MCQ 11 Mark
For a fixed base, if the exponent decreases by $1,$ the number becomes:
- ✓
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.
AnswerCorrect option: A. One-tenth of the previous number.
For a fixed base, if the exponent decreases by $1,$ the number becomes one-tenth of the previous number.
e.g. For $105,$ exponent decreases by $1.$
i.e. $10^{5-1}= 10^4$
$\therefore\ \frac{10^4}{10^5}=\frac{1}{10}$
Note:
Option $(a)$ is possibal only, if we taken base as $10.$
View full question & answer→MCQ 21 Mark
$(-9)^3÷ (-9)^8$ is equal to:
- A
$ (9)^5 $
- B
$ (9)^{-5} $
- C
$ (-9)^5 $
- ✓
$ (-9)^{-5} $
AnswerCorrect option: D. $ (-9)^{-5} $
Given,$(-9)^3\div (-9)^8$
Using law of exponents,
$a^m+ a^n= (a)^{m-n} [\because$ a is non$-$zero integer$]$
$\therefore (-9)^3+ (-9)^8$
$= (-9)^{3-8}$
$ (-9)^{-5} $
View full question & answer→MCQ 31 Mark
$\Big(\frac{1}{10}\Big)^0$ is equal to:
- A
$0$
- B
$\frac{1}{10}$
- ✓
$1$
- D
$10$
AnswerUsing law of exponents, $a^0= 1$ [$\because$ a is non-zero integer]
$\therefore$ $\Big(\frac{1}{10}\Big)^0=1$
View full question & answer→MCQ 41 Mark
Cube of $-\frac{1}{2}$ is:
- A
$\frac{1}{8}$
- B
$\frac{1}{16}$
- ✓
$-\frac{1}{8}$
- D
$-\frac{1}{16}$
AnswerCorrect option: C. $-\frac{1}{8}$
Cube of $a$ is $(a)^3= a × a × a$
Similarly, $\Big(-\frac{1}{2}\Big)^3$
$=\Big(\frac{-1}{2}\Big)\times\Big(\frac{-1}{2}\Big)\times\Big(\frac{-1}{2}\Big)$
$=\Big(-\frac{1}{8}\Big)$
$[$for $(-a)^m$, if $m$ is odd, then $(-a)^m$ is negative$]$
View full question & answer→MCQ 51 Mark
The value of $\frac{1}{4^{-2}}$ is:
- ✓
$16$
- B
$8$
- C
$\frac{1}{16}$
- D
$\frac{1}{8}$
AnswerUsing law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}$ [$\because$ a is non-integer]
$\therefore$ $\frac{1}{4^{-2}}=\frac{1}{\frac{1}{4^2}}$
$=\frac{1}{\frac{1}{16}}=1\times16$
$=16$
View full question & answer→MCQ 61 Mark
The value of $\left(7^{-1}-8^{-1}\right)^{-1}-\left(3^{-1}-4^{-1}\right)^{-1}$ is:
AnswerUsing law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^{\text{m}}}$
$[\therefore$ a is non $-$ zero integer$]$
$\therefore$$\left(7^{-1}-8^{-1}\right)^{-1}-\left(3^{-1}-4^{-1}\right)^{-1}$
$=\Big(\frac{1}{7}-\frac{1}{8}\Big)^{-1}-\Big(\frac{1}{3}-\frac{1}{4}\Big)^{-1}$
$=\Big(\frac{1}{56}\Big)^{-1}-\Big(\frac{1}{12}\Big)^{-1}$
$=56-12=44$
View full question & answer→MCQ 71 Mark
For a non$-$zero rational number $z, (z^{-2})^3$ is equal to:
- A
$ z^6 $
- ✓
$ z^{-6} $
- C
$ z^1 $
- D
$ z^4 $
AnswerCorrect option: B. $ z^{-6} $
Using law of exponents, $(a^m)^n= (a)^{mn} $
$[\because$ a is non$-$zero integer$]$
Similarly, $ \left(z^{-2}\right)^3$
$=(z)^{(-2) \times 3} $
$=(z)^{-6} $
