M.C.Q. [1 Marks Each]

MCQ 11 Mark

Cube of $\big(\frac{-1}{4}\big)$ is:

A
$\frac{-1}{12}$
B
$\frac{1}{16}$
C
$\frac{-1}{64}$
✓
$\frac{1}{64}$

Answer

Correct option: D.

$\frac{1}{64}$

Cube of $\big(\frac{-1}{4}\big)$ is $\big(\frac{-1}{4}\big)^{3}$
So, $\big(\frac{-1}{4}\big)^{3}=\big(\frac{-1}{4}\big)\times\big(\frac{-1}{4}\big)\times\big(\frac{-1}{4}\big)$
$=\frac{(-1)\times(-1)\times(-1)}{4\times4\times4}$
$=\frac{-1}{64}$

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MCQ 21 Mark

For a non-zero rational number $x, x^8 \div x^2$ is equal to:

A
$x^4$
✓
$x^6$
C
$x^{10}$
D
$x^{16}$

Answer

Correct option: B.

$x^6$

We know that, if ' $a$ ' is a rational number, $m$ and $n$ are natural numbers such that $m>n$, then.
$a^m \div a^n$
$=a^{m-n}$
So, $x \div x^2=\frac{x^8}{x^2}$
$=x^{8-2}=x^6$

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MCQ 31 Mark

Which of the follwing is equal to $1$?

A
$2^\circ + 3^\circ + 4^\circ $
✓
$2^\circ \times 3^\circ \times 4^\circ $
C
$(3^\circ - 2^\circ ) \times 4^\circ $
D
$(3^\circ - 2^\circ ) \times (3^\circ +2^\circ)$

Answer

Correct option: B.

$2^\circ \times 3^\circ \times 4^\circ $

Let us solve all the expressions.
For option $(a),$
$2^\circ + 3^\circ + 4^\circ $
$= 1 + 1 + 1$ [$\therefore$ $a^\circ = 1$]
$= 3$
For option $(b),$
$2^\circ \times 3^\circ \times 4^\circ $
$= 1 \times 1 \times 1$ [$\therefore$ $a^\circ = 1$]
$= 1$
Hence, option $(b)$ is the answer.

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MCQ 41 Mark

In standard form, the number $829030000$ is written as $K \times 108$ where $K$ is equal to:

A
$82903$
B
$829.03$
C
$82.903$
✓
$8.2903$

Answer

Correct option: D.

$8.2903$

We have, A number in a standard form is written as $\mathrm{K} \times 10^8$, then $K$ will be a terminating decimal such that $1 \leq \mathrm{K} \leq 10$
So, there is only one option, where $K$
$=8.2903<10$

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MCQ 51 Mark

Which of the following is not equal to $\big(\frac{-5}{4}\big)^{4}$?

A
$\frac{(-5)^{4}}{4^{4}}$
B
$\frac{-5^{4}}{(4)^{4}}$
✓
$-\frac{5^{4}}{4^{4}}$
D
$\big(-\frac{5}{4}\big)\times\big(-\frac{5}{4}\big)\times\big(-\frac{5}{4}\big)\times\big(-\frac{5}{4}\big)$

Answer

Correct option: C.

$-\frac{5^{4}}{4^{4}}$

We know that, $\big(\frac{\text{p}}{\text{q}}\big)^{\text{m}}=\frac{\text{p}^{\text{m}}}{\text{q}^{\text{m}}}$
So, $\big(\frac{-5}{4}\big)^{4}=\frac{(-5)^{4}}{(4)^{4}}$
or $\big(\frac{-5}{4}\big)^{4}=\frac{(5)^{4}}{(-4)^{4}}$
or $\big(\frac{-5}{4}\big)^{4}=\big(\frac{-5}{4}\big)\times\big(\frac{-5}{4}\big)\times\big(\frac{-5}{4}\big)$
Hence, option $(c)$ is not equal to $1$.

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MCQ 61 Mark

In standard form, the number $72105.4$ is written as $7.21054 × 10^n$ where n is equal to:

A
$2$
B
$3$
✓
$4$
D
$5$

Answer

Correct option: C.

$4$

We know that, if the given number is greater than or equal to $10$, then the power of $10$ (i.e. n) is a positive integer equal to the number of places the decimal point has been shifted.
Hence, $72105.4$
$= 7.21054 \times 10^4$

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MCQ 71 Mark

$\left[(-3)^2\right]^3$ is equal to:

A
$(-3)^8$
✓
$(-3)^6$
C
$(-3)^5$
D
$(-3)^{23}$

Answer

Correct option: B.

$(-3)^6$

We know that, if ' $a$ ' is a rational number, $m$ and $n$ are natural numbers, then.
$\left(a^m\right)^n=a^{m \times n}$
$\text { So, }\left[(-3)^2\right]^3$
$=(-3)^{2 \times 3}$
$=(-3)^6$

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MCQ 81 Mark

Square of $\big(\frac{-2}{3}\big)$ is:

A
$\frac{-2}{3}$
B
$\frac{2}{3}$
C
$\frac{-4}{9}$
✓
$\frac{4}{9}$

Answer

Correct option: D.

