MCQ Questions for Class 8 Maths Rational Numbers Quiz 7

Fill in the blank:

$\left [\dfrac {6}{5}\times \left (\dfrac {-2}{12}\right )\right ] + \left [\dfrac {6}{5} \times \dfrac {4}{12}\right ] = \dfrac {6}{5}\times \left [..... + \dfrac {4}{12}\right ]$

0%
$\left (\dfrac {-2}{12}\right )$
0%
$\left (\dfrac {-1}{12}\right )$
0%
$-12$
0%
$\dfrac {4}{12}$

Explanation

Distributive property over addition for rational numbers states that

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

$\therefore \left [\dfrac {6}{5}\times \left (\dfrac {-2}{12}\right )\right ] + \left [\dfrac {6}{5} \times \dfrac {4}{12}\right ]$

$= \dfrac {6}{5}\left [\left (\dfrac {-2}{12}\right ) + \dfrac {4}{12}\right ]$ .

The value of

$\left [\dfrac {3}{4}\times \dfrac {4}{12}\right ] + \left [\dfrac {3}{4} \times \dfrac {-3}{9}\right ]$ is

0%
$\dfrac {3}{4}$
0%
$0$
0%
$1$
0%
$\dfrac {1}{2}$

Explanation

$\left [\dfrac {3}{4}\times \dfrac {4}{12}\right ] + \left [\dfrac {3}{4} \times \dfrac {-3}{9}\right ]$

$= \dfrac {3}{4}\times \left [\dfrac {4}{12} + \dfrac {-3}{9}\right ]$ (

$\because$ distributive property)

$= \dfrac {3}{4} \times 0$

$= 0$ .

$(81\div 9)\div 3=a$ and

$81\div(9\div 3)=b$ , then which of the following is correct?

0%
$a=b$
0%
$a>b$
0%
$b>a$
0%
All of the above

Explanation

Associative property is not satisfied by division.

$(81\div 9)\div 3=a$

$81\div(9\div3)=b$

$\Rightarrow 9\div 3=3=a$

$81\div3 =27 = b$

$\therefore a\neq b$ and

$b > a$

So, option

$C$ is correct.

Which of the following rational numbers lies between

$\dfrac {-4}{5}$ and

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

0%
$\dfrac {-36}{45}$
0%
$\dfrac {-35}{45}$
0%
$\dfrac {-39}{45}$
0%
$\dfrac {-63}{45}$

Explanation

As per the given options, let us multiply both numbers with

$\dfrac {9}{9}$ .

$\therefore \dfrac {-4}{5}\times \dfrac {9}{9} = \dfrac {-36}{45}$ and

$\dfrac {-7}{5}\times \dfrac {9}{9} = \dfrac {-63}{45}$

$\therefore$ Clearly,

$\dfrac {-39}{45}$ lies between

$\dfrac {-4}{5}$ and

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

Find five rational numbers between

$1$ and

$2.$

0%
$\dfrac {1}{10}, \dfrac {2}{10}, \dfrac {3}{10}, \dfrac {4}{10}, \dfrac {5}{10}$
0%
$\dfrac {1}{5}, \dfrac {2}{5}, \dfrac {3}{5}, \dfrac {4}{5}, \dfrac {5}{5}$
0%
$\dfrac {1}{2}, \dfrac {1}{3}, \dfrac {1}{4}, \dfrac {1}{5}, \dfrac {1}{6}$
0%
$\dfrac {8}{7}, \dfrac {9}{7}, \dfrac {10}{7}, \dfrac {11}{7}, \dfrac {12}{7}$

Explanation

$\textbf{Step -1: Find required rational numbers.}$

$\text{We need five rational numbers between }1\text{ and }2.$

$\text{So, multiply and divide the numbers with a natural number greater or equal to } 5+1=6.$

$\text{Lets take }7.$

$1=1\times\dfrac{7}{7}=\dfrac{7}{7}$

$\text{and }2=2\times\dfrac{7}{7}=\dfrac{14}{7}$

$\text{Thus, }1\text{ and }2\text{ becomes }\dfrac{7}{7}\text{ and }\dfrac{14}{7}\text{ respectively.}$

