## Alligation question

by Sue

How many grams of coal tar should be added to 925gm of zinc oxide paste to prepare a 6% coal tar ointment?
Thanks

 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)=6/100cross 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/.06Now 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 = 980.555.5 / 980.5 = .0566The 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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