MCQ 1511 Mark
If $\text{log}^{\text{m}^\text{x}}_{\text{m}^\text{y}}=\frac{3}{4}$ then the value of $8\text{x}- 6 \text{y}+ 1$ is equal to:
View full question & answer→MCQ 1521 Mark
$8848$ is equal to:
- ✓
$8.848 \times 10^3$
- B
$8.848 \times 10^2$
- C
$8.848 \times 10$
- D
$8.848 \times 10^4$
AnswerCorrect option: A. $8.848 \times 10^3$
A. $8.848 \times 10^3$
Solution:
$8848 = 8.848 \times 10^3$
View full question & answer→MCQ 1531 Mark
The value of $\log_2^{2^{\text{xy}+2}}+\log_2^{2^{\text{4}-\text{xy}}}$ is equal to:
View full question & answer→MCQ 1541 Mark
Tick $(\checkmark)$ the correct answer the following:
The value of $(-3)^{-4}$ is:
- A
$12$
- B
$81$
- C
$-\frac{1}{12}$
- ✓
$\frac{1}{81}$
AnswerCorrect option: D. $\frac{1}{81}$
D. $\frac{1}{81}$
Solution:
$(-3)^{-4}$
$=\Big(\frac{1}{-3}\Big)^{-4}$
$=\frac{1}{(-3)\times(-3)\times(-3)\times(-3)}$
$=\frac{1}{81}$
View full question & answer→MCQ 1551 Mark
What is the scientific notation of $0.0023?$
- ✓
$2.3 \times 10^{-3}$
- B
$23 \times 10^{-3}$
- C
$2.3 \times 10^{3}$
- D
$23 \times 10^{3}$
AnswerCorrect option: A. $2.3 \times 10^{-3}$
A. $2.3 \times 10^{-3}$
View full question & answer→MCQ 1561 Mark
$300000000$ is equal to:
- ✓
$3 \times 10^8$
- B
$3 \times 10^7$
- C
$3 \times 10^6$
- D
$3 \times 10^9$
AnswerCorrect option: A. $3 \times 10^8$
A. $3 \times 10^8$
Solution:
$300,000,000 = 3 \times 10^8$
View full question & answer→MCQ 1571 Mark
In $2^n, n$ is known as:
AnswerWe know that an is called the nth power of a; and is also read as a raised to the power n.
The rational number a is called the base and n is called the exponent (power or index). In the same way in $2^n, n$ is known as exponent.
View full question & answer→MCQ 1581 Mark
What is the value of $(-30 + 40 - 50)?$
View full question & answer→MCQ 1591 Mark
The value of $\text{log}\text{m}^\text{n} + \text{log}\text{m}^\text{n+1} +\text{log}\text{m}^\text{1+2n}$ is:
AnswerCorrect option: A. $2\text{log}{\text{m}}$
$2\text{log}{\text{m}}$
View full question & answer→MCQ 1601 Mark
Tick $(\checkmark)$ the correct answer the following : $\Big(\frac{2}{3}\Big)^0=\ ?$
- A
$\frac{3}{2}$
- B
$\frac{2}{3}$
- ✓
$1$
- D
$0$
Answer$\Big(\frac{2}{3}\Big)^0=1\ \big(\because\text{x}^0=1\big)$
View full question & answer→MCQ 1611 Mark
$\frac{5^4}{5^2}$ is equal to:
- A
$5^6$
- B
$5^{-6}$
- C
$5^{-2}$
- ✓
$5^{2}$
AnswerCorrect option: D. $5^{2}$
D. $5^{2}$.
Solution:
By exponent law,
$\frac{\text{a}^\text{m}}{\text{a}^\text{n}}= \text{a}^\text{m-n}$
$\frac{5^4}{5^2}= 5^{4-2}$
$=5^2$
View full question & answer→MCQ 1621 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 1631 Mark
What is the value of ‘m‘ if $(-2)^2 \times (-5)^3 = 50m?$
View full question & answer→MCQ 1641 Mark
$2.5 \times 10^4$ is equal to:
- A
$25$
- B
$250$
- C
$2500$
- ✓
$25000$
AnswerCorrect option: D. $25000$
D. $25000$
Solution:
$2.5 \times 10^4 = 25000$
View full question & answer→MCQ 1651 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. $6.4 \times 10^{-5}$
D. $6.4 \times 10^{-5}$
Solution:
Given,
$0.000064 = 0. 64 \times 10^{-4}$
$= 6.4 \times 10^{-5}$
Hence,
standard form of $0.000064$ is $6.4 \times 10^{-5}$.