$\text{Since, }7<8<9<10<11<12<13<14$

$\therefore 1=\dfrac{7}{7}<\dfrac{8}{7}<\dfrac{9}{7}<\dfrac{10}{7}<\dfrac{11}{7}<\dfrac{12}{7}<\dfrac{13}{7}<\dfrac{14}{7}=2$

$\text{Hence, five rational numbers between }1\text{ and }2\text{ are }\dfrac{8}{7},\dfrac{9}{7},\dfrac{10}{7},\dfrac{11}{7},\text{ and }\dfrac{12}{7}.$

$\textbf{Hence , the correct option is D.}$

Which of the following numbers lies between

$\dfrac {5}{24}$ and

$\dfrac {3}{8}$ ?

0%
$\dfrac {7}{2}$
0%
$1$
0%
$\dfrac {7}{24}$
0%
$0$

Explanation

Mean

$= \dfrac {\dfrac {3}{8} + \dfrac {5}{24}}{2} = \dfrac {\dfrac {9 + 5}{24}}{2} = \dfrac{\left (\dfrac {14}{24}\right )}{2}$

$= \dfrac {7}{12}\times \dfrac {1}{2}$

$= \dfrac {7}{24}$

Mean of two numbers lies between the two numbers.

So,

$\dfrac {7}{24}$ lies between

$\dfrac {3}{8}$ and

$\dfrac {5}{24}.$

The value of

$\left (\dfrac {5}{9}\times \dfrac {6}{11}\right ) + \left (\dfrac {1}{11}\times \dfrac {3}{9}\right )$ is

0%
$\dfrac {1}{5}$
0%
$\dfrac {1}{9}$
0%
$\dfrac {1}{11}$
0%
$\dfrac {1}{3}$

Explanation

$\dfrac {5}{9}\times \dfrac {6}{11} + \dfrac {1}{11}\times \dfrac {3}{9}$

$= \dfrac {1}{11}\left [\dfrac {30}{9} + \dfrac {3}{9}\right ]$ (

$\because$ distributive property)
[Also, commutative property was used to get the terms in line]

$= \dfrac {1}{11} \left [\dfrac {33}{9}\right ]$

$= \dfrac {1}{3}$

Which of the following rational numbers lies between

$\dfrac {3}{4}$ and

$\dfrac {13}{8}$ ?

0%
$\dfrac {11}{16}$
0%
$\dfrac {12}{16}$
0%
$\dfrac {19}{16}$
0%
$\dfrac {26}{16}$

Explanation

$Mean = \dfrac{\left (\dfrac {3}{4} + \dfrac {13}{8}\right )}{2} = \dfrac {6 + 13}{8} \times \dfrac {1}{2} = \dfrac {19}{16}$ .

Mean of two numbers always lies between the two numbers.

So, answer is option

$C.$

Fill in the blank:

$\dfrac {4}{7}\left (\dfrac {7}{15} - \dfrac {21}{4}\right ) = \dfrac {4}{7} \times \dfrac {7}{15} - ....\times \dfrac {4}{7}$

0%
$\dfrac {4}{7}$
0%
$\dfrac {21}{4}$
0%
$\dfrac {-21}{4}$
0%
$\dfrac {7}{15}$

Explanation

Distributive property over subtraction holds true for rational numbers.

Thus,

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

$\therefore \dfrac {4}{7} \left (\dfrac {7}{15} - \dfrac {21}{4}\right ) = \dfrac {4}{7} \times \dfrac {7}{15} - \dfrac {4}{7} \times \dfrac {21}{4}$

$= \dfrac {4}{7} \times \dfrac {7}{15} - \dfrac {21}{4} \times \dfrac {4}{7}$ (

$\because$ commutative property).

