displaystyle\frac{10}{6}$ are

$\displaystyle\frac{-8}{6}$ ,

$\displaystyle\frac{-7}{6}$ ,

$\displaystyle{0}$ ,

$\displaystyle\frac{1}{6}$ ,

$\displaystyle\frac{2}{6},...$

Hence the statement is true.

State true or false

Five rational numbers between

$\dfrac{1}{4}$ and

$\dfrac{1}{2}$ are

$\displaystyle\frac{9}{32},\frac{10}{32},\frac{11}{32},\frac{12}{32},\frac{13}{32}$ .

0%
True
0%
False

Explanation

To get the rational numbers between

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

$\displaystyle\frac{1}{2}$

Take an LCM of these two numbers:

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

$\displaystyle\frac{2}{4}$

Multiply numerator and denominator by 8:

$\displaystyle\frac{8}{32}$ and

$\displaystyle\frac{16}{32}$

All the numbers between

$\displaystyle\frac{8}{32}$ and

$\displaystyle\frac{16}{32}$ form the answer

Some of these numbers are

$\displaystyle\frac{9}{32}$ ,

$\displaystyle\frac{10}{32}$ ,

$\displaystyle\frac{11}{32}$ ,

$\displaystyle\frac{12}{32}$ ,

$\displaystyle\frac{13}{32}$

State true or false:

Ten rational numbers between

$\dfrac{3}{5}$ and

$\dfrac {3}{4}$ are

$\displaystyle\frac{97}{160},\frac{98}{160},\frac{99}{160},\frac{100}{160},\frac{101}{160},\frac{102}{160},\frac{103}{160},\frac{104}{160},\frac{105}{160},\frac{106}{160}$

0%
True
0%
False

Explanation

To get the rational numbers between

$\displaystyle\frac{3}{5}$ and

$\displaystyle\frac{3}{4}$

Take an LCM of these two numbers:

$\displaystyle\frac{12}{20}$ and

$\displaystyle\frac{15}{20}$

So now to make the denominator

$(160)$ as per the question we need to
multiply numerator and denominator by

$8$ :

$\displaystyle\frac{96}{160}$ and

$\displaystyle\frac{120}{160}$

All the numbers between

$\displaystyle\frac{96}{160}$ and

$\displaystyle\frac{120}{160}$ form the answer

Some of these numbers are

$\displaystyle\frac{97}{160}$ ,

$\displaystyle\frac{98}{160}$ ,

$\displaystyle\frac{99}{160}$ ,

$\displaystyle\frac{100}{160}$ ,

$\displaystyle\frac{101}{160}$ ,

$\displaystyle\frac{102}{160}$ ,

$\displaystyle\frac{103}{160}$ ,

$\displaystyle\frac{104}{160}$ ,

$\displaystyle\frac{105}{160}$ ,

$\displaystyle\frac{106}{160}$

State true or false.

Three rational numbers between

$\dfrac{2}{5}$ and

$\dfrac{3}{5}$ is

$\dfrac{9}{20},\dfrac{10}{20},\dfrac{11}{20}$

0%
True
0%
False

Explanation

Given numbers can be written as

$\dfrac { 2\times 5 }{ 5\times 5 } =\dfrac { 10 }{ 25 }\\$

$\dfrac { 3\times 5 }{ 5\times 5 } =\dfrac { 15 }{ 25 }$
therefore 3 rational numbers are

$\dfrac { 11 }{ 25 }, \dfrac { 12 }{ 25 }, \dfrac { 13 }{ 25 }$

Which of the following statements is true?

0%
$\displaystyle \frac{-5}{8}$ lies to the left of $0$ on the number line
0%
$\displaystyle \frac{3}{7}$ lies to the right of $0$ on the number line.
0%
The rational numbers $\displaystyle \frac{1}{3}$ and $\displaystyle \frac{-7}{3}$ are on opposite sides of $0$ on the number line
0%
All the above

Explanation

We know that every positive number occupies right side of the origin while the negative numbers are present on the left side of the origin. Let us learn to represent a rational number on a number line with few steps.

(i) Draw a line and locate the point having the coordinate as

$0$ . This point is known as the origin.

If the given number is positive, locate it on the right side of the origin. If it is a negative number, locate it on the left side of zero.

Divide each unit into the values which are equal to the denominator of the fraction. For example: for representing

$\dfrac {4}{5}$ on the number line you need to divide each unit into

$5$ subunits.

Similarly,

(a)

$\dfrac {-5}{8}$ is a negative rational number and it is known that

$\dfrac {-5}{8}$ is greater than

$-1$ and less than

$0$ .

