Mixture word problems are exercises which involve creating a mixture from two or more different things, and then determining some quantity (such as percentage, price, number of liters, etc) of the resulting mixture. There will always be a "rate" of some sort, such as miles per hour or cost per pound.
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Here is an example:
Your school is holding a family-friendly event this weekend. Students have been pre-selling tickets to the event; adult tickets are $5.00, and child tickets (for kids six years old and under) are $2.50. From past experience, you expect about 13,000 people to attend the event.
But this is the first year in which tickets prices have been reduced for the younger children, so you really don't know how many child tickets and how many adult tickets you can expect to sell. Your boss wants you to estimate the expected ticket revenue. You decide to use the information from the pre-sold tickets to estimate the ratio of adults to children, and compute the expected revenue from this information.
You consult with your student ticket-sellers, and discover that they have not been keeping track of how many child tickets they have sold. The tickets are identical, until the ticket-seller punches a hole in the ticket, indicating that it is a child ticket.
But they don't remember how many holes they've punched. They only know that they've sold 548 tickets for $2460. How much revenue from each of child and adult tickets can you expect?
To solve this, we need to figure out the ratio of tickets that have already been sold. If we work methodically, we can find the answer.
Let A stand for the number of adult tickets pre-sold. Since a total of 548 tickets have been sold, then the number of children tickets pre-sold thus far must be 548 − A.
Since each adult ticket cost $5.00, then 5A stands for the revenue brought in from the adult tickets pre-sold; likewise, 2.5(548 − A) stands for the revenue brought in from the child tickets. (Note: The per-ticket cost is the "rate" for this exercise.)
Organizing this information in a grid, we get:
|tickets sold||$/ticket||total $|
|child||548 − A||2.5||2.5(548 − A)|
From the last column, we get (total $ from the adult tickets) plus (total $ from the child tickets) is (the total $ so far), or, as an equation:
5A + 2.5((548 − A) = 2,460
5A + 1,370 − 2.5A = 2,460
1,370 + 2.5A = 2,460
2.5A = 1,090
So 436 adult tickets were pre-sold, so the number of child tickets pre-sold is:
C = 548 − 436 = 112
So 112 child tickets were pre-sold.
Now we need to figure out how many adult and child tickets we can expect to sell overall. Since 436 out of 548 pre-sold tickets were adult tickets, then we can expect (or roughly 79.6%) of the total tickets sold to be adult tickets.
We expect about 13,000 people in total to attend this event. We have a ratio of adult pre-sold tickets to total pre-sold tickets. Assuming that the pre-sold rate (or ratio) is representative of the total number of adult tickets, we can set up a proportion (of pre-sold adult tickets to total pre-sold tickets), using a variable for the unknown total number of adult tickets expected to be sold:
This works out to about 10,343 adult tickets. The remaining 2,657 tickets, of the expected total of 13,000, will be child tickets. Then the expected total ticket revenue is given by:
5(10,343) + 2.5(2,657) = 58,357.5
So the expected total revenues from ticket sales is $58,357.50.
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Let's try another one. This time, 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 w stand for the number of liters of the weaker 10% solution. Since the total number of liters is going to be 10, then the number of liters left, after the first w liters have been poured, will be 10 − w liters needed of the 30% solution.
(The clear labeling of the variable is important. While I picked w to stand for the weaker acid, I might not remember this by the end of the exercise. If I don't label, I might not be able correctly to interpret my answer in the end.)
For mixture problems, it is often very helpful to do a grid, so let's do that here:
|30% acid||10 − w||0.30||0.30(10 − w)|
|mixture||10||0.15||0.15(10) = 1.5|
When the problem is set up like this, we can usually use the last column to write our equation. In this case, the liters of acid within the 10% solution, plus the liters of acid within the 30% solution, must add up to the liters of acid within the 15% solution. So, adding down the right-hand column, setting the inputs equal to the mixed output, we get:
0.10w + 0.30(10 − w) = 1.5
0.10w + 3 − 0.30w = 1.5
3 − 0.2w = 1.5
1.5 = 0.2w
Looking back and confirming that the variable stands for the number of liters of the weaker acid, we see that we would need 7.5 liters of 10% acid. This means that we would need another 10 − 7.5 = 2.5 liters of the stronger 30% acid.
(If you think about it, this makes sense. Fifteen percent is closer to 10% than it is to 30%, so we ought to need more 10% solution in our mix.)