Блог Анкара

In This Chapter

^ Concentrating on the concentrations of mixtures ^ Using quality times quantity as the standard ^ Taking an interest in interest problems

Classic type of mathematical word problem is the mixture problem. But mixture problems go beyond just mixing up antifreeze with water or chocolate syrup with milk. Mixture problems involve all sorts of situations where you combine so much of one thing that has a certain amount of worth or density with so much of something else that has more worth or more density.

In this chapter, you find problems that multiply numbers of coins times their monetary value, pints of coolant with their concentrations and cartons of goodies with their product count. All these make for some interesting and, yes, even useful problems. So you can put your spoon away and find out what mixture problems are all about.

Standardizing Quality Times Quantity

The common theme in all mixture problems is that you take two or more different Amounts (quantities) of two different Concentrations (or qualities) and mix them together to get an amount (quantity) that’s the sum of the two ingredient quantities and a concentration (quality) that’s somewhere between the concentration of the two ingredients that you started with. Figure 14-1 illustrates this property or theme with containers.

Figure 14-1:

The quantities add up to the resulting amount.

Quantity A

Quantity B

Quantity A + B

Quality (percent, value) A

Quantity A X Quality A

Quality (percent, value) B

Quantity B X Quality B

Quality (percent, value) C

Quantity (A + B) X Quality C where A < C < B or B < C < A

The figure shows that when you add two quantities together, you get an amount or new quantity that’s the sum of the two quantities that are put together. The concentration or quality of the resulting mixture is a blend of the two starting substances — its concentration is somewhere between the two that are being combined.

Mixing It Up with Mixtures

Mixture problems involving actual substances occur when you take two or more different solutions or granular compounds or anything that will combine or Mix And create a new combination that’s no longer purely one or the other. When you pour chocolate syrup into milk, you add volume to the liquid in the glass, and the color of the milk mixture isn’t as dark as the chocolate or as white as the milk. The more chocolate, the darker the mixture. Yum!

Improving the concentration of antifreeze

The radiators of cars serve to cool down the engine with liquid that circulates around and pulls the heat away. The concentration of the fluid in the radiator can be changed to reflect the temperatures of the particular season. The concentration of antifreeze should be greater in the winter or cold months and less in the warmer months.

The Problem: A service station owner wants to mix up some 35 percent antifreeze. He wants to use up his current supply of 100 gallons of 20 percent antifreeze and add enough 40 percent antifreeze to bring the mixture up to 35 percent. How many gallons of 40 percent antifreeze should he add?

Vjj. VLA/V Keep in mind Quality x quantity For the two solutions being added together and the Quality And Quantity Of the resulting solution. Let X Represent the number of gallons of 40 percent antifreeze that needs to be added. The qualities are the percentages of the solutions, and the quantities are the gallons of the respective solutions.

Starting with [20 percent x 100 gallons] + [40 percent x X Gallons] = [35 percent x (100 + x) gallons], you see that the desired concentration, 35 percent, is between 20 percent and 40 percent. Also, the resulting quantity is the sum of the two quantities X And 100. Now rewrite the equation so it can be solved, replacing percentages with the decimal equivalents. Solving for X,

0.20 (100) + 0.40x = 0.35 (100 + X) 20 + 0.40x = 35 + 0.35x

0.05X= 15

15

0.05

300

It’ll take 300 gallons of the 40 percent antifreeze to bring the concentration up to what he wants. Hope he has a large enough container!

The Problem: The same service station owner from the previous problem still wants to make his 100 gallons of 20 percent antifreeze into 35 percent antifreeze. How many gallons of Pure Antifreeze will it take to raise the concentration?

