Activity : Solving Percent Problems Using Percent Converter

ALGEBRA STANDARDS ADDRESSED

This presentation scaffolds material covered in earlier years and takes students from 4th/5th grade math standards to Algebra Standards. It is hope that the visual methods will provide students with strategies that they can make the connection to some of the challenging standards in Algebra. In no way does it state that the material HAS to be covered in this manner or suggest that one has to ignore the normal algebra methods used to address percent problems, solving simultaneous equations and mixture problems. It is our hope that by providing alternate methods to teach algebra standards, students will actually learn the Algebra and teachers will experience success with their teaching.

Activity : Solving percent problems using percent converter

Weight puzzles

Activity : Solving simultaneous equations using unknown weights

Balancing Scales and Teeter- Totter Method to Solve Solution Problems

Steps needed for mastery of Standard 15

WARM-UP

How many ways can you solve….

12.5% of 40

PERCENT PROBLEM TOOL

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Percent Problems

1: Find 40% of 300

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2: 30% of what number is 45 ?

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3: 15 is what percent of 60?

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4: A house is sold for a profit of 20%. If the actual cost of the house is $200,000, determine its sale price?

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5: Principal =$2,000 R = 8% T= 5 years, determine the interest in $.

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Percent Problem Types

TYPE 1: Find 40% of 300

% 100

# 300

% 10 100

# 30 300

% 10 40 100

# 30 120 300

(Explanation: 100% = 300, so 10% =30 and 40% = 120)

TYPE 2: 30% of what number is 45 ?

% 30 100

# 45

% 10 100

# 15

% 10 100

# 15 150

(Explanation: 30% = 45, so 10% =15 and 100% = 150)

TYPE 3: 15 is what percent of 60?

% 100

# 60

% 10 20 30 100

# 6 12 18 60

% 10 20 25 30 100

# 6 12 15 18 60

(Explanation: 100% = 60, so 10% =6 and 20% = 12 and 30% = 18. This is greater than 15. But 15 lies half way between 12 and 18. .So 15 is 25% of 60)

TYPE 4: A house is sold for a profit of 20%. If the actual cost of the house is $200,000, determine its sale price?

% 100

Price $200,000

% 10 20 100

Price $20,000 $40,000 $200,000

% 10 20 100 120%

Price $20,000 $40,000 $200,000 $240,000

(Explanation: 100% =$200,000, so 10% =$20,000 and 20% = $40,000 and 120% = $200,000 + $40,000 or

$240,000)

TYPE 5: Simple Interest

Principal =$2,000 R = 8% T= 5 years, determine the interest in $.

% 100

$ 2,000

% 1 8 100

$ 20 160 2,000

(Explanation: 100% =$2,000, so 1% =$20 and 8% = $160 )

The amount of interest received in 1 year =$160. So in 5 years the interest received is $$160 × 5 =$800

Weight Puzzles

In each of the problems below the scale is balanced.

-Same shapes have same weights

-Different shapes have different weights

-A horizontal bar shows that there is balance

-There is only one correct solution in each problem

-Use the weights given in the problem to determine the other weights

1

Total weight = 40

=______= ______

Write a numerical equation showing that the scale is balanced

______

2.

Total weight = 48

=______=______

Write a numerical equation showing that the scale is balanced

______

(3)

Total weight = 32 and =10

=______=______

Write a numerical equation showing that the scale is balanced

______

(4)

Total weight = 48 and =16

=______= ______

Write a numerical equation showing that the scale is balanced

______

(5)

Total weight = 40 and = 4

=______= ______

Write a numerical equation showing that the scale is balanced

______

(6)

Total weight = 32

=______= ______

Write a numerical equation showing that the scale is balanced

______

(7)

Total weight = 80

=______= ______

Write a numerical equation showing that the scale is balanced

______

(8)

Total weight = 48

=______= ______=______and =______

Write a numerical equation showing that the scale is balanced

______

What is the Weight?

Solve using at least two different methods!

Set A

Set B

Set C

Set D

Using the strategy learned in the previous problem and apply it to solve for x and y in the system of equations:

2x + y = 9

3x + y = 11

DONUTS AND COFFEE

The cost for two donuts and a coffee is $1.35 and the cost of four donuts and three cups of coffee is $3.05. What is the cost of a single donut and a single cup of coffee?