View full question & answer→MCQ 1661 Mark
$16$ is the multiplicative inverse of.
- ✓
$2^{-4}$
- B
$2^8$
- C
$8^2$
- D
$2^4$
AnswerCorrect option: A. $2^{-4}$
A. $2^{-4}$
View full question & answer→MCQ 1671 Mark
Which of the following is not equal to $\Big(\frac{2}{3}\Big)^4?$
- A
$\Big(\frac{3}{2}\Big)^4$
- B
$\Big(\frac{2}{3}\Big)^{-4}$
- ✓
$\Big(\frac{3}{2}\Big)^{-4}$
- D
$\frac{3^4}{2^4}$
AnswerCorrect option: C. $\Big(\frac{3}{2}\Big)^{-4}$
The reciprocal of $\Big(\frac{2}{3}\Big)^{4}$ is $\Big(\frac{3}{2}\Big)^{4}$
Therefore, option $(a)$ is the correct answer.
Option $(b)$ is just re$-$expressing the number with a negative exponent.
Option $(d)$ is obtained by working out the exponent.
Hence,option $(c)$ is not the reciprocal of $\Big(\frac{2}{3}\Big )^4$
View full question & answer→MCQ 1681 Mark
$\Big(\frac{-1}{3}\Big)^3\div\Big(\frac{-1}{5}\Big)^8$ is equal to:
- A
$\Big(\frac{-1}{5}\Big)^{5}$
- B
$\Big(-\frac{1}{5}\Big)^{11}$
- ✓
$(-5)^5 $
- D
$\Big(\frac{1}{5}\Big)^{5}$
AnswerCorrect option: C. $(-5)^5 $
We have:
$\Big(\frac{-1}{5}\Big)^3\div\Big(\frac{-1}{5}\Big)^8$
$=\Big(\frac{-1}{5}\Big)^{3-8}$
$=\Big(\frac{-1}{5}\Big)^{-5}$
$=\frac{1}{\big(\frac{-1}{5}\big)^5}$
$=\frac{1}{\big(\frac{(-1)^5}{5^5}\big)}$
$=\frac{5^5}{(-1)^5}$
$=\Big(\frac{5}{-1}\Big )^5$
$=(-5)^5$
View full question & answer→MCQ 1691 Mark
Find the value of the expression $a^2$ for a $= 10.$
View full question & answer→MCQ 1701 Mark
The number $86,800,000,000,000,000,000,000,000Kg$ is equals to.
- ✓
$8.68 \times 10^{25}K$
- B
$868 \times 10^{23}Kg$
- C
$86.8 \times 10^{-25}Kg$
- D
$868 \times 10^{-23}m$
AnswerCorrect option: A. $8.68 \times 10^{25}K$
A. $8.68 \times 10^{25}K$
View full question & answer→MCQ 1711 Mark
Express $3.657 \times 10^{-7}$ in usual form:
- A
It is already in usual form.
- B
$0.3657 \times 10^{-8}$
- C
$0.00003657$
- ✓
$0.0000003657$
AnswerCorrect option: D. $0.0000003657$
D. $0.0000003657$
Solution:
$3.657 \times 10^{-7} = 3657 \times 10^{-3} \times 10^{-7}$
$= 3657 \times 10^{-10}$
$= 0.0000003657$
View full question & answer→MCQ 1721 Mark
Tick $(\checkmark)$ the correct answer the following : $\Big(\frac{1}{2}\Big)^{-2}+\Big(\frac{1}{3}\Big)^{-2}+\Big(\frac{1}{4}\Big)^{-2}$
- A
$\frac{61}{144}$
- B
$\frac{144}{61}$
- ✓
$29$
- D
$\frac{1}{29}$
Answer$\Big(\frac{1}{2}\Big)^{-2}+\Big(\frac{1}{3}\Big)^{-2}+\Big(\frac{1}{4}\Big)^{-2}$
$=(2)^2+(3)^2+(4)^2$
$=4+9+16\ \because\Big(\frac{1}{\text{x}}\Big)^{-\text{m}}=\text{x}^\text{m}$
$=29$
View full question & answer→MCQ 1731 Mark
$0.0016$ is equal to:
- ✓
$1.6 \times 10^{-3}$
- B
$1.6 \times 10^{-2}$
- C
$1.6 \times 10^{-4}$
- D
$1.6 \times 10^{-5}$
AnswerCorrect option: A. $1.6 \times 10^{-3}$
A. $1.6 \times 10^{-3}$
Solution:
$0.0016 = 1.6 \times 10^{-3}$
View full question & answer→MCQ 1741 Mark
Simplify $4^{-4}\times\Big(\frac{3}{4}\Big)^{-4}$ and write the answer in exponent form:
- ✓
$\frac{1}{3^4}$
- B
$\frac{4^4}{3^4}$
