Answer
Mass of ice = 122,000 T, Mass of rock = 16000 T
Work Step by Step
Please see the attached image first.
We can write,
The iceberg's weight is $W_{iceb}=m_{iceb}g=\rho_{iceb}V_{iceb}g$
The weight of the displaced water is $W_{water}=m_{water}g=\rho_{water}V_{sub}g$
By Archimedes' principle, $W_{water}$ is equal in magnitude to the buoyancy force, which balances gravity when the iceberg is in equilibrium. So,
$\rho_{water}V_{sub}g=\rho_{iceb}V_{iceb}g$
$\rho_{water}\frac{V_{sub}}{V_{iceb}}=\rho_{iceb}$ ; Let's plug known values into this equation.
$1043\space kg/m^{3}\times\frac{95.5}{100}=\rho_{iceb}$
$996\space kg/m^{3}=\rho_{iceb}$
We know, $\rho=\frac{m}{V}\rightarrow V=\frac{m}{\rho}$
$\rho_{iceb}=\frac{138\times10^{6}kg}{V_{iceb}}=>V_{iceb}=\frac{138\times10^{6}kg}{996\space kg/m^{3}}=138554.2\space m^{3}$
We can write, $V_{iceb}=V_{ice}+V_{rock}$
$138554.2\space m^{3}=\frac{m_{ice}}{917kg/m^{3}}+\frac{m_{rock}}{2750kg/m^{3}}$
$138554.2\times917\times2750\space kg=2750m_{ice}+917m_{rock}$
$3.49\times10^{11}kg=2750m_{ice}+917m_{rock}-(1)$
Also, we can write,
$138\times10^{6}kg=m_{ice}+m_{rock}-(2)$
(1)=>(2),
$138\times10^{6}kg=\frac{(3.5\times10^{11}-917m_{rock})}{2750}+m_{rock}$
$138\times10^{6}\times2750kg=3.5\times10^{11}+1833m_{rock}$
$16\times10^{6}kg=m_{rock}$
$16000\space T=m_{rock}$
$m_{ice}=138000\space T-16000\space T=122000\space T$