View full question & answer→MCQ 81 Mark
The value of $\Big(-\frac{2}{3}\Big)^4$ is equal to:
- ✓
$\frac{16}{81}$
- B
$\frac{81}{16}$
- C
$\frac{-16}{81}$
- D
$\frac{81}{-16}$
AnswerCorrect option: A. $\frac{16}{81}$
Given, $\Big(\frac{-2}{3}\Big)^4$
$=\Big(\frac{-2}{3}\Big)\times\Big(\frac{-2}{3}\Big)\times\Big(\frac{-2}{3}\Big)\times\Big(\frac{-2}{3}\Big)$
$=\frac{16}{81}$
$[$for $(-a)^m$, if $m$ is even, then $(-a)^m$ is positive$]$
View full question & answer→MCQ 91 Mark
The usual form for $2.03 \times 10^{-5}$
- A
$0.203$
- B
$0.00203$
- C
$203000$
- ✓
$0.0000203$
AnswerCorrect option: D. $0.0000203$
Given,
$2.03 × 10^{-5}= 0.0000203$
$[\therefore$ placing decimal five digit towards left of original position$]$
View full question & answer→MCQ 101 Mark
If $x$ be any integer different from zero and $m, n$ be any integers, then $(x^m)^n$ is equal to$:$
- A
$\text{x}^{\text{m}+\text{n}}$
- ✓
$\text{x}^{\text{mn}}$
- C
$\text{x}^{\frac{\text{m}}{\text{n}}}$
- D
$\text{x}^{\text{m}-\text{n}}$
AnswerCorrect option: B. $\text{x}^{\text{mn}}$
Using law of exponents, $(\text{a}^{\text{m}})^{\text{n}}=(\text{a})^{\text{m}\times\text{n}}$
$[\because$ a is non$-$zero integer$]$
Similaly, $(\text{x}^{\text{m}})^{\text{n}}=(\text{x})^{\text{m}\times\text{n}}$
$=(\text{x})^\text{mn}$
View full question & answer→MCQ 111 Mark
$\Big(\frac{-7}{5}\Big)^{-1}$ is equal to:
- A
$\frac{5}{7}$
- ✓
$-\frac{5}{7}$
- C
$\frac{7}{5}$
- D
$\frac{7}{5}$
AnswerCorrect option: B. $-\frac{5}{7}$
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^{\text{m}}}$ [$\because$ a is non-zero integer]
$\therefore$ $\Big(\frac{-7}{5}\Big)^{-1}=\frac{1}{\Big(\frac{-7}{5}\Big)}$
$=\Big(-\frac{5}{7}\Big)$
View full question & answer→MCQ 121 Mark
$\Big(-\frac{5}{7}\Big)^{-5}$ is equal to:
- A
$\Big(-\frac{5}{7}\Big)^{-5}$
- B
$\Big(\frac{5}{7}\Big)^5$
- C
$\Big(\frac{7}{5}\Big)^5$
- ✓
$\Big(\frac{-7}{5}\Big)^5$
AnswerCorrect option: D. $\Big(\frac{-7}{5}\Big)^5$
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}$ [$\because$ a is non-zero integer]
$\therefore$ $\Big(-\frac{5}{7}\Big)^5=\frac{1}{\Big(\frac{-5}{7}\Big)^5}$
$=\Big(-\frac{7}{5}\Big)^5$
View full question & answer→MCQ 131 Mark
For a non$-$zero integer $x, (x^4)^{-3}$ is equal to$:$
- A
$ x^{12} $
- ✓
$ x^{-12} $
- C
$ x^{64} $
- D
$ x^{-64} $
AnswerCorrect option: B. $ x^{-12} $
Using law of exponents, $(a^m)^n = (a)^{m\times n}= (a^m)^n$
$[\because$ a is non$-$zero integer$]$
similarly, $(x^4)^{-3}= (x)^{4\times {-3}}$
$= x^{-12} $
View full question & answer→MCQ 141 Mark
For a non$-$zero integer $x, x^7÷ x^{12}$ is equal to$:$
- A
$ x^5 $
- B
$ x^{19} $
- C
$ x^{-5} $
- ✓
$ x^{-19} $
