How many kg of mixed nuts that contain 40% peanuts must Jenny add to 6 kg of mixed nuts that contain 30% peanuts to make a mixture with 34% peanuts?

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How many kg of mixed nuts that contain 40% peanuts must Jenny add to 6 kg of mixed nuts that contain 30% peanuts to make a mixture with 34% peanuts?

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Answer 1 · 2018-06-26T13:17:13

Jenny must add $4$ $4$ $kg$ $"kg"$ .

Explanation:

Let's call the added amount of kg of mixed nuts that contain $40 %$ $40%$ peanuts $m_{a d d e d}$ $m_(added)$ . If the new mixture contains $34 %$ $34%$ peanuts, then:

$\frac{m_{peanuts}}{m_{total}} = 0.34$ $frac(m_("peanuts"))(m_("total"))=0.34$

The bag of $6$ $6$ $kg$ $"kg"$ already contained $0.3 \cdot 6 = 1.8$ $0.3*6=1.8$ $kg$ $"kg"$ of peanuts. Furthermore, $40 %$ $40%$ of $m_{added}$ $m_"added"$ will be peanuts. This means that:

$m_{peanuts} = 0.4 m_{added} + 1.8$ $m_"peanuts"=0.4m_"added"+1.8$

The total mass will be the original $6$ $6$ $kg$ $"kg"$ plus the added mass, so $m_{total} = m_{added} + 6$ $m_"total"=m_"added"+6$ .

This gives us the following equation.

$\frac{0.4 m_{added} + 1.8}{m_{added} + 6} = 0.34$ $frac(0.4m_("added")+1.8)(m_("added")+6)=0.34$

Multiply both sides with $m_{added} + 6$ $m_("added")+6$ .

$0.4 m_{added} + 1.8 = 0.34 (m_{added} + 6)$ $0.4m_("added")+1.8=0.34(m_("added")+6)$

Multiply out the brackets.

$0.4 m_{added} + 1.8 = 0.34 m_{added} + 2.04$ $0.4m_("added")+1.8=0.34m_("added")+2.04$

Subtract $0.34 m_{added}$ $0.34m_("added")$ from both sides.

$0.04 m_{added} + 1.8 = 2.04$ $0.04m_("added")+1.8=2.04$

Subtract $1.8$ $1.8$ from both sides.

$0.06 m_{added} = 0.24$ $0.06m_("added")=0.24$

Divide both sides by $0.06$ $0.06$ .

$m_{added} = 4$ $m_("added")=4$

Therefore, Jenny must add $4$ $4$ $kg$ $"kg"$ .

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How many kg of mixed nuts that contain 40% peanuts must Jenny add to 6 kg of mixed nuts that contain 30% peanuts to make a mixture with 34% peanuts?

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