1 / 24

Ratios, Rates, Proportions and Percent

Ratios, Rates, Proportions and Percent. Grade 7. Mr. Pontrella. Ratios. We use ratios to make comparisons between two things. We are comparing rectangles to triangles. Ratios can be written 3 ways. 1. As a fraction 3 5. 2. Using the word to 3 to 5.

neil-weaver
Télécharger la présentation

Ratios, Rates, Proportions and Percent

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Ratios, Rates, Proportions and Percent Grade 7 Mr. Pontrella

  2. Ratios We use ratios to make comparisons between two things. We are comparing rectangles to triangles. Ratios can be written 3 ways. 1. As a fraction 3 5 2. Using the word to 3 to 5 3. Using a colon 3:5 equivalent ratios Ratios that name the same comparisons To see if to ratios are equivalent 1. Change each to a decimal and compare the decimals. 2. Reduce both ratios and compare. 3. Use cross products.

  3. Rates Rate: is a ratio of 2 measurements with different units Here we are comparing days to inches Example: It rained 4 inches in 30 days The rate is 4 30 We can reduce to 2 15 Unit Rates A rate that has 1 unit as its second term (denominator) If a car travels 325 miles and uses 11 gallons of gas what is the mile per gallon? This is an example of a unit rate. How many miles per 1 gallon? Create a ratio Miles Gallons 325 11 Since every fraction is a division problem we divide 325/ 11 Our Unit Rate is 29. 54 miles per gallon

  4. Camparisons Using Unit Rates Unit rates are often used to make comparisons. Drivers often use Miles Per Gallon(mpg) or Miles Per Hour(mph) Example: the 8th grade parents collect $ 2450 for pizza sales and the 7th grade parents collect $ 2250 for selling Subway. If there were 102 - 8th graders and 90 - 7th graders, which group collect more per student? We have 2 rates. We can make each a unit rate to see which group had more dollars per student. 8th 7th $ 2450 102 $ 2250 90 = $ 24.90/ student = $ 25 per student The 7th grade parents collect $ 0.10 more per student

  5. Shop Rite has 24 oz box of Hulk Cereal for $ 3.99. Path Mark has a 1lb box for$ 2.59. Which is the better buy? Notice in this box the units are different (oz, lbs) Before we solve we need to make the units the same. 1 lb = 16 oz The rates are: We find the unit rates by division. Shop Rite Path Mark $ 3.99 24 oz = about $ 0.17per oz $ 2.59 16 oz = about $ 0.16 per oz Shop Rite has the better buy on Hulk Cereal

  6. Proportions An equation that shows that two ratios are equal We can write proportions in 2 forms. a:b = c:d If 2 ratios are equal then their cross product will be equal. a * d = b * c

  7. Using Proportions to Solve problems A car travels 125 miles in 5 hours. How many miles will the car travel in 8 hours? Solve using proportions. 125 = m 5 8 Proportion Set an equation using cross products 125 * 8 = 5 * m Simplify 1000 = 5m Solve by inverse operation (The opposite of multiplication is division ) 1000 /5 = m 200 = m In 8 hours a car can travel 200 miles

  8. Scale Drawings and Proportions On a map 1.5 inches is equal to 5 miles. If the distance in real life is 22 mile how big will it be on the map? Proportions can help us with this problem. We know 1 ratio is 1.5 in: 5 m. We know 1 part of the second ratio is 22 m. Proportion 1.5 in = X m 5 m 22 m Notice we lined up m to m and in to in Cross Products 1.5 x 22 = 5 x X Simplify 33 = 5X 22 miles is equal to 6.6 inches on the map Inverse Operation 33/5 = X 6.6 = X

  9. The Percent Proportion Dinner cost $75 and you wish to leave a 20% tip. How much will the tip be? We can use the percent proportion to solve. P is the percentage ( a value that is a number for the percent P = R B B is the base or the original amount R is the rate(the percent number over 100) In this problem the Base is $75, the Rate is 20 over 100 and we are solving for the Percentage ( how much money is equal to 20%) Cross products 75 x 20 = P x 100 P = 20 $75 100 Simplify 1500 = 100P Inverse operation 1500/ 100 = P $15 = P The tip will be $15

  10. 3 Types or Percent We have seen 1 type of percent problem. Let’s look at 2 others. If we left a 20% tip which was $25, how much was the bill? We know R is 20% and P is $25. We need to find B. Proportion25 = 20 B 100 Cross products 25 x 100 = 20 x B Simplify 2500 = 20B The dinner bill was $125 Inverse Operation 2500/ 20 = B $125 = B

  11. If Dinner cost $125 and we left a $35 tip what percent of the bill was the tip? P is $35, B is $125. We are trying to find R. Proportion35 = R 125 100 Cross Products 35 x 100 = 125 x R Simplify 3500 = 125 R The tip was 28% of the bill Inverse Operation 3500/ 125 = R 28% = R

  12. The Percent Equation We have used a proportion to solve percent problems for P, R, and B. We can rewrite the proportion to an equation. Let us look at the previous problems using the equation. P = R x B 1. Dinner cost $75 and you wish to leave a 20% tip. How much will the tip be? Formula P = R x B P = 20% x $75 Substitute Solve: use the decimal form of the % P = 0.2 x 75 P = $15 The tip will be $15