$\frac{4}{9}$

Square of $\frac{-2}{3}\text{ is }\big(\frac{-2}{3}\big)^{2}$
So, $\big(\frac{-2}{3}^{2}\big)=\big(\frac{-2}{3}\big)\times\big(\frac{-2}{3}\big)$
$=\frac{4}{9}$
[$\because$ multiplication of two rational number with same sign is always positive]

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MCQ 91 Mark

$a^m \times a^n$ is equal to:

A
$\left(a^2\right)^{m n}$
B
$a^{m-n}$
✓
$a^{m+n}$
D
$a^{m n}$

Answer

Correct option: C.

$a^{m+n}$

We know that, if $‘a’$ is a rational number, $m$ and $n$ are natural numbers, then.
$a^m \times a^n=a^{m+n}$

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MCQ 101 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 111 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 $x=x^2 \times x^3=x^{2+3}\left[\therefore a^m \times a^n=a^{m+n}\right]$
i.e. fifth power of $x$

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MCQ 121 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 $(a)$, $3^2>2^3$
$3 \times 3>2 \times 2 \times 2$
$9>8 \text { (true) }$
For option $(b)$, $4^3=2^6$
$\left(2^2\right)^3=2^6$
$2^6=2^6 \text { (true) }\left[\because\left(a^{\mathrm{m}}\right)^{\mathrm{n}}=a^{\mathrm{m} \times \mathrm{n}}\right]$
For option $(c)$, $3^3=9$
$3 \times 3 \times 3=9$
$29 \neq 9 \text { (not true) }$
For option $(d)$, $2^5>5^2$
$2 \times 2 \times 2 \times 2 \times 2>5 \times 5$
$32>25 \text { (true) }$
Hence, option $(c)$ is not true.

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MCQ 131 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 141 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 151 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 161 Mark

The standard form of the number $12345$ is:

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

Answer

Correct option: D.

$1.2345 \times 10^4$

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

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MCQ 171 Mark

Which of the following is not equal to $1$?

A
$\frac{2^{3}\times3^{3}}{4\times18}$
B
$\big[(-2)^{3}\times(-2)^{4}\big]\div(27)^{7}$
C
$\frac{2^{0}\times5^{3}}{5\times25}$
✓
$\frac{2^{4}}{(7^{0}+3^{0})}$

Answer

Correct option: D.

$\frac{2^{4}}{(7^{0}+3^{0})}$

Let option (a) $\frac{2^{3}\times3^{3}}{4\times18}=\frac{2\times2\times2\times3\times3}{4\times18}$
$=\frac{4\times18}{4\times18}=1$
For option (b), $[(-2)^{3}\times(-2)^{4}]\div(-2)^{7}=\frac{(-2)^{3}\times(-2)^{}4}{(-2)^{7}}$
$=\frac{(-2)^{3+4}}{(-2)^{7}}$ $\big[\because\text{a}^{\text{m}}\times\text{a}^{\text{n}}=\text{a}^{\text{m+n}}\big]$
$=\frac{(-2)^{7}}{(-2)^{7}}=1$ $\big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{n}}=\text{a}^{\text{m-n}}\big]$
For option (c), $=\frac{3^{0}\times5^{3}}{5\times25}=\frac{1\times5\times5\times5}{5\times25}$
$=\frac{5\times25}{5\times25}=1$ $\big[\because\text{a}^{0}=1\big]$
For option (d), $\frac{2^{4}}{(7^{0}+3^{0})^{3}}=\frac{2^{4}}{(1+1)^{3}}$ $\big[\because\text{a}^{0}=1\big]$
$=\frac{2^{4}}{2^{3}}=2^{4-3}$ $\big[\because\frac{\text{a}^{m}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}},\text{m>n}\big]$
$=-2^{1}=2$
Hence, option $(d)$ is not equal to $1$

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MCQ 181 Mark

In standard form $72$ crore is written as:

A
$72 \times 10^7$
B
$72 \times 10^8$
✓
$7.2 \times 10^8$
D
$7.2 \times 10^7$

Answer

Correct option: C.

$7.2 \times 10^8$

We know that, A number in standard form is written as $a \times 10^{\mathrm{k}}$, where a is the terminating decimal such that $1 \leq \mathrm{a} \leq$ $10$ and $k$ is any integer.
$\text { So, } 72 \text { crore }$
$=720000000$
$=7.2 \times 10^8$
Note Here, power of $10 ($i.e. $k)$ is a positive integer equal to the number of places the decimal point has been shifted.

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MCQ 191 Mark

Which power of $8$ is equal to $2^6$?

A
$3$
✓
$2$
C
$1$
D
$4$

Answer

Correct option: B.

$2$

Let power of $8$ be $x$.
According to the question,
$8^{\text{x}=2^{6}}$
$\Rightarrow(2)^{3\text{x}}=2^{6}$ $\big[\because8=2\times2\times2=2^{3}\big]$
$\Rightarrow2^{3\text{x}}=2^{6}$ $\big[\because(\text{a}^{\text{m}})^{\text{n}}=\text{a}^{\text{m}\times\text{n}}\big]$
Since, base equal, then by eqyating their exponents, we get
$3\text{x}=6$
$\Rightarrow\frac{3\text{x}}{3}=\frac{6}{3}$ [dividing both sides by $3$]
$\Rightarrow \text{x}=2$
Hence, the power of $8$ is $2$, which is equal to $2^6$

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

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