$5\mathrm{.}9,46\mathrm{.}5kg(102\mathrm{.}3lb)$ of fine aggregate is mixed with $72\mathrm{.}3kg(159\mathrm{.}1lb)$ of coarse

aggregate. The fine aggregate has a molsture content of $2\mathrm{.}0\%$ and absorption

of $3\mathrm{.}4\%,$ whereas the coarse aggregate has a molsture content of $1\mathrm{.}3\%$ and

TABLE P $5\mathrm{.}7$

absorption of $3\mathrm{.}8\%.$ What is the amount of water required to increase the

molsture contents of both fine and coarse aggregates to reach absorption$?$

Why is it important to determine the amount of water required to increase the

molsture content of aggregate to reach absorption$?$ Use SI units