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CBSE Questions for Class 9 Maths Number Systems Quiz 4 - MCQExams.com
CBSE
Class 9 Maths
Number Systems
Quiz 4
The rational number between $$\cfrac{1}{2}$$ and $$\cfrac{6}{10}$$ is
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$$\cfrac{1}{4}$$
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$$\cfrac{3}{4}$$
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$$\cfrac{21}{40}$$
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$$\cfrac{33}{100}$$
Explanation
$$\dfrac{1}{2}$$ =
$$\dfrac{5}{10}$$
$$\dfrac{5}{10}$$ 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 $$\dfrac{21}{40}$$
Write five rational numbers which are smaller than $$2$$.
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$$1,\,\displaystyle\frac{1}{2},\,0,\,-1,\,-\displaystyle\frac{1}{2}$$
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$$0, 1 , 1.414, \sqrt3, -1$$
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$$0, 1 , \sqrt2, \sqrt3, -1$$
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$$0, 1 , 1.732, \sqrt2, -1$$
Explanation
Five rational numbers less than $$2$$ may be taken $$1,\,\displaystyle\frac{1}{2},\,0,\,-1,\,-\displaystyle\frac{1}{2}$$
(There can be many more such rational numbers).
Simplification of $$\displaystyle \left ( \frac{3}{5} \right )^{3} \times \left ( \frac{15}{2} \right )^{3}$$ is $$\dfrac{729}{8}$$
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True
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False
Explanation
$$\displaystyle \left ( \frac{3}{5} \right )^{3} \times \left ( \frac{15}{2} \right )^{3}$$
$$=\displaystyle \left ( \frac{3}{5} \right )^{3} \times \left ( \frac{3\times 5}{2} \right )^{3}$$
$$=\displaystyle \left ( \frac{3^3}{5^3} \right ) \times \left ( \frac{3^3\times 5^3}{2^3} \right )$$
$$=\displaystyle \left ( \frac{3^3}1 \right ) \times \left ( \frac{3^3}{2^3} \right )$$
$$=\dfrac{3^6}{2^3}$$
$$=\dfrac{729}{8}$$
The rationalising factor of $$ \displaystyle \left ( a+\sqrt{b} \right ) $$ is
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$$ \displaystyle a-\sqrt{b} $$
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$$ \displaystyle \sqrt{a}-b $$
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$$ \displaystyle \sqrt{a}-\sqrt{b} $$
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None of this
Explanation
Rationalizing factor of $$a+\sqrt{b}$$ is $$a-\sqrt{b}$$
$$(a+\sqrt{b})\times (a-\sqrt{b})=a^2-b^2$$
Which of the following expresses zero law of exponents?
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$$0^{x} = 0$$
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$$x^{0} = 1$$
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$$x^{1} = x$$
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None of the above
Explanation
According to the zero law of exponents,
any non - zero number raised to the power of zero is equal to $$1$$.
$$\therefore x^{0} = 1$$, where $$x\neq 0$$
So, option $$B$$ is correct.
$$4\times 4^{10}$$ is represented as:
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$$4^{40}$$
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$$4^{10}$$
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$$4^{11}$$
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$$16^{10}$$
Explanation
$$4\times 4^{10} = 4^{1 + 10}$$
$$= 4^{11}$$
So, option $$C$$ is correct.
In simplified form, $$((3)^{0} + (5)^{0})^{0}$$ is equal to:
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$$2$$
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$$1$$
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$$0$$
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$$8$$
Explanation
$$(3^{0} + 5^{0})^{0} = (1 + 1)^{0}$$
$$= (2)^{0}$$
$$= 1$$
So, option $$B$$ is correct.
$$(2+\sqrt 3)$$ is
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rational
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irrational
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an integer
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not real
Explanation
$$\sqrt{3}$$ is an irrational number. So any irrational number added to a rational number gives an irrational number.
