Anyone know how to solve?

Question

How many gallons of a 80% antifreeze solution must be mixed with 70% gallons of 15% antifreeze to get a mixture that is 70% antifreeze?

Answer 1

let's say that V₁ is the volume of the 80% antifreeze solution. So, in this solution we will have 0.8V₁ quantity of the antifreeze substance.

 

the total volume of the mixture would be:

(V₁+70) gallons, if we assume that the percentage of the antifreeze is 70%, then in (V1+70) gallons we will have:

(V1+70)0.7 antifreeze regardless of the unit we use.

 we know that in V₁ gallons we have 0.8V₁ and in 70 gallons of the other solution we have 70*0.15=10.5 (regardless of the unit). this let us solve for V₁:

0.8v₁+10.5=(V1+70)0.7  this leads to the following:

0.8v₁-0.7v₁=49-10.5=38.5

then V₁=38.5/0.1=385 gallons

We need 385 gallons of the 80% solution to mix with 70 gallons of 15% to obtain a mixture of 70%

 

I hope this will help

 

Regards,