Teeter Totter Problems

1. Determine where the yellow figure should sit on the teeter totter if the fulcrum (or pivot) is at the 50 cm mark

and the teeter totter is to be balanced.

2. Determine where the yellow figure should sit on the teeter totter if the fulcrum (or pivot) is at the 35 cm mark

and the teeter totter is to be balanced.

3. Determine the weight of the yellow figureif the fulcrum (or pivot) is at the 35 cm mark

and the teeter totter is to be balanced.

Mixtures –Finding the Balance

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MIXTURE PROBLEMS

THE JELLY BEAN PROBLEM

White jelly beans cost $0.25 a pound and blue jelly beans cost $0.85 a pound. How many pounds of white jelly beans must be added to 25 pounds of blue jelly beans to arrive at a mixture worth $0.45 a pound?

SALT SOLUTION

A mixture containing 6% salt is to be mixed with 2 ounces of a mixture which is 15% salt, in obtaining a solution which is 12% salt. How much of the first solution must be used?

A SOLUTION

How many liters of a 70% acid solution must be added to 50 liters of a 40% acid solution to produce a 50% acid solution?

COFFEE ANYONE?

Find the selling price per pound of a coffee mixture made from 8 pounds of coffee that sells for $9.20 per pound and 12 pounds of coffee that costs $5.50 per pound.

A VEGETABLE MEDLEY

How many pounds of lima beans that cost $0.90 per pound must be mixed with 16 pounds of corn that costs $0.50 per pound to make a mixture of vegetables that costs $0.65 per pound?

A LITE PUNCH

Two hundred liters of a punch that contains 35% fruit juice is mixed with 300 L of another punch. The resulting fruit punch is 20% fruit juice. Find the percent of fruit juice in the 300 liters of punch.

THE CEREAL PROBLEM

Ten grams of sugar are added to a 40-g serving of a breakfast cereal that is 30% sugar. What is the percent concentration of sugar in the resulting mixture?

How many liters of water must be added to 50 L of 30% acid solution in order to produce a 20% acid solution?

ADDITIONAL MIXTURE PROBLEMS

(Composed by Glenda Griffin)

1. Hal is doing a chemistry experiment that calls for a 30% solution of copper sulfate. Hal has 40 mL of 25% solution. How many milliliters of a 60% solution should Hal add to obtain the required 30% solution?

1. How much whipping cream (9% butterfat) should be added to 1 gallon of milk (4% butterfat) to obtain a 6% butterfat mixture?

1. A chemist had 2.5 liters of a solution which is 70% acid. How much water should be added to obtain a 50% acid solution?

1. How much pure copper must be added to 50 kg of an alloy containing 12% copper to raise the copper content to 21%?

1. A liter of cream has 9.2% butterfat. How much skim milk containing 2% butterfat should be added to the cream to obtain a mixture with 6.4% butterfat?

1. How much coffee costing $3 a pound should be mixed with 5 pounds of coffee costing $3.50 a pound to obtain a mixture costing $3.25 a pound?

1. Ground chuck sells for $1.75 a pound. How many pounds of ground round selling for $2.45 a pound should be mixed with 20 pounds of ground chuck to obtain a mixture that sells for $2.05 a pound?

1. An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice?

1. A pharmacist had 150 dL of a 25% solution of peroxide in water. How many deciliters of pure peroxide should be added to obtain a 40% solution?

1. A health food store sells a mixture of raisins and roasted nuts. Raisins sell for $3.50/kg and nuts sell for $4.75/kg. How many kilograms of each should be mixed to make 20 kg of this snack worth $4.00 kg?

1. An auto mechanic has 300 mL of battery acid solution that is 60% acid. He must add water to this solution to dilute it so that it is only 45% acid. How much water should he add?

1. A chemist has 40 mL of a solution that is 50% acid. How much water should he add to make a solution that is 10% acid?

1. If 800 mL of a juice drink is 15% grape juice, how much grape juice should be added to make a drink that is 20% grape juice?

1. How many liters of water must be added to 50 L of a 30% acid solution in order to produce a 20% acid solution?

1. How many milliliters of water must be added to 60 mL of a 15% iodine solution in order to dilute it to a 10% iodine solution?

1. A spice mixture is 25% thyme. How many grams of thyme must be added to 12 g of the mixture to increase the thyme content 40%?

1. A grocer mixes two kinds of nuts. One kind cost $5.00/kg and the other $5.80/kg. How many kilograms of each type are needed to make 40 kg of a blend worth $5.50/kg?

