Answer
The first stage burns for $90$ seconds and the second stage burns for $62$ seconds.
Work Step by Step
Let $t$ be the time that is needed for the second stage of the rocket.
It is given that the first stage needs $28$ more seconds than the second stage.
Therefore, the time needed for the first stage would be $t+28$.
Since the total burning time is $152$ seconds, algebraically it can be expressed as : $(28+t)+t=152$
Combining like terms yield
$28+t+t=152$
$28+2t=152$
Subtract $28$ from each side to obtain:
$28+2t-28=152-28$
$2t=124$
Divide both sides by $2$:
$\dfrac{2t}{2}=\dfrac{124}{2}$
$t=62$
Thus, the second stage burns for $62$ seconds while the first stage burns for $62+28=90$ seconds.
