MCQ 11 Mark
$\Big\{\Big(\frac{1}{3}\Big)^{-3}-\Big(\frac{1}{2}\Big)^{-3}+\Big(\frac{1}{4}\Big)^{-3}\Big\}$
- ✓
$\frac{19}{64}$
- B
$\frac{64}{19}$
- C
$\frac{27}{16}$
- D
$\frac{-19}{64}$
AnswerCorrect option: A. $\frac{19}{64}$
$\Big\{\Big(\frac{1}{3}\Big)^{-3}-\Big(\frac{1}{2}\Big)^{-3}\Big\}\div\Big(\frac{1}{4}\Big)^{-3}$
$=\Big\{\Big(\frac{3}{1}\Big)^{3}-\Big(\frac{2}{1}\Big)^{3}\Big\}\div\Big(\frac{4}{1}\Big)^{3}$ $\text{As, }\text{x}^{-1}=\frac{1}{\text{x}}$
$=\{3^3-2^3\}\div4^3$
$=\{27-8\}\div64$
$=19\div64$
$=\frac{19}{64}$
Hence, the correct alternative is option $(a).$
View full question & answer→MCQ 21 Mark
$(1+3+5+7+9+11)^{\frac{3}{2}}=$
Answer$(1+3+5+7+9+11)^{\frac{3}{2}}$
$=(36)^{\frac{3}{2}}$
$=(6^2)^{\frac{3}{2}}$
$=6^{2\times\frac{3}{2}}$ $[\text{As,}(\text{x}^{\text{m}})=\text{x}^{\text{mn}}]$
$=6^3$
$=216$
Hence, the correct alternative is option $(b)$.
View full question & answer→MCQ 31 Mark
If $\Big(\frac{5}{3}\Big)^{-5}\times\Big(\frac{5}{3}\Big)^{11}=\Big(\frac{5}{3}\Big)^{8\text{x}}, \text{then x}=?$
- A
$-\frac{1}{2}$
- B
$-\frac{3}{4}$
- ✓
$\frac{3}{4}$
- D
$\frac{4}{3}$
AnswerCorrect option: C. $\frac{3}{4}$
$\Big(\frac{5}{3}\Big)^{-5}\times\Big(\frac{5}{3}\Big)^{11}=\Big(\frac{5}{3}\Big)^{8\text{x}}$
$\Rightarrow\Big(\frac{5}{3}\Big)^{-5+11}=\Big(\frac{5}{3}\Big)^{8\text{x}}$ $\Big(\text{As, }\text{x}^{\text{m}\times\text{n}}=\text{x}^{\text{m+n}}\Big)$
$\Rightarrow\Big(\frac{5}{3}\Big)^{6}=\Big(\frac{5}{3}\Big)^{8\text{x}}$
Comparing the exponent of both the sides, we get:
$8\text{x}=6$
$\Rightarrow\text{x}=\frac{6}{8}$
$\therefore\text{x}=\frac{3}{4}$
Hence, the correct alternative is option $(c)$.
View full question & answer→MCQ 41 Mark
If $3^x=6561$, then $3^{x-3}=$
Answer$3^x=6561$
$\Rightarrow 3^x=3^8$
Comparing the exponent of both the sides, we get:
$x=8$
Now, $3^{x-3}=3^{8-3}$
$=3^5=243$
Hence, the correct alternative is option $(b)$.
View full question & answer→MCQ 51 Mark
If $x y z=0$, then find the value of $\left(a^x\right)^{y z}+\left(a^y\right)^{z x}+\left(a^z\right)^{x y}=$
AnswerSince,
$\left(a^x\right)^{y z}+\left(a^y\right)^{z x}+\left(a^z\right)^{x y}$
$=a^{x y z}+a^{y z x}+a^{z x y}\left[\text { As, }\left(x^m\right)^n=x^{m n}\right]$
$=a^{x y z}+a^{x y z}+a^{x y z}$
$=a^0+a^0+a^0$
$=1+1+1\left(\text { As, } x^0=1\right)$
$=3$
Hence, the correct option is $(a)$.
View full question & answer→MCQ 61 Mark
$2^{3^{2}}=$
AnswerSince, $2^{3^{2}}=2^9=512$
Hence, the correct alternative is option $(d)$.