The value of

$\dfrac {1}{5} \left (\dfrac {15}{7} - \dfrac {20}{14}\right )$ is

0%
$\dfrac {2}{7}$
0%
$\dfrac {3}{7}$
0%
$\dfrac {1}{7}$
0%
$7$

Explanation

$\dfrac {1}{5} \left (\dfrac {15}{7} - \dfrac {20}{14}\right )$

$= \dfrac {1}{5} \left (\dfrac {15}{7} - \dfrac {10}{7}\right )$

$= \dfrac {1}{5} \times \dfrac {15}{7} - \dfrac {1}{5}\times \dfrac {10}{7}$ (

$\because$ distributive property)

$= \dfrac {3}{7} - \dfrac {2}{7}$

$= \dfrac {1}{7}$

The value of

$8\times 2\dfrac {4}{5} + 8\times 3\dfrac {1}{5}$ is

0%
$\dfrac {8}{5}$
0%
$24$
0%
$48$
0%
$-48$

Explanation

$8\times 2\dfrac {4}{5} + 8\times 3\dfrac {1}{5}$

$= 8\times \dfrac {14}{5} + 8\times \dfrac {16}{5}$

$= \dfrac {8}{5} \times 14 + \dfrac {8}{5} \times 16$

$= \dfrac {8}{5} \times (14 + 16)$ (

$\because$ distributive property)

$= \dfrac {8}{5}\times 30$

$= 8\times 6$

$= 48$ .

The value of

$\dfrac {3}{5} \times \dfrac {35}{24} + \dfrac {10}{1}\times \dfrac {3}{5}$ is

0%
$\dfrac {55}{8}$
0%
$\dfrac {8}{55}$
0%
$\dfrac {3}{5}$
0%
$\dfrac {35}{8}$

Explanation

$\dfrac {3}{5} \times \dfrac {35}{24} + \dfrac {10}{1}\times \dfrac {3}{5}$

$= \dfrac {3}{5} \times \dfrac {35}{24} + \dfrac {10}{1}\times \dfrac {3}{5}$

$= \dfrac {3}{5} \times \dfrac {35}{24} + \dfrac {3}{5}\times \dfrac {10}{1}$ (

$\because$ commutative property)

$= \dfrac {3}{5} \left (\dfrac {35}{24} + 10\right )$ (

$\because$ distributive property)

$= \dfrac {3}{5} \times \dfrac {35 + 240}{24}$

$=\dfrac {275}{8\times 5}$

$= \dfrac {55}{8}$ .

The value of

$2\dfrac {1}{2} \times 10 - 4\dfrac {1}{3} \times 10$ is

0%
$\dfrac {3}{55}$
0%
$\dfrac {-3}{55}$
0%
$\dfrac {55}{3}$
0%
$\dfrac {-55}{3}$

Explanation

$2\dfrac {1}{2} \times 10 - 4\dfrac {1}{3} \times 10$

$= \dfrac {5}{2}\times 10 - \dfrac {13}{3} \times 10$

$= 10\times \dfrac {5}{2} - 10\times \dfrac {13}{3}$ (

$\because$ commutative property)

$= 10 \left (\dfrac {5}{2} - \dfrac {13}{3}\right )$ (

$\because$ distributive property)

$= 10\times \left (\dfrac {15 - 26}{6}\right )$

$= 10\times \left (\dfrac {-11}{6}\right )$

$= \dfrac {-55}{3}$ .

The points

$A, B$ and

$C$ on the given number line represents

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

Explanation

The given number line can be completed as:

Thus,

$A = \dfrac {-3}{10}, B = \dfrac {-1}{10}$ and

$C = \dfrac {3}{4}$ .

A rational number between

$\dfrac {-9}{10}$ and

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

0%
$\left (\dfrac {-9}{10} + \dfrac {4}{5}\right ) \times \dfrac {1}{2}$
0%
$\left (\dfrac {-9}{10} - \dfrac {4}{5}\right ) + \dfrac {1}{2}$
0%
$\left (\dfrac {-9}{10} + \dfrac {4}{5}\right ) \times 2$
0%
All above are correct

Explanation

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

$A.$

Which of the following numbers lies between

$-1$ and

$-2$ ?

0%
$\dfrac {-1}{2}$
0%
$\dfrac {-3}{2}$
0%
$\dfrac {1}{2}$
0%
$\dfrac {3}{2}$

Explanation

Mean

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

Mean of two numbers always lies between the two numbers.

So, answer is option

$B.$

Which of the following represents a rational number between

$-6$ and

$-7$ ?