  13. 2. If we left a 20% tip which was $25, how much was the bill? P = R x B Formula Substitute $ 25 = 20% x B Solve using inverse operation 25 0.2 = B 125 = B The dinner bill was $125

  14. 3. If Dinner cost $125 and we left a $35 tip what percent of the bill was the tip? P = R x B Formula $ 35 = R x $125 Substitute 35 125 = R Solve using inverse operation Note: We are dividing by numbers not the rate so we use the numbers. 0.28 = R Since our answer is a decimal we convert that decimal to get a percent 28% = R The tip was 28% of the bill

  15. Using Percent Simple Interest When money is borrowed, interest is charged for the use of that money for a certain period of time. When the money is paid back, theprincipal(amount of money that was borrowed) and the interest is paid back. The amount to interest depends on the interest rate, the amount of money borrowed (principal) and the length of time that the money is borrowed. Interest = Principal * Rate * Time The formula for finding simple interest is: How much will the interest be if we borrow 20,000 for 2 years at 6%? I = P x R x T Note: we use the decimal form to multiply and the length of time is based on 1 year. I = 20000 x 6% x 2 I = 20000 x 0.06 x 2 The interest will be $ 2400 I = 2400

  16. Simple Interest If we put 20000 in the bank at 5.5% for 18 months, How much will we have at the end of 18 months? There are differences in this problem 1. The interest rate has a decimal 2. The time is not if full years 3. We are asked for a total not just the interest. I = P x R x T I = 20000 x 5.5% 18 months Note: since 1 year = 12 months we use 12 as a denominator. We could reduce (1 1 ) or use the decimal form(1.5) 2 I = 20000 x 0.055 x 18 12 This is the interest. We now add that to the principle of 20000 I = 1650 1650 + 20000 = 21650 The total at the end of the time period is $21650

  17. Using Percent Tax Many of the things we buy, the money we earn and the places we live come with a tax. .The tax is found by finding a percentage of the purchase or income called the tax rate. By finding the percentage we can calculate tax. In much the same way we did with simple interest except we eliminate the time component of the formula. The formula can be written: Tax = Principle * tax rate When a total is asked for we add the percentage to the original amount as we did with simple interest.

  18. Tax We purchased a car for $28,568. If the tax rate is 6%, how much tax did we pay? Tax = Principle * tax rate T = $28568 * 6% The tax on our car is $ 1714.08 T = 28568 * 0.06 T = 1714.08 What is the total cost of the car? This problem requires us to just add the tax and the price of the car much the same as we did with the total for simple interest. $28568 + $ 1714.08 = $30282.08 $30282.08 is the total cost of the car

  19. Using Percent Discounts Retailers often offer sales. They are usually in percents, We can solve these problems in the same way using percents. A DVD is on sale for 20% off. It originally sells for $275. How much will we save? We can put our % equation in this form. D = P * R D = $275 * 20% D = 275 * 0.2 D = 55 The discount on the DVD is $ 55 What is the sale price of the DVD? To solve we just subtract our discount from the original price $ 275 - $55 = $ 225 The sale price of the DVD is $ 225

  20. Discounts Macy’s is having a 20% off sale. If you buy it today you receive an additional 15% of the sale price.If you buy a $45 sweater today how much will it cost? In this problem it looks as if you will get 35% off. But we will only get 15% off the sale price not the original price. D = $ 45 *20 % D = 45 * 0.2 D = $ 9 $ 45 - $ 9 = $36 The sale price is $ 36. Now we take off the 15%. D = $ 36 *15 % D = 36 * 0.15 D = $ 5.40 $ 36 - $ 5.40 = $31.60 The final price is $ 31.60 But how much would it be if there was a 6% sales tax???

  21. Using Percent Appreciation and Depreciation Often times when we buy things they can either increase in value (appreciate) or decrease in value (depreciate). The differences can looked at as Percent of Change. We refer to these situations as either the percent of increase (appreciation) or the percent of decrease (depreciation) Usually, the homes we buy appreciate. When we sell our homes we often get more than we paid for them. We also buy stocks in the hope that they will also go up (not always the case) On the other hand, cars often go down in value as they get older.

  22. Joe Smith bought his home in 1999 for $ 325,000 and sold it in 2003 for $ 545,000. What was his percent of increase? 1. Determine whether this is an increase or a decrease. ( If the new price is higher it is an increase. If the new price is lower it is a decrease. In this case the price is an increase). 2. Subtract the higher price from the lower price to find the difference. $545,000 - $325,000 = $220,000 difference $ 220,000 $ 325,000 3. Make a fraction by placing the difference over the original price. Original price 4. Change the fraction to a decimal by division. Then to a percent by moving the decimal point. The percent of increase is 67.7% (Answer rounded to the nearest tenth of a percent).

  23. Mrs. Princing bought stock in the I.O.U company worth $5500 in May. In June the stock was worth $3000 . Find the percent of change in the stock. Since the new price is lower we will be finding the percent of decrease. $5500 - $3000 = $2500 Find the difference. Make a fraction Change to decimal Change to percent $ 2500 $ 5500 0.4545 45.5% The percent of change (decrease) for Mrs. Princing’s stock is 45.5 % ( rounded to the nearest tenth).

More Related