Conquering Percent Mixture Problems by the “Bucket Method”
Many textbooks ask our students to arrange the information given in a percent mixture
problem in a chart. Charts can be great, but I found that my students had a hard time
remembering what to write in the headings and on the sides… as a result, the chart that was
designed to help organize their thinking was actually getting in the way…
I have found a way to conquer percent mixture problems by using the “Bucket Method”.
The “Bucket Method” simply arranges the “chart” in a more conceptual way. Here it is:
+
=
These boxes represent the concept that we are mixing one solution with another solution
to obtain a resulting solution that contains both mixed together.
There are three basic formats for percent mixture problems:
1. Mixing two solutions together.
2. Mixing one solution with pure substance. (key: pure substance means 100% is being added)
3. Mixing one solution with water. (key: water means 0% of that substance is being added)
Each of the formats can be handled beautifully by the “Buckets”.
1. Mixing two solutions together: (Two examples to illustrate the two ways the amounts may be given):
a) A chemist mixed some 20% acid solution with some 40% acid solution to obtain
100 mL of 30% acid solution. How much 20% acid solution did he use in the mixture?
Here’s how you fill in the boxes:
Amount:
x
+
Percent mixture:
Here’s how you get the equation:
100 – x
100 mL
=
20%
40%
20x
+ 40(100 – x) =
20x + 4000 – 40x =
-20x =
x =
30%
30(100)
3000
-1000
50 mL 20% solution
b) A pharmacist mixed 25 mL of 10% saline solution with some 30% saline solution to
obtain a 20% solution. How much 30% solution did he use?
Amount:
Percent mixture:
25 mL
+
10%
10(25) +
x
=
30%
30x
250 + 30x
10x
x
25 + x
20%
=
=
=
=
20(25 + x)
500 + 20x
250
25 mL of 30% solution
2. Mixing a solution with a pure substance: (key: pure substance means 100% is being added)
Example: How much pure acid needs to be added to 100mL of 50% acid solution to make it
a 75% acid solution?
Amount:
Percent mixture:
x
+
=
50%
100%
100x
100 mL
+
50(100)
100x + 5000
25x
x
100 + x
75%
=
=
=
=
75(100 + x)
7500 + 75x
2500
100 mL of pure acid
3. Mixing a solution with water: (key: water means 0% of that substance is being added)
Example: How much water needs to be added to 50mL of a 70% saline solution to dilute it to
a 20% saline solution?
Amount:
Percent mixture:
x
+
50 mL
=
70%
0%
0
+
70(50)
3500
2500
x
50 + x
20%
=
=
=
=
20(50 + x)
1000 + 20x
20x
125 mL of water
Last Type of Mixture Problem:
Dry Mixtures: These can be put in “Buckets” as well. : )
Example: Peanuts cost $2 per pound and almonds cost $5 per pound. How many pounds of each
should be used to make a 30-lb mixture that costs $4 per pound?
Amount:
Cost per pound:
x
+
$2
2x
Cathy Ruddell
Beckman High School
30 – x
=
$5
+ 5(30 – x)
2x + 150 – 5x
–3x + 150
–3x
x
30 – x
30 lbs
$4
=
=
=
=
=
=
4(30)
120
120
–30
10 pounds of Peanuts
20 pounds of Almonds