r = k [A] for a reaction, 50 % of A is decomposed in 120 minutes....

$r = k\left[A\right]$

So, order of reaction $= 1$

${t}_{1/2} = 120min$

$k = \dfrac{0 . 693}{{t}_{1/2}}$

For $90$ completion of reaction,

$⇒ k = \dfrac{2 .303}{t}\log \left(\dfrac{a}{a - x}\right)a = 100, ( a - x ) = 10$

By putting the values in above equation,

$⇒ \dfrac{0 .693}{{t}_{1/2}} = \dfrac{2 .303}{t}log\dfrac{100}{10}$

$∴ t = 399min$