^VLAiV You can use the same basic equation of Quality x quantity And multiply the unknown amount, X, By 100 percent. The equation takes the form of [20 percent x 100 gallons] + [100 percent x X Gallons] = [35 percent x (100 + x) gallons]. Writing this in a form to be solved,

0.20 (100) + 1.00x = 0.35 (100 + X) 20 + 1.00x = 35 + 0.35x 0^5x = 15 15

X = 7TFF . 23.077 0.65

It’ll take just a little more than 23 gallons of pure antifreeze to raise the concentration to 35 percent.

If you want to add pure antifreeze to what’s in your radiator right now, to increase the concentration, you have to take some of the mixture out of the radiator, first. The radiator holds only so much fluid. So, an even more interesting problem involves removing a certain amount of what’s in the radiator now and replacing it with pure antifreeze to achieve the level of concentration that you want.

The Problem: A man has a 16-quart radiator that now contains 16 quarts of 20 percent antifreeze. How much of the current mixture has to be removed and replaced with pure antifreeze to raise the level of concentration to 35 percent?

The basic structure for the mixture problem will have one more term in it. You’ll have a Quality x quantity Term that’s subtracted from the original amount. Let X Represent the number of quarts taken out and put back into the radiator. The end quantity will have to be the original 16 quarts.

[20 percent x 16 quarts] – [20 percent x X Quarts] + [100 percent x X Quarts] = [35 percent x 16 quarts]

Solving this equation,

0.20 (16) – 0.20x + 1.00x = 0.35 (16) 3.2 + 0.80x = 5.6 0.80x = 2.4

2.4

X = AOA = 3

Take out 3 quarts of the 20 percent solution and replace it with pure antifreeze.

Coffee, tea, or not

Two friends were at a restaurant and ordered coffee. The waiter came back with another cup. coffee to have with their dessert. One of the But the friend became very upset and returned it friends found a fly in his coffee and called for the again, saying that the second cup was the same waiter to take it away and bring a fresh cup of as the first cup. How did he know?

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Watering down the wine

Just as concentrations of solutions are increased by adding a pure substance, the concentrations can also be decreased by adding pure water. You perform this type of operation all the time, such as when mixing up orange juice from frozen concentrate or when fixing mixed drinks such as bourbon and water.

The Problem: How much water must be added to 8 ounces of 40 percent alcohol to produce a mixture that’s 7 percent alcohol?

Water is 0 percent alcohol, so use the Quality x quantity Setup to solve for X, The number of ounces of water needed.

[40 percent X 8 ounces] + [0 percent X X Ounces] = [7 percent X (8 + x) ounces]

Writing an equation and solving for X,

0.40 (8) + 0.00x = 0.07 (8 + X) 3.2 + 0 = 0.56 + 0.07x

2.64= 0.07X

2.64

0 07 = X x. 37.71

It’ll take about 38 ounces of water to get the alcohol down to the 7 percent level.

Mixing up insecticide

Farmers know how important it is to keep destructive insects out of their crop-producing fields. They apply enough insecticide to keep the bugs away, but they don’t want to overdo it and poison the field for the future. Mixing the insecticides and applying them where necessary is a complicated problem.

The Problem: To get rid of a nasty strain of root worm, an insecticide mixture should have a concentration that’s 9 percent insecticide. A farmer has two mixtures on hand — one with 5 percent insecticide and the other with 15 percent insecticide. What is the ratio of the 5 percent to 15 percent mixtures that should be combined to get a 9 percent mixture? If the farmer needs 40 gallons of mixture, how many gallons of each should she use?

^VLA/V First solve for the ratio of the different mixtures and then apply it to the quantity of 40 gallons. Use the Quality X Quantity Setup. Let X Represent the fraction of 5 percent solution and 1 – X Represent the fraction of 15 percent solution. Adding X + 1 – X You get 1, which is All Of the final mixture.

