Free Pre-Algebra Lesson 44 page 1 Lesson 44 Percents in Mixtures Chemists may specify the strength of a solution by using a percent. For example, you can buy isopropyl alcohol at the drug store in a 91% solution, which means that it contains 91% alcohol and 9% something else (probably water). The proportions of oil and vinegar in salad dressing might be specified as two parts oil to one part vinegar, a ratio which allows you to mix the exact amount you need. It s sometimes tricky to think about percents in solutions and mixtures, but these problems are designed to help you be prepared for the similar, but more advanced problems you will encounter in your algebra course. Three Examples of Mixtures The examples we ll do use these mixtures: A bin of trail mix, consisting of nuts and raisins. A bottle of salad dressing, consisting of oil and vinegar. A chemical solution, consisting of water and alcohol. Although we could have more than two different ingredients in a mixture it s slightly easier if we limit ourselves to two. Key points: The entire mixture, with the ingredients combined, is the base of the percents. The amounts of each of the two ingredients are added together to make the mixture. The percents of the two ingredients should add up to 100%. 100% is the entire mixture. Example: Find the total amount in the mixture. The bin of mixed nuts and raisins contained 4 lb nuts and 3 lb raisins. How much is in the bin? 4 lb + 3 lb = 7 lb A salad dressing consists of 2/3 cup oil and 1/2 cup vinegar. How much salad dressing is there? 2/3 cup + 1/2 cup = 4/6 cup + 3/6 cup = 7/6 cups An alcohol solution consists of 50 ml alcohol and 250 ml water. How many ml of solution are there? 50 ml + 250 ml = 300 ml Example: Find the percent of the other ingredient. The bin held 55% nuts, by weight. What percent was raisins? 100% 55% = 45% The salad dressing was 60% oil. What percent was vinegar? 100% 60% = 40% A solution was 20% alcohol. What percent was water? 100% 20% = 80%
Free Pre-Algebra Lesson 44 page 2 Finding the Percent of Ingredients To find the percent of each ingredient, divide the amount of that ingredient by the total amount in the mixture. Example: Find the percent of the ingredient, rounded to the nearest whole percent. The bin of mixed nuts and raisins contained 4 lb nuts and 3 lb raisins. What percent is nuts? 4 lb + 3 lb = 7 lb 4 lb / 7 lb = 0.5714 about 57% nuts A salad dressing consists of 2/3 cup oil and 1/2 cup vinegar. What percent is vinegar? 2/3 cup + 1/2 cup = 4/6 cup + 3/6 cup = 7/6 cups 1/2 cup / 7/6 cups = 0.4285 about 43% vinegar An alcohol solution consists of 50 ml alcohol and 250 ml water. What percent is alcohol? 50 ml + 250 ml = 300 ml 50 ml / 300 ml = 0.1666 about 17% alcohol About what percent is raisins? 100% 57% = 43% about 43% raisins About what percent is oil? 100% 43% = 57% about 57% oil About what percent is water? 100% 17% = 83% about 83% water Finding the Amount of Ingredients These are just percent problems. The percent of the base is the amount. Example: Find the amount of the ingredient, rounded to the nearest tenth. There were 20 lb of mixed nuts and raisins, of which 30% was nuts. How many pounds of nuts? 30% of 20 lb is 0.3 20 = 6 There were 6 lb nuts. There were 1 1 / 2 cups of salad dressing. The dressing was 60% oil. How much oil was in the dressing? 60% of 1 1 /2 cups is 0.6 1.5 = 0.9 9/10 cup of oil There were 1800 ml of a 20% alcohol solution. How many ml of alcohol were there? 20% of 1800 ml is 0.2 1800 = 360 360 ml alcohol How many pounds of raisins? 20 lb 6 lb = 14 lb 14 pounds of raisins How many cups of vinegar? 11/2 9/10 = 15/10 9/10 = 6/10 = 3/5 3/5 cup of vinegar How many ml water? 1800 ml 360 ml = 1440 ml 1440 ml water
Free Pre-Algebra Lesson 44 page 3 Finding the Base The base of the percents is the total amount of the mixture or solution. In these problems we solve the percent equation to find the base Example: Find the total amount in the mixture or solution. When they dumped in 18 lb nuts, the mixture was 75% nuts. How many lbs in the entire mixture? 75% of entire = 18 lb entire = 18 lb/0.75 = 24 lb The entire mixture was 24 lb. How many pounds of raisins? 24 lb 18 lb = 6 lb 6 pounds of raisins He added 3/4 cup of oil to the vinegar, which made it 60% oil salad dressing. How many cups of salad dressing were there? 60% of dressing = 3/4 dressing = 0.75 / 0.60 = 1.25 There were 1 1 /4 cups of salad dressing. How many cups of vinegar? 1 1 /4 3/4 = 2/4 = 1/2 1/2 cup of vinegar Adding 80 ml alcohol to the water made a 40% alcohol solution. How much liquid was in the entire solution? 40% of solution = 80 ml solution = 80 ml / 0.4 = 200 There were 200 ml of solution. How many ml water? 200 ml 80 ml = 120 ml 120 ml water Turn the page. Keep going. Ignore the kitten.
