Answer
The fraction of the body's volume that is above the quicksand surface is $~~0.41$
Work Step by Step
The weight of the displaced quicksand is equal to the weight of the body.
Then the mass of the displaced quicksand is equal to the mass of the body.
Let $V_b$ be the volume of the body.
We can find an expression for $V_q$, the volume of displaced quicksand:
$\rho_q~V_q = \rho_b~V_b$
$V_q = \frac{\rho_b~V_b}{\rho_q}$
$V_q = \frac{0.95~\rho_w~V_b}{1.6~\rho_w}$
$V_q = 0.59~V_b$
Therefore, the fraction of the body's volume that is above the quicksand surface is $~~0.41$
