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Alligation question

by Sue
(BC, Canada)

Can you please show me how to figure out the answer.
How many grams of coal tar should be added to 925gm of zinc oxide paste to prepare a 6% coal tar ointment?
Thanks

Comments for
Alligation question

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Mar 30, 2011
A little bit more algebra involved...
by: Mike C

Set up so .06x = coal tar added.
Set up so 925 + .06x = Total Weight.

Coal tar Added/Total needs to equal 6%.

So
.06x/(925+.06x)=6/100

cross multiply and solve for x.

Take x you found and put it into .06x equation.

This should give you the amount of coal tar you need to add.

To check your answer, plug your x into
.06x/(925+.06x) and it will equal .06 .

Good luck!

Dec 23, 2010
Not an alligation problem
by: Anonymous

This is not an alligation problem, it is a concentration problem--and a very enlightening one at that!

Oct 18, 2010
Math correction?
by: Feintheart

I believe the previous answer would only hold true if the final quantity of ointment were exactly 925. However, the question asks us to add tar until the final balance is 6%. This means we will have more than 925g final product.


94% of that product will be the zinc oxide. 6% will be the tar.

x/925 = .94/.06

Now cross multiply, etc. as above.

Works out to 59g or so. You need slightly more than 6% of 925g (55.5g) to give the patient their money's worth.

The final ointment will weigh 984g, not 980.5g.

The previous answer only gives the patient 5.66% coal tar.

925 + 55.5 = 980.5
55.5 / 980.5 = .0566

The above answer would be correct if you scooped/dumped out 55.5g of zinc oxide first before displacing its weight with coal tar.

Sep 14, 2010
answer to this Alligation Problem
by: Anonymous

you have 6 g in 100 g so you need to find how many g you need in 925 g

6/100 = X/925

cross multiply to get 5550 = 100X

55.5 = X

so you will need 55.5g to make 925g of 6% ointment

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