M.C.Q. [1 Marks Each]

MCQ 1011 Mark

Fill in the blanks
$(0.000729)^{\frac{-3}{4}}\times(0.09)^{\frac{-3}{4}}=\text{_____.}$

A
$\frac{10^3}{3^3}$
B
$\frac{10^5}{3^5}$
C
$\frac{10^2}{3^2}$
✓
$\frac{10^6}{3^6}$

Answer

Correct option: D.

$\frac{10^6}{3^6}$

$(0.000729)^{\frac{-3}{4}}\times(0.09)^{\frac{-3}{4}}$
$\Rightarrow(3^8\times10^{-8})^{\frac{-3}{4}}$
$\Rightarrow\frac{10^6}{3^6}$

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MCQ 1021 Mark

If $.0000058 = 5.8 \times 10^n, n =$

A
$-4$
B
$-5$
✓
$-6$
D
$-7$

Answer

Correct option: C.

$-6$

Given,$ 0.0000058 = 5.8 \times 10^{-6}$
$\Rightarrow 5.8 \times 10^{-6}= 5.8 \times 10^n$
$\Rightarrow n = -6$

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MCQ 1031 Mark

$3.268 \times 104 =$

A
$3268 \times 10^1$
B
$0.003268 \times 10^7$
C
$326800 \times 10^{-1}$
✓
All of these

Answer

Correct option: D.

All of these

$3.268 \times 10^4= 3268 \times 10^1= .003268 \times 10^7= 326800 \times 10^{-1}$
Hence all of the options are correct

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MCQ 1041 Mark

The value of $(101)^3$is:

A
$10303001$
B
$10030301$
✓
$1030301$
D
$10300301$

Answer

Correct option: C.

$1030301$

$(101)^3 = (100 + 1)^3$
$= (100)^3 + 3 x (100)^2 x 1 + 3 x 100 x (1)^2+ (1)^3 ...........$ {formula $: a^3+ 3a^2b + 3ab^2 + b^3$}
$= 1000000 + 3 x 10000 x 1 + 300 x 1 + 1$
$= 1000000 + 30000 + 300 + 1$
$= 1030301$

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MCQ 1051 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
None of these.

Answer

Correct option: B.

$29$

$\Big(\frac{1}{2}\Big)^{-2}+\Big(\frac{1}{3}\Big)^{-2}+\Big(\frac{1}{4}\Big)^{-2}$
$=(2)^2+(3)^2+(4)^2$
$=4+9+16=29$

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MCQ 1061 Mark

For non-zero numbers a and b, $\big(\frac{\text{a}}{\text{b}}\big)^{\text{m}}\div\big(\frac{\text{a}}{\text{b}}\big)^{\text{n}}$ where $m > n$, is equal to:

A
$\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{mn}}$
B
$\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{m+n}}$
✓
$\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{m-n}}$
D
$\bigg(\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{m}}\bigg)^{\text{n}}$

Answer

Correct option: C.

$\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{m-n}}$

We know that,
$\text{a}^{\text{m}}+\text{a}^{\text{n}}=\text{a}^{\text{m-n}}(\text{m}>\text{n})$
So, $\Big(\frac{\text{a}}{\text{b}}\Big)+\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{n}}=\Big(\frac{\text{a}{}}{\text{b}}\Big)^{\text{m-n}}$

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MCQ 1071 Mark

Simplify : $\big(\frac{1}{2}\big)^{-2}+\big(\frac{1}{3}\big)^{-2}+\big(\frac{1}{4}\big)^{-2}\text{_____}$

A
$19$
✓
$29$
C
$39$
D
$49$

Answer

Correct option: B.

$29$

We have $\big(\frac{1}{2}\big)^{-2}+\big(\frac{1}{3}\big)^{-2}+\big(\frac{1}{4}\big)^{-2}$
$=\big(\frac{2}{1}\big)^2+\big(\frac{3}{1}\big)^2+\big(\frac{4}{1}\big)^2$
$=(2^2+3^2+4^2)$
$=4+9+16=29$

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MCQ 1081 Mark

$6^8, 8$ is:

A
Base
B
Number
✓
Power
D
None of these

Answer

Correct option: C.

Power

$8$ is power.

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MCQ 1091 Mark

Which one of the following is the value of $(101)^\circ?$

A
$0$
B
$101$
C
$1010$
✓
$1$

Answer

Correct option: D.

