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# Problems Involving Mixtures, Coins, Moving Objects

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Multi-step word problems using a single variable

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### Problems Involving Mixtures, Coins, Moving Objects

1. 1. Multi-step Word problems<br />For complicated problems, it is helpful to spend more time in the fact finding phase. This presentation includes some of the common types of problems involving mixtures, coins and moving objects and we will show how to organize the facts into a table . Once we have the facts in order, the solution proceeds just like any other problem.<br />
2. 2. Mixture Problems<br />Aunt Flossie’s trail mix is made up of mixed nuts, dried cherries and chocolate chunks. She forgot to give you the recipe, but she did say that it contains equal parts of chocolate and cherries, and that you can make 5 pounds of it for \$23.50. You check the cost of each ingredient and find that the price per pound of the nuts is \$5, for the cherries \$6, and for the chocolate \$3. How much of each ingredient will you buy? <br />FIND: # of pounds of cherries, chocolate and nuts<br />FACTS: There is a lot of information to gather. It will work best to organize it in a table.<br />The recipe contains equal parts of cherries and chocolate, so we can represent the amount of each of them with x. <br />Also, we know the price per lb of each ingredient<br />
3. 3. Mixture Problems<br />Aunt Flossie’s trail mix is made up of mixed nuts, dried cherries and chocolate chunks. She forgot to give you the recipe, but she did say that it contains equal parts of chocolate and cherries, and that you can make 5 pounds of it for \$23.50. You check the cost of each ingredient and find that the price per pound of the nuts is \$5, for the cherries \$6, and for the chocolate \$3. How much of each ingredient will you buy? <br />FACTS: Continue to build the table<br />We know that the total amount is 5 lbs, so the amount of nuts will be 5 – (amt of cherries and chocolate)<br />The cost for each item is the amount times the unit price<br />
4. 4. Mixture Problems<br />FORMULA: Create equation from the information in the table. The sum of the cost of all items is \$23.50, so<br /> 6x + 3x + 5(5 – 2x) = 23.50<br />SOLVE: 6x + 3x + 25 – 10x = 23.50 (distributive property)<br /> 6x + 3x – 10x + 25 = 23.50 (commutative property)<br /> – x + 25 = 23.50 (combine like terms)<br /> – x + 25 – 25 = 23.50 -25 (subtract 25 from both sides)<br /> - x = - 1.5 <br /> x = 1.5 (divide both sides by -1) <br />ANSWER: The value of x answers the amount of both cherries and chocolate. To get the amount of nuts, we need to substitute the value of x into the expression that represents the amount of nuts : 5 – 2x = 5 – 2(1.5) = 5 -3 = 2. So our answer is 1.5 lbs of cherries, 1.5 lbs of chocolate and 2 lbs of nuts. To check, multiply the amount by the cost of each ingredient: <br />(6)(1.5) + (3)(1.5) + (5)(2) = 9 + 4.5 + 10 = 23.5 (our answer checks out)<br />
5. 5. Coin Problem<br />Sam took \$10.00 in quarters, dimes and nickels from his jar of spare change. He had twice as many quarters as nickels and 10 more dimes than quarters. How many coins of each denomination did he have?<br />FIND: number of dimes, nickels, quarters<br />FACTS: we will gather the facts in a table<br /><ul><li>We’ll use x to represent the number of nickels. There are twice as many quarters, so that is represented by 2x; and 10 more dimes than quarters gives us 2x+10
6. 6. Next we show the value of each coin and compute the value for the number of coins we have of each.
7. 7. We know that the total value of all coins is \$10 </li></li></ul><li>Coin Problem<br />FORMULA: The total value of all of the coins add up to \$10, so we can use the sum of their values to create our equation. <br />.05x + .10(2x+10) + .25(2x) = 10<br />SOLVE: .05x + .20x + 1 + .5x = 10 (use distributive law to simplify) <br /> .75x + 1 = 10 (combine like terms)<br />.75 x + 1 - 1 = 10 – 1 (subtract 1 from each side)<br /> .75x = 9<br /> .75x ÷ .75 = 9 ÷ .75 (divide both sides by .75)<br /> x = 12<br />ANSWER: x gives us the number of nickels. We’ll substitute that value into the expressions representing quarters and dimes: <br /> 2x = 2•12 = 24 quarters<br /> 2x + 10 = 2• 12 + 10 = 24 + 10 = 34 dimes<br />Check the answer by multiplying the number of each coin by its value<br />(.05)(12) + (.10)(34) + (.25)(24) = .6 + 3.4 + 6 = 10 (we get the total value of 10)<br />
8. 8. Moving objects<br />George is practicing for a marathon. An hour after he leaves for his practice run, his friend Fred leaves on his bicycle to check on his progress. It takes Fred 40 minutes to catch up with him. If George maintains a running pace of 7 mph, how fast does Fred ride?<br />FIND: Fred’s rate in mph<br />FACTS: Gather the facts in a table. Think about what you know about the distance formula: d=rt.<br />George has traveled 1 hr 40 min. We need to express that in hours since our rate is in miles per hour. <br />FORMULA: Both travelers have gone the same distance, so we can create our equation by setting the two expressions for the travelers’ distances equal to one another.<br />
9. 9. Moving objects<br />SOLVE: <br />(Multiply both sides by 3)<br />(Divide both sides by 2)<br />ANSWER: Our solution indicates that Fred has traveled 17.5 mph on his bicycle. We can substitute that value into the original equation<br />(2/3) • 17.5 = 35/3 (the result checks out)<br />