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
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