$1$

$101^\circ= 1$

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MCQ 1101 Mark

How can $9^{10} ÷ 9^1$ be represented as?

A
$9$
✓
$9^9$
C
$9^{-9}$
D
$9^{10}$

Answer

Correct option: B.

$9^9$

$9^{10}$÷ $9^{1}$ = $9^{10-1}$= $9^{9}$

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MCQ 1111 Mark

Mark $(\checkmark)$ against the correct answer in the following:
$\Big(\frac{-2}{3}\Big)^{10}\div\Big(\frac{-2}{3}\Big)^8=?$

✓
$\frac{4}{9}$
B
$\frac{-4}{9}$
C
$\Big(\frac{-2}{3}\Big)^{18}$
D
None of these.

Answer

Correct option: A.

$\frac{4}{9}$

$\big[\text{Since a}^\text{m}\div\text{a}^\text{n}=\text{a}^{\text{m}-\text{n}}]$
$\Big(\frac{-2}{3}\Big)^{10}\div\Big(\frac{-2}{3}\Big)^8=\Big(\frac{-2}{3}\Big)^{10-8}$
$=\Big(\frac{-2}{3}\Big)^2$
$=\frac{(-2)^2}{3^2}=\frac{4}{9}$

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MCQ 1121 Mark

If $3^{\text{n}-2}=\frac{1}{81},\text{n}=$

A
$0$
B
$1$
C
$2$
✓
$-2$

Answer

Correct option: D.

$-2$

Given, $3^{\text{n}-2}=\frac{1}{81}$
$\Rightarrow3^{\text{n}-2}=\frac{1}{3^4}$
$\Rightarrow3^{\text{n}-2}=3^{-4}$
$\Rightarrow\text{n}-2=-4$
$\Rightarrow\text{n}=-2$

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MCQ 1131 Mark

$\text{If abc} = 0,$ $\text{then}\frac{\Big\{(\text{x}^{\text{a}})^{\text{b}}\Big\}^{\text{c}}}{\Big\{(\text{x}^{\text{b}})^{\text{c}}\Big\}}=$

A
$3$
B
$0$
C
$-1$
✓
$1$

Answer

Correct option: D.

$1$

$\frac{\Big\{(\text{x}^{\text{a}})^{\text{b}}\Big\}^{\text{c}}}{\Big\{(\text{x}^{\text{b}})^{\text{c}}\Big\}}$
$=\frac{(\text{x}^{\text{a}})^{\text{bc}}}{(\text{x}^{\text{b}})^{\text{ac}}}$ $[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=\frac{\text{x}^{\text{abc}}}{\text{x}^{\text{abc}}}$ $[\text{As,}(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=\frac{\text{x}^{0}}{\text{x}^{0}}$ $[\text{As, abc}=0]$
$=\frac{1}{1}$ $[\text{As, }\text{x}^{0}=1]$
$=1$
Hence, the correct alternative is option $(d)$.

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MCQ 1141 Mark

$x$ is a non-zero rational number. Product of the square of $x$ with the cube of $x$ is equal to the:

A
Second power of $x$
B
Third power of $x$
✓
Fifth power of $x$
D
Sixth power of $x$

Answer

Correct option: C.

Fifth power of $x$

Square of $x = x^2$
Cube of $x = x^3$
Product of square with the cube of
$x = x^2 × x^3 = x^{2+3}$ [$\therefore$ $a^m \times a^n =a^{m+n}$]
i.e. fifth power of $x$

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MCQ 1151 Mark

$(3^{-1}\times5^{-1})^{-1}=$

A
$\frac{1}{15}$
B
$-\frac{1}{15}$
✓
$15$
D
$-15$

Answer

Correct option: C.

$15$

$(3^{-1}\times5^{-1})^{-1}$
$=\Big(\frac{1}{3}\times\frac{1}{5}\Big)^{-1}$ $\Big(\text{As},\text{x}^{-1}=\frac{1}{\text{x}}\Big)$
$=\Big(\frac{1}{15}\Big)^{-1}$
$=\frac{15}{1}$ $\Big(\text{As},\text{x}^{-1}=\frac{1}{\text{x}}\Big)$
$=15$
Hence, the correct alternative is option $(c)$.

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MCQ 1161 Mark

Value of $62^{2}$is:

A
$124$
B
$-124$
C
$3244$
✓
$3844$

Answer

Correct option: D.