[5 percent x X ] + [15 percent x (1 - x)] = [9 percent x 1]

Solving for X In the related equation,

0.05 (X) + 0.15 (1 – X) = 0.09 (1) 0.05x + 0.15 – 0.15x = 0.09

0.15-0.10X=0.09 -0.10X=-0.06

_ A A/? C O

X = — = T0 = 5

The farmer needs to let 3, or 60 percent, of the insecticide be made up of 5

The 5 percent solution and 5, or 40 percent, of the mixture be 15 percent

Solution. In a 40-gallon situations, 60 percent is 0.60 x 40 = 24 gallons and the other 40 percent is 0.40 x 40 = 16 gallons.

Counting on the Money

Problems concerning money — no, I’m talking about Math Problems, not just any problem over money — involve the value of the currency or coin or vending unit along with the number of each of these monetary values. The theme of the Quality X Quantity Applies very well here, where the Quality Is the coin value, bill value, or commodity value in the problem.

Determining how many of each denomination

Coins and paper money come in many different denominations, allowing people to carry a lot of money with just a few bills or to make change for purchases using smaller bills and coins. What’s consistent for all money in all countries is that you determine the total amount that you have by multiplying the value of the coin or bill times the number of them that you’re carrying.

The Problem: Cassie has a total of $4.10 in dimes and quarters. If she has twenty coins in all, how many of them are quarters?

^VLA/V Multiply the value (quality) of a quarter, $0.25, times the number of quarters and the value of a dime, $0.10, times the number of dimes. The sum of the two results is set equal to $4.10. You know that the number of quarters and dimes equals 20. Also, you want to solve for quarters. So let Q Represent the number of quarters, leaving 20 – Q To represent the number of dimes. Setting up the equation and solving,

0.25 (q ) + 0.10 (20 — Q ) = 4.10 0.25q + 2 — 0.10q = 4.10 0.15q + 2 = 4.10

0.15Q = 2.10

2.10

Q = 015 = 14

Cassie has 14 quarters. You check your answer by determining the number of dimes, 20 – Q = 20 – 14 = 6. Multiply 14 x 0.25 = 3.50 and 6 x 0.10 = 0.60. The sum 3.50 + 0.60 = 4.10. So the solution checks.

The Problem: Grace has twice as many nickels as quarters and five less than three times as many dimes as quarters. If she has a total of $6, then how many of each coin does she have?

^VLA/rV Because both the nickels and dimes are compared to quarters, let Q Represent the number of quarters and write expressions for the number of nickels and dimes using the relationships. The number of nickels is written as 2q, and the number of dimes is 3Q - 5. Multiply each quantity of coin by its quality and add up the products, setting the sum equal to 6.

0.25 (q) + 0.05 (2q) + 0.10 (3q — 5) = 6.00 0.25q + 0.10q + 0.30q — 0.50 = 6.00 0.65q — 0.50 = 6.00

0.65Q = 6.50 Q 6.50

Grace has 10 quarters. If she has twice as many nickels as quarters, that makes it 20 nickels. And 5 less than 3 times 10 is 30 minus 5 or 25 dimes. Checking on the total, 0.25(10) + 0.05(20) + 0.10(25) = 2.50 + 1 + 2.50 = 6.

It isn’t always clear-cut which coin or piece of currency you want to use to have their amount represent the variable. You almost always have more than one choice for X. Your goal is to pick the money amount that makes writing the problem as easy as possible.

The Problem: A cash drawer contains singles, fives, tens, and twenties. There are two more tens than fives, eight less than twice as many twenties as tens, and ten more than twice as many singles than twenties. If the total amount of money in the cash drawer is $650, then how many of each bill is there?

Several different comparisons are going on here, each with the number of bills of a particular denomination being compared to the number of bills of another denomination. You need to pick a variable to represent how many you have of one type of bill and work from that amount.

One suggestion is to start by letting the number of fives be represented by F. Then the number of tens is represented by F + 2. Comparing twenties to tens, you write Eight less than twice as many twenties as tens By subtracting 8 from 2 times the number of tens, or 2(f + 2) – 8. Now take that number of twenties and write the number of singles (ten more than twice as many singles as twenties) As 2[2(f + 2) - 8] + 10.