Therefore,

$\dfrac {-5}{8}$ lies between

$0$ and

$-1$ which is on the left side of

$0$ on the number line.

(b)

$\dfrac {3}{7}$ is a positive rational number and it is known that

$\dfrac {3}{7}$ is less than

$1$ and greater than

$0$ .

Therefore,

$\dfrac {3}{7}$ lies between

$0$ and

$1$ which is on the right side of

$0$ on the number line.

(c)

$\dfrac {-7}{3}$ is a negative rational number lies between

$-2$ and

$-3$ and on the left side of

$0$ on the number line, whereas,

$\dfrac {1}{3}$ is a positive rational number lies between

$0$ and

$1$ and on the right side of

$0$ on the number line

Hence, the rational numbers

$\dfrac {1}{3}$ and

$\dfrac {-7}{3}$ are on opposite sides of

$0$ on the number line.

All of the statements are true.

__________ are rational numbers between between 5 and -2.

0%
$\displaystyle 5, \frac{33}{4},\ \frac{3}{2}, -\frac{1}{4}, -2$
0%
$\displaystyle \frac{13}{4},\ \frac{3}{2}, -\frac{1}{4}$
0%
$\displaystyle 5, \frac{13}{4},\ \frac{13}{2}, -\frac{1}{4}, -2$
0%
none of the above

Explanation

A rational number between two numbers

$a$ and

$b = \dfrac {(a + b)}{2}$

So, a rational number between

$5$ and

$- 2 = \dfrac {( 5 - 2 )}{2} = \dfrac {3}{2}$

Now, another rational number between

$5$ and

$\dfrac {3}{2} = \dfrac {( 5 + \dfrac {3}{2})}{2} = \dfrac {13}{4}$

Another rational number between

$\dfrac {3}{2}$ and

$-2 = \dfrac {( \dfrac {3}{2}) - 2}{2} = -\dfrac {1}{4}$

$\therefore \dfrac {13}{4}, \dfrac {3}{2}, - \dfrac {1}{4}$ are the rational numbers between

$5$ and

$-2$

Which of the following order is correct for rational numbers between

$\displaystyle \frac{4}{11}$ and

$\displaystyle \frac{9}{16}?$

0%
$\dfrac{4}{11}$ , $\dfrac{67}{176}$ , $\dfrac{79}{176}$ , $\dfrac{9}{16}$
0%
$\dfrac{4}{11}$ , $\dfrac{67}{276}$ , $\dfrac{79}{176}$ , $\dfrac{9}{16}$
0%
$\displaystyle \frac{4}{11}, \frac{17}{38}, \frac{13}{27}, \frac{22}{43}, \frac{9}{16}$
0%
$\displaystyle \frac{4}{11}, \frac{27}{38}, \frac{13}{27}, \frac{22}{43}, \frac{9}{16}$

Explanation

$L.C.M$ of

$11$ and

$16$ is

$11\times 16 =176$

$\dfrac{4}{11}=\dfrac{4\times 16}{11\times 16}=\dfrac{64}{176}$

$\dfrac{9}{16}=\dfrac{9\times 11}{16\times 11}=\dfrac{99}{176}$

rational numbers between

$\dfrac{4}{11}$ and

$\dfrac{9}{16}$ are

$\dfrac{67}{176}$ ,

$\dfrac{79}{176}$

Which property of multiplication is illustrated by :

$\displaystyle {\frac{-4}{3}\, \times\, \left (\frac{-6}{5}\, +\, \frac{8}{7} \right )\, =\, \left (\frac{-4}{3}\, \times\, \frac{-6}{5} \right )\, +\, \left (\frac{-4}{3}\, \times\, \frac{8}{7} \right )}$

0%
Associative property
0%
Commutative property
0%
Distributive property
0%
None of these

Explanation

We know that distributive property of multiplication with respect to addition is defined as:

For any real number

$a, b, c$ ,

$a\times (b + c) = a\times b + a\times c$

Let us take

$a=\dfrac {-4}{3},b=\dfrac {-6}{5}$ and

$c=\dfrac {8}{7}$ and apply the distributive property stated above:

$\dfrac { -4 }{ 3 } \times \left( \dfrac { -6 }{ 5 } +\dfrac { 8 }{ 7 } \right) =\left( \dfrac { -4 }{ 3 } \times \dfrac { -6 }{ 5 } \right) +\left( \dfrac { -4 }{ 3 } \times \dfrac { 8 }{ 7 } \right)$

Hence, it is the distributive property of multiplication.

CBSE Questions for Class 8 Maths Rational Numbers Quiz 5 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Rational Numbers Quiz 5 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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