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CBSE Questions for Class 8 Maths Exponents And Powers Quiz 3 - MCQExams.com
CBSE
Class 8 Maths
Exponents And Powers
Quiz 3
State true or false.
$$10^{-2}=\dfrac{1}{100}$$
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0%
True
0%
False
Explanation
True.
Using $$x^{-m}=\dfrac{1}{x^{m}}$$
$$10^{-2}=\dfrac{1}{10^{2}}=\dfrac{1}{10\times 10}=\dfrac{1}{100}$$
$$3^{-2}$$ can be written as
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0%
$$3^2$$
0%
$$\dfrac {1}{3^2}$$
0%
$$\dfrac {1}{3^{-2}}$$
0%
$$-\dfrac {2}{3}$$
Explanation
We will use, law of exponent $$a^{-m}=\dfrac {1}{a^m}$$, where $$a=$$ non zero integer
Therefore, $$3^{-2}=\dfrac {1}{3^2}$$
Mark tick against the correct answer of the following:
$$\left(\dfrac23\right)^{-5}=$$?
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0%
$$\dfrac{32}{243}$$
0%
$$\dfrac{243}{32}$$
0%
$$-\dfrac{32}{243}$$
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$$-\dfrac{243}{32}$$
Explanation
We know that,
$$ \left(\dfrac ab\right)^{-n}=\left(\dfrac ba\right)^{n}$$
$$\therefore\left(\dfrac23\right)^{-5}\\=\left(\dfrac32\right)^5$$
$$=\dfrac{3^5 }{2^5}$$ $$\left[\because \left(\dfrac ab\right)^{n}=\left(\dfrac{a^n}{b^n}\right)\right]$$
$$=\dfrac{243}{32}$$
Hence the correct answer is $$Opt:[B]$$
The value of $$\dfrac {1}{4^{-2}}$$ is
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0%
$$16$$
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$$8$$
0%
$$\dfrac {1}{16}$$
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$$\dfrac {1}{8}$$
Explanation
We will use, law of exponent $$a^{-m}=\dfrac {1}{a^m}$$,
Therefore, $$\dfrac {1}{4^{-2}}=\dfrac {\dfrac {1}{1}}{4^{2}}=\dfrac {\dfrac {1}{1}}{16} =1\times 16=16$$
The value of $$\left(\dfrac 25 \right)^{-2}$$ is
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0%
$$\dfrac 45$$
0%
$$\dfrac {4}{25}$$
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$$\dfrac {25}{4}$$
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$$\dfrac {5}{2}$$
Explanation
We will use, law of exponent $$a^{-m}=\dfrac {1}{a^m}$$ where $$a=$$ non-zero integer
$$\left(\dfrac {2}{5}\right)^{-2}=\dfrac {1}{\left(\dfrac {2}{5}\right)^2} \\=\dfrac {1}{\dfrac {4}{25}}\\=\dfrac {25}{4}$$
The value of $$(-2)^{2\times 3-1}$$ is
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0%
$$32$$
0%
$$64$$
0%
$$-32$$
0%
$$-64$$
Explanation
To find the value of $$(-2)^{2\times 3-1}$$
$$(-2)^{2\times 3-1}$$
$$=(-2)^{6-1}$$
$$=(-2)^5$$
$$=(-2)\times (-2)\times (-2)\times (-2)\times (-2)$$
$$=(4)\times (4)\times (-2)$$
$$=(16)\times (-2)$$
$$=-32$$
Hence, the result $$=-32$$
The value of $$(7^{-1}-8^{-1})^{-1} -(3^{-1}-4^{-1})$$ is
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0%
$$44$$
0%
$$56$$
0%
$$68$$
0%
$$12$$
Explanation
Using law of exponents, $$a^{-m}=\dfrac {1}{a^m}$$
$$\Rightarrow \ (7^{-1}-8^{-1})^{-1}-(3^{-1} -4^{-1})^{-1}=\left(\dfrac {1}{7}-\dfrac {1}{8}\right)^{-1}-\left(\dfrac {1}{3}-\dfrac {1}{4}\right)^{-1}$$
$$\Rightarrow \ \left(\dfrac {1\times 8-1\times 7}{56}\right)^{-1}- \left(\dfrac {1\times 4-1\times 3}{12}\right)^{-1}$$
$$=\left(\dfrac {8-7}{56}\right)^{-1} -\left(\dfrac {4-3}{12}\right)^{-1}$$
$$=\left(\dfrac {1}{56}\right)^{-1} -\left(\dfrac {1}{12}\right)^{-1}$$
Again we will use $$a^{-m} =\dfrac {1}{a^m}$$
$$\Rightarrow (7^{-1} -8^{-1})^{-1} -(3^{-1} -4^{-1})^{-1}=\left(\dfrac {1}{56}\right)^{-1}-\left(\dfrac {1}{12}\right)^{-1}$$
$$=56-12=44$$.
