|
Mixture problems are commonly considered to be those in which someone is mixing together ingredients of various types - and the ingredients have different concentrations of some key item (like salt, gold, copper, etc.) When solving mixture problems I usually concentrate on the part of the mixture that is stable - that is, the amount of salt, the amount of gold and form my equation from that information. As in many of the other applications, I will organize the information in the form of a chart.
Example 1: How much of a 10% gold alloy should be mixed with 32 grams of a 24% gold alloy to make a 22% gold alloy?
Step 1: Identify which column you can complete from the information given in the problem (without having to use a variable) and fill it out. In this case, we know the percent of gold in each alloy so we will fill that part in.
Step 2: Identify what you are looking for and define your variable. Here we are looking for the number of grams of the 10% gold alloy.Let x represent the number of grams of the 10% gold alloy.
Step 3: Fill out the last column by using the first two columns and the relationship that
Step 4: The equation that will model this problem will now be drawn from the last column# of grams of pure gold in the10 % gold alloy
Now using the first example as a template, let's try a problem that changes just slightly.
Example 2: How much of each a 10% gold alloy and a 24% gold alloy should be mixed together to make 40 grams of a 22% gold alloy? Step 1: Identify which column you can complete from the information given in the problem without having to use a variable and fill it out. In this case, we know the percent of gold in each alloy so we will fill that part in.
So far this is the same as Example 1.
Step 2: Identify what you are looking for and define your variable. Here we are looking for the number of grams of both the 10% gold alloy AND the 24% gold alloy. We cannot assume that the amounts are the same so we cannot simply allow the varibale 'x' to represent them both. I will arbitrarily choose x to represent one of them.Let x represent the number of grams of the 10% gold alloy.
Step 3: Fill out the last column by using the first two columns and following the relationship
Step 4: The equation that will model this problem will now be drawn from the last column# of grams of pure gold in the10 % gold alloy
Now the final example is one of my favorites and, truth be known, I probably created this entire page just so that I could throw this one in. I first discovered it in a newpaper column by Marilyn vos Savant and it defies intuitive reasoning. I call it the cucumber paradox.
Example 3: Suppose that you go out and buy 100 lbs. of cucumbers that are 99% water. Unfortunatly, you do not observe the rules of caring for your cucumbers so after some time they dehydrate until they are only 98% water. How much do they then weigh? Now this problem is almost deliberately misleading as it begs you to concentrate on the weight of the water (which changes as it evaporates) instead of what remains constant (the weight of the pure cucumber solid- no water). As before, however, the equation will describe what will not change - that is, the amount of the pure cucumber solid. Since 99% of the 100lb. of cucumber is water, that leaves only 1% to be cucumber solid. Or, 1 lb. That will remain unchanged even as the water evaporates.
Step 1: Identify which column you can complete from the information given in the problem without having to use a variable and fill it out. In this case, we know the percent of solid cucumbers in both the original and the evaporated piles.
Step 2: Identify what you are looking for and define your variable. Here we are looking for how much the cucumbers weigh after evaporation.
Step 3: Fill out the last column by using the first two columns and following the relationship(# of lbs. of cucumber) X (% of cucumber solid) = (# of lbs of cucumber solid)
Step 4: The equation that will model this problem will now be drawn from the last column# of lbs of cucumber solid originally (before evaporation) Answer to the original question:After evaporation, the cucumbers only weigh 50 lbs. I told you that this one was totally counter-intuitive. Try it out on your friends. I'll bet no one is able to guess the answer.
© 1999 Jo Steig
|