View full question & answer→MCQ 71 Mark
$(8^4+8^2)^{\frac{1}{2}}=$
- A
$84$
- B
$8\sqrt{77}$
- C
$72$
- ✓
$8\sqrt{65}$
AnswerCorrect option: D. $8\sqrt{65}$
$(8^4+8^2)^{\frac{1}{2}}$
$=(8^{2+2}+8^2)^{\frac{1}{2}}$
$=(8^2\times8^2+8^2)^{\frac{1}{2}}$ $(\text{As},\text{x}^{\text{m+n}}=\text{x}^{\text{m}}\times\text{x}^{\text{n}})$
$=[8^2\times(8^2+1)]^{\frac{1}{2}}$ $[\text{As}, \text{ab+ac}=\text{a}\times(\text{a+c})]$
$=(8^2)^{\frac{1}{2}}\times(8^2+1)^{\frac{1}{2}}$ $[\text{As, }(\text{ab})^{\text{m}}=\text{a}^{\text{m}}\times\text{b}^{\text{m}}]$
$=8^{2\times\frac{1}{2}}\times(64+1)^{\frac{1}{2}}$
$=8\times65^{\frac{1}{2}}$
$=8\sqrt{65}$
Hence, the correct alternative is option $(d)$.
View full question & answer→MCQ 81 Mark
The number 4,70,394 is standard form is written as:
- ✓
$4.70394 \times 10^5$
- B
$4.70394 \times 10^4$
- C
$47.0394 \times 10^4$
- D
$4703.94 \times 10^2$
AnswerCorrect option: A. $4.70394 \times 10^5$
Since, $4,70,394=4.70394 \times 100000=4.70394 \times 10^5$
So, the number $4,70,394$ in standard form is written as $4.70394 \times 10^5$
Hence, the correct alternative is option $(a).$
View full question & answer→MCQ 91 Mark
$(6^{-1} - 8^{-1})^{-1 }=$
- A
$\frac{1}{24}$
- ✓
$24$
- C
$-24$
- D
$-\frac{1}{24}$
Answer$(6^{-1}-8^{-1})^{-1}=$
$=\Big(\frac{1}{6}-\frac{1}{8}\Big)^{-1}$ $\Big(\text{As},\text{x}^{-1}=\frac{1}{\text{x}}\Big)$
$=\Big(\frac{4-3}{24}\Big)^{-1}$
$=\Big(\frac{1}{24}\Big)^{-1}$
$=\frac{24}{1}$ $\Big(\text{As},\text{x}^{-1}=\frac{1}{\text{x}})$
$=24$
Hence, the correct alternative is option $(b).$
View full question & answer→MCQ 101 Mark
If $2^n=4096$, then $2^{n-5}=$
AnswerSince,
$\text { As, } 2^n=4096$
$\Rightarrow 2^n=2^{12}$
Comparing the exponent of both the sides, we get:
$\mathrm{n}=12$
Now, $2^{n-5}=2^{12-5}$
$=2^7$
$=128$
Hence, the correct alternative is option $(a)$.
View full question & answer→MCQ 111 Mark
$(2^{3^{4}})=$
- A
$2^{4^{3}}$
- B
$2^{3^{4}}$
- ✓
$\big(2^4\big)^3$
- D
AnswerCorrect option: C. $\big(2^4\big)^3$
$\big(2^3\big)^4$
$=2^{3\times4}$ $[\text{As}, (\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=2^{4\times3}$
$=(2^{4})^{3}$
Hence, the correct alternative is option $(c).$
View full question & answer→MCQ 121 Mark
If $a = 25,$ then $\text{a}^{25^{0}}+\text{a}^{0^{25}}$
Answer$26$
$\text{a}^{25^{0}}+\text{a}^{0^{25}}$
$=\text{a}^{1}+\text{a}^{0}$ $\Big(\text{As, }25^{0}=1\text{ and }0^{25}=0\Big)$
$=\text{a}+1$ $\Big(\text{As, }\text{a}^{0}=1\Big)$
$=25+1$
$=26$
Hence, the correct alternative is option $(b)$.