On multiplying ________ by $$2^{-5}$$ we get $$2^5$$
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0%
$$1$$
0%
$$2$$
0%
$$-1$$
0%
$$0$$
Explanation
$$2^{-5} \times 2^{5}= \dfrac{1}{2^5} \times 2^{5} = 1$$
$$\left(-\dfrac 57 \right)^{-5}$$ is equal to
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0%
$$\left(\dfrac 57 \right)^{-5}$$
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$$\left(\dfrac 57 \right)^{5}$$
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$$\left(\dfrac 75 \right)^{5}$$
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$$\left(-\dfrac 75 \right)^{5}$$
Explanation
Using law of exponent $$a^{-m}=\dfrac {1}{a^m}$$
[Where, $$a$$ is non-zero integer]
$$\left(\dfrac {-5}{7}\right)^{-5}=\dfrac {1}{\left(\dfrac {-5}{7}\right)^{5}}=\left(-\dfrac {7}{5}\right)^{5}$$
If $$x$$ be any integer different from zero and $$m$$ be any integers then $$x^{-m}$$ is equal to
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$$x^m$$
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$$-x^m$$
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$$\dfrac {1}{x^m}$$
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$$\dfrac {-1}{x^m}$$
Explanation
Using law of exponents,
$$a^{-m} =\dfrac {1}{a^m}$$
[Where,$$a$$ is non-zero integer]
Similarly, $$x^{-m}=\dfrac {1}{x^m}$$
$$\left(\dfrac {-7}{5} \right)^{-1}$$ is equal
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0%
$$\dfrac 57$$
0%
$$-\dfrac 57$$
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$$\dfrac 75$$
0%
$$\dfrac {-7}{5}$$
Explanation
Using law of exponent $$a^{-m}=\dfrac {1}{a^m}$$
[Where, $$a$$ is non-zero integer]
$$\left(\dfrac {-7}{5}\right)^{-1}=\dfrac {1}{\left(\dfrac {-7}{5}\right)}=\left(-\dfrac {5}{7}\right)$$
The expression for $$3^5$$ with a negative exponent is _________.
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0%
$$\dfrac{1}{3^{-5}}$$
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$$-\dfrac{1}{3^{-5}}$$
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$${3^{-5}}$$
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$$\dfrac{1}{3^{5}}$$
Explanation
Using $$a^m=\dfrac{1}{a^{-m}}$$
$$3^5 = \dfrac{1}{3^{-5}}$$
The value of $$[2^{-1} \times\ 3^{-1}]^{2}$$ is ______.
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0%
$$6^2$$
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$$\dfrac{1}{6^2}$$
0%
$$-6^2$$
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$$-\dfrac{1}{6^2}$$
Explanation
$$[2^{-1} \times\ 3^{-1}]^{2} =\dfrac{1}{(2 \times 3)^{2}}=\dfrac{1}{6^{2}}$$
State whether the following statement is true (T) or false (F):
The value of $$(-2)^{-3}$$ is $$\dfrac{-1}{8}$$
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0%
True
0%
False
Explanation
$$(-2)^{-3} = \dfrac{1}{(-2)^3} = \dfrac{1}{-8} = -\dfrac{1}{8}$$
Choose the correct answer from the given four options:
The value of $$(6^{-1}-8^{-1})^{-1}$$ is
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0%
$$\dfrac{-1}{2}$$
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$$-2$$
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$$\dfrac{1}{24}$$
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$$24$$
Explanation
$$(6^{-1}-8^{-1})^{-1}$$
$$=\bigg(\dfrac{1}{6}-\dfrac{1}{8}\bigg)^{-1}$$
$$=\bigg(\dfrac{4-3}{24}\bigg)^{-1}$$
$$=\bigg(\dfrac{1}{24}\bigg)^{-1}$$
$$=24$$
Choose the correct answer from the given four options:
$$\bigg(-\dfrac{3}{2}\bigg)^{-1}$$ is equal
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0%
$$\dfrac{2}{3}$$
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$$-\dfrac{2}{3}$$
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$$\dfrac{3}{2}$$
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$$\dfrac{4}{9}$$
Explanation
$$\bigg(-\dfrac{3}{2}\bigg)^{-1} =-\dfrac{2}{3}$$
If the coefficients of $$
x^{7} \& x^{8}
$$, in the expansion of $$
\left[2+\frac{x}{3}\right]^{n}
$$ are equal, then the value of n is:
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0%
15
0%
45
0%
55
0%
56
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