Free Pre-Algebra Lesson 44 page 4 Combining Mixtures My bin of trail mix gets poured into your bin of trail mix, etc. We use the amounts and bases from the old mixtures to find the facts about the new mixture. Example: Mixture C is created by combining Mixture A with Mixture B. Answer the questions about each mixture. a. You have a bin of Mixture A, which consists of 12 lb nuts and 8 lb raisins. b. You have another bin of Mixture B, which weighs 40 lb. It is 75% nuts. c. Mixture C is created by pouring Mixtures A and B together. How much does Mixture A weigh? 8 lb + 12 lb = 20 lb What percent is nuts? 12/20 = 0.6 60% What percent is raisins? 100% 60% = 40% How much do the nuts in Mixture B weigh? 0.75 40 lb = 30 lb What percent of Mixture B is raisins? 100% 75% = 25% How much do the raisins weigh? 40 lb 30 lb = 10 lb How much does Mixture C weigh? Wt A + Wt B = Wt C 20 lb + 40 lb = 60 lb How much do the raisins in Mixture C weigh? Wt A + Wt B = Wt C 8 lb + 10 lb = 18 lb What percent of Mixture C is raisins? 18/60 = 0.3 30% It s VERY IMPORTANT to notice that the percents do not add when the mixtures are added together. The weights or volumes of the old mixtures will add to make the new mixture. But the percents in the new mixture come from a new base, the total amount in the new mixture. So Mixture A is 40% raisins, and Mixture B is 25% raisins, but Mixture C is 30% raisins. The percent in the new mixture will always be BETWEEN the two percents from the old mixtures. You can see why by looking at the picture of Mixture C. We can organize the information from the problem into a table: Raisins + Nuts = Total Mixture A 8 lb 12 lb 20 lb Mixture B 10 lb 30 lb 40 lb Mixture C 18 lb 42 lb 60 lb The columns add across, because the total amount in mixture A is the sum of the amount of raisins plus the amount of nuts. The columns also add down, since Mixture C comes from adding Mixture A and Mixture B together.
Free Pre-Algebra Lesson 44 page 5 Example: Solution C is created by combining Solution A with Solution B. Answer the questions about each solution. a. Solution A has 90 ml alcohol and is 450 ml total. Fill in the table for Solution A. 450 90 = 360 Alcohol + Water = Total Solution A 90 ml 360 ml 450 ml c. Solution B is 30% alcohol and is 300 ml total. Fill in the table for Solution B. alcohol is 30% of 300 ml alcohol = 0.3 300 ml = 90 ml 300 90 = 210 Alcohol + Water = Total Solution B 90 ml 210 ml 300 ml b. What percent alcohol is Solution A? 90 / 450 = 0.2 = 20% d. Solution C is made by pouring Solutions A and B together. Fill in the table for Solution C. Solution C Alcohol + Water = Total 90 ml 360 ml 450 ml + 90 ml + 210 ml + 300 ml 180 ml What percent alcohol is Solution C? 570 ml 180 / 750 = 0.24 24% 750 ml
Free Pre-Algebra Lesson 44: Percents in Mixtures Worksheet Lesson 44 page 6a Name Round to the nearest whole percent or the nearest tenth of a unit, if rounding is necessary. 1. If a mixture of oil and vinegar is 24% vinegar, what percent is oil? 2. If a mixture of nuts and raisins weighs 18 lb, and 7 lb is raisins, what percent is raisins? If a mixture of oil and vinegar is 14 fluid ounces, and 5 ounces are vinegar, how many ounces are oil? What percent is nuts? 3. If a solution of alcohol and water is 900 ml and 78% alcohol, how many ml of alcohol are present? 4. A piece of metal was composed of 45% nickel. The amount of nickel was 4g. How much did the piece of metal weigh? How many ml of water? 5. Mixture C is formed by combining Mixtures A and B. Mixture A: 2 cubic meters of soil mixture, 35% sand Mixture B: 6 cubic meters of soil mixture, 15% sand c. How many cubic meters of sand are in Mixture A? d. How many cubic meters of sand are in Mixture B? a. The percent of sand in Mixture C is between % and %. b. The amount of soil mixture in Mixture C is cubic meters. e. How many cubic meters of sand are in Mixture C? f. What percent of Mixture C is sand?
