Iodine Solutions

Iodine is a great material for disinfection. It kills bacteria very well, but it's a little burning when put on a wound. Mathematic calculations of iodine solutions are not as burning, but they definitely require some more thought.

A pharmacist has two iodine solutions: one with a concentration of 30% and one with a concentration of 80%. How much of each solution the pharmacist needs to create 8 liters of a 50% solution?

First, we need to understand the situation? Why would a pharmacist need 8 liters of iodine solution? After you answer that in your head, go back to math.

We need two conditions to happen at once: the solution must have 8 liters of solution and 50% iodine, which means 4 liters of iodine. How do we solve that? A system of equations!

Let x by the number of liters of the 80% solution
Let y be the number of liters of the 30% solution

The amount of iodine in each solution is the percent times the volume. In this case, the amount of iodine from the 80% solution is 0.8x and the amount of iodine from the 30% solution is 0.3y. Now let's create those equations:
0.8x + 0.3y = 4
x + y = 8

The 4 comes from the explanation above: 50% of 8 is 4, which means there are 4 liters of iodine in that solution (ouch). Let's solve this system:
0.8x + 0.3y = 4 /*10
x + y = 8

8x + 3y = 40
x + y = 8 /*3

8x + 3y = 40
3x + 3y = 24

Subtract the equations:
5x = 16
x = 16/5 = 3.2 liters

3.2 + y = 8
y = 4.8 liters

Check:
80% * 3.2 = 2.56 liters of iodine
30% * 4.8 = 1.44 liters of iodine
Combined: 4 liters of iodine out of 8 liters of liquid. A 50% solution.

The pharmacist needs 3.2 liters of the 80% solution and 4.8 liters of the 30% solution. That's a lot of iodine.

Have a great weekend,
Nadav

nadavs

Percentages and Copy Machines

Today's question is relatively easy, but very unique. It does not require complex mathematical calculations, but rather a simple mathematical thinking.

A copy machine has 7 buttons to increase or decrease the area of the photocopied area. They are 50%, 75%, 80%, 100%, 120%, 125%, and 150%. Which buttons can be eliminated from the machine in a way that saves the machines original area choices. Notice: when two buttons are pushed consecutively, the effect of the second button is on the area that remains after the first button (for example, two pushes on 50% will photocopy 25%).

It seems easy, and it really is. All you have to do is turn these percents into fractions and see how you can multiply them to get the others.

50% = 1/2
75% = 3/4
80% = 4/5
100% = 1
120% = 6/5
125% = 5/4
150% = 3/2

50% cannot be created by any combination, so it must remain in the machine.
75% = 3/4 = 1/2 * 3/2
80% cannot be created by any combination, so it also stays in the machine.
100% = 1 = 5/4 * 4/5
120% = 4/5 * 3/2
125% is another number that cannot be created from others.
150% is a component in 75% and 120%, so it can't go away.

However, this question has another answer, where 150% = 6/5 * 5/4, and then 120% stays in the machine and 150% goes out.

Easy, isn't it?
Nadav

nadavs