- C
$3^4$
- D
$\frac{1}{3^3}$
AnswerCorrect option: A. $\frac{1}{3^4}$
$4^{-4}\times\Big(\frac{3}{4}\Big)^{-4}=\frac{1}{4^4}\times\frac{3^{-4}}{4^{-4}}$
$\frac{1}{4^4}\times\frac{3^{-4}}{4^{-4}}=\frac{1}{4^4}\times\frac{4^{4}}{3^{4}}$
$=\frac{1}{3^4}$
View full question & answer→MCQ 1751 Mark
$0.000003$ is equal to:
- ✓
$3 \times 10^{-6}$
- B
$3 \times 10^6$
- C
$3 \times 10^5$
- D
$3 \times 10^{-5}$
AnswerCorrect option: A. $3 \times 10^{-6}$
A. $3 \times 10^{-6}$
Solution:
$0.000003 = 3 \times 10^{-6}$
View full question & answer→MCQ 1761 Mark
$(-2)^{-5} \times (-2)^6$ is equal to:
AnswerB. $-2$
Solution:
$(-2)^{-5} \times (-2)^6 = (-2)^{-5+6} = (-2)^1 = -2$
View full question & answer→MCQ 1771 Mark
Simplify $(-3)^{-5} \times (7)^{-5} \times (4)^{-5}$ and write the answer in exponent form:
- A
$(- 84)^5$
- B
$(84)^5$
- C
$\frac{1}{(84)^5}$
- ✓
$\frac{1}{(-84)^5}$
AnswerCorrect option: D. $\frac{1}{(-84)^5}$
D. $\frac{1}{(-84)^5}$
Solution:
$(-3)^-5 × (7)^-5 × (4)^-5 = [(-3) × 7 × 4]^-5$
$= (-84)^-5$
$=\frac{1}{(-84)^5}$
View full question & answer→MCQ 1781 Mark
$(\frac{1}{2})^{-4}$ is equal to:
- A
$2$
- ✓
$2^{-4}$
- C
$1$
- D
$2^{-6}$
AnswerCorrect option: B. $2^{-4}$
B. $2^{-4}$
Solution:
$(\frac{1}{2})^{-4}=\frac{(1)^{-4}}{(2)^{-4}}= \frac{1}{(2)^{-4}}$
$=2^4$
View full question & answer→MCQ 1791 Mark
For which of the following is $m = 8?$
- ✓
$\frac{\Big(5^{\text{m}}-5^{-3}\Big)}{5^2=5^3}$
- B
$\frac{\Big(5^{\text{m}}-5^{-3}\Big)}{5^3=5^2}$
- C
$\frac{\Big(5^{\text{m}}-5^{3}\Big)}{5^2=5^3}$
- D
$\frac{\Big(5-5^{-2}\Big)}{5^2=5^3}$
AnswerCorrect option: A. $\frac{\Big(5^{\text{m}}-5^{-3}\Big)}{5^2=5^3}$
$\text{m}=8$
$=\frac{5^2-5^{-3}}{5^2}=5^3$
$=\frac{5^{[8+(-3)]}}{5^2}$
$=\frac{5^5}{5^2}$
$=5^3=5^3$
Hence Prove.
View full question & answer→MCQ 1801 Mark
Simplify $2^{-7} \times\left(2^5 \div 2^9\right)$ and write the answer in exponent form:
- A
$\frac{1}{2^9}$
- B
$\frac{1}{2^5}$
- ✓
$\frac{1}{2^{11}}$
- D
$2^{11}$
AnswerCorrect option: C. $\frac{1}{2^{11}}$
C. $\frac{1}{2^{11}}$
Solution:
$2^{-7} \times\left(2^5 \div 2^9\right)=2^{-7} \times\left(2^{5 \cdot 9}\right)\left(a^m \div a^n=a^{m-n}\right) $
$=2^{-7} \times 2^{-4} $
$=2^{-7+(-4)}\left(a^m \times a^n=a^{m+n}\right) $
$=2^{-11} $
$=\frac{1}{2^{11}}$
View full question & answer→MCQ 1811 Mark
$\Big(\frac{-2}{5}\Big)^7\div\Big(\frac{-2}{5}\Big)^5$ is equal to:
AnswerCorrect option: A. $\frac{4}{25}$
We have:
$\Big(\frac{-2}{5}\Big)^7\div\Big(\frac{-2}{5}\Big)^5=\Big(\frac{-2}{5}\Big)^{7-5}$
$=\Big(\frac{-2}{5}\Big)^{2}$
$=\frac{(-2)^2}{5^2}$
$=\frac{4}{25}$
View full question & answer→MCQ 1821 Mark
$2{\text{log}^4_2} + 2{\text{log}^8_2} + 2{\text{log}^{16}_2}$ is equal to:
View full question & answer→MCQ 1831 Mark
Cube of $\frac{-1}{2}$ is:
- A
$-\frac{1}{3}$
- B
$\frac{1}{16}$
- ✓
$-\frac{1}{8}$
- D
$\frac{-1}{16}$
AnswerCorrect option: C. $-\frac{1}{8}$
The cube of a number is the number raised to the power of $3$.