AnswerCorrect option: D. $ x^{-19} $
$ x^{-5} $
View full question & answer→MCQ 151 Mark
If $x$ be any non$-$zero integer, then $x^{-1}$ is equal to$:$
- A
$\text{x}$
- ✓
$\frac{1}{\text{x}}$
- C
$-\text{x}$
- D
$\frac{-1}{\text{x}}$
AnswerCorrect option: B. $\frac{1}{\text{x}}$
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{a^{\text{m}}} [\because a$ is non$-$zero integer$]$
Similarly, $\text{x}^{-1}=\frac{1}{\text{x}}$
View full question & answer→MCQ 161 Mark
The multiplicative inverse of $10^{-100}$ is:
- A
$10$
- B
$100$
- ✓
$10^{100}$
- D
$10^{-100}$
AnswerCorrect option: C. $10^{100}$
For multiplicative inverse, let a be the multiplicative inverse of $10^{-100}$.
so, $a × b = 1$
$\therefore$ $a × 10^{100} = 1$
$\Rightarrow\text{a}=\frac{1}{10^{-100}}\times\frac{1}{\frac{1}{10^{100}}}\ \Big[\because\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}\Big]$
$=10^{100}$
View full question & answer→MCQ 171 Mark
If $x$ be any integer different from zero and $m$ be any positive integer, then $x^{-m}$ is equal to:
AnswerCorrect option: C. $\frac{1}{\text{x}^\text{m}}$
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}$ [$\because a$ is non$-$zero integer$]$
Similarly, $\text{x}^{-\text{m}}=\frac{1}{\text{x}^\text{m}}$
View full question & answer→MCQ 181 Mark
For any two non$-$zero rational numbers $x$ and $y, x^4 ÷ y^4$ is equal to:
- A
$ (x \div y)^0 $
- B
$ (x \div y)^1 $
- ✓
$ (x \div y)^4 $
- D
$ (x \div y)^8 $
AnswerCorrect option: C. $ (x \div y)^4 $
Using law of exponents, $\frac{\text{a}^\text{m}}{\text{b}^\text{m}}=\Big(\frac{\text{a}}{\text{b}}\Big)^\text{m} $
$[\because a$ and $b$ are non$-$zero integers$]$
Similarly,
$x^4÷ y^4$
$=\Big(\frac{\text{x}}{\text{y}}\Big)^4$
$=(\text{x}\div\text{y})^4$
View full question & answer→MCQ 191 Mark
Which of the following is not the reciprocal of $\Big(\frac{2}{3}\Big)^4$ ?
- A
$\Big(\frac{3}{2}\Big)^4$
- ✓
$\Big(\frac{3}{2}\Big)^{-4}$
- C
$\Big(\frac{2}{3}\Big)^{-4}$
- D
$\frac{3^4}{2^4}$
AnswerCorrect option: B. $\Big(\frac{3}{2}\Big)^{-4}$
Reciprocal of a is $\frac{1}{\text{a}}$.
Similarly,
$\Big(\frac{2}{3}\Big)^4$
$=\Big(\frac{3}{2}\Big)^4=\frac{3^4}{2^4}$
$=\Big(\frac{2}{3}\Big)^{-4}$
Hence,
option $(b)$ is not the reciprocal of $\Big(\frac{2}{3}\Big)^4$
View full question & answer→MCQ 201 Mark
The multiplicative inverse of $\Big(-\frac{5}{9}\Big)^{-99}$ is:
- ✓
$\Big(-\frac{5}{9}\Big)^{99}$
- B
$\Big(\frac{5}{9}\Big)^{99}$
- C
$\Big(\frac{9}{-5}\Big)^{99}$
- D
$\Big(\frac{9}{5}\Big)^{99}$
AnswerCorrect option: A. $\Big(-\frac{5}{9}\Big)^{99}$
For multiplicative inverse, a is called multiplicative inverse of b, if $a × b = 1$.
