Mixture Problem

Mixture Problem

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.

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)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.

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:

liters sol’n percent acid liters acid
10% sol’n 10 – y 0.10 0.10(10 – y)
30% sol’n y 0.30 0.30 y
mixture x + y = 10 0.15 (0.15)(10) = 1.5

Example 1 :

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.)

Usually, these exercises are fairly easy to solve once you’ve found the equations. To help you see how to set up these problems, below are a few more problems with their grids (but not solutions).

Example 2 :

How many liters of a 70% alcohol solution must be added to 50 liters of a 40% alcohol solution to produce a 50% alcohol solution?

liters sol’n % alcohol total liters alcohol
70% sol’n x 0.70 0.70x
40% sol’n 50 0.40 (0.40)(50) = 20
50% mix 50 + x 0.50 0.50(50 + x)

From the last column, you get the equation 0.7x + 20 = 0.5(50 + x). Solve for x.

Example 3 :

How many ounces of pure water must be added to 50 ounces of a 15% saline solution to make a saline solution that is 10% salt?

ounces liquid % salt total ounces salt
water x 0 0
15% sol’n 50 0.15 (50)(0.15) = 7.5
10% mix 50 + x 0.10 0.10(50 + x)

From the last column, you get the equation 7.5 = 0.1(50 + x). Solve for x.

(Note the percentage for water. “Pure water” contains no salt, so the percent of salt is zero. If, on the other hand, you were trying to increase the salt content by adding pure salt, the percent would have been one hundred.)

Example 4 :

Find the selling price per pound of a coffee mixture made from 8 pounds of coffee that sells for $9.20 per pound and 12 pounds of coffee that costs $5.50 per pound.

pounds coffee $/pound total $ for coffee
pricey 8 $9.20 (8)($9.20) = $73.60
cheapo 12 $5.50 (12)($5.50) = $66
mix 8 + 12 = 20 ? $73.60 + 66 = $139.60

From the last row, you see that you have 20 pounds for $139.60, or $139.60/(20 pounds). Simplify the division to find the unit rate.

Example 5 :

How many pounds of lima beans that cost $0.90 per pound must be mixed with 16 pounds of corn that costs $0.50 per pound to make a mixture of vegetables that costs $0.65 per pound?

pounds $/pound total $ for veggies
lima beans x $0.90 $0.90x
corn 16 $0.50 (16)($0.50) = $8
mix 16 + x $0.65 (16 + x)($0.65)

From the last column, you get the equation $0.90x + $8 = (16 + x)($0.65). Solve for x.

Example 6 :

Two hundred liters of a punch that contains 35% fruit juice is mixed with 300 liters (L) of another punch. The resulting fruit punch is 20% fruit juice. Find the percent of fruit juice in the 300 liters of punch.

liters punch % juice total liters juice
35% juice 200 0.35 (200)(0.35) = 70
other punch 300 x 300x
mix 200 + 300 = 500 0.20 (500)(0.20) = 100

From the last column, you get the equation 70 + 300x = 100. Solve for x, and then convert the decimal answer to a percentage.

Example 7 :

Ten grams of sugar are added to a 40-g serving of a breakfast cereal that is 30% sugar. What is the percent concentration of sugar in the resulting mixture?

grams in bowl % sugar total grams sugar
sugar 10 1.00 10
cereal 40 0.30 (40)(0.30) = 12
mix 50 ? 10 + 12 = 22

From the last row, you see that there are 22 grams of sugar in the 50 grams in the bowl, or 22/50. Simplify, and then convert to a percentage.