$3844$

$62 \times 62 = 3844$

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MCQ 1171 Mark

$\Big(-\frac{3}{5}\Big)^{-1}=$

A
$\frac{3}{5}$
B
$\frac{5}{3}$
✓
$-\frac{5}{3}$
D
$-\frac{3}{5}$

Answer

Correct 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 $(c)$.

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MCQ 1181 Mark

Simplify the following using law of exponents. $\frac{5^7}{5^2}$

✓
$5^{5}$
B
$5^{20}$
C
$5^{50}$
D
$5^{15}$

Answer

Correct option: A.

$5^{5}$

we know, $\frac{\text{a}^\text{m}}{\text{a}^\text{n}}=\text{a}^{\text{m}-\text{n}}$
so, $\frac{5^7}{5^2}=5^{7-2}$
$=5^5$

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MCQ 1191 Mark

Solve for n in $4^n + 4^n + 4^n + 4^n = 2^{2010}$

A
$1005$
B
$2010$
✓
$1004$
D
$1003$

Answer

Correct option: C.

$1004$

Given, $4^n + 4^n + 4^n + 4^n = 2^{2010}$
$\Rightarrow 4 \times 4^n = 2^{2010}$
$\Rightarrow 2^{2n} = 2{2008}$
On equating, we get
$2^n = 2008$
$\Rightarrow n = 1004$

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MCQ 1201 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
$\frac{-27}{64}$
✓
$\frac{-64}{27}$

Answer

Correct option: D.

$\frac{-64}{27}$

$\Big(\frac{-3}{4}\Big)^{-3}=\Big(\frac{4}{-3}\Big)^3$ $\Big[\text{Since}\Big(\frac{\text{a}}{\text{b}}\Big)^{-\text{n}}=\Big(\frac{\text{b}}{\text{a}}\Big)^\text{n}\Big]$
$=\frac{4^3}{(-3)^3}$
$=\frac{4\times4\times4}{(-3)\times(-3)\times(-3)}=\frac{64}{(-27)}$
$=\frac{64\times-1}{-27\times-1}=\frac{-64}{27}$

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MCQ 1211 Mark

$(-1)^8=$ value is -

A
$-1$
B
$0$
✓
$+1$
D
None of these

Answer

Correct option: C.

$+1$

$(-1)^{2n} = 1$
$\therefore (-1)^8= 1$

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MCQ 1221 Mark

Which of the following is not true?

A
$3^2 > 2^3$
B
$4^3 = 26$
✓
$3^3 = 9$
D
$2^5 > 5^2$

Answer

Correct option: C.

$3^3 = 9$

For option $(a)$,$ 3^2> 2^3$
$3 \times 3 > 2 \times 2 \times 2$
$9 > 8 $(true)
For option $(b)$, $4^3 = 2^6$
$(2^2)^3 = 2^6$
$2^6 = 2^6$ (true) [$\because(\text{a}^{\text{m}})^{\text{n}}=\text{a}^{\text{m}\times\text{n}}$]
For option $(c)$, $3^3 = 9$
$3 \times 3 \times 3 = 9$
$29\neq9$ (not true)
For option $(d)$, $2^5 > 5^2$
$2 \times 2 \times 2 \times 2 \times 2 > 5 \times 5$
$32 > 25$ (true)
Hence, option $(c)$ is not true.

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MCQ 1231 Mark

If $\sqrt{\sqrt{2500}+\sqrt{961}}=(\text{x})^2,$ then x equals.

A
$81$
✓
$3$
C
$6561$
D
$9$

Answer

Correct option: B.

$3$

$\sqrt{\sqrt{2500}+\sqrt{961}}=(\text{x})^2....$ Given
$\Rightarrow\sqrt{50+31}=(\text{x})^2$
$\Rightarrow\sqrt{81}=(\text{x})^2$
$\therefore\text{x}^2=9$
$\therefore\text{x}=3$

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MCQ 1241 Mark

Value of $\frac{10^{22}+10^{20}}{10^{20}}$ is:

A
$10$
B
$10^{42}$
✓
$101$
D
$10^{22}$

Answer

Correct option: C.

