Once you have the volume of the full cylinder, how do you adjust that volume to account for the drum being only $80\%$ full?

After finding the liquid volume in cubic meters, use the given conversion to milliliters, then use the density to get grams, and finally convert grams to pounds with the given relationship.

At each step, write down the units (m$^3$, mL, g, lb) to make sure you are multiplying and dividing in a way that cancels units correctly and ends in pounds.

In Desmos, type 3.14*(0.45^2)*1.2 to get the full cylinder volume in cubic meters. Then multiply that result by 0.8 (for 80% full). You can type it all at once as 0.8*3.14*(0.45^2)*1.2 and note the output (in m^3).

Take your previous expression and multiply by 1000000*0.85 to convert from cubic meters to milliliters and then to grams (since $1\text{ m}^3 = 1{,}000{,}000\text{ mL}$ and the density is $0.85\text{ g/mL}$). For example: 0.8*3.14*(0.45^2)*1.2*1000000*0.85.

Now divide that entire expression by 453.6 to convert grams to pounds: 0.8*3.14*(0.45^2)*1.2*1000000*0.85/453.6. The resulting value is the approximate number of pounds; compare this number to the answer choices and pick the closest one.

Use the volume formula for a cylinder: $V = \pi r^2 h$.

The interior diameter is $0.9\text{ m}$, so the radius is half of that:

• $r = 0.9/2 = 0.45\text{ m}$
• $h = 1.2\text{ m}$

Now compute the full interior volume:

• $V_{\text{full}} = 3.14 \times (0.45)^2 \times 1.2$.

Calculate $(0.45)^2 = 0.2025$, so

• $V_{\text{full}} \approx 3.14 \times 0.2025 \times 1.2 \approx 0.763\text{ m}^3$ (keeping a few decimal places).

The drum is only $80\%$ full, so the liquid volume is $0.8$ times the full volume.

• $V_{\text{liquid}} = 0.8 \times V_{\text{full}}$
• $V_{\text{liquid}} \approx 0.8 \times 0.763 \approx 0.610\text{ m}^3$

So the liquid occupies about $0.61\text{ m}^3$.

We are given $1\text{ m}^3 = 1{,}000{,}000\text{ mL}$.

Multiply the volume in cubic meters by $1{,}000{,}000$ to get milliliters:

• $V_{\text{liquid}} \approx 0.610 \times 1{,}000{,}000 \approx 610{,}000\text{ mL}$.

(More precisely, you might get around $610{,}000$ to $610{,}500$ mL depending on rounding; any close value is fine.)

Density is mass per unit volume. We are given a density of $0.85\text{ g/mL}$.

Mass in grams is

• $m_{\text{grams}} = (\text{volume in mL}) \times (\text{density in g/mL})$.

Using our volume approximation:

• $m_{\text{grams}} \approx 610{,}000 \times 0.85 \approx 518{,}500\text{ g}$.

(Again, a value near $519{,}000\text{ g}$ is fine depending on your earlier rounding.)

We are given $1\text{ lb} = 453.6\text{ g}$. To convert grams to pounds, divide by $453.6$:

• $m_{\text{pounds}} = \dfrac{m_{\text{grams}}}{453.6}$
• Using $m_{\text{grams}} \approx 518{,}500\text{ g}$, we get $m_{\text{pounds}} \approx \dfrac{518{,}500}{453.6}$.

This quotient is a bit more than $1{,}100$ and comes out to about $1{,}140$ when computed carefully. So the best answer choice is 1,140.