There are many examples of mixture word problems. The goal is to really help you understand deeply a few.Then, you will be able to tackle similar problems on your own

I offer two solution for the mixture word problems here

A store owner wants to mix cashews and almonds. Cashews cost 2 dollars per pound and almonds cost 5 dollars per pound. He plans to sell 150 pounds of a mixture. How many pounds of each type of nuts should be mixed if the mixture will cost 3 dollars?

Let x be the number of pounds of cashews

So, 150 - x will represent the number of almonds

Since each pound of the mixture cost 3 dollars, 150 pounds will cost 3 × 150 = 450 dollars

Cost of cashews + cost of almonds = 450

2 × x + (150 - x) × 5 = 450

2x + 150 × 5 - x × 5 = 450

2x + 750 - 5x = 450

2x - 5x + 750 = 450

-3x + 750 = 450

-3x + 750 - 750 = 450 - 750

-3x = -300

-3x/-3 = -300/-3

x = 100

150 - x = 150 - 100 = 50.

The store owner should mix 100 pounds of cashews with 50 pounds of almonds

Let x represent the number of pounds of cashews

Let y be the number of pounds of almonds

x + y = 150

2x + 5y = 450

Solve by elimination

Multiply x + y = 150 by -2

-2(x + y) = 150 × -2

-2x + -2y = -300

Put the two equations together again.

-2x + -2y = -300

2x + 5y = 450

Add the left sides and the right sides to get:

3y = 150

3y/3 = 150/3

y = 50

Example 2 and 3 are more challenging mixture word problems

Suppose a car can run on ethanol and gas and you have a 15 gallons tank to fill. You can buy fuel that is either 30 percent ethanol or 80 percent ethanol. How much of each type of fuel should you mix so that the mixture is 40 percent ethanol?

Le x represent number of gallons of gas that contain 30 percent ethanol

Let 15 - x be number of gallons of gas that contain 80 percent ethanol

Since the mixture contains 40 percent ethanol, only 40% of the 15 gallons will be ethanol

40% of 15 = (40/100) times 15 = (40/100) times 15/1 = (40 × 15) / (100 × 1) = 600 / 100 = 6

In order for x gallons of gas to contain 30% ethanol, we must take 30% of x or 30% times x

In order for 15 - x gallons of gas to contain 80% ethanol, we must take 80% of 15 - x or 80% times 15 - x

0.30 × x + 0.80 × (15 - x) = 6

0.30x + 0.80 × 15 - 0.80 × x = 6

0.30x + 12 - 0.80x = 6

0.30x - 0.80x + 12 = 6

-0.50x + 12 = 6

-0.50x = -6

x = 12

So 12 gallons of gass contain 30 percent ethanol and 15 - 12 = 3 gallons contain 80 percent ethanol

Therefore, mix 12 gallons of a 30% ethanol with 3 gallons of an 80% ethanol

Indeed 30% of 12 = 0.30 × 12= 3.6 and 80% of 3 = 0.80 × 3 = 2.4

3.6 + 2.4 = 6

Le x be number of gallons of gas that contain 30 percent ethanol

Let y represent number of gallons of gas that contain 80 percent ethanol

x + y = 15

0.30 × x + 0.80 × y = 0.40 × 15

x + y = 15

0.30x + 0.80y = 6

Solve by substitution

Use x + y = 15 to get x ( subtract y from both sides) and replace x into the other equation

x = 15 - y

0.30 × (15 - y) + 0.80y = 6

0.30 × 15 - 0.30 × y + 0.80y = 6

4.5 + 0.80y - 0.30y = 6

4.5 + 0.50y = 6

4.5 - 4.5 + 0.50y = 6 - 4.5

0.50y = 1.5

0.50y/0.50 = 1.5/0.50

y = 3

Again, x = 15 - y = 15 - 3 = 12

So far, you probably noticed that mixture word problems can be quite challenging!

You have 6 liters of water that have 20 percent strawberry juice. How many liters of a 80 percent strawberry juice should be added to the mixture to make 75 percent strawberry juice?

Le x be number of liters of water that contain 20 percent strawberry juice

Let y represent number of liters of water that contain 80 percent strawberry juice

x + y = 6

0.20x + 0.80y = 0.75 × 6

Solve by substitution

x = 6 - y

0.20 × (6 - y) + 0.80y = 4.5

0.20 × 6 - 0.20 × y + 0.80y = 4.5

1.2 + 0.80y - 0.20y = 4.5

1.2 + 0.60y = 4.5

1.2 - 1.2 + 0.60y = 4.5 - 1.2

0.60y = 3.3

0.60y / 0.60 = 3.3 / 0.60

y = 5.5

x = 6 - 5.5 = 0.5

Therefore, if you want your juice to contain 75% strawberry juice, do the following:

Mix 5.5 liters of water that has 80% strawberry juice with 0.5 liter of water that has 20% strawberry juice

Your juice shoud taste better now!

The mixture word problems I solved above are typical questions. You may not encounter these word problems a lot in algebra. However, it is good idea to know how to solve these mixture word problems

When solving mixture word problems, I suggest you do it with a system of linear equations