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"Mixture" Word Problems (page 1 of 2)


Mixture problems involve creating a mixture from two or more things, and then determining some quantity (percentage, price, etc) of the resulting mixture. For instance:

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 figure the expected revenue from this information.Bigbucks University Presents!

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, and C stand for the child tickets pre-sold. Then A + C = 548. Also, since each adult ticket cost $5.00, then ($5.00)A stands for the revenue brought in from the adult tickets pre-sold; likewise, ($2.50)C stands for the revenue brought in from the child tickets. Then the total income so far is given by ($5.00)A + ($2.50)C = $2460. But we can only solve an equation with one variable, not two. So look again at that first equation. If A + C = 548, then A = 548 – C (or C = 548 – A; it doesn't matter which variable you solve for). Organizing this information in a grid, we get:

      tickets sold $/ticket total $
    adult 548 – C $5 $5(548 – C)
    child C $2.50 $2.50C
    total 548 --- $2460

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:

    ($5.00)(548 – C) + ($2.50)C = $2460
    $2740 – ($5.00)C + ($2.50)C = $2460

    $2740 – ($2.50)C = $2460

    –($2.50)C = –$280

    C = –$280/–$2.50 = 112

Then 112 child tickets were pre-sold, so A = 548 – 112 = 436 adult tickets were sold. (Using "A" and "C" for our variables, instead of "x" and "y", was helpful, because the variables suggested what they stood for. We knew instantly that "C = 112" meant "112 child tickets". This is a useful technique.)

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 436/548, or about 79.6%, of the total tickets sold to be adult tickets. Since we expect about 13,000 people, this works out to about 10,343 adult tickets. (You can find this value by using proportions, by the way.) The remaining 2657 tickets will be child tickets. Then the expected total ticket revenue totals to $58,357.50, of which ($5.00)(10,343) = $51,715 will come from adult tickets, and ($2.50)(2,657) = $6,642.50 will come from child tickets.


 

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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 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:

      liters sol'n percent acid total liters acid
    10% sol'n x 0.10 0.10x
    30% sol'n y 0.30 0.30y
    mixture x + y = 10 0.15 (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'n percent acid liters acid
    10% sol'n 10 – y 0.10 0.10(10 – y)
    30% sol'n y 0.30 0.30y
    mixture x + y = 10 0.15 (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
    1
    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.)

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Cite this article as:

Stapel, Elizabeth. "'Mixture' Word Problems." Purplemath. Available from
    http://www.purplemath.com/modules/mixture.htm. Accessed
 

 

 

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