MCQ 11 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:
$\bigg\{6^{-1}+\Big(\frac{3}{2}\Big)^{-1}\bigg\}=?$
- A
$\frac{2}{3}$
- B
$\frac{5}{6}$
- ✓
$\frac{6}{5}$
- D
AnswerCorrect option: C. $\frac{6}{5}$
$\bigg\{6^{-1}+\Big(\frac{3}{2}\Big)^{-1}\bigg\}=\bigg[\Big(\frac{1}{6}\Big)+\Big(\frac{2}{3}\Big)\bigg]^{-1}$
$=\Big(\frac{1+4}{6}\Big)^{-1}=\Big(\frac{5}{6}\Big)^{-1}$
$=\frac{6}{5}$
View full question & answer→MCQ 21 Mark
Mark $(\checkmark)$ against the correct answer in the following:
$\Big(\frac{3}{4}\Big)^0=?$
- A
$0$
- B
$\Big(\frac{4}{3}\Big)$
- ✓
$1$
- D
Answer$(\text{a})^0=1$
$\therefore \Big(\frac{3}{4}\Big)^0=1$
View full question & answer→MCQ 31 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:
If $\Big(\frac{5}{3}\Big)^{-5}\times \Big(\frac{5}{3}\Big)^{11}=\Big(\frac{5}{3}\Big)^{8\text{x}},$ then $x = ?$
- A
$\frac{-1}{2}$
- B
$\frac{-3}{4}$
- ✓
$\frac{3}{4}$
- D
$\frac{4}{3}$
AnswerCorrect option: C. $\frac{3}{4}$
$\because \Big(\frac{5}{3}\Big)^{-5}\times \Big(\frac{5}{3}\Big)^{11}=\Big(\frac{5}{3}\Big)^{8\text{x}},$
$\Rightarrow \Big(\frac{5}{3}\big)^{-5+11}=\big(\frac{5}{3}\Big)^{8\text{x}}$
$\Rightarrow \Big(\frac{5}{3}\Big)^6=\Big(\frac{5}{3}\Big)^{8\text{x}}$
Comparing $8x = 6$
$\Rightarrow \text{x}=\frac{6}{8}=\frac{3}{4}$
View full question & answer→MCQ 41 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:
$\Big(\frac{-3}{2}\Big)^{-1}=?$
- A
$\frac{2}{3}$
- ✓
$\frac{-2}{3}$
- C
$\frac{3}{2}$
- D
AnswerCorrect option: B. $\frac{-2}{3}$
$\because \Big(\frac{-3}{2}\Big)^{-1}=\Big(\frac{2}{-3}\Big)^1$
$=\frac{2\times(-1)}{-3\times(-1)}=\frac{-2}{3}$
View full question & answer→MCQ 51 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:
$\Big(\frac{2}{3}\Big)^{-5}=?$
- A
$\frac{32}{243}$
- ✓
$\frac{243}{32}$
- C
$\frac{-32}{243}$
- D
$\frac{-243}{32}$
AnswerCorrect option: B. $\frac{243}{32}$
$\because \Big(\frac{2}{3}\Big)^{-5}=\Big(\frac{3}{2}\Big)^{5}$
$=\frac{3}{2}\times\frac{3}{2}\times\frac{3}{2}\times\frac{3}{2}\times\frac{3}{2}$
$=\frac{243}{32}$
View full question & answer→MCQ 61 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:
$\Big(\frac{-2}{5}\Big)^7\div\Big(\frac{-2}{5}\Big)^5=?$
AnswerCorrect option: A. $\frac{4}{25}$
$\because \Big(\frac{-2}{5}\Big)^7\div\Big(\frac{-2}{5}\Big)^5$
$=\Big(\frac{-2}{5}\Big)^{7-5}=\Big(\frac{-2}{5}\Big)^2$
$=\frac{-2}{5}\times\frac{-2}{5}=\frac{4}{25}$
View full question & answer→MCQ 71 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:
$(5^{-1}\times3^{-1})^{-1}=?$
- A
$\frac{1}{15}$
- B
$\frac{-1}{15}$
- ✓
$15$
- D
$-15$
Answer$\because (5{-1}\times3^{-1})^{-1}=\Big(\frac{1}{5}\times\frac{1}{3}\Big)^{-1}$
$\Big(\frac{1}{15}\Big)^{-1}=15$
View full question & answer→MCQ 81 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:
$(6^{-1}-8^{-1})^{-1}=?$
- A
$-\frac{1}{2}$
- B
$-2$
- C
$\frac{1}{24}$
- ✓
$24$
Answer$\because (6^{-1}-8^{-1})=\Big(\frac{1}{6}+-\frac{1}{8}\Big)^{-1}$
$=\Big(\frac{4-3}{24}\Big)^{-1}$
$=\Big(\frac{1}{24}\Big)^{-1}=24$
View full question & answer→MCQ 91 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:
$\Big(\frac{-1}{2}\Big)^{-6}=?$
- A
$-64$
- ✓
$64$
- C
$\frac{1}{64}$
- D
$\frac{-1}{64}$
Answer$\because \Big(\frac{-1}{2}\Big)^{-6}=(-2)^6$
$=(-2)\times(-2)\times(-2)\times(-2)\times(-2)\times(-2)$
$=64$
View full question & answer→MCQ 101 Mark
Mark $(\checkmark)$ against the correct answer in the following:
Which of the following numbers is in standard form?