1. Joanne makes a mixture of dried fruits by mixing dried apples costing $6.00/kg with dried apricots costing $8.00/kg. How many kilograms of each are needed to make 20 kg of a mixture worth $7.20/kg?

Answers:

1. 6.7 mL of the 60% solution)

1. 0.67 gal

1. 1 L

1. 5.7 kg

1. 0.64 L

1. 5 lbs

1. 15 lbs

1. 2.5 qts

1. 37.5 dL

1. 12 kg of raisins and 8 kg of nuts

1. 100 mL of water

1. 160 mL water

1. 50 mL juice

1. 25 L

15. 30 L

16. 3 g

17. (15 kg of $5.00/kg, 25 kg of $5.80/kg

18. 8 kg dried apples, 12 kg dried apricots

MIXTURE PROBLES- A MEDLEY OF METHODS

(Written by Ed D’Souza, Ph.D.)

Method 1 : Intuitive Method

White jelly beans cost $0.25 a pound and blue jelly beans cost $0.85 a pound. How many pounds of white jelly beans must be added to 25 pounds of blue jelly beans to arrive at a mixture worth $0.45 a pound?

With an intuitive approach, we would begin by reasoning that if the mixture contained the same amount of white and blue jelly beans, it would cost an amount that is right in the middle. A $0.60 difference exists between the two types. A middle price would be $0.25 + $0.30 (half the difference), or $0.55. But the problemstates that the new mixture must be worth $0.45. Therefore, the mixture must contain more white jelly beans than blue. Also, it must be greater than 25 pounds.

Then we would reason that the price difference between the white jelly beans and the mixture is $.20. The price difference between the blue jelly beans and the mixture is $.40. As a ratio, the price differences compare 1:2. It makes sense that the amounts will follow this ratio. Since more white jelly beans are needed, the

ratio of white to blue must be 2:1. Therefore, since the mix has 25 pounds of blue, it must have 50 pounds of white.

METHOD 2:Teeter-Totter Method

White jelly beans cost $0.25 a pound and blue jelly beans cost $0.85 a pound. How many pounds of white jelly beans must be added to 25 pounds of blue jelly beans to arrive at a mixture worth $0.45 a pound?

Using the tug-of-war approach, imagine the white jelly beans having a tug-of-war game with the blue jelly beans. Both want the final price to be closer to their price.

1. Draw a "rope" (straight line) and label the three numbers you know.

W=25 M =45 B= 85

1. The left point on the rope is the price of the white jelly beans (25), the right

point is the price of the blue jelly beans (85), and the "fulcrum" will be

located at the final price (45), and closer to the left side. Try to approximate

where 45 would land between 25 and 85.

Price (cents) W= 25 M= 45 B= 85

1. Label what you know - the pull to the right is 25. Label what you don't know the pull to the left is x.

Amount (lbs) 20 40

Price(cents) W= 25 M=45 B= 85

Amount (lbs) x 25

5. Calculate the gap in price between the knot and the left side (20cents) and between the knot and the right side (40 cents- see bullet 3.). Since the gap on the left is half as big, the left side must bepulling twice as hard. Therefore, we must have 50 pounds of white jelly beans.

Students will eventually discover the rule: (gap) x (pull) = (gap) x (pull). Multiply the gap on the left (20) by the pull to the left (the unknown). Set this quantity to the gap on the right (40) times the pull to the right (25). Solving for the unknown yields 50 pounds.

(20)(x) = (40)(25)

x = (40)(25) = 50lbs

20

So , 50 lbs of white jelly beans need to be added to the mixture.

3rd Method: Column Method

White jelly beans cost $0.25 a pound and blue jelly beans cost $0.85 a pound. How many pounds of white jelly beans must be added to 25 pounds of blue jelly beans to arrive at a mixture worth $0.45 a pound?

Use only one unknown

Equation: 0.25x + (0.85)(25) =(x+25)(0.45)

Multiply throughout by 100 (this eliminates decimals)

(100) (0.25x) + (100) (0.85)(25) = (x+25)(0.45)(100)

25 x + (85)(25) = 45(x+25)

25 x + 2125 = 45x + 1125

2125 – 1125 = 45x – 25 x

1000 = 20 x

x =50 So, 50 white jelly beans need to be added to the mixture.

Strategies for Solving Problems

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