View full question & answer→MCQ 131 Mark
$\bigg[\Big\{\big(-\frac{1}{3}\big)\Big\}^{-2}\bigg]^{-1}=$
- ✓
$\frac{1}{81}$
- B
$\frac{1}{9}$
- C
$-\frac{1}{81}$
- D
$-\frac{1}{9}$
AnswerCorrect option: A. $\frac{1}{81}$
$\bigg[\Big\{\big(-\frac{1}{3}\big)\Big\}^{-2}\bigg]^{-1}$
$\bigg[\Big\{\big(-\frac{1}{3}\big)^{2}\Big\}^{(-2)\times(-1)}\bigg]$ $[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}\big]$
$=\Big(\frac{-1}{3}\Big)^{2\times2}$ $[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}\big]$
$=\Big(\frac{-1}{3}\Big)^4$
$=\frac{(-1)^4}{3^4}$ $\Big[\text{As, }\Big(\frac{\text{x}}{\text{y}}\Big)^{\text{m}}=\frac{\text{x}^{\text{m}}}{\text{y}^{\text{m}}}\Big]$
$=\frac{1}{81}$
Hence, the correct alternative is option $(a).$
View full question & answer→MCQ 141 Mark
$\text{If abc} = 0,$ $\text{then}\frac{\Big\{(\text{x}^{\text{a}})^{\text{b}}\Big\}^{\text{c}}}{\Big\{(\text{x}^{\text{b}})^{\text{c}}\Big\}}=$
Answer$\frac{\Big\{(\text{x}^{\text{a}})^{\text{b}}\Big\}^{\text{c}}}{\Big\{(\text{x}^{\text{b}})^{\text{c}}\Big\}}$
$=\frac{(\text{x}^{\text{a}})^{\text{bc}}}{(\text{x}^{\text{b}})^{\text{ac}}}$ $[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=\frac{\text{x}^{\text{abc}}}{\text{x}^{\text{abc}}}$ $[\text{As,}(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=\frac{\text{x}^{0}}{\text{x}^{0}}$ $[\text{As, abc}=0]$
$=\frac{1}{1}$ $[\text{As, }\text{x}^{0}=1]$
$=1$
Hence, the correct alternative is option $(d)$.
View full question & answer→MCQ 151 Mark
$(3^{-1}\times5^{-1})^{-1}=$
- A
$\frac{1}{15}$
- B
$-\frac{1}{15}$
- ✓
$15$
- D
$-15$
Answer$(3^{-1}\times5^{-1})^{-1}$
$=\Big(\frac{1}{3}\times\frac{1}{5}\Big)^{-1}$ $\Big(\text{As},\text{x}^{-1}=\frac{1}{\text{x}}\Big)$
$=\Big(\frac{1}{15}\Big)^{-1}$
$=\frac{15}{1}$ $\Big(\text{As},\text{x}^{-1}=\frac{1}{\text{x}}\Big)$
$=15$
Hence, the correct alternative is option $(c).$
View full question & answer→MCQ 161 Mark
$\Big(-\frac{3}{5}\Big)^{-1}=$
- A
$\frac{3}{5}$
- B
$\frac{5}{3}$
- ✓
$-\frac{5}{3}$
- D
$-\frac{3}{5}$
AnswerCorrect option: C. $-\frac{5}{3}$
Since,
$\Big(-\frac{3}{5}\Big)^{-1}=-\frac{5}{3}$ $\Big(\text{As, }\text{x}^{-1}=\frac{\text{1}}{\text{x}}\Big)$
Hence, the correct alternative is option $(c).$
View full question & answer→MCQ 171 Mark
If If $a=3^{-3}-3^3$ and $b=3^3-3^{-3}$, then $\frac{\text{a}}{\text{b}}-\frac{\text{b}}{\text{a}}=$
AnswerSince,
$\text{a}=3^{-3}-3^3$
$=\frac{1}{3^3}-3^3$ $\Big(\text{As},\text{x}^{-\text{m}}=\frac{1}{\text{x}^{\text{m}}}\Big)$
$=\frac{1}{27}-\frac{27}{1}$
$=\frac{1}{27}-\frac{27\times27}{1\times27}$
$=\frac{1}{27}-\frac{729}{27}$
$=\frac{1-729}{27}$
$=\frac{-728}{27}$
Also, $\text{b}=3^3-3^{-3}$
$=3^3-\frac{1}{3^3}$ $\Big(\text{As, }\text{x}^{-\text{m}}=\frac{1}{\text{x}^{\text{m}}}\Big)$
$=\frac{27}{1}-\frac{1}{27}$
$=\frac{27\times27}{1\times27}-\frac{1}{27}$
$=\frac{729}{27}-\frac{1}{27}$
$=\frac{729-1}{27}$
$=\frac{728}{27}$
Now,
$\frac{\text{a}}{\text{b}}-\frac{\text{b}}{\text{a}}$
$=\frac{\Big(\frac{-728}{27}\Big)}{\Big(\frac{728}{27}\Big)}-\frac{\Big(\frac{728}{27}\Big)}{\Big(\frac{-728}{27}\Big)}$
$=(-1)-(-1)$
$=-1+1$
$=0$
Hence, the correct alternative is option $(a)$.
