Thread: mixture problem

I have two cans of paint. Can A has 9 parts of blue paint to one part of yellow paint. Can B is 20% blue and the rest is yellow. How much paint should I use from each can to obtain 4 liters of paint which is half blue and half yellow.

Answers in liters.

Hello, math34a!

I have two cans of paint.
Can A has 9 parts of blue paint to one part of yellow paint.
Can B is 20% blue and the rest is yellow.
How much paint should I use from each can to obtain
4 liters of paint which is half blue and half yellow.
Answers in liters.

This is a standard "mixture problem".


Can A is 90% blue.
Can B is 20% blue.
. . We want 4 liters of mixture which contains 2 liters of blue.

Let = number of liters from can A.
. . It contains: . liters of blue.

Then = number of liters from can B.
. . It contains: . liters of blue.

Together, this mixture will contains 2 liters of blue.


There is our equation: .

Originally Posted by math34a

I have two cans of paint. Can A has 9 parts of blue paint to one part of yellow paint. Can B is 20% blue and the rest is yellow. How much paint should I use from each can to obtain 4 liters of paint which is half blue and half yellow.

Answers in liters.

Here is one way.

Let A = liters of paint to get from can A.
And B = liters of paint to get fropm can B.

For blue parts,
A(9) +B(2) = 4(5)
9A +2B = 20 --------------(1)

For yellow parts,
A(1) +B(8) = 4(5)
A +8B = 20 -----------------(2)

20 = 20,
9A +2B = A +8B
9A -A = 8B -2B
8A = 6B
A = 6B /8 = (3/4)B ------(i)
Substitute that into, say, (2),
(3/4)B +8B = 20
8.75B = 20
B = 20/8.75) = 2.286 liters

So,
A = (3/4)B = 0.75(2.286) = 1.714 liters

Check,
A +B = 4
1.714 +2.286 =? 4
4 =? 4
Yes, so, OK.