- A
$32.63 \times 10^4$
- B
$326.3 \times 10^3$
- ✓
$3.263 \times 10^5$
- D
AnswerCorrect option: C. $3.263 \times 10^5$
A given number is said to be in standarf form if it can be expressed as $k \times 10^n$,
where $k$ is a real number such that $1 \leq k <10$ and $n$ is a positive integer.
For example: $3.263 \times 10^5$
View full question & answer→MCQ 111 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:
$\bigg\{\Big(\frac{3}{4}\Big)^{-1}-\Big(\frac{1}{4}\Big)^{-1}\bigg\}^{-1}=?$
- A
$\frac{3}{8}$
- ✓
$\frac{-3}{8}$
- C
$\frac{8}{3}$
- D
$\frac{-8}{3}$
AnswerCorrect option: B. $\frac{-3}{8}$
$\bigg\{\Big(\frac{3}{4}\Big)^{-1}-\Big(\frac{1}{4}\Big)^{-1}\bigg\}^{-1}=\Big(\frac{4}{3}-\frac{4}{1}\Big)^{-1}$
$=\Big(\frac{4-12}{3}\Big)^{-1}=\Big(\frac{-8}{3}\Big)^{-1}$
$=\frac{-3}{8}$
View full question & answer→MCQ 121 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:
$\bigg\{\Big(\frac{1}{2}\Big)^{-3}-\Big(\frac{1}{2}\Big)^{-3}\bigg\}\div\Big(\frac{1}{4}\Big)^{-3}=?$
- ✓
$\frac{19}{64}$
- B
$\frac{64}{19}$
- C
$\frac{27}{16}$
- D
AnswerCorrect option: A. $\frac{19}{64}$
$\because \bigg\{\Big(\frac{1}{2}\Big)^{-3}-\Big(\frac{1}{2}\Big)^{-3}\bigg\}\div\Big(\frac{1}{4}\Big)^{-3}$
$=[(3)^3-(2)^3]\div(4)^3$
$=(27-8)\div64$
$=19\div64=\frac{19}{64}$
View full question & answer→MCQ 131 Mark
Mark $(\checkmark)$ tick against the correct answer in the following: $\bigg\{\Big(\frac{1}{3}\Big)^2\bigg\}^4=?$
- A
$\Big(\frac{1}{3}\Big)^6$
- ✓
$\Big(\frac{1}{3}\Big)^8$
- C
$\Big(\frac{1}{3}\Big)^{16}$
- D
$\Big(\frac{1}{3}\Big)^{24}$
AnswerCorrect option: B. $\Big(\frac{1}{3}\Big)^8$
$\because \bigg\{\Big(\frac{1}{3}\Big)^2\bigg\}^4=\Big(\frac{1}{3}\Big)^{2\times4}$
$=\Big(\frac{1}{3}\Big)^8$
View full question & answer→MCQ 141 Mark
Mark $(\checkmark)$ against the correct answer in the following: $\Big(\frac{-5}{4}\Big)^{-1}=?$
- A
$\frac{3}{5}$
- ✓
$\frac{-3}{5}$
- C
$\frac{5}{3}$
- D
$\frac{-5}{3}$
AnswerCorrect option: B. $\frac{-3}{5}$
$\Big(\frac{-5}{3}\Big)^{-1}=\Big(\frac{3}{-5}\Big)^1$ $\Big[\text{Since}\Big(\frac{\text{a}}{\text{b}}\Big)^{-\text{n}}=\Big(\frac{\text{b}}{\text{a}}\Big)^\text{n}\Big]$
$=\Big(\frac{3}{-5}\times\frac{-1}{-1}\Big)=\frac{-3}{5}$
View full question & answer→MCQ 151 Mark
Mark $(\checkmark)$ tick against the correct answer in the following: $(2^{-1}-4^{-1})^2=?$
- ✓
$4$
- B
$-4$
- C
$\frac{1}{16}$
- D
$\frac{-1}{16}$
Answer$\because(2^{-1}-4^{-1})^2=\Big(\frac{1}{2}-\frac{1}{4}\Big)^2$