Put $b = \Big(-\frac{5}{9}\Big)^{-99}$
$\Rightarrow\text{a}\times\Big(\frac{-5}{9}\Big)^{-99}=1$
$\Rightarrow\text{a}=\frac{1}{\frac{-5}{9}}^{-99}$
$\Rightarrow\text{a}=\Big(-\frac{5}{9}\Big)^{99}\ \Big[\because\text{a}^{-\text{m}}=\frac{1}{\text{a}^{\text{m}}}\Big]$
View full question & answer→MCQ 211 Mark
The value of $35 \div 3^{-6}$ is:
- A
$3^5$
- B
$3^{-6}$
- ✓
$3^{11}$
- D
$3^{-11}$
AnswerCorrect option: C. $3^{11}$
Using law of exponents, $a^m+a^n=a^{m-n} [\because a$ is non$-$integer$]$
$3^5 \div 3^{-6}=3^{5-(-6)} $
$ =3^{5+6} $
$=3^{11}$
View full question & answer→MCQ 221 Mark
For a non$-$zero rational number $p, p^{13} \div p^8$ is equal to:
- ✓
$\mathrm{p}^5$
- B
$\mathrm{p}^{21}$
- C
$\mathrm{p}^{-5}$
- D
$\mathrm{p}^{-19}$
AnswerCorrect option: A. $\mathrm{p}^5$
Using law of exponents, $a^m \div a^n=(a)^{m-n} [\because a$ is non$-$zero integer$]$
Similarly $p^{13} \div p^8$
$=(p)^{13-8}$
$=\mathrm{p}^{5}$
View full question & answer→MCQ 231 Mark
In $2^n, n$ is known as:
View full question & answer→MCQ 241 Mark
Which of the following is equal to $\Big(-\frac{3}{4}\Big)^{-3}$ ?
- A
$\Big(\frac{3}{4}\Big)^{-3}$
- B
$-\Big(\frac{3}{4}\Big)^{-3}$
- C
$\Big(\frac{4}{3}\Big)^{3}$
- ✓
$\Big(-\frac{4}{3}\Big)^{3}$
AnswerCorrect option: D. $\Big(-\frac{4}{3}\Big)^{3}$
Given,
$\Big(\frac{-3}{4}\Big)^{-3}$
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{m}}$ [$\because$ a is non-zero integer]
$\therefore$ $\Big(\frac{-3}{4}\Big)^{-3}$
$=\frac{1}{\Big(\frac{-3}{4}\Big)^4}$
$=\Big(-\frac{4}{3}\Big)^{3}$
View full question & answer→MCQ 251 Mark
The reciprocal of $\Big(\frac{2}{5}\Big)^{-1}$ is
- A
$\frac{2}{5}$
- ✓
$\frac{5}{2}$
- C
$-\frac{5}{2}$
- D
$-\frac{2}{5}$
AnswerCorrect option: B. $\frac{5}{2}$
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^{\text{m}}}$ [$\because$ a is non-integer]
$\therefore$ $\Big(\frac{2}{5}\Big)^{-1}=\frac{1}{\Big(\frac{2}{5}\Big)^{1}}$
$=\frac{5}{2}$
View full question & answer→MCQ 261 Mark
The standard form for $0.000064$ is:
- A
$ 64 \times 10^4 $
- B
$ 64 \times 10^{-4} $
- C
$ 6.4 \times 10^5 $
- ✓
$ 6.4 \times 10^{-5} $
AnswerCorrect option: D. $ 6.4 \times 10^{-5} $
Given, $ 0.000064 $
$=6.4 \times 10^{-4} $
$ =6.4 \times 10^{-5} $
Hence, standard form of $0.000064$ is $ 6.4 \times 10^{-5} $.
