Question

In the group $(Q^+ ,\star )$ of positive rational numbers w.r.t. the binary operation $\star$ defined by $a^{\star} b=\frac {ab}{3} , \forall a,b \in Q^{+}$ the solution of the equation $5 ^{\star} x=4^{-1}$ in $Q^+$ is

• A. $\frac {27}{20}$
• B. $\frac {20}{27}$
• C. $\frac {1}{20}$
• D. $20$

Answer 1

Answer: (A)

$\frac {27}{20}$

Answer 2

Let $e$ is the identity element of $Q^{+}$
then, $e^{*} a=a, \forall a \in Q^{+}$
$\frac{e a}{3} =a$
$left[\because a^{*} b=\frac{a b}{3}]$
$\therefore\, e=3$
Again, let $b$ be the inverse of $a$, then $a^{*} b =e$
$\frac{ab}{3}=3$
$[\because e=4]$
$b=\frac{9}{a}$
$\therefore$ Inverse of 4 is $\frac{9}{4}$.
From the given condition, $5^{*}x=4^{-1}$
$\Rightarrow\, \frac{5 x}{3}=\frac{9}{4}$
$\Rightarrow \,x=\frac{27}{20}$