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Category results for 'percent'.

Question: "If I’ve used ninety percent of my hand sanitizer which was seventy percent alcohol, and I refill the bottle with new hand sanitizer which is eighty percent alcohol, what is the new alcohol content of the bottle?"

This is a fairly straightfoward mixture problem. One way of approaching this problem is to begin by writing everything as a percent:

Initailly, the bottle of hand sanitizer was 100% full, then 90% was removed. Thus, we had 100% - 90% = 10%. And 70% of that was alcohol.

Then we filled the container back up, meaning we added (100% - 10%) = 90%. And 80% of that was alcohol.

Thus, the amount of alcohol in the mixture is
10% x 70% + 90% x 80% =
0.10 x 0.70 + 0.90 x 0.80 =
0.07 + 0.72 = 0.79 =

NOTE: If you want to practice doing word problems, the lesson plans section of the site contains a page on mixture word problems. There are worksheets available, and answer keys for pro members.

Akash from Surat asks, "There were 200 fishes in an aquarium, 99% of which were red. How many red fishes must be removed to make the percentage of red fishes 98%?"

Hi Akash, let's see if we can set up a solution for this one.

First of all, 99% of 200 is 198 red fish. (Because 200 x 0.99 is 198).

If we remove x red fish from the tank, we have 198 - x red fish left out of 200 - x total.

Thus, the equation we need is (198 - x)/(200 - x) = 0.98.

198 - x = 0.98(200 - x)
198 - x = 196 - 0.98x
2 = 0.02x

x = 100 red fish removed.

Let's check our answer. If we remove 100 fish, and they're all red, we'll have 198 - 100 = 98 red fish, and a total of 200 - 100 = 100 fish.

Sure enough, 98 out of 100 is 98%.

It's tempting to think that if 198 fish is 99%, then 196 must be 98%, but that reasoning forgets that we aren't just decreasing the number of red fish; we're also decreasing the total number of fish!

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