Volume of water in rectangular tanks

Tank X has a rectangular base 25 cm by 12 cm. Tank Y has a rectangular base 15 cm by 10 cm. The height of Y$Y$ is 14 cm. At first, the water in Y is 8 cm high.

Mei poured one-fifth of the water from X into Y until Y was completely filled. Kai then poured all the water from Y into X. 300 cm³ of water overflowed from X.

(a) Find the volume of water in X at first.

(b) Find the height of X$X$.

(a) Pictorial — Look at Y. Water sits 8 cm high in a 14 cm tank, so a 6 cm empty band sits on top. Mei poured from X until that empty band vanished. So the water she poured equals the empty volume in Y.

Empty height = 14 - 8 = 6$14-8=6$ cm. Empty volume = 15 × 10 × 6 = 900$15\times 10\times 6=900$ cm³.

That empty volume is exactly one-fifth of the water in X. So water in X at first = 900 × 5 = 4500$900\times 5=4500$ cm³.

Move: empty volume, hence multiply by 5. Why: one-fifth of X$X$ filled the empty part, so five-fifths is all of X$X$.

(b) Pictorial — Now pour every drop from Y into X. The water that was already in Y joins the water that was in X. X cannot hold it all: 300 cm³ spills over the rim. What stays behind is a full tank X.

Water already in Y = 15 × 10 × 8 = 1200$Y=15\times 10\times 8=1200$ cm³. Total water = 4500 + 1200 = 5700$4500+1200=5700$ cm³.

After 300 cm³ overflows, X is full with 5700 - 300 = 5400$5700-300=5400$ cm³. Height of X = 5400 ÷ (25 × 12) = 18$X=5400÷(25\times 12)=18$ cm.

Move: total − overflow, hence divide by base area. Why: full volume ÷ base gives the height.

The empty part of Y$Y$ is 900 cm³. That empty part is one-fifth of the water in X. What was the volume of water in X at first?

The total volume of water is 5700 cm³. After 300 cm³ overflows, X is full. The base of X$X$ is 25 cm by 12 cm. What is the height of X$X$?