Free Pre-Algebra Lesson 44 page 7a 6. Mixture C is formed by combining Mixtures A and B. Mixture A: 9 lb mixed nuts, 55% peanuts Mixture B: 6 lb mixed nuts, 40% peanuts c. How many lbs peanuts are in Mixture A? d. How many lbs peanuts are in Mixture B? a. The percent of peanuts in Mixture C is between % and %. e. How many lbs peanuts are in Mixture C? b. The amount of mixed nuts in Mixture C is lbs. f. What percent of Mixture C is peanuts? 7. Solution C is formed by combining Solutions A and B. Solution A: 80 liters of 20% alcohol Solution B: 40 liters of 50% alcohol c. How many liters of alcohol are in Solution A? d. How many liters of alcohol are in Solution B? a. The percent of alcohol in Solution C is between % and %. e. How many liters of alcohol are in Solution C? b. Solution C is liters. f. What percent of Solution C is alcohol?
Free Pre-Algebra Lesson 44 page 8 Lesson 44: Percents in Mixtures Homework 44A Name 1. 2/3 of the 504 callers wanted directions. How many callers wanted directions? 2. 66 of the 88 sandwiches contained meat. What fraction of the sandwiches contained meat? 3. 80% of the 340 guests stayed past midnight. How many guests stayed past midnight? 4. 218 of the 545 votes were opposed. What percent were opposed? 5. Find equivalent fractions with a common denominator. 5 14 6. Add 8 5 14 + 1119 28 19 28 7. Convert 8 months to years (12 months = 1 year). 8. Convert 43 centimeters to meters (100 cm = 1 m). 9. Simplify 0.5x + 0.2( 12 x ) 10. Simplify 20 3x 10 + 1 $ 4%
Free Pre-Algebra Lesson 44 page 9 11. The sales tax rate was 9.25%. If the sales tax for the item was $11.47, what was the price? 12. If Lexi and Brandy left a $6 tip for the meal that cost $32.40, what percent did they tip? 13. The sale advertised 20% off a $158 carpet. What was the sale price? 14. The loan was for $700 at 6% simple interest. If you pay back the loan plus two years interest, what will the total you pay be? 15. $25,000 was invested in two accounts. $7,500 was invested at 2% simple interest and the rest at 4%. What is the total interest on the two accounts after one year? 16. $25,000 was invested at 3% interest compounded twice each year. How much was in the account after 10 years? A = P 1+ r $ n % nt 19. Mixture C is made by combining Mixtures A and B. Mixture A: 30 lbs of concrete mixture contains 30% sand. Mixture B: 50 lbs of concrete mixture contains 40% sand. a. The percent of sand in Mixture C is between % and %. b. How many lbs does Mixture C contain? lb. c. How many lbs of sand are in Mixture A? d. How many lbs of sand are in Mixture B? e. How many lbs of sand are in Mixture C? f. What percent of Mixture C is sand?