You then write an equation taking the number of each type of bill times its monetary value, adding up all these products, and setting the sum equal to $650. But first, it’s a good idea to simplify the expressions for the numbers of

Each bill. The number of fives is F, Which is fine. The number of tens, F+ 2, is also as simple as it can get. But the number of twenties can be simplified. 2(f + 2) – 8 = 2f + 4 – 8 = 2f – 4. And the number of singles has a long way to go, because 2[2(f + 2) - 8] + 10 = 2[2f + 4 - 8] + 10 = 2[2f - 4] + 10 = 4f – 8 + 10 = 4f + 2. Now you’re set to go. Write the equation.

5 (F) + 10 (F + 2) + 20 (2f — 4) + 1 (4f + 2) = 650 5f + 10f + 20 + 40f — 80 + 4f + 2 = 650 59f — 58 = 650 59f= 708

F = l08 = 12

There are 12 five-dollar bills, two more tens than fives (or 14 ten-dollar bills), eight less than twice as many twenties as tens (or 20 twenties), and ten more than twice as many singles as twenties (or 50 singles). Checking this out for the total: 5(12) + 10(14) + 20(20) + 1(50) = 60 + 140 + 400 + 50 = 650.

Making a marketable mixture of candy

Some people prefer dark chocolate. Others go for the gooey chocolate-covered cherries. Many don’t care for coconut in their sweets. And most people are happy with any mixture of any kind. Different kinds of candies have different prices, depending on the ingredients. When you combine different candies in packages, the Quality Or price of each type is multiplied by the Quantity Or weight to determine the price of the mixture.

The Problem: Malted milk balls sell for $3 per pound, and chocolate-covered peanuts cost $4 per pound. How many pounds of each type of candy should be used to create a 5-pound box of candy that costs $3.60 per pound?

^VLA* Let M Represent the number of pounds of malted milk balls. Because the total amount of candy is to be 5 pounds, then 5 – M Can represent the number of pounds of chocolate-covered peanuts. Multiply $3 times M And add it to the product of $4 and 5 – M. Set that sum equal to $3.60 times 5. Each term has the price times the weight. Solve for M.

3 (m) + 4 (5 — m) = 3.60 (5) 3m + 20 — 4m = 18 20 —m =18 2 = M

If you use 2 pounds of malted milk balls, you’ll need 3 pounds of the peanuts to make 5 pounds. Checking, 3(2) + 4(3) = 6 + 12 = 18. If you divide the total price of $18 by 5 pounds, you get $3.60, the cost per pound.

The Problem: A box of candy is to contain chocolate covered cherries that cost $5 per pound, nougats that cost $3 per pound, assorted cremes that cost $2 per pound, and caramels that cost $6 per pound. There should be an equal amount (by weight) of the caramels and cherries. You want twice as many nougats as caramels and twice as many cremes as nougats. How much of each type of candy should be used if the box is to cost $4.25?

V£.?LA/V First, you have to stop drooling and think about the mathematics, not the candy. Because the caramels and cherries will have equal weight, it makes sense to assign a variable to represent the weights of those two candies. Let C Be the number of pounds of caramels and C Be the number of pounds of chocolate-covered cherries. For the nougat, you want twice as much as there are caramels, so that’s represented by 2C. And twice as much of the nougat is 2(2C) = 4C, For the weight of the cremes. Now multiply each weight of candy by the respective cost, add them together, and set the sum equal to 4.25.

5 (C) + 6 (C) + 3 (2c) + 2 (4c) = 4.25 5c + 6c + 6c + 8c = 4.25 25C = 4.25

4.25

25

0.17

So the mixture will contain 0.17 pound of chocolate covered cherries, 0.17 pound of caramels, 0.34 pound of nougat, and 0.68 pound of assorted cremes.

