A cylindrical bucket of internal radius 15 cm holds water standing 28 cm deep. All of it is emptied into a rectangular trough whose base measures 45 cm by 22 cm. How deep does the water stand in the trough, in cm? (Use π = 22/7.)

The water keeps its volume: (22/7) $\times 15^{2}x28=22\times 225\times$ 4 = 19800 cubic cm. The trough's base is $45\times 22=990$ sq cm, so the depth = 19800/990 = 20 cm. The trap 5 reads the bucket's 15 cm as a diameter, which quarters the volume; 28 assumes the depth travels with the water, though the trough is a different width; 10 doubles the trough's 22 cm width to 44 cm, halving the depth.