Hence the cube of $-\frac{1}{2}$ is $\frac{(-1)^3}{2^3}$
$=-\frac{1}{8}$
View full question & answer→MCQ 1841 Mark
$100^0 + 20^0 + 5^0$ is equal to:
- A
$125$
- B
$25$
- C
$\frac{1}{125}$
- ✓
$3$
AnswerD. $3$
Solution:
By exponent law we know:
$a^0 = 1$
$100^0 + 20^0 + 5^0 = 1 + 1 + 1 = 3$
View full question & answer→MCQ 1851 Mark
$1\text{ micron}=\frac{1}{1000000}\text{m}$ which of the following is its standard form?
- A
$1.1 \times 10^{-5}$
- B
$1.6 \times 10^{-5}$
- C
$0.1 \times 10^{-6}$
- ✓
$1.0 \times 10^{-6}$
AnswerCorrect option: D. $1.0 \times 10^{-6}$
D. $1.0 \times 10^{-6}$
Solution:
In the scientific notation, a "micro" is always represented by the number $10$ raised to the power of $-6.$ Thus, in the scientific notation, a micron will be represented in a similar manner.
Therefore, a micron will be:
$1\text{ micron}=\frac{1}{1000000}\text{m}$
$=\frac{1}{106}\text{m}\begin{Bmatrix}\text{a}^{-\text{m}}=\frac{1}{\text{a}^{\text{m}}}\end{Bmatrix}$
$1$ micron $= 10^{-6}m$ [Staudand from]
Or, $1.0 \times 10^{-6}$
View full question & answer→MCQ 1861 Mark
Tick $(\checkmark)$ the correct answer the following : The value of $\Big(\frac{2}{5}\Big)^{-3}$ is:
- A
$-\frac{8}{125}$
- B
$\frac{25}{4}$
- ✓
$\frac{125}{8}$
- D
$-\frac{2}{5}$
AnswerCorrect option: C. $\frac{125}{8}$
$\Big(\frac{2}{5}\Big)^{-3}$
$=\Big(\frac{5}{2}\Big)^3$
$=\frac{5\times5\times5}{2\times2\times2}$
$=\frac{125}{8}$
View full question & answer→MCQ 1871 Mark
The value of $4\text{log}^8_{16}$ is equal to:
AnswerCorrect option: D. $2\sqrt{2}$
$2\sqrt{2}$
View full question & answer→MCQ 1881 Mark
$\Big[\Big(\frac{1}{2}\Big)^{-1}+\big(\frac{2}{3}\big)^2-\big(\frac{3}{4}\big)^0\Big]^{-2}$ is equal to:
- A
$\frac{81}{484}$
- ✓
$\frac{81}{169}$
- C
$\frac{179}{81}$
- D
$\frac{16}{81}$
AnswerCorrect option: B. $\frac{81}{169}$
Given $\Big[\Big(\frac{1}{2}\Big)^{-1}+\Big(\frac{2}{3}\Big)^2-\Big(\frac{3}{4}\Big)^0\Big]^{-2}$
Solving the power,
$=\Big[\frac{2}{1}+\frac{4}{9}-1\Big]^{-2}$
$=\Big[\frac{18+4-9}{9}\Big]^{-2}$
$=\Big[\frac{13}{9}\Big]^{-2}$
$=\Big[\frac{9}{13}\Big]^2$
$=\frac{81}{169}$
Therefore $\Big[\Big(\frac{1}{2}\Big)^{-1}+\Big(\frac{2}{3}\Big)^2-\Big(\frac{3}{4}\Big)^0\Big]^{-2}=\frac{81}{169}$
View full question & answer→MCQ 1891 Mark
$ 2^2 \times 2^3 \times 2^4 $ is equal to:
- A
$2^{24}$
- B
$2^{-5}$
- ✓
$2^9$
- D
$2^{-9}$
AnswerC. $2^9$
Solution:
By laws of exponents:
$a^m \times a^n=a^{m+n} $
$ 2^2 \times 2^3 \times 2^4=2^{2+3+4}=2^9$
View full question & answer→MCQ 1901 Mark
Which one of the following is the value of $1^{15}.$
View full question & answer→MCQ 1911 Mark
The approximate distance of moon from the earth is $384,467,000\ m$ and in exponential form. This distance can be written as.
