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Course 1 7-9 Percent Problems 7-9 Percent Problems Course 1 Warm Up Warm Up Lesson Presentation Lesson Presentation Problem of the Day Problem of the Day.

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Presentation on theme: "Course 1 7-9 Percent Problems 7-9 Percent Problems Course 1 Warm Up Warm Up Lesson Presentation Lesson Presentation Problem of the Day Problem of the Day."— Presentation transcript:

1 Course 1 7-9 Percent Problems 7-9 Percent Problems Course 1 Warm Up Warm Up Lesson Presentation Lesson Presentation Problem of the Day Problem of the Day

2 Course 1 7-9 Percent Problems Warm Up Write each decimal as a percent and fraction. 1. 0.38 2. 0.06 3. 0.2 38%, 6%, 20%, 19 50 __ 3 50 __ 1 5

3 Course 1 7-9 Percent Problems Problem of the Day Lucky Jim won $16,000,000 in a lottery. Every year for 10 years he spent 50% of what was left. How much did Lucky Jim have after 10 years? $15,625

4 Course 1 7-9 Percent Problems Learn to find the missing value in a percent problem.

5 Course 1 7-9 Percent Problems To find the percent one number is of another, use this proportion: % 100 = is of If you are looking for 45% of 420, 45 replaces the percent sign and 420 replaces “of.” The first denominator, 100, always stays the same. The “is” part is what you have been asked to find.

6 Course 1 7-9 Percent Problems Additional Example 1: Application First estimate your answer. Think: 35% =, which is close to, and 560 is close to 600. So about of the students participate in after-school sports. 35 100 ___ 1 3 __ 1 3 1 3 This is the estimate. 600 = 200 Think: “35 out of 100 is how many out of 560?” Helpful Hint There are 560 students in Ella’s school. If 35% of the students participate in after-school sports, how many students participate in after- school sports?

7 Course 1 7-9 Percent Problems Additional Example 1 Continued Now solve: s 560 ___ 35 100 ___ = 100 s = 35 560 100s = 19,600 100s 100 ____ 19,600 100 _____ = s = 196 Let s represent the number of students who participate in after- school sports. The cross products are equal. s is multiplied by 100. Divide both sides of the equation by 100 to undo multiplication. Since 196 is close to your estimate of 200, 196 is a reasonable answer. 196 students participate in after-school sports.

8 Course 1 7-9 Percent Problems Check It Out: Example 1 There are 480 students in Tisha’s school. If 70% of the students participate in the fundraising program, how many students participate in the fundraising program? First estimate your answer. Think: 70% =, which is close to, and 480 is close to 500. So about of the students participate in after school sports. 70 100 ___ 3 4 __ 3 4 This is the estimate. 3 4 __ 500 = 375

9 Course 1 7-9 Percent Problems Check It Out: Example 1 Continued Now solve: s 480 ___ 70 100 ___ = 100 s = 70 480 100s = 33,600 100s 100 ____ 33,600 100 _____ = s = 336 Let s represent the number of students who participate in the fundraising program. The cross products are equal. s is multiplied by 100. Divide both sides of the equation by 100 to undo multiplication. Since 336 is close to your estimate of 375, 336 is a reasonable answer. 336 students participate in the fundraising program.

10 Course 1 7-9 Percent Problems Additional Example 2: Application Johan is 25% of the way through his exercises. If he has exercised for 20 minutes so far, how much longer does he have to work out? is of __ % 100 ___ = 100 20 = 25 m 25% of the exercises are completed, so 20 minutes is 25% of the total time needed. The cross products are equal. 20 m __ 25 100 ___ = Set up a proportion. The “of” part is what you have been asked to find.

11 Course 1 7-9 Percent Problems Additional Example 2 Continued 2,000 = 25m 2,000 25 _____ 25m 25 ____ = 80 = m m is multiplied by 25. Divide both sides by 25 to undo multiplication. The time needed for the exercises is 80 min. So far, the exercises have taken 20 min. Because 80 – 20 = 60, Johan will be finished in 60 min.

12 Course 1 7-9 Percent Problems Check It Out: Example 2 Phil is 30% of the way through his homework. If he has worked for 15 minutes so far, how much longer does he have to work? is of __ % 100 ___ = 100 15 = 30 m 30% of the exercises are completed, so 15 minutes is 30% of the total time needed. The cross products are equal. 15 m __ 30 100 ___ = Set up a proportion. The “of” part is what you have been asked to find.

13 Course 1 7-9 Percent Problems Check It Out: Example 2 Continued 1,500 = 30m 1,500 30 _____ 30m 30 ____ = 50 = m m is multiplied by 30. Divide both sides by 30 to undo multiplication. The time needed for the homework is 50 min. So far, the homework has taken 15 min. Because 50 – 15 = 35, Phil will be finished in 35 min.

14 Course 1 7-9 Percent Problems Instead of using proportions, you can also multiply to find a percent of a number.

15 Course 1 7-9 Percent Problems Additional Example 3: Multiplying to Find a Percent of a Number 36% = 0.36 0.36 50 Write the percent as a decimal. Multiply using the decimal. 18 So 18 is 36% of 50. Find 36% of 50. 0 5 10 15 20 25 30 35 40 45 50 0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100% Check Use a model to check the answer.

16 Course 1 7-9 Percent Problems Check It Out: Example 3 20% = 0.20 0.20 70 Write the percent as a decimal. Multiply using the decimal. 14 So 14 is 20% of 70. Find 20% of 70. 0 7 14 21 28 35 42 49 56 63 70 0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100% Check Use a model to check the answer.

17 Course 1 7-9 Percent Problems Lesson Quiz 1. Find 28% of 310. 2. Find 70% of 542. 3. Martha is taking a 100-question test. She has completed 60% of the test in 45 minutes. How much longer will it take her to finish the test? 4. Crystal has a collection of 72 pennies. If 25% of them are Canadian, how many Canadian pennies does she have? 379.4 86.8 30 min 18


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