you want to make 500kg to of 28% alcohol solution by mixing 40% alcohol solution and 20% alcohol solution. how many kg of each substance does he need?

Question

i have the anwser but i dont know how to get there plz help

Answer 1

You can set this up as a system of two equations, using the two things you know about mass, in kg, of the 40% alcohol solution you need (which we can label m₄₀) and the mass of the 20% solution you need (m₂₀).

 

The first thing you know is that the total mass is 500 kg, which we can represent in equation form as:

 

m₄₀ + m₂₀ = 500

 

Then, we know that 40% of m₄₀ and 20% of m₂₀ must equal 28% of 500 kg, which can be written as

 

0.4m₄₀ + 0.2m₂₀ = 0.28*500 = 140

 

We can solve this system in several ways, of which I will choose substitution: using the first equation above to write the second equation in terms of one variable, and then solve for that variable, like this:

 

m₄₀ = 500 - m₂₀  (this is just a rearrangement of the first equation)

 

Then the second equation can be written as:  0.4(500-m₂₀) + 0.2m₂₀ = 140

This can now be solved algebraically for m₂₀:

 

200 - 0.4m₂₀ + 0.2m₂₀ = 140 --> 200 - 0.2m₂₀ = 140 --> 0.2m₂₀ = 200 - 140 = 60 --> m₂₀ = 60/.2 = 300 kg

 

So 300 kg of the 20% solution is required.  Putting this value back into the first equation lets us solve for m₄₀, which is 200 kg.