View full question & answer→MCQ 181 Mark
The number $2.35 \times 10^4$ in the usual form is written as:
- A
$2.35 \times 10^3$
- ✓
$23500$
- C
$2350000$
- D
$235 \times 10^4$
AnswerCorrect option: B. $23500$
Since, $2.35 \times 10^4=2.35 \times 10000=23500$
So, the number $2.35 \times 10^4$ in the usual form is written as $23500$ .
Hence, the correct alternative is option $(b)$.
View full question & answer→MCQ 191 Mark
$\frac{(144)^{\frac{1}{2}}+(256)^{\frac{1}{2}}}{3^2-2}=$
Answer$\frac{(144)^{\frac{1}{2}}+(256)^{\frac{1}{2}}}{3^2-2}$
$=\frac{(12^2)^{\frac{1}{2}}+(16^2)^{\frac{1}{2}}}{9-2}$
$=\frac{12^{2\times\frac{1}{2}}+16^{2\times\frac{1}{2}}}{7}$ $[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=\frac{12+16}{7}$
$=\frac{28}{7}$
$=4$
Hence, the correct alternative is option $(b).$
View full question & answer→MCQ 201 Mark
$(-1)^{301}+(-1)^{302}+(-1)^{303}+\dots+(-1)^{400}$
AnswerSince,
$(-1)^{301}+(-1)^{302}+(-1)^{303}+\dots+(-1)^{400}$
$=(-1)+(1)+(-1)+\dots+(1)$ $\Big[\text{As, }(-1)^{\text{odd}}=-1\text{ and }(-1)^{\text{even}}=1\Big]$
$=-50+50$ $[\text{As, there are} 50(-1)'\text{s and 50 (1)}'\text{s}]$
$=0$
Hence, the correct alternative is option is $(d)$.
View full question & answer→MCQ 211 Mark
If $2^n = 1024$, then $2^{(\frac{\text{n}}{2}+2)}=$
Answer$\text { As, } 2^n=1024$
$\Rightarrow 2^n=2^{10}$
Comparing the exponent of both the sides, we get:
$n = 10$
Now,
$2^{\Big(\frac{\text{n}}{2}+2\Big)}$
$=2^{\Big(\frac{\text{10}}{2}+2\Big)}$
$=2^{(5+2)}$
$=2^7$
$=128$
Hence, the correct alternative is option $(b)$.
View full question & answer→MCQ 221 Mark
What should be multiplied to $6^{-2}$ so that the product may be equal to $216$?
AnswerSince,
Let $6^{\mathrm{m}}$ should be multiplied to $6^{-2}$
$A.T.Q.$
$6^{-2} \times 6^m=216$
$\Rightarrow 6^{-2+m}=6^3$
Comparing th exponent of both the sides, we get
$-2+m=3$
$\Rightarrow m=3+2$
$\therefore m=5$
So, $6^5$ should be multiplied.
Hence, the correct alternative is option $(b)$.
View full question & answer→MCQ 231 Mark
If $abc = 0$, then find the value of $\Big\{(\text{x}^{\text{a}})^{\text{b}}\Big\}^{\text{c}}$
AnswerSince,
$\Big\{(\text{x}^{\text{a}})^{\text{b}}\Big\}^{\text{c}}$
$=(\text{x}^{\text{a}})^{\text{bc}}$ $[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=\text{x}^{\text{abc}}$ $[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=\text{x}^{0}$ $[\text{As, abc}=0]$
$=1$ $[\text{As, }\text{x}^{0}=1]$
Hence, the correct alternative is option $(a).$
View full question & answer→MCQ 241 Mark
$\Big\{(33)^2-(31)^2\Big\}^{\frac{5}{7}}$
AnswerSince,
$\Big\{(33)^2-(31)^2\Big\}^{\frac{5}{7}}$
$=\{1089-961\}^{\frac{5}{7}}$
$=\{128\}^{\frac{5}{7}}$
$=\{2^7\}^{\frac{5}{7}}$
$=2^{7\times\frac{5}{7}}$$[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=2^5$
$=32$
Hence, the correct alternative is option $(c)$.