$=\Big(\frac{2-1}{4}\Big)^2$
$=\Big(\frac{1}{4}\Big)^2=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}$
View full question & answer→MCQ 161 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:
$(3^2-2^2)\times\Big(\frac{2}{3}\Big)^{-3}=?$
- A
$\frac{45}{8}$
- B
$\frac{8}{45}$
- C
$\frac{8}{135}$
- ✓
$\frac{135}{8}$
AnswerCorrect option: D. $\frac{135}{8}$
$\because(3^2-2^2)\times\Big(\frac{2}{3}\Big)^{-3}=(9-4)\Big(\frac{3}{2}\Big)^3$
$=5\times\frac{3}{2}\times\frac{3}{2}\times\frac{3}{2}=\frac{135}{8}$
View full question & answer→MCQ 171 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:
$\Bigg[\bigg\{\Big(-\frac{1}{2}\Big)^2\bigg\}^{-2}\Bigg]^{-1}=?$
- ✓
$\frac{1}{16}$
- B
$16$
- C
$\frac{-1}{16}$
- D
$-16$
AnswerCorrect option: A. $\frac{1}{16}$
$\Bigg[\bigg\{\Big(-\frac{1}{2}\Big)^2\bigg\}^{-2}\Bigg]^{-1}=\Big(\frac{-1}{2}\Big)^{2\times(-2)\times(-1)}$
$=\Big(\frac{-1}{2}\Big)^4=\Big(\frac{-1}{2}\Big)\Big(\frac{-1}{2}\Big)\Big(\frac{-1}{2}\Big)\Big(\frac{-1}{2}\Big)$
$=\frac{1}{16}$
View full question & answer→MCQ 181 Mark
Mark $(\checkmark)$ against the correct answer in the following: $\bigg\{\Big(\frac{1}{3}\Big)^{-3}-\Big(\frac{1}{2}\Big)^{-3}\bigg\}=?$
- ✓
$19$
- B
$\frac{1}{19}$
- C
$-19$
- D
$\frac{-1}{19}$
Answer$\Big[\text{Since}\Big(\frac{\text{a}}{\text{b}}\Big)^{-1}=\Big(\frac{\text{b}}{\text{a}}\Big)^1\Big]$
$\bigg\{\Big(\frac{1}{3}\Big)^{-3}-\Big(\frac{1}{2}\Big)^{-3}\bigg\}=\bigg\{\Big(\frac{3}{1}\Big)^3-\Big(\frac{2}{1}\Big)^3\bigg\}$
$=\big\{(3)^3-(2)^3\big\}$
$=(27-8)=19$
View full question & answer→MCQ 191 Mark
Mark $(\checkmark)$ tick against the correct answer in the following: $\Big(\frac{1}{2}\Big)^{-2}+\Big(\frac{1}{3}\Big)^{-2}+\Big(\frac{1}{4}\Big)^{-2}=?$
- A
$\frac{61}{144}$
- ✓
$29$
- C
$\frac{144}{61}$
- D
Answer$\Big(\frac{1}{2}\Big)^{-2}+\Big(\frac{1}{3}\Big)^{-2}+\Big(\frac{1}{4}\Big)^{-2}$
$=(2)^2+(3)^2+(4)^2$
$=4+9+16=29$
View full question & answer→MCQ 201 Mark
Mark $(\checkmark)$ tick against the correct answer in the following: $\Big(\frac{-1}{2}\Big)^3=?$
- A
$\frac{-3}{2}$
- ✓
$\frac{-1}{8}$
- C
$\frac{-1}{6}$
- D
AnswerCorrect option: B. $\frac{-1}{8}$
$=\Big(\frac{-1}{2}\Big)\times \Big(\frac{-1}{2}\Big)\times\Big(\frac{-1}{2}\Big)$
$=\frac{-1}{8}$
View full question & answer→MCQ 211 Mark
Mark $(\checkmark)$ against the correct answer in the following: $\Big(\frac{-2}{3}\Big)^{10}\div\Big(\frac{-2}{3}\Big)^8=?$
AnswerCorrect option: A. $\frac{4}{9}$