$101$

We can write the given expression as
$\frac{10^{22}}{10^{20}}+\frac{10^{20}}{10^{20}}=10^{22-20}+1$$\big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^\text{m-n}\text{m}>\text{n}\big]$
$= 10^2 + 1$
$= 10 \times 10 + 1$
$= 100 + 1 = 101$

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MCQ 1251 Mark

Exponential form of $\Big(\sqrt{\sqrt{\text{a}}\times\sqrt{\text{b}}}\Big)^2$ is:

A
$\text{ab}$
✓
$(\text{ab})^{\frac{1}{2}}$
C
$(\text{ab})^{\frac{1}{4}}$
D
$(\text{ab})^2$

Answer

Correct option: B.

$(\text{ab})^{\frac{1}{2}}$

Consider $\Big(\sqrt{\sqrt{\text{a}}\times\sqrt{\text{b}}}\Big)^2$ and simplify it as follows:
$\Big(\sqrt{\sqrt{\text{a}}\times\sqrt{\text{b}}}\Big)^2=(\sqrt{\text{a}}\times\sqrt{\text{b}})^{\frac{1}{2}\times2}$
$=\sqrt{\text{ab}}=(\text{ab})^{\frac{1}{2}}$
Hence, $\Big(\sqrt{\sqrt{\text{a}}\times\sqrt{\text{b}}}\Big)^2=(\text{ab})^{\frac{1}{2}}.$

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MCQ 1261 Mark

If $a = 3^{-3} - 3^3$ and $b = 3^3 - 3^{-3},$ then $\frac{\text{a}}{\text{b}}-\frac{\text{b}}{\text{a}}=$

✓
$0$
B
$1$
C
$-1$
D
$2$

Answer

Correct option: A.

$0$

Since,
$\text{a}=3^{-3}-3^3$
$=\frac{1}{3^3}-3^3$ $\Big(\text{As},\text{x}^{-\text{m}}=\frac{1}{\text{x}^{\text{m}}}\Big)$
$=\frac{1}{27}-\frac{27}{1}$
$=\frac{1}{27}-\frac{27\times27}{1\times27}$
$=\frac{1}{27}-\frac{729}{27}$
$=\frac{1-729}{27}$
$=\frac{-728}{27}$
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 $(a)$.

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MCQ 1271 Mark

Simplify : $\frac{2^{2009}-2^{2007}}{2^{2006}-2^{2008}}$

A
$-4$
✓
$-2$
C
$-1$
D
$2$

Answer

Correct option: B.

$-2$

$\frac{2^{2009}-2^{2007}}{2^{2006}-2^{2008}}=\frac{2^{2007}(2^2-1)}{-(2)^{2006}(2^2-1)}$
$=-2^{2007-2006}$
$=-2$

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MCQ 1281 Mark

If $x$ and $y$ are related by $x - y - 10 = 0$ and mode of $x$ is known to be $23$, then the mode of $y$ is:

A
$20$
✓
$13$
C
$3$
D
$23$

Answer

Correct option: B.

$13$

$x - y - 10 = 0$
$x = 23$
$23 - y - 10 = 0$
$y = 23 - 10$
$y = 13$

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MCQ 1291 Mark

The solution of the equation $\frac{2^3}{\log3\text{x}}=\frac{1}{64}$ is?

A
$3$
B
$\frac{1}{3}$
✓
$\frac{1}{\sqrt{(3)}}$
D
$\text{None of these}$

Answer

Correct option: C.

$\frac{1}{\sqrt{(3)}}$

$\frac{3}{\log_3\text{x}}=-6$
$\therefore64=2^6$
$\therefore\log_3\text{x}=-\frac{1}{2}$ or $\text{x}=3^{-\frac{1}{2}}=\frac{1}{\sqrt3}$

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MCQ 1301 Mark

Find the value of m for which: $100^m ÷ 100^4 = 100^8:$

A
$1$
B
$2$
C
$3$
✓
$12$

Answer

Correct option: D.

$12$

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MCQ 1311 Mark

If $2^m+ 2^{1 + m}= 24,$ then what is value of $m$?

A
$0$
B
$\frac{1}{3}$
✓
$3$
D
$6$

Answer

Correct option: C.

$3$

Given, $ 2m + 2{1 + m} = 24$
$\therefore, = 2^m+ 2 \times 2^m= 24$
$\Rightarrow 3 \times 2^m = 24$
$\Rightarrow 2^m = 8$
$\Rightarrow 2^m = 2^3$
$\Rightarrow m = 3$

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MCQ 1321 Mark

The number $2.35 \times 10^4$ in the usual form is written as:

A
$2.35 × 10^3$
✓
$23500$
C
$2350000$
D
$235 × 10^4$

Answer

Correct 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 $(b)$.

