190 likes | 369 Views
Advanced Bar Modeling. Gregg Velatini Dianna Spence 2009 GCTM. 8. 11. 1. 1. 1. Solving an Algebraic Equation. Three more than twice a number is eleven. What is the number ?. 2x + 3 = 11 2x = 8 x = 8/2 x = 4. 4. The number is 4. ?. Solving an Algebraic Equation.
E N D
Advanced Bar Modeling Gregg Velatini Dianna Spence 2009 GCTM
8 11 1 1 1 Solving an Algebraic Equation • Three more than twice a number is eleven. What is the number? 2x + 3 = 11 2x = 8 x = 8/2 x = 4 4 The number is 4
? Solving an Algebraic Equation • Mrs. Spence has a bag of treats for her cats. If she gives each cat 7 treats, she will have 17 treats left. If she gives each cat 4 treats, she will have 50 treats left. How many cats does Mrs. Spence own? 7 treats n n n n n 17 n n 4 treats n n n n 50 Let n = Number of Cats 7n + 17 = 4n + 50
11 Solving an Algebraic Equation • Mrs. Spence has a bag of treats for her cats. If she gives each cat 7 treats, she will have 17 treats left. If she gives each cat 4 treats, she will have 50 treats left. How many cats does Mrs. Spence own? 7 treats n n n n n 17 n n 7n + 17 = 4n + 33 + 17 3n = 33 n=11 4 treats 17 n n n n 33 50 • Mrs. Spence has 11 cats.
80 x 3 = 240 lbs 10 lbs Solving an Algebraic Equation • The average weight of Peter, Paul and Mary is 80 lbs. Paul is twice as heavy as Mary. Peter is 10 lbs lighter than Paul. Find Paul’s weight. Let M = Mary’s weight W = Paul’s weight P = Peter’s weight W = 2M P = W – 10 P+W+M = 240 Paul Mary Peter Re-Draw
80 x 3 = 240 lbs 10 lbs 10 lbs • The average weight of Peter, Paul and Mary is 80 lbs. Paul is twice as heavy as Mary. Peter is 10 lbs lighter than Paul. Find Paul’s weight. Paul Let M = Mary’s weight W = Paul’s weight P = Peter’s weight P+W+M = 240 W = 2M P = W – 10 Mary Peter 100 lbs ? 5M -10 =240 5M = 250 M=50 W=2M Paul 50 lbs Mary 80 x 3 = 240 lbs + 10 = 250 lbs Peter 250/5 = 50 lbs Paul weighs 100 lbs.
Solving Fraction Equations • Maria won a cash prize. She spent 3/8 of it on washing machine, 2/5 of it on a refrigerator, and $180 on a fan. If she had $270 left, how much was the prize? Maria’s Prize
1/8 Solving Fraction Equations • Maria won a cash prize. She spent 3/8 of it on washing machine, 2/5 of it on a refrigerator, and $180 on a fan. If she had $270 left, how much was the prize? Maria’s Prize
1/5 1/8 Solving Fraction Equations • Maria won a cash prize. She spent 3/8 of it on washing machine, 2/5 of it on a refrigerator, and $180 on a fan. If she had $270 left, how much was the prize? Maria’s Prize
1/5 1/8 Solving Fraction Equations • Maria won a cash prize. She spent 3/8 of it on washing machine, 2/5 of it on a refrigerator, and $180 on a fan. If she had $270 left, how much was the prize? Refrigerator Washing Machine Maria’s Prize $450 $50 $450/9 =$50 $180 + $270 = $450 Total prize = 40 x $50 = $2000
126 Red, Green, and Yellow 1/3 Red 2/5 Blue Rest Green and Yellow 56 4 Yellow 3 Green Solving Fraction Equations • In a jar filled with beads, 2/5 of the beads are blue, 1/3 of them are red, and the rest are green and yellow. The total number of red, green and yellow beads is 126. There are ¾ as many green beads as there are yellow beads. How many yellow beads are there? Beads 14 126/9 = 14 There are 4 x 14 = 56 Green and Yellow Beads 56/7 = 8 “There are ¾ as many green beads as there are yellow beads.” There are 8 x 4 = 32 Yellow Beads 8
3 Parts 4 Parts Ratios and Proportions • The ratio of Clinton’s baseball cards to Jesse’s baseball cards was 3:4. After Clinton bought another 40 baseball cards, he had twice as many baseball cards as Jesse. How many baseball cards did Clinton have at first? Clinton Jesse
40 Cards 2 Parts 1 Part Ratios and Proportions • Set this up as a “Before and After” problem. 8 8 8 Clinton 8 x 3 = 24 Before 3 Parts Clinton had 24 cards to begin with Jesse 4 Parts 40/5 = 8 8 Clinton After Jesse
Ratios and Proportions • If you mix 1 gal of 40% acid solution with 2 gal of 60% acid solution, what is the resulting acid concentration?. 1 gal 2 gal 3 gal ? % 40 % + 60 % = 16/30 = 53 1/3 % The final concentration is 53 1/3 % acid.
Ratios and Proportions • What amount and concentration of acid solution must be added to 1 gal of 60% acid solution in order to get 3 gal of 80% acid solution? 3 gal -1 gal = 2 gal 2 gal ? gal 1 gal 3 gal ? % 80 % 60 % + = There are 24 shaded units here. 6 come from the first bucket. 18 must come from the second bucket. Shading each gallon equally to get 18 total shaded units results in each gallon with 9 of 10 shaded units 2 gal of 90% acid solution must be added to 1 gal of 60 % acid solution to yield 3 gal of 80% acid solution.
$92 Percentages • A toy plane which originally cost $92 was sold at a loss of 25%. For how much was it sold? Plane Cost
$92 $92 Percentages • A toy plane which originally cost $92 was sold at a loss of 25%. For how much was it sold? $69 ? 0.75 x $92 = $69 75% Plane Cost 25% The plane was sold for $69. 100%-25% = 75% Alternative: Tie percentages to fractions. $69 ? 25% = 1/4 ¼ of $92 = $23 So ¾ of $92 = $69 Plane Cost 1/4
375 mi. 150 mi 3 hrs Time, Rate, Distance Lydia and Nicholas left Atlanta, GA at the same time. Both were headed for Panama City, FL and each drove at a uniform speed for the entire trip. When Lydia reached Panama City, Nicholas was still 150 miles away. After 3 hours, Nicholas reached Panama City. If the two cities are 375 mi. apart, what speed was Lydia traveling? Atlanta Panama City Lydia ? Speed Nicholas
375 mi. 150 mi, 3 hrs The bar model here has become more abstract. It is used more as an organizational and conceptual tool, and less as a computational tool. Atlanta Panama City 83.1/3 mph Lydia ? Speed Nicholas Lydia’s was traveling at a rate of 83 1/3 mph Nicholas’s Speed = 150 mi ÷ 3hrs= 50 mph Nicholas’s time = 375 mi ÷ 50 mph = 7.5 hrs Lydia’s time = 7.5 hrs – 3 hrs = 4.5 hrs Lydia’s speed = 375 mi ÷ 4.5 hrs = 83 1/3 mph