SOLUTION: A gasoline refinery mixes different types of gasoline. How many gallons of grade A worth 17.60 per gallon must be mixed with grade B worth 19.20 per gallon and grade C at 22.40 per

Click here to see ALL problems on Mixture Word Problems

Question 1180566: A gasoline refinery mixes different types of gasoline. How many gallons of grade A worth 17.60 per gallon must be mixed with grade B worth 19.20 per gallon and grade C at 22.40 per gallon to produce 10 gallons of gas that will sell for 19.52 per gallon if equal amount of grade A and C are mixed?

Please help me🙏 this is for my final project. Thank you so much🙏😇
Found 3 solutions by josgarithmetic, Edwin McCravy, ikleyn:
Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website!
very typical three part mixture problem

Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website!

|Price | | per |Money | |Gallons|Gallon|Value | ------------------------------- Grade A | x |17.60 |17.60x| Grade B | y |19.20 |19.20y| Grade C | x |22.40 |22.40x| ------------------------------- Mixture | 10 |19.52 |195.20| Equations: Solve by substitution or elimination. Answer: 2 gallons each of grades A and C, and 6 gallons of grade B Edwin

Answer by ikleyn(52787)   (Show Source):
You can put this solution on YOUR website!
.
A gasoline refinery mixes different types of gasoline. How many gallons of grade A worth 17.60 per gallon
must be mixed with grade B worth 19.20 per gallon and grade C at 22.40 per gallon to produce 10 gallons of gas
that will sell for 19.52 per gallon if equal amount of grade A and C are mixed?
~~~~~~~~~~~~~~~~~~~~~~

            From the first glance, this problem is for 3 unknowns and 3 equations.

            Edwin managed to reduce it to two unknowns in two equations.

            I will go FURTHER and will show you even MORE SIMPLE way to solve the problem using only ONE unknown and ONE equation.

            Watch attentively every my step.

            You may feel yourself shocked. Do not worry - it is very natural feeling for those who see this trick for the first time.

Let x be the amount (the volume in gallons) of the grade A. The amount of the grade C is the same x gallons as the grade A, according to the condition. Hence, the amount of the grade B is the rest (10-2x) gallons. Now we write the total cost equation 17.60x + 19.20*(10-2x) + 22.40x = 10*19.52. I think that you do not need long explanations regarding this equation: you write it as you read the problem. The only thing to explain is the right side, which is the cost of 10 gallons of the mixture. Now simplify and solve this equation 17.60x - 19.20*10 - 19.20*2x + 22.40x = 10*19.52 1.6x = 10*19.52 - 19.20*10 1.6x = 3.2 x = 3.2/1.6 = 2. So, the problem is just solved: you need 2 gallons of the grade A, 2 gallons of the grade C and the rest (10 - 2 - 2) = 6 gallons is the grade B.

Solved, answered and carefully explained.

---------------

Now estimate the advantages of my method:

- you do not need to draw this table of the components (which I hate to spend my time preparing it); - you do not need solve NEITHER system of three equations NOR system of two equations; - and even 6-th or 7-th grade student can solve it easily . . . if he or she knows the method.

Now I will tell you what other people do not know:

            1)   This problem is DESIGNED and is INTENDED to be solved by this method.

            2)   And this method is the EXPECTED WAY to solve and expected way TO TEACH students solving such and similar problems.

Happy learning (!)

/////////////

Do not forget to post your "THANKS" to me for my teaching.