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Practice each question, then open the answer or worked solution.
In 2n, n is known as
(a) Base
(b) Constant
(c) Exponent
(d) Variable
Ans: Option (c) is correct.
2 is a rational number as base in expression, and n is the power of that base which is called an exponent or power or index.
For a fixed base, if the exponent decreases by 1, the number becomes
(a) One-tenth of the previous number.
(b) Ten times of the previous number.
(c) Hundredth of the previous number.
(d) Hundred times of the previous number.
Ans: Option (a) is correct.
If the exponent is decreased by 1, then for the
fixed base, the number becomes one-tenth of the previous
number.
e.g.- For 104, exponent decreases by 1
104-1 = 103
$\Rightarrow \dfrac{{{{10}^3}}}{{{{10}^4}}} = \dfrac{1}{{10}}$
3–2 can be written as
(a) 32
(b) $\dfrac{1}{{{3^2}}}$
(c) $\dfrac{1}{{{3^{ - 2}}}}$
(d) $ - \dfrac{2}{3}$
Ans: Option (b) is correct.
Use, law of exponent ${a^{ - m}} = \dfrac{1}{{{a^m}}}$ where a is a non zero integer.
Thus, ${3^{ - 2}} = \dfrac{1}{{{3^2}}}$
The value of $\dfrac{1}{{{4^{ - 2}}}}$ is
(a) 16
(b) 8
(c) $\dfrac{1}{{16}}$
(d) $\dfrac{1}{8}$
Ans: Option (a) is correct.
Use, law of exponent ${a^{ - m}} = \dfrac{1}{{{a^m}}}$ where a is a non zero integer.
Thus, $\dfrac{1}{{{4^{ - 2}}}} = \dfrac{1}{{\dfrac{1}{{{4^2}}}}} = {4^2} = 16$
The value of 35 ÷ 3–6 is
(a) 35
(b) 3–6
(c) 311
(d) 3–11
Ans: Option () is correct.
Use, law of exponent ${a^m} \div {a^n} = {a^{m - n}}$ where a is a non zero integer.
Thus, ${3^5} \div {3^{ - 6}} = {3^{5 - \left( { - 6} \right)}} = {3^{5 + 6}} = {3^{11}}$.
The value of ${\left( {\dfrac{{\mathbf{2}}}{{\mathbf{5}}}} \right)^{{\mathbf{ - 2}}}}$ is
(a) $\dfrac{4}{5}$
(b) $\dfrac{4}{{25}}$
(c) $\dfrac{{25}}{4}$
(d) $\dfrac{5}{2}$
Ans: Option (c) is correct.
Use, law of exponent ${a^{ - m}} = \dfrac{1}{{{a^m}}}$ where a is a non zero integer.
Thus, ${\left( {\dfrac{2}{5}} \right)^{ - 2}} = \dfrac{1}{{{{\left( {\dfrac{2}{5}} \right)}^2}}} = \dfrac{1}{{\dfrac{4}{{25}}}} = \dfrac{{25}}{4}$
The value of ${\left( {\dfrac{{\mathbf{2}}}{{\mathbf{5}}}} \right)^{{\mathbf{ - }}1}}$ is
(a) $\dfrac{2}{5}$
(b) $\dfrac{5}{2}$
(c) $ - \dfrac{5}{2}$
(d) $ - \dfrac{2}{5}$
Ans: Option (b) is correct.
Use the law of exponent ${a^{ - m}} = \dfrac{1}{{{a^m}}}$, where a is a non zero integer.
Thus, ${\left( {\dfrac{2}{5}} \right)^{ - 1}} = \dfrac{1}{{{{\left( {\dfrac{2}{5}} \right)}^1}}} = \dfrac{1}{{\dfrac{2}{5}}} = \dfrac{5}{2}$
The multiplicative inverse of 10-100 is
(a) 10
(b) 100
(c) 10100
(d) 10-100
Ans: Option (c) is correct.
For multiplicative inverse, let a be the multiplicative of 10-100
So,
$a \times {10^{ - 100}} = 1$
$\Rightarrow a = \dfrac{1}{{{{10}^{ - 100}}}} = {10^{100}}$
The value of (-2)2×3-1 is
(a) 32
(b) 64
(c) -32
(d) -64
Ans: Option (c) is correct.
For (-a)m, if m is odd, then (-a)m is negative
(-2)2×3-1 = (-2)6–1 = (-2)5
= (-2) × (-2) × (-2) × (-2) × (-2) = - 32
The value of ${\left( {\dfrac{{ - 2}}{3}} \right)^4}$ is equal to
(a) $\left( {\dfrac{{16}}{{81}}} \right)$
(b) $\left( {\dfrac{{81}}{{16}}} \right)$
(c) $\left( {\dfrac{{ - 16}}{{81}}} \right)$
(d) $\left( {\dfrac{{81}}{{ - 16}}} \right)$
Ans: Option (a) is correct.
For (-a)m, if m is even, then (-a)m is positive.
${\left( { - \dfrac{2}{3}} \right)^4} = \left( { - \dfrac{2}{3}} \right) \times \left( { - \dfrac{2}{3}} \right) \times \left( { - \dfrac{2}{3}} \right) \times \left( { - \dfrac{2}{3}} \right) = \dfrac{{16}}{{81}}$