$\big[\text{Since a}^\text{m}\div\text{a}^\text{n}=\text{a}^{\text{m}-\text{n}}]$
$\Big(\frac{-2}{3}\Big)^{10}\div\Big(\frac{-2}{3}\Big)^8=\Big(\frac{-2}{3}\Big)^{10-8}$
$=\Big(\frac{-2}{3}\Big)^2$
$=\frac{(-2)^2}{3^2}=\frac{4}{9}$
View full question & answer→MCQ 221 Mark
Mark $(\checkmark)$ against the correct answer in the following:$\Big(\frac{-3}{4}\Big)^{-3}=?$
- A
$\frac{27}{64}$
- B
$\frac{64}{27}$
- C
$\frac{-27}{64}$
- ✓
$\frac{-64}{27}$
AnswerCorrect option: D. $\frac{-64}{27}$
$\Big(\frac{-3}{4}\Big)^{-3}=\Big(\frac{4}{-3}\Big)^3$ $\Big[\text{Since}\Big(\frac{\text{a}}{\text{b}}\Big)^{-\text{n}}=\Big(\frac{\text{b}}{\text{a}}\Big)^\text{n}\Big]$
$=\frac{4^3}{(-3)^3}$
$=\frac{4\times4\times4}{(-3)\times(-3)\times(-3)}=\frac{64}{(-27)}$
$=\frac{64\times-1}{-27\times-1}=\frac{-64}{27}$
View full question & answer→MCQ 231 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:$\Big(\frac{-2}{3}\Big)^2=?$
- A
$\frac{4}{3}$
- B
$\frac{-2}{9}$
- ✓
$\frac{4}{9}$
- D
$\frac{-4}{9}$
AnswerCorrect option: C. $\frac{4}{9}$
$\because \Big(\frac{-2}{3}\Big)^2=\frac{-2}{3}\times\frac{-2}{3}$
$=\frac{4}{9}$
View full question & answer→MCQ 241 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:
Which of the following numbers is in standard form?
- A
$21.56 \times 10^5$
- B
$215.6 \times 10^4$
- ✓
$2.156 \times 10^6$
- D
AnswerCorrect option: C. $2.156 \times 10^6$
The number which is in standard form is $2.156 \times 10^6$
View full question & answer→MCQ 251 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:$\Big(\frac{5}{6}\Big)^0=?$
Answer$\because\Big(\frac{5}{6}\Big)^0=1$ $\bigg\{\because\Big(\frac{\text{a}}{\text{b}}\Big)^0=1\bigg\}$
View full question & answer→MCQ 261 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:
$\Big(\frac{-1}{5}\Big)^3\div\Big(\frac{-1}{5}\Big)^8=?$
- 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. $(-5)^5$
$\because\Big(\frac{-1}{5}\Big)^3\div\Big(\frac{-1}{5}\Big)^8$
$=\Big(-\frac{1}{5}\Big)^{3-8}=\Big(-\frac{1}{5}\Big)^{-5}$
$=(-5)^5$
View full question & answer→MCQ 271 Mark
Mark $(\checkmark)$ tick against the correct answer in the following:
By what number should $(-8)^{-1}$ be multiplied to get $10^{-1}$ ?
- A
$\frac{4}{5}$
- B
$\frac{-5}{4}$
- ✓
$\frac{-4}{5}$
- D
AnswerCorrect option: C. $\frac{-4}{5}$
$\text { Required number }=10^{-1} \div(-8)^{-1}$
$=\frac{1}{10} \div\left(\frac{1}{-8}\right)=\frac{1}{10} \times \frac{-8}{1}$
$=\frac{4}{5}$
View full question & answer→