# The cube

The cube has an edge of 12 dm. The second cube has an edge exactly 20% longer. How many % is more water in the second cube than in the first cube, if the first cube is full to 3/4 and the second to 3/8?

Correct result:

p =  -13.6 %

#### Solution:

$a_{1}=12 \ \text{dm} \ \\ a_{2}=(1+\dfrac{ 20 }{ 100 } ) \cdot \ a_{1}=(1+\dfrac{ 20 }{ 100 } ) \cdot \ 12=\dfrac{ 72 }{ 5 }=14.4 \ \text{dm} \ \\ \ \\ V_{1}=\dfrac{ 3 }{ 4 } \cdot \ a_{1}^3=\dfrac{ 3 }{ 4 } \cdot \ 12^3=1296 \ \text{dm}^3 \ \\ V_{2}=\dfrac{ 3 }{ 8 } \cdot \ a_{2}^3=\dfrac{ 3 }{ 8 } \cdot \ 14.4^3=1119.744 \ \text{dm}^3 \ \\ \ \\ p=100 \cdot \ \dfrac{ V_{2}-V_{1} }{ V_{1} }=100 \cdot \ \dfrac{ 1119.744-1296 }{ 1296 }=- \dfrac{ 68 }{ 5 }=-13.6 \%$

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