0%
$\dfrac {-6 - 7}{2}$
0%
$\dfrac {-6 + 7}{2}$
0%
$\dfrac {6 + 7}{2}$
0%
$-6 - 7$

Explanation

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

$A.$

The values of

$A$ and

$B$ represented on the number line are

0%
$1, 1$
0%
$-1, 1$
0%
$1, -1$
0%
$-1, -1$

Explanation

The denominators of all the numbers represented on the number line here , are

$1$ . Hence,

$A = 1$ and

$B = 1$ .

Which of the following rational number lies between

$\dfrac {4}{9}$ and

$\dfrac {4}{5}$ ?

0%
$-1$
0%
$\dfrac {28}{45}$
0%
$0$
0%
$1$

Explanation

$Mean = \dfrac{\left (\dfrac {4}{9} + \dfrac {4}{5}\right )}{2} = \left (\dfrac {20 + 36}{45}\right ) \times \dfrac {1}{2} = \dfrac {56}{45}\times \dfrac {1}{2}$

$= \dfrac {28}{45}$

Mean of two numbers always lies between the two numbers.

So, answer is option

$B.$

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

$\dfrac{7}{-4}$ lies to the right zero on the number line.

0%
True
0%
False

Explanation

The given rational number

$\dfrac{7}{-4}$ is a negative rational number that lies on the left side of zero in the number line.

Hence the given statement is false

Which number is represented by

$A$ , in the following number line?

0%
$13$
0%
$\dfrac {6}{13}$
0%
$6$
0%
$\dfrac {0}{13}$

Explanation

In the given number line, every number increases by

$\dfrac {1}{13}$ as we move right.

$\therefore \dfrac {5}{13} + \dfrac {1}{13} = \dfrac {6}{13} \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \dfrac {A}{13}$

$\therefore A = 6$ .

Which number is represented by

$x$ , in the given number line?

0%
$-1$
0%
$1$
0%
$0$
0%
$3$

Explanation

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,

$x = \dfrac {-1}{3} + \dfrac {1}{3} = 0$ .

What rational number lies exactly halfway between

$\dfrac{2}{3}$ and

$\dfrac{3}{4}$ ?

0%
$\dfrac{3}{5}$
0%
$\dfrac{5}{6}$
0%
$\dfrac{7}{12}$
0%
$\dfrac{9}{16}$
0%
$\dfrac{17}{24}$

Explanation

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.

Identify a rational number between

$\dfrac {1}{3}$ and

$\dfrac {4}{5}$

0%
$\dfrac {1}{4}$
0%
$\dfrac {9}{10}$
0%
$\dfrac {17}{30}$
0%
$\dfrac {7}{10}$

Explanation

$\frac{1}{3}\,and\,\frac{4}{5}\text{ }$

$To\text{ find rational number between to two number make the denominator }$

$\text{same and then check the numbers in between numerators}$

$\frac{6}{30}\,and\,\frac{24}{30}$

$now\,\text{ the possible numbers between 1 and 4 are 7,8,9,10,11,12,13,14,15...23}$

$\therefore \,\text{ the possible rational number between}\frac{1}{3}and\frac{4}{5\,}is \frac{17}{30}\,and\,\frac{7}{10}$

Subtract the additive inverse of

$\dfrac {5}{6}$ from the multiplicative inverse of

$\dfrac {-5}{7}\times \dfrac {14}{15}$ .

0%
$\dfrac {3}{2}$
0%
$\dfrac {-2}{3}$
0%
$\dfrac {-3}{2}$
0%
$\dfrac {2}{3}$

Explanation

$Additive\quad inverse\quad of\quad \frac { 5 }{ 6 } =-\frac { 5 }{ 6 } \\ Now\quad \frac { -5 }{ 7 } \times \frac { 14 }{ 15 } =-\frac { 2 }{ 3 } \\ Multiplicative\quad inverse\quad of\quad -\frac { 2 }{ 3 } =-\frac { 3 }{ 2 } \\ now\quad subtracting\quad addtive\quad inverse\quad from\quad multiplicative\quad inverse\\ -\frac { 3 }{ 2 } -(-\frac { 5 }{ 6 } )=-\frac { 2 }{ 3 }$

So correct answer will be option B

An illustration of the associative law for multiplication is given by

0%
$\left (\dfrac{1}{3}\times 5\right)\times 8=\dfrac{1}{3}\times (5\times 8)$
0%
$\dfrac{1}{3}\times 5\times 8=\dfrac{1}{3}\times 8\times 5$
0%
$\dfrac{1}{3}\times 5+\dfrac{1}{3}\times 8=\dfrac{1}{3}\times 13$
0%
$\dfrac{1}{3}\times 5\times 8=\left (\dfrac{1}{3}\times 5\right)\times \left (\dfrac{1}{3}\times 8\right)$
0%
none of these

Explanation

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.