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MCQ 1331 Mark

$\frac{2^{\text{n}+4}-2(2^\text{n}}{2(2^{\text{n}+3})}+2^{-3}$ is equal to

A
$2^{\text{n}+1}$
B
$-2^{\text{n}+1}+\frac{1}{8}$
C
$\frac{9}{8}-2^\text{n}$
✓
$1$

Answer

Correct option: D.

$1$

$\frac{2^{\text{n}+4}-2(2^\text{n}}{2(2^{\text{n}+3})}+2^{-3}$
$=\frac{2^{\text{n}+4}-(2^{\text{n}+4})}{(2^{\text{n}4})}+2^{-3}$
$=1-2^{-3}+2^{-3}=1$

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MCQ 1341 Mark

If $3^n = 27,$ then $3^n - 2$ is:

✓
$3$
B
$\frac{1}{3}$
C
$\frac{1}{9}$
D
$9$

Answer

Correct option: A.

$3$

Given, $3^n = 27$
$\Rightarrow n = 3$
$\Rightarrow 3^n - 2 = 3^{3 - 2} = 3$

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MCQ 1351 Mark

The value of $(0.125)^{\frac{2}{3}}$ is:

A
$2.5$
✓
$0.25$
C
$0.025$
D
$0.0025$

Answer

Correct option: B.

$0.25$

$(0.125)^{\frac{2}{3}}$
$0.125$ could be written as $(0.5)^3$
So, $(0.5^3)^{\frac{2}{3}}$
$=(0.5)^2$
$=0.25$

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MCQ 1361 Mark

$\frac{(144)^{\frac{1}{2}}+(256)^{\frac{1}{2}}}{3^2-2}=$

A
$8$
✓
$4$
C
$-4$
D
$-8$

Answer

Correct option: B.

$4$

$\frac{(144)^{\frac{1}{2}}+(256)^{\frac{1}{2}}}{3^2-2}$
$=\frac{(12^2)^{\frac{1}{2}}+(16^2)^{\frac{1}{2}}}{9-2}$
$=\frac{12^{2\times\frac{1}{2}}+16^{2\times\frac{1}{2}}}{7}$ $[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=\frac{12+16}{7}$
$=\frac{28}{7}$
$=4$
Hence, the correct alternative is option $(b)$.

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MCQ 1371 Mark

Simplify the following using law of exponents. $9^{2}$\times $9^{18}$\times $9^{10}$ :

✓
$9^{30}$
B
$9^{25}$
C
$9^{15}$
D
$9^{10}$

Answer

Correct option: A.

$9^{30}$

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MCQ 1381 Mark

The value of $(1^2+ 2^2+ 3^2)^2$ is:

A
$236$
✓
$196$
C
$189$
D
$97$

Answer

Correct option: B.

$196$

$(1^2 + 2^2 +3^2)^2 = (1 + 4 + 9)^2$
$= 14^2$
$= 196$

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MCQ 1391 Mark

$(-1)^{301}+(-1)^{302}+(-1)^{303}+\dots+(-1)^{400}$

A
$1$
B
$101$
C
$100$
✓
$0$

Answer

Correct option: D.

$0$

Since,
$(-1)^{301}+(-1)^{302}+(-1)^{303}+\dots+(-1)^{400}$
$=(-1)+(1)+(-1)+\dots+(1)$ $\Big[\text{As, }(-1)^{\text{odd}}=-1\text{ and }(-1)^{\text{even}}=1\Big]$
$=-50+50$ $[\text{As, there are} 50(-1)'\text{s and 50 (1)}'\text{s}]$
$=0$
Hence, the correct alternative is option is $(d)$.

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MCQ 1401 Mark

In simplified form, $((3)^\circ + (5)^\circ)^\circ$ is equal to:

A
$2$
✓
$1$
C
$0$
D
$8$

Answer

Correct option: B.

$1$

$((3)^\circ + (5)^\circ)^\circ$ $= (1 + 1)^\circ$
$= (2)^\circ =$ $1$

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MCQ 1411 Mark

By what number should $3^{-4}$ be multiplied, so that the product is $729?$

A
$3^8$
B
$3^9$
✓
$3^{10}$
D
$3^{12}$

Answer

Correct option: C.