Running a concession stand

Picture this: It’s game night, and you’ve made a dash to the food vendor to get a quick snack. So did half the people in the stadium. Now you have time to stand in line and wait and ponder the problem of combining different foods and drinks available in the concession stand so that you can spend every penny of the money in your pocket — no more, no less.

The Problem: Stan had exactly $20.40 in his pocket and managed to spend it all at the concession stand. He bought three hot dogs, two servings of cheese fries, and one large drink for that amount of money. One hot dog costs $1.80 more than one serving of cheese fries. And a drink costs $1.20 less than a hot dog. How much did each item cost?

The number of items have to be multiplied by their respective prices, and the costs totaled. Set that total equal to $20.40 to solve for the individual prices. Let the cost of cheese fries be represented by C. Then a hot dog costs C + 1.80. A drink is $1.20 less than a hot dog, so it costs C + 1.80 – 1.20 = C + 0.60.

2 (C) + 3 (C + 1.80) + 1 (C + 0.60) = 20.40 2c + 3c + 5.40 + C + 0.60 = 20.40 6c + 6.00 = 20.40 6c = 14.40

C = 1440 = 2.40

6

So cheese fries cost $2.40, the hot dogs are $1.80 more than that (or $4.20), and the drink is $0.60 more than the fries (or $3).

The Problem: You’re buying supplies for the concession stand Friday night and need to purchase candy and pretzels and gum in bulk. Twix candy bars are 35<t each if you buy a case of 36. M&M’s bags are 36<t each if you buy a case of 48. Pretzels are 19<t a bag if you buy a case of 30. Cookies are 24<t a package if you buy a case of 33. And gum costs 18<t a pack if you buy a case of 40. You buy the same number of cases of cookies and gum. You buy twice as many cases of M&M’s as cookies. You buy two less than twice as many cases of pretzels as cookies. And you buy five fewer cases of Twix than M&M’s. If you spend $270.72 on all these supplies, then how many individual items do you have on hand to sell Friday night?

This problem just begs to be organized. A spreadsheet would be a big help here. But the next best thing is a table showing the item name, the number of items in a case, the cost per case, and a representation of the number of cases ordered. Because the number of cases of cookies and gum is the same, let the number of cases of these items be represented by G. The number of cases of M&M’s is 2g. The number of cases of pretzels is 2g – 2; and the number of cases of Twix is 2g – 5. Table 14-1 shows all the entries.

The equation consists of the sum of the products of the costs and the number of cases set equal to the total of $270.72.

7.92 _g) + 7.20_g) + 17.28 _2g) + 5.70_2g – 2) + 12.60_2g – 5) = 7.92g + 7.20g + 34.56g + 11.40g – 11.40 + 25.50g – 63 =

86.28G-74.40 = 86.28G=

G=

G=

You bought 4 cases of cookies and 4 cases of gum. You bought 8 cases of M&M’s, 6 cases of pretzels and 3 cases of Twix. Multiplying each number of cases by the number of items in that case, you have 4(33) + 4(40) + 8(48) + 6(30) + 3(36) = 964 items. Have fun!

270.72 270.72 270.72 345.12

345.12 86.28 4

Being Interested in Earning Interest

Earning interest on money invested is of utmost importance to the wise investor. Some funds pay a higher rate of interest but may be a bit risky. To offset the risk, the shrewd investor puts some money in a high-risk fund and the rest in a fund that doesn’t pay as well but one that can be trusted to give a return and not lose any of the investment.

Making your investment Work for you

The problems in this section assume the use of the Simple-interest formula, Compounded annually. In actuality, financial institutions use compound interest and computer programs to figure out these problems. But you get a good idea of how it works — and a pretty good estimate of the actual answer using the less complex simple-interest formula.

The simple-interest formula says that the interest earned, /,is equal to the amount invested (principal), P, Times the percentage rate, R, Written as a decimal, times the number of years (time), T. The formula is: I = Prt.