So, $$2+\sqrt{3}$$ is an irrational number
Answer is Option B
$$a^0=1$$ is true for all $$a$$ except
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$$-1$$
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negative integers
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0
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1
Explanation
The correct answer is C
A. $${ (-1 )}^{ 0 }=1$$
B. $${ (-2) }^{ 0 }=1$$
C. $${ 0 }^{ 0 }$$ is not defined.
D. $${ 1 }^{ 0 }=1$$
$$ \displaystyle \frac{-3}{0} $$ is a-
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Negative rational number
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Positive rational number
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Either positive or negative rational number
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None of these
Explanation
-3/0 is undefined. Which means that it is neither a negative rational number nor a positive rational number.
So option D is the correct answer.
The value of $$21^{0}$$ is _____.
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$$0$$
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$$21$$
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$$1$$
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$$-21$$
Explanation
$$21^{0} = 1$$
$$\because a^{0} = 1$$
This is the law for zero exponent.
So, option $$C$$ is correct.
$$ \displaystyle \frac{-2}{-19} $$ is a -
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Negative rational number
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positive rational number
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neither positive nor negative rational number
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None of these
Explanation
Both the negative signs of the numerator and denominator will cancel each other out. So the given fraction is a positive rational number.
So option B is the correct answer.
Which of the following expresses product law of exponents?
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$$(a^{m})^{n} = a^{mn}$$
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$$a^{m}\times a^{n} = a^{m+n}$$
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$$a^{m} \div a^{n} = a^{m-n}$$
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$$a^{m}\div b^{m} = \left (\dfrac {a}{b}\right )^{m}$$
Explanation
The product law of exponents states that,
when multiplying two powers that have the same base,
we can add their exponents.
$$\therefore a^{m}\times a^{n} = a^{m+n}$$
So, option $$B$$ is correct.
Which one of the following is the value of $$(101)^{0}$$?
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$$0$$
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$$101$$
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$$1010$$
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$$1$$
Explanation
$$101^{0} = 1$$
So, option $$D$$ is correct.
The value of $$(10)^{0} \times (20)^{0}\times (30)^{0}$$ is:
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$$1$$
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$$3$$
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$$60$$
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$$0$$
Explanation
$$(10)^{0} \times (20)^{0} \times (30)^{0} = 1\times 1\times 1 = 1$$
So, option $$A$$ is correct.
Find the value of $$(6)^{0} - (10)^{0}$$
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$$-4$$
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$$2$$
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$$1$$
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$$0$$
Explanation
$$(6)^0 - (10)^0 = 1-1 = 0$$
So, option $$D$$ is correct.
The expression $$((2)^{0} + (3)^{0} + (5)^{0})^{0}$$ is equal to _____
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$$3$$
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$$1$$
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$$10$$
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$$0$$
Explanation
$$((2)^{0} + (3)^{0} + (5)^{0})^{0} = (1 + 1 + 1)^{0}$$
$$= 3^{0}$$
$$= 1$$
So, option $$B$$ is the correct answer.
Which of the following is equal to $$1$$?
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$$(3)^{0} + (3)^{0}$$
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$$(3)^{0} - (3)^{0}$$
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$$(3)^{0}$$
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$$(3^{3})^{0} + (3^{0})^{3}$$
Explanation
$$\textbf{Step 1: Simplify the following expressions.}$$
$$(a)\ 3^0+3^0=1+1=2\neq 1$$
$$(b)\ 3^0-3^0=1-1=0\neq 1$$
$$(c)\ 3^0=1$$
$$(d)\ (3^3)^0+(3^0)^3=9^0+1^3=1+1=2\neq 1$$
$$\textbf{Hence, Option C is correct.}$$
Simplify: $$\left((9)^{0} + (11)^{0} + (13)^{0}\right) \div (23)^{0}$$
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$$1$$
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$$3$$
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$$0$$
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$$2$$
Explanation
$$\left((9)^{0} + (11)^{0} + (13)^{0}\right) \div (23)^{0}$$
$$= (1 + 1 + 1) \div 1$$
$$= 3\div 1$$
$$= 3$$
So, option $$B$$ is correct.
The value of $$((6)^{0} + (16)^{0}) \div ((7)^{0} + (17)^{0})$$ is _____
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$$2$$
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$$4$$
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$$1$$
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$$0$$
Explanation
$$((6)^{0} + (16)^{0}) \div ((7)^{0} + (17)^{0})$$
$$ = (1 + 1)\div (1 + 1)$$
$$= 2\div 2$$
$$= 1$$
So, option $$C$$ is correct.
