Problema Solution
A decorator can buy pink paint from two manufacturers.
• Paint A is made up from red and white paint in the ratio 1:3
• Paint B is made up from red and white paint in the ratio 1:7
He can mix the paints to produce a different shade of pink.
If Paint A and Paint B come in same size cans, what is the least number he would need of each type in order to produce pink paint containing red and white in the following ratios:
• 1:4
• 1:5
• 1:6
Answer provided by our tutors
Let
A = r1 + w1 where r1 = red, w1 = white and r1: w1 = 1:3 that is w1=3r1
B = r2 + w2 where r2 = red, w2 = white and r2: w2 = 1:7 that is w2=7r2
We know that A and B are of the same size so A = B that is
r1 + w1 = r2 + w2
r1 + 3r1 = r2 + 7r2
4r1 = 8r2
r1 = 2r2
Lets thake s cans of A and t cans of B.
s cans of A have: s*r1 = s*2r2 = 2*s*r2 red color
t cans of B have: t*r2 red color
or total sA + tB have (2*s*r2 + t*r2) = r2(2s + t) red color
s cans of A have: s*w1 = s*3r1 = s*3*2*r2 = 6*s*r2 white color
t cans of B have: t*w2 = 7*t*r2 white color
or total sA + tB have ( 6*s*r2 + 7*t*r2 ) = r2(6s + 7t) white color
The rations of red and white color for sA + tB is
r2(2s + t)/(r2(6s + 7t) = (2s+t)/(6s+7t)
We need to find minimum s and t so that they are non-negative integers and the ratio of red/white color be
1) 1:4
that is (2s+t)/(6s+7t) = 1/4
s/t = 3/2
We need at least 3 cans of A and 2 cans of B to have red/white = 1/4.
2) 1:5
(2s+t)/(6s+7t) = 1/5
s/t = 1/2
We need at least 1 can of A and 2 cans of B to have red/white = 1/5.
3) 1:6
(2s+t)/(6s+7t) = 1/6
s/t = 1/6
We need at least 1 can of A and 6 cans of B to have red/white = 1/6.