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?

Result

p =  -13.6 %

Solution:

a1=12 dm a2=(1+20100)⋅ a1=(1+20100)⋅ 12=725=14.4 dm  V1=34⋅ a13=34⋅ 123=1296 dm3 V2=38⋅ a23=38⋅ 14.43=1119.744 dm3  p=100⋅ V2−V1V1=100⋅ 1119.744−12961296=−685=−13.6=−13.6%a_{ 1 } = 12 \ dm \ \\ a_{ 2 } = (1+\dfrac{ 20 }{ 100 } ) \cdot \ a_{ 1 } = (1+\dfrac{ 20 }{ 100 } ) \cdot \ 12 = \dfrac{ 72 }{ 5 } = 14.4 \ dm \ \\ \ \\ V_{ 1 } = \dfrac{ 3 }{ 4 } \cdot \ a_{ 1 }^3 = \dfrac{ 3 }{ 4 } \cdot \ 12^3 = 1296 \ dm^3 \ \\ V_{ 2 } = \dfrac{ 3 }{ 8 } \cdot \ a_{ 2 }^3 = \dfrac{ 3 }{ 8 } \cdot \ 14.4^3 = 1119.744 \ dm^3 \ \\ \ \\ p = 100 \cdot \ \dfrac{ V_{ 2 }-V_{ 1 } }{ V_{ 1 } } = 100 \cdot \ \dfrac{ 1119.744-1296 }{ 1296 } = - \dfrac{ 68 }{ 5 } = -13.6 = -13.6 \%

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