Irregular solid A is immersed in a graduated cylinder filled with water. The water level is at the 200 mL mark. Then, irregular solid B is immersed in the same cylinder and the water level rises to 320mL. If solid A is taken out of the cylinder, the water level will become 240mL. Which solid has the greater than volume, and how much greater over the other solid?
Visual Illustration: 
Solution:
Based from the illustration above
200mL = Vw + VA
320 mL = Vw + VA + VB
240 mL = Vw + VB
Where:
Vw = Volume of water inside the graduated cylinder
VA = Volume Irregular Solid A
VB = Volume Irregular Solid B
so,
200mL = Vw + VA (eq. 1)
320 mL = Vw + VA + VB (eq. 2)
240 mL = Vw + VB (eq. 3)
from eq. 1
VA = 200mL – VW (eq.4)
from eq. 3
VB = 240mL – VW (eq.5)
Substitute eq. 4 and eq. 5 to eq. 3
320 mL = Vw + (200mL – Vw) + (240mL – Vw)
320mL - 200mL – 240mL = Vw – Vw – Vw
-120mL = - Vw
Vw = 120mL
substitute the value of Vw to eq. 4 and eq. 5
VA = 200mL – 120mL = 80mL
VB = 240mL – 120mL = 120mL
Therefore;
VB is 40mL greater than VA (answer)