- ✓
$3.84,467 \times 10^8m$
- B
$384,467 \times 10^{-8}m$
- C
$384,467 \times 10^8m$
- D
$3.844,67 \times 10^{-13}m$
AnswerCorrect option: A. $3.84,467 \times 10^8m$
A. $3.84,467 \times 10^8m$
View full question & answer→MCQ 1921 Mark
$100^\circ + 20^\circ+ 5^\circ$ is equal to:
- A
$125$
- B
$25$
- C
$\frac{1}{125}$
- ✓
$3$
AnswerD. $3$
Solution:
By exponent law we know:
$a^\circ = 1$
$100^\circ + 20^\circ + 5^\circ$
$= 1 + 1 + 1$
$= 3$
View full question & answer→MCQ 1931 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. $2.34 \times 10^8$
A. $2.34 \times 10^8$
Solution:
Given,
$234000000 = 234 \times 10^6$
$= 2.34 \times 10^6$
$= 2.34 \times 10^8$
Hence,
standard form of $234000000$ is $2.34 \times 10^8.$
View full question & answer→MCQ 1941 Mark
$149600000000$ is equal to:
- ✓
$1.496 \times 10^{11}$
- B
$1.496 \times 10^{10}$
- C
$1.496 \times 10^{12}$
- D
$1.496 \times 10^{8}$
AnswerCorrect option: A. $1.496 \times 10^{11}$
A. $1.496 \times 10^{11}$
Solution:
$149,600,000,000 = 1.496 \times 10^{11}$
View full question & answer→MCQ 1951 Mark
The value of $(-2)^{2 \times 3-1}$ is:
AnswerC. $-32$
Solution:
Given,
$(-2)^{2 \times 3-1}=(-2)^{6-1} $
$=(-2)^5 $
$=(-2) \times(-2) \times(-2) \times(-2) \times(-2) $
$=-32$
[for $(-a)^m$, if $m$ is odd, then $(-a)^m$ is negative]
View full question & answer→MCQ 1961 Mark
Size of a bacteria is $0.00000063m$ Express it into standard form:
- A
$6.3 \times 10^{-6}$
- B
$6.3 \times 10^{-10}$
- ✓
$6.3 \times 10^{-7}$
- D
$6.3 \times 10^{-8}$
AnswerCorrect option: C. $6.3 \times 10^{-7}$
C. $6.3 \times 10^{-7}$
Solution:
$0.00000063 = 6.3 \times 10^{-8}$
$= 6.3 \times 10 \times 10^{-8}$
$= 6.3 \times 10^{-7}$
View full question & answer→MCQ 1971 Mark
Which of the following is the value of $\frac{\Big(\frac{4}{5}\Big)^{-9}}{\Big(\frac{4}{5}\Big)^{-9}} ?$
Answer$\frac{\Big(\frac{4}{5}\Big)^{-9}}{\Big(\frac{4}{5}\Big)^{-9}}$
$=\frac{\Big(\frac{\not{4}}{\not{5}}\Big)^{\not{9}}}{\Big(\frac{\not{4}}{\not{5}}\Big)^{\not{9}}}=1$
View full question & answer→MCQ 1981 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. $x^{m+n}$
B. $x^{m+n}$
Solution:
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 1991 Mark
The value of $\text{log}_\text{x}(\text{x+y+z)}$ is equal to:
- A
$\text{log}_\text{x}^\text{x}+\text{log}_\text{x}^\text{y}+\text{log}_\text{x}^\text{z}$
- ✓
$\frac{\text{log}(\text{x+y+z)}}{\text{log}\text{x}}$
- C
$\text{log}^\text{xyz}_\text{x}$
- D
$\text{None of these}$
AnswerCorrect option: B. $\frac{\text{log}(\text{x+y+z)}}{\text{log}\text{x}}$
$\frac{\text{log}(\text{x+y+z)}}{\text{log}\text{x}}$
View full question & answer→MCQ 2001 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. $\frac{1}{3^2}$
B. $\frac{1}{3^2}$
Solution:
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→