# Pebble

The aquarium with internal dimensions of the bottom 40 cm × 35 cm and a height of 30 cm is filled with two-thirds of water. Calculate how many millimeters the water level in the aquarium rises by dipping a pebble-shaped sphere with a diameter of 18 cm.

Result

x =  21.812 mm

#### Solution:

$a=40 \ \text{cm} \ \\ b=35 \ \text{cm} \ \\ c=30 \ \text{cm} \ \\ \ \\ c_{1}=\dfrac{ 2 }{ 3 } \cdot \ c=\dfrac{ 2 }{ 3 } \cdot \ 30=20 \ \text{cm} \ \\ \ \\ D=18 \ \text{cm} \ \\ r=D/2=18/2=9 \ \text{cm} \ \\ \ \\ V=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r^3=\dfrac{ 4 }{ 3 } \cdot \ 3.1416 \cdot \ 9^3 \doteq 3053.6281 \ \text{cm}^3 \ \\ V=abd \ \\ \ \\ d=V/(a \cdot \ b)=3053.6281/(40 \cdot \ 35) \doteq 2.1812 \ \text{cm} \ \\ \ \\ d < c-c_{1} \ \\ \ \\ x=d \rightarrow mm=d \cdot \ 10 \ mm=21.81163 \ mm=21.812 \ \text{mm}$

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Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):

Math student
where did we get the small d?

Dr Math
small d = height of water level rise. c is occupied yet, thus the next variable is d.

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