Mobile Math Teachers’ Circle
Paint Problem
June 15, 2016
Problem
Anne decides to repaint her house in purple. She can buy two different types of purple paint.
• Paint A is made up from blue and red paint in the ratio 1 : 2.
• Paint B is made up from blue and red paint in the ratio 1 : 8.
She can mix the paints to produce different shades of purple. If Paint A and Paint B come
in same size cans, what is the least number she would need of each type in order to produce
purple paint containing blue and red in the ratio 1 : 3.
Source: NRICH enriching mathematics http://nrich.maths.org
Outline of Activity and Solution:
1. Ask each group to present their solution to the problem on a poster.
2.
Sort the solutions (same answer, similar methods, type of exposition etc.) and critique other groups solutions.
3. Compare different solutions and find the correct one.
4.
See if we can find a solution if we use the ratios part to whole. 1/3 of Paint A is
blue and 1/9 of Paint B is blue. We want to mix in such a way that exactly 1/4 of the mixture
is blue paint.
1
An Important Observation: Sometimes a picture helps.
Paint A
Paint B
red
red
red
red
red
red
red
red
red
red
red
red
blue
red
blue
red
blue
blue
Solution 1: Table, Trial and Error
From the picture above we can see that one can of Paint A contains 3 parts of blue and 6
parts of red while one can of Paint B contains 1 part of blue and 8 parts of red. It might be a
good idea to imagine that one can contains 9 ounces of paint. Note that combining one can of
Paint A with one can of Paint B yields a mixture that contains 4 ounces of blue paint and 14
ounces of red paint, ratio of 4 : 14. The table below lists the number of ounces of blue and the
number of ounces of red paint for the corresponding mixtures of cans of A and B. Note that 5
cans of A and 3 cans of B yield 18 ounces of blue and 54 ounces of red, a ratio of 18:54, which
equals the desired ratio of 1:3.
Paint A
1
2
Paint B
3
4
5
1
4:14
5:22
6:30
7:38
8:46
2
7:20
8:28
9:36
10:44
11:52
2
3
10:26
11:34
12:42
13:50
14:58
4
13:32
14:40
15:48
16:56
17:64
5
16:38
17:46
18:54
19:62
20:70
Solution 2: Comparing Ratios This solution is more efficient.
Let’s start with a can of Paint A.
What we want
1:3
1:3
1:3
1:3
1:3
1:3
1:3
1:3
What we have
3:6
4:14
7:20
8:28
11:34
14:40
15:48
18:54
Too much blue or too little blue?
too much blue
too little blue
too much blue
too little blue
too little blue
too much blue
too little blue
just right
Correction
add B
add A
add B
add A
add A
add B
add A
done!
We started with one can of A and added 4 more cans of A and 3 cans of B.
We used 5 cans of A and 3 cans of B.
Does this algorithm always work?
Solution 3: Ratio of Part to Whole, Algebra
Note that this argument uses the ratio of part to whole.
1/3 of a can of Paint A is blue and 1/9 of one can of Paint B is blue. Both cans are of
the same size. We want to mix x cans of Paint A and y cans of paint B in order to get the
equivalent of x + y cans of a certain mixed paint, 1/4 of which is blue.
If we total the amount of red paint in x cans of A and y cans of B, we obtain
x·
1
1
+y· .
3
9
On the other hand x + y cans containing 1/4 of red paint each would give the following total
of red paint
1
(x + y) · .
4
Setting the two sides equal to each other gives us
x·
1
1
1
+ y · = (x + y) · .
3
9
4
Multiplying both sides with the common denominator of 36 yields
12x + 4y = 9(x + y) = 9x + 9y
Collecting terms results in
3x = 5y
or
x
5
=
y
3
Hence the ratio of number of cans of Paint A to the number of cans of Paint B should be 5 : 3
in order to obtain the desired mixture. We need at least 5 cans of A and 3 cans of B.
3
Your Solution:
4