Showing posts with label Mixture Problem in Mathematics. Show all posts
Showing posts with label Mixture Problem in Mathematics. Show all posts

Tuesday, May 17, 2011

Mixture Problem in Mathematics

(Mula sa http://www.purplemath.com/modules/mixture.htm ang halimbawa)


Ang mixture problem sa Math ay itinuturing na mahirap na sagutin ng mga mag-aaral. Ngunit kung pag-aaralan lamang mabuti at maraming pagsasanay, napakadali ng problemang ito. Tunghayan ang halimbawa sa ibaba.


Suppose you work in a lab. You need a 15% acid solution for a certain test, but your supplier only ships a 10% solution and a 30% solution. Rather than pay the hefty surcharge to have the supplier make a 15% solution, you decide to mix 10% solution with 30%solution, to make your own 15% solution. You need 10 liters of the 15% acid solution. How many liters of 10% solution and 30% solution should you use?

Let x stand for the number of liters of 10% solution, and let y stand for the number of liters of 30%solution. (The labeling of variables is, in this case, very important, because "x" and "y" are not at all suggestive of what they stand for. If we don't label, we won't be able to interpret our answer in the end.) For mixture problems, it is often very helpful to do a grid:


Nakatutulong na mabuti ang pagsasaayos ng mga elemento sa isang grid tulad nito:
    liters sol'npercent acidtotal liters acid
    10% sol'nx0.100.10x
    30% sol'ny0.300.30y
    mixturex + y = 100.15(0.15)(10) = 1.5
Since x + y = 10, then x = 10 – y. Using this, we can substitute for x in our grid, and eliminate one of the variables:   Copyright © Elizabeth Stapel 1999-2011 All Rights Reserved
    liters sol'npercent acidliters acid
    10% sol'n10 – y0.100.10(10 – y)
    30% sol'ny0.300.30y
    mixturex + y = 100.15(0.15)(10) = 1.5
When the problem is set up like this, you can usually use the last column to write your equation: The liters of acid from the 10% solution, plus the liters of acid in the 30% solution, add up to the liters of acid in the 15% solution. Then:
    0.10(10  y) + 0.30y = 1.5  0.10y + 0.30y = 1.5  1 + 0.20y = 1.5  0.20y = 0.5  y = 0.5/0.20 = 2.5
Then we need 2.5 liters of the 30% solution, and x = 10 – y = 10 – 2.5 = 7.5 liters of the 10%solution. (If you think about it, this makes sense. Fifteen percent is closer to 10% than to 30%, so we ought to need more 10% solution in our mix.)