1. Percent – Base and rate Problems You might know the “base / rate” problems as “is / of” problems. For example, 23 is what percent of 50 ?

2. Percent – Base and rate Problems You might know the “base / rate” problems as “is / of” problems. For example, 23 is what percent of 50 ? Every one of these problems can be set up with the same format.

3. Percent – Base and rate Problems You might know the “base / rate” problems as “is / of” problems. For example, 23 is what percent of 50 ? Every one of these problems can be set up with the same format : To set up the problem, go thru the given information and insert known values into the format…

4. Percent – Base and rate Problems You might know the “base / rate” problems as “is / of” problems. For example, 23 is what percent of 50 ? Every one of these problems can be set up with the same format : To set up the problem, go thru the given information and insert known values into the format… In our example above…23 is what percent of 50

5. Percent – Base and rate Problems You might know the “base / rate” problems as “is / of” problems. For example, 23 is what percent of 50 ? Every one of these problems can be set up with the same format : To set up the problem, go thru the given information and insert known values into the format… In our example above…23 is what percent of 50

6. Percent – Base and rate Problems You might know the “base / rate” problems as “is / of” problems. For example, 23 is what percent of 50 ? Every one of these problems can be set up with the same format : To set up the problem, go thru the given information and insert known values into the format… In our example above…23 is what percent of 50 It’s now a basic proportion and you find what is missing…in this case %

7. Percent – Base and rate Problems You might know the “base / rate” problems as “is / of” problems. For example, 23 is what percent of 50 ? A short cut for solving proportions, one that gets all the Algebra OUT is this : 1. Begin with the partner of the unknown ( red box ) 2. Move around the outside the proportion ( green box ) 3. Divide, then multiply to get the unknown Divide Multiply

8. Percent – Base and rate Problems You might know the “base / rate” problems as “is / of” problems. For example, 23 is what percent of 50 ? A short cut for solving proportions, one that gets all the Algebra OUT is this : 1. Begin with the partner of the unknown ( red box ) 2. Move around the outside the proportion ( green box ) 3. Divide, then multiply to get the unknown Divide Multiply

9. Percent – Base and rate Problems EXAMPLE # 2 : What is 35 % of 80

10. Percent – Base and rate Problems EXAMPLE # 2 : What is 35 % of 80

11. Percent – Base and rate Problems EXAMPLE # 2 : What is 35 % of 80

12. Percent – Base and rate Problems EXAMPLE # 2 : What is 35 % of 80 Divide Multiply

13. Percent – Base and rate Problems EXAMPLE # 3 : 15 is 20 % of what number

14. Percent – Base and rate Problems EXAMPLE # 3 : 15 is 20 % of what number Divide Multiply

15. APPLICATION PROBLEMS In most percent application problems, you have to determine what the problem is asking for and then use the is / of proportion to solve.

16. APPLICATION PROBLEMS In most percent application problems, you have to determine what the problem is asking for and then use the is / of proportion to solve. EXAMPLE : And inspector rejects 12 out of 250 parts due to being out of tolerance. What percentage of parts were rejected ?

17. What is the problem asking us to find ?

18. The problem is asking for the % of rejects, so the “is” and “of” part is in the wording somewhere.

19. What is the problem asking us to find ? The problem is asking for the % of rejects, so the “is” and “of” part is in the wording somewhere. I can see the “of” easily…

20. What is the problem asking us to find ? The problem is asking for the % of rejects, so the “is” and “of” part is in the wording somewhere. I can see the “of” easily… Which leaves the 12 as the “is”…

21. What is the problem asking us to find ? The problem is asking for the % of rejects, so the “is” and “of” part is in the wording somewhere. I can see the “of” easily… Which leaves the 12 as the “is”… Now just solve like we did before… Divide Multiply

22. APPLICATION PROBLEMS EXAMPLE #2 : A motor is said to be 80% efficient if the output power delivered is 80% of the input power received. How many horsepower does a motor have if it is 80% efficient with a 6.20 horsepower output.

23. Again we can quickly identify one part of the proportion… 80 % goes in the “%” spot.

24. Since the motor IS 80% efficient, the known output ( 6.20 hp ) goes in the “is” spot…

26. APPLICATION PROBLEMS EXAMPLE #3 : By replacing high – speed cutters with carbide cutters, a machinist increases productivity by 35%. Using carbide cutters, 270 pieces per day are produced. How many pieces per day were produced by the steel cutters ?

27. APPLICATION PROBLEMS EXAMPLE #3 : By replacing high – speed cutters with carbide cutters, a machinist increases productivity by 35%. Using carbide cutters, 270 pieces per day are produced. How many pieces per day were produced by the steel cutters ? SOLUTION :

28. APPLICATION PROBLEMS EXAMPLE #3 : By replacing high – speed cutters with carbide cutters, a machinist increases productivity by 35%. Using carbide cutters, 270 pieces per day are produced. How many pieces per day were produced by the steel cutters ? SOLUTION : When production increases or decreases, the beginning percent is always 100%. An increase will add to the 100%, a decrease will subtract from the 100%. Since production increased by 35%, our new production rate is 100% + 35% = 135%.

29. APPLICATION PROBLEMS EXAMPLE #3 : By replacing high – speed cutters with carbide cutters, a machinist increases productivity by 35%. Using carbide cutters, 270 pieces per day are produced. How many pieces per day were produced by the steel cutters ? SOLUTION : When production increases or decreases, the beginning percent is always 100%. An increase will add to the 100%, a decrease will subtract from the 100%. Since production increased by 35%, our new production rate is 100% + 35% = 135%. The new number of pieces is a result of the increase or decrease in production and must be placed in the proportion relative to the increase or decrease.

30. APPLICATION PROBLEMS EXAMPLE #3 : By replacing high – speed cutters with carbide cutters, a machinist increases productivity by 35%. Using carbide cutters, 270 pieces per day are produced. How many pieces per day were produced by the steel cutters ? SOLUTION : When production increases or decreases, the beginning percent is always 100%. An increase will add to the 100%, a decrease will subtract from the 100%. Since production increased by 35%, our new production rate is 100% + 35% = 135%. The new number of pieces is a result of the increase or decrease in production and must be placed in the proportion relative to the increase or decrease.