Individual

To prepare 14 kg of cheese with 45% dry matter, 3 types of cheese mixtures were mixed. The first was 5 kg, the second was 3 kg and had 14% more dry matter (14 percentage points more) than the first. The third mixture, which had 2 times as much dry matter as the first mixture, was 6 kg. What % dry matter was in each cheese mixture?

Correct answer:

p₁ =  29.4 %
p₂ =  43.4 %
p₃ =  58.8 %

Step-by-step explanation:

$$\begin{gathered} r_1=14\%=\frac{14}{100}=\frac{7}{50}=0.14 \\ {r}_2=45\%=\frac{45}{100}=\frac{9}{20}=0.45 \\ {m}_1=5\text{kg} \\ {m}_2=3\text{kg} \\ {m}_3=6\text{kg} \\ m=m_1+m_2+m_3=5+3+6=14\text{kg} \\ S=m\cdot r_2=14\cdot 0.45=\frac{63}{10}=6.3\text{kg} \\ {m}_1\cdot r+m_2\cdot (r+r_1)+m_3\cdot 2\cdot r=S \\ 5\cdot r+3\cdot (r+0.14)+6\cdot 2\cdot r=6.3 \\ 20r=5.88 \\ r=\frac{5.88}{20}=0.294 \\ r=\frac{147}{500}=0.294 \\ {p}_1=100\cdot r=100\cdot 0.294=29.4\% \end{gathered}$$

$p_2=100\cdot (r+r_1)=100\cdot (0.294+0.14)=43.4\%$

$p_{3} = 100 \cdot \ 2 \cdot \ r = 100 \cdot \ 2 \cdot \ 0.294 = \dfrac{ 294 }{ 5 } = 58.8 = 58.8 \% \ \\ \ \\ \text{ Verifying Solution: } \ \\ s_{1} = m_{1} \cdot \ \dfrac{ p_{1} }{ 100 } = 5 \cdot \ \dfrac{ 29.4 }{ 100 } = \dfrac{ 147 }{ 100 } = 1.47 \ \text{kg} \ \\ s_{2} = m_{2} \cdot \ \dfrac{ p_{2} }{ 100 } = 3 \cdot \ \dfrac{ 43.4 }{ 100 } = \dfrac{ 651 }{ 500 } = 1.302 \ \text{kg} \ \\ s_{3} = m_{3} \cdot \ \dfrac{ p_{3} }{ 100 } = 6 \cdot \ \dfrac{ 58.8 }{ 100 } = \dfrac{ 441 }{ 125 } = 3.528 \ \text{kg} \ \\ s_{4} = s_{1}+s_{2}+s_{3} = 1.47+1.302+3.528 = \dfrac{ 63 }{ 10 } = 6.3 \ \text{kg} \ \\ \ \\ S = s_{4}$

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