The Problem: Robert has $50,000 to invest. He wants to put some of this money in an account that earns 8 percent interest and the rest in a riskier account that promises to earn at the rate of 12 percent. He needs yearly earnings of a total of $4,500. How much of his $50,000 should he invest in each account?

^VLA* Write an equation in which the First interest Plus the Second interest Is equal to $4,500. The First interest And Second interest Are expressions that you write using the interest formula. This is another Quality Times Quantity Situation. Let X Represent the amount of money invested in the first account and 50,000 – X Be the amount of money in the second account. Assume that the time is one year. Then solve for X.

Robert needs to invest $37,500 at 8 percent and the rest, $50,000 – $37,500 = $12,500, at 12 percent.

The Problem: Stella has been watching several investment funds and has decided that she’ll deposit some of her $100,000 in each. She will put twice as much in the fund earning 5 percent as that earning 4 percent, $5,000 more in the fund earning 10 percent than in the fund earning 4 percent, and $5,000 less in the fund earning 6 percent as in the fund earning 4 percent. How much will she get in total interest earnings at the end of the year? How much did she invest in each fund?

.cVLA/v First, determine how much she’s investing in each fund by letting X Represent the amount in the 4 percent fund, let 2x represent the amount in the 5 percent fund, let X + 5,000 represent the amount in the 10 percent fund, and let X - 5,000 represent the amount in the 6 percent fund. Add up all the fund amounts and set them equal to 100,000 to solve for X.

X + 2x + x + 5000 + X – 500 = 100,000

5X = 100,000

100,000 20 000 X =-f-= 20,000

If X = 20,000, then the amounts in the 4 percent, 5 percent, 10 percent, and 6 percent funds are $20,000, $40,000, $25,000 and $15,000, respectively. Multiply each fund amount and percentage together to get the total interest.

$20,000(0.04) + $40,000(0.05) + $25,000(0.10) + $15,000(0.06) = $800 + $2,000 + $2,500 + $900 = $6,200. Stella should earn $6,200 in interest.

Determining how much is needed for the future

It’s all well and good if you have $100,000 or some such amount of money to invest. Wouldn’t we all just love that? Frequently, though, the question is more like, "How much do I need to invest in order to get that kind of interest?" Nonprofit organizations like to have Endowment funds (money put aside that’s never touched, with only the interest used to pay expenses). They want to know how much is needed in the endowment fund in order to have a particular income every year from the interest.

The Problem: The local Kiwanis Club needs $20,000 in interest annually in order to adequately fund its benevolence programs. It currently has $120,000 invested — one-third of it in a mutual fund earning 5 percent interest and the rest in a money-market fund earning 10 percent interest. This division of the funds is mandated by its endowment agreement. How much more money does the Kiwanis Club need, and how much does it have to have in each fund in order to earn that $20,000 annual interest?

First determine the total amount of money needed to generate the $20,000 interest each year when one-third goes into the 5 percent fund and two-thirds into the 10 percent fund. Then, after you find out how much is needed, you can see what the difference is between that and what the Kiwanis Club already has.

Think of the division of money as being divided into three parts — one part in the 5 percent fund and two-thirds in the 10 percent fund. Rather than use fractions, let the whole three-thirds be represented by 3X. Then let X Represent the amount of money invested at 5 percent and 2x represent the amount invested at 10 percent. Write the two interest terms, add them up, and set them equal to 20,000.

X (0.05) + 2x (0.10) = 20,000 0.05x + 0.20x = 20,000

0.25X= 20,000

20,000 80 000 X = AOR= 80,000 0.25

It needs to have $80,000 invested at 5 percent and twice that, $160,000 invested at 10 percent. So a total of $240,000 is needed in the two funds. Subtracting what it already has, $240,000 – $120,000 = $120,000. The Kiwanis Club is halfway to its goal. Looks like it’ll have to host a few more pancake breakfasts to get there!

Chapter 15