Pharmacy Tech Math - Alligation tutorial
By Ron Aylor, CPhT
(Cont'd) - Back to previous page
On the previous pages, you learned some Alligation Math Basics. On this page, we'll look at another, more complex example.
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Alligation Example #2 |
How many mL of water must be added to 300 mL of 70% alcohol solution to make a 40% alcohol solution? A. 225mL |
*Important consideration - Before the matrix can be formed, the problem has to be analyzed to see which method is best for working the problem. The key words which indicate that this is an alligation problem are "must be added to." Other key words indicating alligation as the best method are "Must be combined" and "must be mixed."
In this problem, 40% is the desired strength and must be placed in the center of the matrix. The next step is to see if a higher or lower strength is given. 70% is a higher strength and must be placed in the upper left-hand corner. If no lower strength is given, it can be assumed to be 0%.
The relationship of parts of each percent strength to their
combined final volume may be used as the first ratio of a proportion. To
formulate the complete equation, place the known factors in the proper
position on the matrix. Assign the X value first: The question asks,
"How many milliliters of water?"; the X value is placed on the extended
line opposite the percentage denoted by water (0%). The other known
factor is that the water will be added to 300 ml of 70%. The 300mL
pertains to the 70% so it is placed on the line opposite the 70% on the
matrix.
These values form the
proportion: 4 /3 = 300 / x ; cross multiply: 4 x = 900; x = 225 mL of
water (0%). Take note that when distilled water, ointment bases, or
normal saline are used as diluents, they will contain zero percent (0%)
active ingredient.
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Alligation Example #3 |
A prescription calls for an elixir that contains 45% alcohol. On hand, you have 10% alcoholic elixir and a 75% alcoholic elixir. How many mL of the 75% elixir must be combined to make 1000 mL of 45% alcoholic elixir? A. 538mL |
To solve for the higher (75%), place the X on the line
opposite the 75% (see above). The only other known factor is that 1000
mL of the 45% must be prepared. The 1000 mL must be placed on the line
opposite the 45%. The arrow indicates that the bottom line is the 45%
line (see above). With these figures in place, construct you proportion
and solve for X.
The volume of the lower (45%) ingredient may be found by subtracting the known volume from the final volume. In this case we just found the volume of the 75% elixir to be 538.46 mL. Subtract this figure from the final volume of 1000 mL to get 461.54 mL; or, using the same matrix place the X opposite the 10% on the matrix and construct another proportion (see above)
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