RATE PROBLEMS

INTRODUCTION Several types of problems fall into the category known as “rate problems”: Distance Work Percent problems Mixture problems

Distance Problems Rate x Time = Distance Within the category of distance problems, we encounter 3 types of questions: Opposite direction Same direction Round-trip

Opposite Direction Key Words: opposite direction, towards each other, away from each other Example: Two trains start from the same point and travel in opposite directions. The northbound train averages 45 mph and starts 2 hours before the southbound train, which averages 50 mph. How long after the northbound train starts will the trains be 470 miles apart?

Opposite Direction (Cont.) What do we know? Organize the information:   RATE TIME DISTANCE N 45 X 45x S 50 X-2 50(x – 2) Northbound– 45 mph Southbound– 50 mph Northbound train starts 2 hours earlier. What do we want to find out? “How long…” indicates time.

Opposite Direction Practice Two boats leave at 7:00 AM from ports that are 252 miles apart and cruise toward each other at speeds of 30 mph and 26 mph. At what time will they pass each other? Answer: 11:30

Same Direction Key words: same direction, pass, overtake, catch up to Example: In a cross country race, Adam left the starting line at 7:00 AM and drove an average speed of 75 km/h. At 8:30, Manuel left the starting point and drove the same route, averaging 90 km/h. At what time did Manuel overtake Adam?

Same Direction (Cont.) What do we know? Organize the information:   RATE TIME DISTANCE A 75 x+1.5 75(x+1.5) M 90 X 90x Adam– 75 km/h Manuel– 90 km/h Manuel left 1.5 hours later. What do we want to find out? “What time…” indicates time.

Same Direction Practice In a cycling race, Mary left the starting line at 6:00 AM averaging 25 mph. At 6:30 AM, Peter left the starting point and traveled the same route at 30 mph. At what time did Peter pass Mary? Answer: 9:00

Round Trip Key words: round trip, to and from, there and back Example: A ski lift carried Maria up a slope at the rate of 6 km/h and she skied back down parallel to the lift at 34 km/h. The round trip took 30 min. How far did she ski?

Round Trip (Cont.) What do we know? Organize the information:   RATE TIME DISTANCE U 6 ½ - x 6(1/2 – x) D 34 x 34x Uphill– 6 km/h Downhill (ski) – 34 km/h Round trip took 30 min. What do we want to find out? “How far…” indicates distance. “How long…” indicates time

Practice 1) At noon, a plane leaves Hawaii for California at 555 km/h. At the same time, a plane leaves California for Hawaii at 400 km/h. The distance between the two cities is about 3800 km. At what time do the planes meet? SOLUTION:

Practice 2) Maria and Tom start at the park. They drive in opposite directions for 3 hours. The are then 510 km apart. Maria’s speed is 80 km/h. What is Tom’s speed? SOLUTION:

Practice 3) Kara floats from A to B at 6 km/h. She returns by motorboat at 18 km/h. The float trip is 4 hours longer than the motorboat trip. How far is it from A to B? SOLUTION: The question asks us to find distance. Since distance = rate x time, D = 18(2) = 36 km

Head Wind- Tail Wind Some distance problems add the variable of wind speed or water current. Tail Wind– Wind blowing in the same direction as the vehicle (increases speed) Head Wind– Wind blowing in the opposite direction (decreases speed) Upstream– Craft is traveling against the current (decreases speed) Downstream– Craft is traveling with the current (increases speed)

Example # 1 Up x – y 10 10(x-y) Down x + y 8 8(x+ y) A boat traveled downstream 160 km in 8 hours. It took the boat 10 hours to travel the same distance upstream. Find the speed of the boat and the current. Let x = speed of the boat y = speed of the current Rate Time Distance Up x – y 10 10(x-y) Down x + y 8 8(x+ y)

Example # 2 x – y 4 4(x-y) x + y 3 3(x+y) When flying with a tailwind, a plane traveled 1200 miles in 3 hours. Flying with a headwind, the plane traveled the same distance in 4 hours. Find the plane’s speed and the speed of the wind. Let x = speed of the plane y = speed of the wind Rate Time Distance Headwind x – y 4 4(x-y) Tailwind x + y 3 3(x+y)

Work Rate Work Rate problems are directly related to distance problems: Rate x Time = Work Done When setting up the table, Rate = amount of work completed per unit of time Time = amount of time working together

Example #1 An office has two envelope stuffing machines.  Machine A can stuff a batch of envelopes in 5 hours, while Machine B can stuff a batch of envelopes in 3 hours.  How long would it take the two machines working together to stuff a batch of envelopes?

Example # 1 Cont. What do we know? Machine A takes 5 hours. Machine B takes 3 hours. Rate (Per hr.) Time (together) Work Done A t B

Example #2 Mary can clean an office complex in 5 hours.  Working together John and Mary can clean the office complex in 3.5 hours.  How long would it take John to clean the office complex by himself?

Example # 2 Cont. What do we know? Mary can clean the office in 5 hours. Working together, it takes 3.5 hours. Rate (Per hr.) Time (together) Work Done Mary 3.5 John

Mixture Problems Mixture problems fall into 2 different categories: Concentration/ percent Money/ Cost There are 2 different strategies for solving mixture problems: Table Diagram

Example #1 CONCENTRATION A solution is 10% acid. How much pure acid must be added to 8 L of solution to get a solution that is 20% acid?

Example # 1 (Cont.) Strategy: TABLE Amount of solution % of acid Amount of acid Original solution 8 .10 .8 Acid added X 1 x New solution 8+x .20 .20(8+x) Use the last column in the table to write an equation. What is the resulting Concentration? You need to add 1 L of acid.

Multiply the percent on Example # 1 (Cont.) Strategy: DIAGRAM Start Add Result 8L x 8+x + = .10 1 .20 To write the equation, Multiply the percent on The outside with the Amount on the inside!

Example #2 Cost/ Money A mixture of raisins and peanuts sells for $4 a kilogram. The raisins sell for $3.60 a kilogram. The peanuts sell for $4.20 a kilogram. How many kilograms of each are used in 12 kilograms of the mixture?

Example # 2 (Cont.) Strategy: TABLE Amount of ingredient Price Cost Raisins R 3.60 3.60R Peanuts 12 – R 4.20 4.20(12-R) Mixture 12 4 48 Use the last column in the table to write an equation. You need 4 kg of raisins, and 8 kg of peanuts.

Example # 2 (Cont.) Strategy: DIAGRAM Raisins Peanuts Mixture R 12-R + = 3.60 4.20 4 To write the equation, Multiply the cost on The outside with the Amount on the inside!

Example #3 CONCENTRATION You have 40g of a 50% solution of acid in water. How much water must you add to make a 10% acid solution?

Multiply the percent on Example # 3 (Cont.) Strategy: DIAGRAM Start Add Result 40g x 40+x + = .50 .10 To write the equation, Multiply the percent on The outside with the Amount on the inside! We need 160g of water!

Example 4-- Investment Mervin Invested an amount of money at 5% interest and another amount at 8.25% interest. The amount invested at 5% was $1000 more than the amount at 8.25%. If his total income from simple interest for 1 year was $315, how much did he invest at each rate?

Example 4 Cont. STRATEGY: TABLE Amount Invested % Invested Income 5% X .05 .05x 8.25% X-1000 .0825 .0825(x-1000) $3000 invested at 5%. $2000 invested at 8.25%