Identity the rational number that does not lie between

$\cfrac{3}{5}$ and

$\cfrac{2}{3}$ .

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

Given that

$Q$ is a rational number:

(i) Difference of two

$Q$ s is

$Q$ .

(ii) Subtraction is commutative on

$Q$ .

(iii) Addition is not commutative on

$Q$ .

Which option is wrong?

0%
(ii) and (iii)
0%
(i) only
0%
(i) and (iii)
0%
All the above

The property under addition used in each of the following are
(i)

$\left (\dfrac {-3}{7}\right ) + \dfrac {1}{9} = \dfrac {1}{9} + \left (\dfrac {-3}{7}\right )$
(ii)

$\dfrac {4}{9} + \left (\dfrac {7}{8} + \dfrac {1}{2}\right ) = \left (\dfrac {4}{9} + \dfrac {7}{8}\right ) + \dfrac {1}{2}$
(iii)

$8 + \dfrac {7}{10} = \dfrac {7}{10} + 8$
(iv)

$\left (\dfrac {-7}{15}\right ) + 0 = \dfrac {-7}{15} = 0 + \left (\dfrac {-7}{15}\right )$
(v)

$\dfrac {2}{5} + \left (\dfrac {-2}{5}\right ) = 0$

(i) Commutative
(ii) Associative
(iii) Commutative
(iv) Additive identity
(v) Additive inverse.

0%
True
0%
False

Explanation

1) A binary operation is commutative if changing the order of the operands does not change the result.

$\left (\dfrac {-3}{7}\right ) + \dfrac {1}{9} = \dfrac {1}{9} + \left (\dfrac {-3}{7}\right )$

2) The associative property states that you can add or multiply regardless of how the numbers are grouped.

$\dfrac {4}{9} + \left (\dfrac {7}{8} + \dfrac {1}{2}\right ) = \left (\dfrac {4}{9} + \dfrac {7}{8}\right ) + \dfrac {1}{2}$

3) A binary operation is commutative if changing the order of the operands does not change the result.

$8 + \dfrac {7}{10} = \dfrac {7}{10} + 8$

4) The additive identity of a set that is equipped with the operation of addition is an element which, when added to any element x in the set, yields x.

$\left (\dfrac {-7}{15}\right ) + 0 = \dfrac {-7}{15} = 0 + \left (\dfrac {-7}{15}\right )$

5) the additive inverse of a number a is the number that, when added to a, yields zero

$\dfrac {2}{5} + \left (\dfrac {-2}{5}\right ) = 0$

A rational number lie between

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

$\displaystyle\frac{1}{3}$ is _________.

0%
$\displaystyle\frac{7}{24}$
0%
$\displaystyle\frac{15}{48}$
0%
$\displaystyle\frac{13}{48}$
0%
All of these

Explanation

So, a rational number between

$\dfrac {1}{4}$ and

$\dfrac {1}{3}$

$= \dfrac {\dfrac {1}{4} + \dfrac {1}{3}}{2} = \dfrac {7}{24}$

Now, another rational number between

$\dfrac {1}{4}$ and

$\dfrac {7}{24}$

$= \dfrac {\dfrac {1}{4} + \dfrac {7}{24}}{2} = \dfrac {13}{48}$

Now, another rational number between

$\dfrac {1}{3}$ and

$\dfrac {7}{24}$

$= \dfrac {\dfrac {1}{3} + \dfrac {7}{24}}{2} = \dfrac {15}{48}$

Hence, required two rational numbers between

$\dfrac {1}{4}$ and

$\dfrac {1}{3}$ are

$\dfrac {7}{24} ,\dfrac {13}{48}, \dfrac {15}{48}$

CBSE Questions for Class 8 Maths Rational Numbers Quiz 7 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Rational Numbers Quiz 7 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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