Maths MCQ

$1\div\frac{1}{3}$
$=1\times3$ $\Big(\text{x}\div\text{y}=\text{x}\times\frac{1}{\text{y}}\Big)$
$=3$
Hence, the correct answer is option $(b).$

Sum of the given number and the required number $= 2$
Given number $=\frac{-7}{9}$
$\therefore$ Required number $=$ Sum of the numbers $-$ Given number
$=2-\Big(\frac{-7}{9}\Big)$
$=\frac{2}{1}+\frac{7}{9}$
$=\frac{2\times9+7\times1}{9}$
$=\frac{18+7}{9}$
$=\frac{25}{9}$
Hence, the correct answer is option $(d).$

$​​\frac{\text{x}}{2}+\frac{1}{3}=1$
$\Rightarrow\frac{\text{x}}{2}=1-\frac{1}{3}$
$\Rightarrow\frac{\text{x}}{2}=\frac{3\times1-1}{3}$
$\Rightarrow\frac{\text{x}}{2}=\frac{3-1}{3}$
$\Rightarrow\frac{\text{x}}{2}=\frac{2}{3}$
$\Rightarrow\frac{2\text{x}}{2}=\frac{2\times2}{3} ($Multiplying both sides by $2)$
$\Rightarrow\text{x}=\frac{4}{3}$
Hence, the correct answer is option $(b).$

$3\frac{1}{7}\times1\frac{5}{6}\times1\frac{2}{5}\times1\frac{1}{11}$
$=\frac{22}{7}\times\frac{11}{6}\times\frac{7}5{}\times\frac{12}{11}$
$=\frac{22\times11\times7\times2}{7\times6\times5\times11}$ $\Big(\frac{\text{a}}{\text{b}}\times\frac{\text{c}}{\text{d}}=\frac{\text{a}\times\text{c}}{\text{b}\times\text{d}}\Big)$
$=\frac{44}{5}$
$=\frac{8\times5+4}{5}$
$=8\frac{4}{5}$
Hence, the correct answer is option $(c).$

For every non$-$zero rational number $\frac{\text{a}}{\text{b}}$ there exists a rational number $\frac{\text{b}}{\text{a}}$ such that:
$\frac{\text{a}}{\text{b}}\times\frac{\text{b}}{\text{a}}=1$
Here, the rational number $\frac{\text{b}}{\text{a}}$ is called the multiplicative inverse or reciprocal of $\frac{\text{a}}{\text{b}}$
Thus, if the product of two non$-$zero rational numbers is $1$, then they are multiplicative inverse or reciprocal of each other.
Hence, the correct answer is option $(d).$

$(-2)\div\Big(​​-\frac{5}{3}\Big)$
$=-2\times\Big(-\frac{3}{5}\Big)$ $\text{x}\div\text{y}=\text{x}\times\frac{1}{\text{y}}$
$=\frac{-2}{1}\times\frac{(-3)}{5}$
$=\frac{-2\times(-3)}{1\times5}$ $\Big(\frac{\text{a}}{\text{b}}\times\frac{\text{c}}{\text{d}}=\frac{\text{a}\times\text{c}}{\text{b}\times\text{d}}\Big)$
$=\frac{6}{5}$
Hence, the correct answer is option $(c).$

Difference of the given number and required number $=\frac{4}{5}$
Given number $=\frac{-2}{3}$
$\therefore$ Required number $=$ Given number $-$ Difference of the numbers
$=\frac{-2}{3}-\frac{4}{5}$
$=\frac{-2}{3}+\Big(\frac{-4}{5}\Big)$
$=\frac{-2\times5+(-4)\times3}{15}$
$=\frac{-10+(-12)}{15}$
$=\frac{-22}{15}$
Hence, the correct answer is option $(b).$

We know that the multiplicative inverse of the rational number $\frac{\text{a}}{\text{b}}\text{ is }\frac{\text{b}}{\text{a}}$
$\therefore$ Multiplicative inverse of $\frac{4}{-5}=\frac{-5}{4}=\frac{5}{-4}$
Hence, the correct answer is option $(c).$

We know that $0$ divided by any non$-$zero rational number is always $0.$
$\therefore0\div\frac{3}{5}=0$
$\Big(0\div\frac{\text{a}}{\text{b}}=0\Big)$
Hence, the correct answer is option $(a).$

We know that the reciprocal of the rational number $\frac{\text{a}}{\text{b}}\text{ is }\Big(\frac{\text{a}}{\text{b}}\Big)^{-1}=\frac{\text{b}}{\text{a}}$
$\therefore$ Reciprocal of $\frac{-3}{4}$
$=\Big(\frac{-3}{4}\Big)^{-1}$
$=\frac{4}{-3}$
$=\frac{4\times(-1)}{-3\times(-1)}$
$=\frac{-4}{3}$
Hence, the correct answer is option $(c).$

$\frac{5}{4}-\frac{7}{6}-\frac{-2}{3}$
$=\frac{5}{4}+\Big(\frac{-7}{6}\Big)+\frac{2}{3}$ $\Big[-\Big(\frac{-2}{3}\Big)=\frac{2}{3}\Big]$
$=\frac{5\times3+(-7)\times2+2\times4}{12} (\text{LCM}$ of $3, 4$ and $6 = 12)$
$=\frac{15-14+8}{12}$
$=\frac{9}{12}$
$=\frac{9\div3}{12\div3} ($Dividing numerator and denominator by $3)$
$=\frac{3}{4}$
Hence, the correct answer is option $(a).$