$3^{10}$

we know,
$729 = 3^6$
also,
$a^m ∗ a^n = a^{m + n}$
let the number be $x$
So, $ 3^{-4+x} = 3^6$
$x - 4 = 6$
$x = 10$
$\Rightarrow 3^{10}$

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MCQ 1421 Mark

$(10 + 20 + 30)$ is equal to:

A
$0$
B
$1$
✓
$3$
D
$6$

Answer

Correct option: C.

$3$

As we know, $a = 1$
$1^\circ + 2^\circ + 3^\circ $
$= 1 + 1 + 1$
$= 3$

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MCQ 1431 Mark

State true or false: $\bigg[\Big(\frac{3}{7}\Big)\bigg]^3=\Big(\frac{3}{7}\Big)^5$

A
True
✓
False
C
Ambiguous
D
Data insufficient

Answer

Correct option: B.

False

$\bigg[\Big(\frac{3}{7}\Big)\bigg]^3=\Big(\frac{3}{7}\Big)^{2\times3}=\Big(\frac{3}{7}\Big)^6$

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MCQ 1441 Mark

which number is equivalent to $\frac{3^4}{3^2}$

A
$2$
✓
$9$
C
$81$
D
$729$

Answer

Correct option: B.

$9$

Number equivalent to $\frac{3^4}{3^2}\frac{3^4}{3^2}=(3)^{4-2}=3^2=3\times3=9.$

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MCQ 1451 Mark

The value of $[10^{150} ÷ 10^{146}]$ is:

A
$1000$
✓
$10000$
C
$100000$
D
$105$

Answer

Correct option: B.

$10000$

$(10)^{150}\div(10)^{146}$
$=\frac{(10)^{150}}{(10)^{146}}$
$=10^{(150-146)}=10^4$
$=10000$

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MCQ 1461 Mark

If $2^{1998}-2^{1997}-2^{1995}+2^{1995}=\text{K.2}^{1995}$ then the value of $K$ is:

A
$1$
B
$2$
✓
$3$
D
$4$

Answer

Correct option: C.

$3$

Given,
$2^{1998}-2^{1997}-2^{1995}+2^{1995}=\text{K.2}^{1995}$
$\Rightarrow2^{1995+3}-2^{1995+2}-2^{1995+1}+2^{1995}\times1=\text{K.2}^{1995}$
$\Rightarrow2^{1995}[2^{3}-2^{2}-2^{1}+1]=\text{K.2}^{1995}$ $\big[\because\text{a}^{\text{m+n}}=\text{a}^{\text{m}}\times\text{a}^{\text{n}}\big]$
$\Rightarrow2^{1995}[8-4-2+1]=\text{K.2}^{1995}$
$\Rightarrow3=\frac{\text{K.2}^{1995}}{2^{1995}}$
$3=\text{K}\text{ or }\text{K}=3$
So, the value of $k$ is $3$

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MCQ 1471 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
$\frac{-4}{9}$

Answer

Correct option: C.

$\frac{4}{9}$

$\because \Big(\frac{-2}{3}\Big)^2=\frac{-2}{3}\times\frac{-2}{3}$
$=\frac{4}{9}$

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MCQ 1481 Mark

The standard form of the number $12345$ is:

A
$1234.5 × 10^1$
B
$123.45 × 10^2$
C
$12.345 × 10^3$
✓
$1.2345 × 10^4$

Answer

Correct option: D.

$1.2345 × 10^4$

A number in standard form is written as $a × 10^k,$ where a is a terminating decimal such that $1 ≤ a ≤ 10$ and k is any integer.
So, $12345$
$= 1.2345 \times 10^4$

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MCQ 1491 Mark

The exponential form of $3 × 3 × 3 × 3$ is:

A
$4^3$
B
$3^3$
✓
$3^4$
D
$3^5$

Answer

Correct option: C.

$3^4$

$= 3^1 \times 3^1 \times 3^1 \times 3^1 = 3^{1 + 1 + 1 + 1} (a^x\times a^y= a^{x + y}) = 3^4$

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MCQ 1501 Mark

Value of $\frac{2^{100}}{2}$ is-

A
$1$
B
$50^{100}$
C
$2^{50}$
✓
$2^{99}$

Answer

Correct option: D.

$2^{99}$

$\Rightarrow\frac{2^\text{n}}{2}=2^{\text{n}-1}$
$\Rightarrow\frac{2^{100}}{2}=2^{100-1}=2^{99}$

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Maths M.C.Q. [1 Marks Each]