View full question & answer→MCQ 271 Mark
The standard form for $234000000$ is:
- ✓
$ 2.34 \times 10^8 $
- B
$ 0.234 \times 10^9 $
- C
$ 2.34 \times 10^{-8}$
- D
$ 0.234 \times 10^{-9} $
AnswerCorrect option: A. $ 2.34 \times 10^8 $
Given, $234000000 $
$= 234 × 10^6$
$ = 2.34 \times 10^6 $
$= 2.34 \times 10^8 $
Hence, standard form of $234000000$ is $ 2.34 \times 10^8 $
View full question & answer→MCQ 281 Mark
The value of $(-2)^{2\times 3-1}$ is:
AnswerGiven, $ (-2)^{2 \times 3-1}$
$=(-2)^{6-1} $
$ =(-2)^5 $
$= (-2) × (-2) × (-2) × (-2) × (-2)$
$= -32$
$[$for $(-a)^m,$ if $m$ is odd, then $(-a)^m$ is negative$]$
View full question & answer→MCQ 291 Mark
If $x$ be any non$-$zero integer and $m, n$ be negative integers, then $x^m \times x^n$ is equal to:
- A
$x^m$
- ✓
$x^{m+n}$
- C
$x^n$
- D
$x^{m-n}$
AnswerCorrect option: B. $x^{m+n}$
Using law of exponents,
$a^m \times a^n=(a)^{m+n}[\because a$ is non$-$zero integer$]$
Similarly, $x^m \times x^n=(x)^{m+n}$
View full question & answer→MCQ 301 Mark
$3^{-2}$ can be written as$:$
- A
$3^2$
- ✓
$\frac{1}{3^2}$
- C
$\frac{1}{3^{-2}}$
- D
$-\frac{2}{3}$
AnswerCorrect option: B. $\frac{1}{3^2}$
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}} [ \because a$ is non$-$zero integer$]$
So, we can write $3^{-2}$ as $\frac{1}{3^2}$.
View full question & answer→MCQ 311 Mark
$\Big(\frac{3}{4}\Big)^5\div\Big(\frac{5}{3}\Big)^5$ is equal to:
- ✓
$\Big(\frac{3}{4}\div\frac{5}{3}\Big)^5$
- B
$\Big(\frac{3}{4}\div\frac{5}{3}\Big)^1$
- C
$\Big(\frac{3}{4}\div\frac{5}{3}\Big)^0$
- D
$\Big(\frac{3}{4}\div\frac{5}{3}\Big)^{10}$
AnswerCorrect option: A. $\Big(\frac{3}{4}\div\frac{5}{3}\Big)^5$
Using law of exponents, $a^m+b^m$
$=(a+b)^m [\because a$ and $b$ are non$-$zero integers$]$
$\therefore \Big(\frac{3}{4}\Big)^5\div\Big(\frac{5}{3}\Big)^5$
$=\Big(\frac{3}{4}\div\frac{5}{3}\Big)^5$
View full question & answer→MCQ 321 Mark
The value of $\Big(\frac{2}{5}\Big)^{-2}$ is:
- A
$\frac{4}{5}$
- B
$\frac{4}{25}$
- ✓
$\frac{25}{4}$
- D
$\frac{5}{2}$
AnswerCorrect option: C. $\frac{25}{4}$
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}$ [$\because$ a is non-integer]
$\therefore$ $\Big(\frac{2}{5}\Big)^{-2}=\frac{1}{\Big(\frac{2}{5}\Big)^{2}}$
$=\frac{1}{\frac{4}{25}}=\frac{25}{4}$
View full question & answer→MCQ 331 Mark
If $y$ be any non$-$zero integer, then $y^0$ is equal to:
AnswerUsing law of exponents,
$a^0= 1 [\because a$ is non$-$zero integer$]$
Similarly,
$y^0= 1$
View full question & answer→