View full question & answer→MCQ 251 Mark
$\left(8^4+8^2\right)^{1 / 2}=$
- A
- B
$8 \sqrt{77}$
- C
- ✓
$8 \sqrt{65}$
AnswerCorrect option: D. $8 \sqrt{65}$
View full question & answer→MCQ 261 Mark
If $2^n=1024$, then $2^{\frac{n}{2}+2}=$
View full question & answer→MCQ 271 Mark
If $3^x=6561$, then $3^{x-3}=$
View full question & answer→MCQ 281 Mark
The number $2.35 \times 10^4$ in the usual form is written as
- A
$2.35 \times 10^3$
- ✓
- C
- D
$235 \times 10^4$
View full question & answer→MCQ 291 Mark
The number $4,70,394$ is standard form is written as
- ✓
$4.70394 \times 10^5$
- B
$4.70394 \times 10^4$
- C
$47.0394 \times 10^4$
- D
$4703.94 \times 10^2$
AnswerCorrect option: A. $4.70394 \times 10^5$
View full question & answer→MCQ 301 Mark
If $2^n=4096$, then $2^{n-5}=$
View full question & answer→MCQ 311 Mark
If $x y z=0$, then find the value of $\left(a^x\right)^{y z}+\left(a^y\right)^{z x}+\left(a^z\right)^{x y}=$
View full question & answer→MCQ 321 Mark
What should be multiplied to $6^{-2}$ so that the product may be equal to 216 ?
View full question & answer→MCQ 331 Mark
If $a=3^{-3}-3^3$ and $b=3^3-3^{-3}$, then $\frac{a}{b}-\frac{b}{a}$
View full question & answer→MCQ 341 Mark
If $a b c=0$, then find the value of $\left\{\left(x^a\right)^b\right\}^c=$
View full question & answer→MCQ 351 Mark
$\left\{(33)^2-(31)^2\right\}^{5 / 7}=$
View full question & answer→MCQ 361 Mark
$\left(2^3\right)^4=$
- A
$2^{4^3}$
- B
$2^{3^4}$
- ✓
$\left(2^4\right)^3$
- D
AnswerCorrect option: C. $\left(2^4\right)^3$
View full question & answer→MCQ 371 Mark
If $a b c=0$, then $\frac{\left\{\left(x^a\right)^b\right\}^c}{\left\{\left(x^b\right)^c\right\}^a}=$
View full question & answer→MCQ 381 Mark
$(1+3+5+7+9+11)^{3 / 2}=$
View full question & answer→MCQ 391 Mark
$\frac{(144)^{1 / 2}+(256)^{1 / 2}}{3^2-2}=$
View full question & answer→MCQ 401 Mark
$\left[\left\{\left(-\frac{1}{3}\right)^2\right\}^{-2}\right]^{-1}=$
- ✓
$\frac{1}{81}$
- B
$\frac{1}{9}$
- C
$-\frac{1}{81}$
- D
$-\frac{1}{9}$
AnswerCorrect option: A. $\frac{1}{81}$
View full question & answer→MCQ 411 Mark
If $\left(\frac{5}{3}\right)^{-5} \times\left(\frac{5}{3}\right)^{11}=\left(\frac{5}{3}\right)^{8 x}$, then $x=$ ?
- A
$-\frac{1}{2}$
- B
$-\frac{3}{4}$
- ✓
$\frac{3}{4}$
- D
$\frac{4}{3}$
AnswerCorrect option: C. $\frac{3}{4}$
View full question & answer→MCQ 421 Mark
$\left(25^2-15^2\right)^{3 / 2}=$
View full question & answer→MCQ 431 Mark
$\left\{\left(\frac{1}{3}\right)^{-3}-\left(\frac{1}{2}\right)^{-3} \div\left(\frac{1}{4}\right)^{-3}\right\}=$
- ✓
$\frac{19}{64}$
- B
$\frac{64}{19}$
- C
$\frac{27}{16}$
- D
$\frac{-19}{64}$
AnswerCorrect option: A. $\frac{19}{64}$
View full question & answer→MCQ 441 Mark
If $a=25$, then $a^{25^0}+a^{0^{25}}=$
View full question & answer→MCQ 451 Mark
$(-1)^{301}+(-1)^{302}+(-1)^{303}+\ldots+(-1)^{400}$
View full question & answer→MCQ 461 Mark
$\left(-\frac{3}{5}\right)^{-1}=$
- A
$\frac{3}{5}$
- B
$\frac{5}{3}$
- ✓
$-\frac{5}{3}$
- D
$-\frac{3}{5}$
AnswerCorrect option: C. $-\frac{5}{3}$
View full question & answer→MCQ 471 Mark
$\left(3^{-1} \times 5^{-1}\right)^{-1}=$
- A
$\frac{1}{15}$
- B
$-\frac{1}{15}$
- ✓
- D
View full question & answer→MCQ 481 Mark
$2^{3^2}$
View full question & answer→MCQ 491 Mark
$\left(6^{-1}-8^{-1}\right)^{-1}=$
- A
$\frac{1}{24}$
- ✓
- C
- D
$-\frac{1}{24}$
View full question & answer→