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Blueroses's avatar

Would you please help me with this math problem?

Asked by Blueroses (18256points) December 27th, 2011

If you know me, you know I’m not looking for homework answers.

I know the answer to this but I got it by a convoluted method of working out which multiple choice answer was correct.

The question:
How many grams of yellow mercuric oxide must be added to 30g of 1% yellow mercuric oxide ointment to prepare a 5% ointment?

What I did figure: yellow mercuric oxide = 100%
To get 5%, I need 4 parts of 100% plus 95 parts of 1%
therefore; 30g(1%) = 95 parts/99 parts

And there I freeze on how to get to grams needed of 100%.

Is it impossible without knowing the final volume? Or am I just stuck?

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8 Answers

Response moderated (Unhelpful)
Response moderated (Unhelpful)
gasman's avatar

I’m not sure where you got “4 parts of 100% plus 95 parts of 1%,” but that adds up to only .04 + .95 = 99%.

Let’s say you add x grams of yellow mercuric oxide (ymo).
I’m assuming percentages are by mass, so volume is irrelevant.

The final total mass of the mixture is 30 + x (because you added x)
The initial mass of ymo was (30g)(1%) = 0.3 grams
The final mass of ymo is therefore x + 0.3 (because you added x) ***

The final concentration is 5%, so final mass of ymo is (30+x)(5%)
= (30 + x)(.05)
= 1.5 + .05x ***

Equating the two starred equations above *** for mass of ymo:
x + 0.3 = 1.5 + .05x
.95x = 1.2
x = 1.2/.95 = 1.263… grams of ymo added.

So the final ointment contains 1.563… g of ymo in a total mass of 31.263… grams, which is exactly 5%.

Blueroses's avatar

Perfect @gasman. I completely understand where you got that 1.26g answer, because you explained it so well. I’ll learn that and apply it to the next problem I have and I’d give you 1000 GAs for your answer.

However it doesn’t fit my textbook and I can’t reconcile it with the methods we are supposed to be learning (and what I am supposed to be tutoring to my fellow students).

We are being taught alligation. To end up with 5%, you have the 5 in the middle of a box; the 100 at the top left and the 1 at the bottom left. That means you have 99 total parts; 4 parts are of the 100% and 95 parts are of the 1%. It doesn’t have to add up to 100: most times the alligation adds up to much less than 100, but from there you can figure amounts IF you have a final volume.

gasman's avatar

@Blueroses The truth is I never heard of alligation until you mentioned it. I just like to work algebra problems, I guess, but you should learn the methods they teach you. Good luck!

Blueroses's avatar

@gasman The alligation for my field looks like this

From there, I would set up a ratio/proportion or a dimensional analysis and solve, but I can’t use either of those methods without a total volume. Your algebra method works for me, and again I thank you for that, but is there any way to derive that answer using one of the methods we are being taught? If there is not, then it is a bad question and should be discarded from the test.

Response moderated (Unhelpful)
Blueroses's avatar

After letting this problem simmer for 24 hours, the very simple solution came to me out of the blue. Thank you @gasman for explaining an alternate method. It put my brain on the right track.

If anybody’s interested, this is the simple proportion I came up with:

Xg (100%)/4 parts = 30g (1%)/95 parts
Cross multiply to get 95x = 120
Divide both sides by 95
X=1.26

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