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

$512^\frac{2}{9}=(2^9)^{-\frac{2}{9}}=2^{-2}=\frac{1}{4}$

$\bigg\{6^{-1}+\Big(\frac{3}{2}\Big)^{-1}\bigg\}=\bigg[\Big(\frac{1}{6}\Big)+\Big(\frac{2}{3}\Big)\bigg]^{-1}$
$=\Big(\frac{1+4}{6}\Big)^{-1}=\Big(\frac{5}{6}\Big)^{-1}$
$=\frac{6}{5}$

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

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

$\frac{1}{2^3}+\frac{1}{3^2}$
$=\frac{1}{8}+\frac{1}{9}$
$=\frac{9+8}{72}$
$=\frac{17}{72}$

$−2 × −2 × −2 = (−2)1 + 1 + 1 = −23$

Let us put $n = 1$ in equation $3^{2n + 1} + 2^{2n + 1}$
Therefore, $3^3 + 2^3 = 27 + 8 = 35$
Therefore, digit in the unit place is $5.$

$ac = b^2$
$\Rightarrow < br > < br > 2^x. 8^z = (4^y)^2$
$\Rightarrow < br > < br > 2^x. (2^3)^z= ((2^2)^y)^2$
$\Rightarrow < br > < br > 2^x. 2^{3z}= 2^{4y}$
$\Rightarrow 2^x + 3^z = 2^4y$
$\Rightarrow x + 3z = 4y$

Given, $4^{2b - 3}= 4^{1 - b}$As, bases are equal powers are equal
$⟹ 2b - 3 = 1 - b$
$⟹ 3b = 4$
$\Rightarrow\text{b}=\frac{4}{3}$

$\text { Given, } 4^{ x } \times n ^2=4^{ x +1} $
$\Rightarrow 4^{ x } \times n ^2=4^{ x } . $
$4^1=4^{ x } .$
$4$ Since $x$ is a positive integer, we can divide both sides by $4^x$, and get $n^2=4$
$\Rightarrow n = \pm$
$2$ Slnce $n$ is also a positive integer, $n=2$

We know that, if $a$, $b$ and $m$ are rational numbers, then
$\text{a}^{\text{m}}\times\text{b}^{\text{m}}=\text{ab}^{\text{m}}$
So, $\big(\frac{2}{3}\big)^{3}\times\big(\frac{5}{7}\big)^{3}=\big(\frac{2}{3}\times\frac{5}{7}\big)^{3}$

$\Big(\frac{5}{3}\Big)^{-5}\times\Big(\frac{5}{3}\Big)^{11}=\Big(\frac{5}{3}\Big)^{8\text{x}}$
$\Rightarrow\Big(\frac{5}{3}\Big)^{-5+11}=\Big(\frac{5}{3}\Big)^{8\text{x}}$ $\Big(\text{As,}\text{x}^{\text{m}\times\text{n}}=\text{x}^{\text{m+n}}\Big)$
$\Rightarrow\Big(\frac{5}{3}\Big)^{6}=\Big(\frac{5}{3}\Big)^{8\text{x}}$
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 $(c)$.

$(\text{a})^0=1$
$\therefore \Big(\frac{3}{4}\Big)^0=1$

$\because \Big(\frac{5}{3}\Big)^{-5}\times \Big(\frac{5}{3}\Big)^{11}=\Big(\frac{5}{3}\Big)^{8\text{x}},$
$\Rightarrow \Big(\frac{5}{3}\big)^{-5+11}=\big(\frac{5}{3}\Big)^{8\text{x}}$
$\Rightarrow \Big(\frac{5}{3}\Big)^6=\Big(\frac{5}{3}\Big)^{8\text{x}}$
Comparing $8x = 6$
$\Rightarrow \text{x}=\frac{6}{8}=\frac{3}{4}$

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

$3^x = 6561$
$\Rightarrow 3^x = 3^8$
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 $(b)$.

$10^8$means $10$ is to be multiplied $8$ times with itself.
Therefore, it means that when we expand $10^8$, there will be $8$ zeros after $1.$

$\Big(\frac{\text{a}}{\text{b}}\Big)^{2\text{x}-1}=\Big(\frac{\text{b}}{\text{a}}\Big)^{\text{x}-2}$
$\Rightarrow\Big(\frac{\text{a}}{\text{b}}\Big)^{2\text{x}-1}=\Big(\frac{\text{a}}{\text{b}}\Big)^{2-\text{x}}$
$\therefore2\text{x}-1=2-\text{x}$
or $\text{x}=1$

$4 \times 4^{10} = 4^{1+10} = 4^{11}$

As we can see that $p$ is raised to power $0.$
It means that there is no term of $p.$
So, $p^0 = 1$

$7 - 7 = 0,$
anything raised to the power of $0$ equals $1$

$2 ∗ 2 ∗ 2 ∗ 2 = 2^4$

$(64)^\frac{-2}{3}\times\Big(\frac{1}{4}\Big)^{-3}$
$=(4^3)^{-\frac{2}{3}}\times\Big(\frac{1}{4}\Big)^{-3}$
$\Rightarrow4^{-2}\times\frac{1}{4^{-3}}$
$\Rightarrow4^{-2+3}=4^1=4$

$\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 $(a)$.

$\because \Big(\frac{-2}{5}\Big)^7\div\Big(\frac{-2}{5}\Big)^5$
$=\Big(\frac{-2}{5}\Big)^{7-5}=\Big(\frac{-2}{5}\Big)^2$
$=\frac{-2}{5}\times\frac{-2}{5}=\frac{4}{25}$

$\because (5{-1}\times3^{-1})^{-1}=\Big(\frac{1}{5}\times\frac{1}{3}\Big)^{-1}$
$\Big(\frac{1}{15}\Big)^{-1}=15$

Given, $\text{5}^{5}\div\text{y}^{5}=\frac{\text{x}^{5}}{\text{y}^{5}}$
As we know, $\frac{\text{p}^{\text{n}}}{\text{q}^{\text{n}}}=\big(\frac{\text{p}}{\text{q}}\big)^\text{n}$
Thus, $\frac{\text{x}^{5}}{\text{y}^{5}}=\big(\frac{\text{x}^{5}}{\text{y}^{5}}\big)^{5}=(\text{x}+\text{y})^{5}$

we know,
$a^x ∗ a^y ∗ a^z = a^{x + y + z}$
$\therefore$ $2^{1 + 1 + 1}$
$= 2^3$
$= 8$

$(8^4+8^2)^{\frac{1}{2}}$
$=(8^{2+2}+8^2)^{\frac{1}{2}}$
$=(8^2\times8^2+8^2)^{\frac{1}{2}}$ $(\text{As},\text{x}^{\text{m+n}}=\text{x}^{\text{m}}\times\text{x}^{\text{n}})$
$=[8^2\times(8^2+1)]^{\frac{1}{2}}$ $[\text{As}, \text{ab+ac}=\text{a}\times(\text{a+c})]$
$=(8^2)^{\frac{1}{2}}\times(8^2+1)^{\frac{1}{2}}$ $[\text{As, }(\text{ab})^{\text{m}}=\text{a}^{\text{m}}\times\text{b}^{\text{m}}]$
$=8^{2\times\frac{1}{2}}\times(64+1)^{\frac{1}{2}}$
$=8\times65^{\frac{1}{2}}$
$=8\sqrt{65}$
Hence, the correct alternative is option $(d)$.