Sea water

Mixing 62 kg of sea water with 84 kg rainwater is created water containing 3.1% salt. How many percent sea water contains salt?

Result

x =  7.3 %

Solution:

m1=62 kg m2=84 kg  p=3.1 %  p=100 x100 m1m1+m2  p m1+m2100=x100 m1  p (m1+m2)=x m1  x=p m1+m2m1=3.1 62+8462=7310=7.3=7.3%m_{ 1 } = 62 \ kg \ \\ m_{ 2 } = 84 \ kg \ \\ \ \\ p = 3.1 \ \% \ \\ \ \\ p = 100 \cdot \ \dfrac{ \dfrac{ x }{ 100 } \cdot \ m_{ 1 } }{ m_{ 1 }+m_{ 2 } } \ \\ \ \\ p \cdot \ \dfrac{ m_{ 1 }+m_{ 2 } }{ 100 } = \dfrac{ x }{ 100 } \cdot \ m_{ 1 } \ \\ \ \\ p \cdot \ (m_{ 1 }+m_{ 2 }) = x \cdot \ m_{ 1 } \ \\ \ \\ x = p \cdot \ \dfrac{ m_{ 1 }+m_{ 2 } }{ m_{ 1 } } = 3.1 \cdot \ \dfrac{ 62+84 }{ 62 } = \dfrac{ 73 }{ 10 } = 7.3 = 7.3 \%



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