The value of $$(100)^{0}$$ is _____.
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$$1$$
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$$0$$
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$$100$$
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$$1000$$
Explanation
According to the law of power of $$0$$,
$$(100)^{0} = 1$$.
So, option $$A$$ is correct.
Evaluate: $$({(10)^0} + (12)^{0})\times (18)^{0}$$
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$$1$$
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$$0$$
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$$2$$
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$$18$$
Explanation
$$((10)^{0} + (12)^{0})\times (18)^{0} = (1 + 1) \times 1$$
$$= (2)\times 1$$
$$= 2$$
So. option $$C$$ is correct.
What are rational numbers?
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A number of the form $$\dfrac {p}{q}$$ where $$p$$ and $$q$$ are any positive integers and $$q\neq 0$$ is called a rational number.
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A number of the form $$\dfrac {p}{q}$$ where $$p$$ and $$q$$ are any negative integers and $$q\neq 0$$ is called a rational number.
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A number of the form $$\dfrac {p}{q}$$ where $$p$$ and $$q$$ are integers is called a rational number.
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A number of the form $$\dfrac {p}{q}$$ where $$p$$ and $$q$$ are any integers and $$q\neq 0$$ is called a rational number.
Explanation
A number of the form $$\dfrac {p}{q}$$ where $$p$$ and $$q$$ are any integers and $$q\neq 0$$ is called a rational number.
Ex:- Consider $$\dfrac{2}{3}$$ here $$2,3$$ are co-primes so the number $$\dfrac{2}{3}$$ is a rational number
$$\dfrac {7}{9}$$ is a/an _______ number.
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rational
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composite
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irrational
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prime
Explanation
$$\dfrac{7}{9}$$ is of the form $$\dfrac {p}{q}$$ form , hence it is rational no.
Which of the following is not a rational number?
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$$\dfrac {0}{1}$$
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$$\dfrac {1}{0}$$
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$$\dfrac {0}{3}$$
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$$\dfrac {0}{2}$$
Explanation
A number of the form $$\dfrac {p}{q}$$, where $$'p'$$ and $$'q'$$ are any integers and $$q\neq 0$$ is called a rational number.
$$\therefore \dfrac {1}{0}$$ is not a rational number because it is not defined.
$$\sqrt {23}$$ is not a ...... number.
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irrational
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co-prime
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composite
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rational
Explanation
As per the theorem, the square root of any prime number is irrational. $$\sqrt {23}$$ is a prime number, so is not a rational number. It is irrational.
Therefore, $$D$$ is the correct answer.
Irrational number is defined as
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a real number that cannot be made by dividing two integers.
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a real number that can be made by dividing two integer.
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a number that can be made derived after multiplying two integers.
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a real number that can be written as whole number.
Explanation
An irrational is any real number that cannot be expressed as a ratio of integers.
Therefore, $$A$$ is the correct answer.
There exists ....... number of rational numbers between $$\dfrac {2}{5}$$ and $$\dfrac {4}{5}.$$
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$$0$$
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$$1$$
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$$5$$
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infinite
Explanation
There exists infinite number of rational numbers between any two rational numbers. i.e. in this case between $$\dfrac {2}{5}$$ and $$\dfrac {4}{5}$$.
If the quotient is terminating decimal, the division is complete only when ...............
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we get the remainder $$1$$
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we get the remainder zero
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we get the remainder as the repeated numbers
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All of the above
Explanation
Any division is complete and terminates only when we get the remainder zero.
e.g. $$\dfrac 2 {10}=0.2$$
we can see the remainder becomes zero in this division.
Therefore, $$B$$ is the correct answer.
Which of the following rational numbers lies between $$0$$ and $$-1$$?
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$$0$$
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$$-1$$
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$$\dfrac {-1}{4}$$
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$$\dfrac {1}{4}$$
Explanation
Clearly, $$0$$ and $$-1$$ cannot lie between $$0$$ and $$-1$$.
Also,
$$0=\dfrac{0}{4}$$ and
$$-1=\dfrac{-4}{4}$$
We can clearly see that $$\dfrac {-1}{4}$$ lies between $$0$$ and $$-1$$.
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