Distance = Rate x Time

Transcription

1 Math 233A Chapter 2.4: More Application Problems Objectives: Finding distance, rate, and time for going in the same direction Finding distance, rate, and time for going in towards each other Finding distance rate, and time for going in opposite directions Finding distance, rate, and time for going slow, then going faster Solve mixture problems Chapter 2.4 Steps for Solving Application Problems: 1. Read, throw out nonsense numbers 2. Assign a variable (What is it asking for?) 3. Write an equation 4. Solve the equation 5. Answer the question! 6. Check, does it make sense? Distance = Rate x Time Finding Distance, Rate, and Time For Going in the Same Direction Ex: Two people start walking at the same time in the same direction. One person walks 4 mph and the other person walks 6 mph. How long until they are 0.5 miles apart. Slower Faster faster's dislance slower's distance total distance apart Answer: 0.25 hours = time it takes to be 0.5 miles apart Finding Distance, Rate, and Time For Going in the Same Direction and Catching Up Ex: Say you and your friend are hanging out. Your friend takes off going 55 mph down the freeway. Five minutes later, you decide to catch up with her and take off going 60 mph down the same freeway. How long does it take you to catch up to her, and how long has she been driving? Your friend friend's distance your distance You Answer: 55 minutes = time it takes took you to catch up 1 hour = time she has been driving

2 Finding Distance, Rate, and Time For Going Towards Each Other Ex: The distance between Gilroy and Watsonville is 24 miles. If you left Gilroy going 40 mph at the same time as your friend leaves Watsonville, going 45 mph, how long will it take to meet up? you friend your distance friend's distance total distance traveled Answer: 0.28 hours (about 17 minutes)«time it takes for both to travel 24 miles 1. Two cars are traveling toward each other from two cities 270 miles apart. One car is going 5 mph slower than the other. They meet up after 2 hours. How fast are they each going? = = Faster Slower faster's distance slower's distance total distance traveled Answer: 70 mph = faster car's rate 65 mph = slower car's rate

3 Finding Distance, Rate, and Time For Going In Opposite Directions Ex: Two cars are going in opposite directions and one car is going 10 mph slower than the other. After 1.5 hours they are 150 miles apart. How fast are they each going? _^ Faster Slower faster's distance slower's distance total distance traveled Answer: 55 mph = faster car's rate 45 mph = slower car's rate 1. A freight train and a passenger train start from the same place and travel in opposite directions. The passenger train travels 15 mph faster than the freight train. After 2 hours, they are 278 miles apart. How fast are they each going? = = Freight Passenger freight's distance passenger's distance total distance traveled Answer: 62 mph = freight train's rate 77 mph = passenger train's rate

4 Solving Mixture Problems Ex: Three pounds of type A sunflower seed, which sells for $3.60 per pound, are mixed with 2 pounds of type B sunflower seeds, which sells for $4.80 per pound. What should the mixture cost per pound? $3.60 $4.80 = $x 3 pounds 2 pounds 5 pounds 3 pounds for S3.60 each 2 pounds for $4.80 each 5 pounds for Sx each Answer: $4.08 per pound = cost of mixture 1. A nurse mixes 3.3 quarts of a 12% saline solution and 6.7 quarts of a 9% saline solution. What percent of saline solution will the mixture be? (round to the nearest percent) = quarts quarts quarts = V V V quarts of % solution quarts of % solution quarts of % solution Answers: 10% = percent of saline solution in the 10 quart mixture

5 Ex: How many ounces of 16% and 30% iodine solution should be mixed together to get 20 ounces of an iodine solution that is 25%? _ 16% 30% = 25% X ounces 20-.x ounces 20 ounces X ounces of 16% solution 20-x ounces of 30% solution 20 ounces of 25% solution Answer: 7.14 ounces = number of ounces of 16% iodine solution ounces = number of ounces of 30% iodine solution Grass seed A sells for $2.65 per pound and grass seed B sells for $2.80 per pound. How many pounds of each type should be used to get a 12-pound mixture that sells for $2.70 per pound? $ $ $ = pounds pounds pounds pounds for each pounds for each pounds for each Answer: 8 pounds = the number of pounds of grass seed A, that sells for $2.65/lb 4 pounds = the number of pounds of grass seed B, that sells for $2.80/lb

6 Ex: How many liter of an 8% sulfuric acid solution must be mixed with 6 liters of a 15 % solution to get a solution that is 12% sulfuric acid? 8% 15% = 12% X liters 6 liters (A:6) liters X liters of 8% solution 6 liters of 15%solulion x6 liters of 12% solution Answer: 4.5 = number of liters of 8% sulfuric acid solution 10.5 = number of liters of 12% sulfuric acid solution 1. How many liter of a 12% sulfuric acid solution must be mixed with 2 liters of a 9% solution to get a solution that is 11% sulfuric acid? = == liters liters liters ^ V ' > V ' > V ' liters of % solution liters of % solution liters of % solution Answer: 4 liters = number of liters of 12% sulfuric acid solution 6 liters = number of liters of 11% sulfuric acid solution