Free Pre-Algebra Lesson 44 page 10 Lesson 44: Percents in Mixtures Homework 44A Answers 1. 2/3 of the 504 callers wanted directions. How many callers wanted directions? 168 2 3 504 = 336 336 callers wanted directions. 3. 80% of the 340 guests stayed past midnight. How many guests stayed past midnight? 80% of 340 guests is 0.8 340 = 272 272 guests stayed past midnight. 2. 66 of the 88 sandwiches contained meat. What fraction of the sandwiches contained meat? 66 88 = 2 3 11 2 2 2 11 = 3 4 3/4 of the sandwiches contained meat. 4. 218 of the 545 votes were opposed. What percent were opposed? 218 / 545 = 0.4 40% were opposed. 5. Find equivalent fractions with a common denominator. 5 14 = 5 2 7 2 2 = 10 28 19 19 = 28 2 2 7 = 19 28 7. Convert 8 months to years (12 months = 1 year). 6. Add 8 5 14 + 1119 28 ( ) + 10 = 8 + 11 = 8 10 28 + 1119 28 28 + 19 $ 28% = 19 + 1 1 28 = 20 1 28 = 19 + 29 28 8. Convert 43 centimeters to meters (100 cm = 1 m). 8 months 1 1 year 12 months = 0.6 years 43 cm 1 1 m 100 cm = 0.43 m 9. Simplify 0.5x + 0.2( 12 x ) = 0.5x + 0.2 12 0.2x = 0.5x 0.2x + 2.4 = 0.3x + 2.4 10. Simplify 20 3x 10 + 1 $ 4% 2 20 3x 5 10 + 20 1 4 = 6x + 5
Free Pre-Algebra Lesson 44 page 11 11. The sales tax rate was 9.25%. If the sales tax for the item was $11.47, what was the price? 9.25% of the price is $11.47 the price is $11.47 / 0.0925 = $124.00 The item cost $124. 12. If Lexi and Brandy left a $6 tip for the meal that cost $32.40, what percent did they tip? $6 / $32.40 = 0.185185 They tipped about 18.5%. 13. The sale advertised 20% off a $158 carpet. What was the sale price? 100% 20% = 80% 0.8 158 = 126.4 The sale price was $126.40. 14. The loan was for $700 at 6% simple interest. If you pay back the loan plus two years interest, what will the total you pay be? I = Prt = ($700)(0.06)(2) = $84 interest You will pay $784 in all. 15. $25,000 was invested in two accounts. $7,500 was invested at 2% simple interest and the rest at 4%. What is the total interest on the two accounts after one year? 25,000 7,500 = 17,500 2% of 7,500 + 4% of 17,500 = 0.02 7,500 + 0.04 17,500 = 150 + 700 = 850 The total interest is $850. 19. Mixture C is made by combining Mixtures A and B. Mixture A: 30 lbs of concrete mixture contains 30% sand. Mixture B: 50 lb of concrete mixture contains 40% sand. a. The percent of sand in Mixture C is between 30 % and 40 %. b. The amount of Mixture C is 80 lb. 16. $25,000 was invested at 3% interest compounded twice each year. How much was in the account after 10 years? A = P 1+ r $ n % nt A = 25000 1+ 0.03 $ 2 % A = 25,000 1.015 ( ) 20 ( ) = 25,000 1.346855007 = 33671.37516 There was $33,671.38 after 10 years. c. How many lbs of sand are in Mixture A? 330% of 30 lb is 0.3 30 lb = 9 lb d. How many lbs of sand are in Mixture B? 40% of 50 lb = 20 lb e. How many lbs of sand are in Mixture C? 9 lb + 20 lb = 29 lb f. What percent of Mixture C is sand? 29 / 80 = 0.3625 36.25% or about 36% 2 10
Free Pre-Algebra Lesson 44 page 12 Lesson 44: Percents in Mixtures Homework 44B Name 1. 7/8 of the 504 callers wanted directions. How many callers wanted directions? 2. 52 of the 65 sandwiches contained meat. What fraction of the sandwiches contained meat? 3. 80% of the 540 guests stayed past midnight. How many guests stayed past midnight? 4. 163 of the 815 votes were opposed. What percent were opposed? 5. Find equivalent fractions with a common denominator. 17 30 6. Add 15 17 30 + 111 18 11 18 7. Convert 5 months to years (12 months = 1 year). 8. Convert 3 centimeters to meters (100 cm = 1 m). 9. Simplify 0.8x + 0.3( 20 x ) 10. Simplify 30 3x 15 + 1 $ 2%
Free Pre-Algebra Lesson 44 page 13 11. The sales tax rate was 9.75%. If the sales tax for the item was $15.60, what was the price? 12. If Lexi and Brandy left a $5 tip for the meal that cost $32.40, what percent did they tip? 13. The sale advertised 25% off a $168 carpet. What was the sale price? 14. The loan was for $800 at 6.5% simple interest. If you pay back the loan plus 6 months interest, what will the total you pay be? 15. $25,000 was invested in two accounts. $9,500 was invested at 2% simple interest and the rest at 6%. What is the total interest on the two accounts after one year? 16. $85,000 was invested at 4.5% interest compounded weekly. How much was in the account after 5 years? A = P 1+ r $ n % nt 19. Mixture C is made by combining Mixtures A and B. Mixture A: 80 lbs of concrete mixture contains 30% sand. Mixture B: 20 lb of concrete mixture contains 40% sand. a. The percent of sand in Mixture C is between % and %. b. How many lbs is Mixture C? lb. c. How many lbs of sand are in Mixture A? d. How many lbs of sand are in Mixture B? e. How many lbs of sand are in Mixture C? f. What percent of Mixture C is sand?