Download presentation
Presentation is loading. Please wait.
Published byKelley Wheeler Modified over 8 years ago
1
Aim # 15: How Do We Create Mixtures and Solutions?
2
Do Now 1) If a restaurant purchases 45 lbs of butter costing $72.45, how much is the restaurant paying per pound of butter? Answer: $1.16 cents per lb. [Known as the unit cost] 2) If there is a 60% chance of rain, what is the chance of “no rain?” Answer: 40% chance it won’t rain.
3
Minilesson: Mixture and solution problems are extremely similar to each other—and very similar to coin and investment problems: Ex: Notice: Units cancel like common factors.
4
Minilesson (cont’d): EUREKA MOMENT?
5
Minilesson (cont’d):
6
35 lb bag Of cashews 55 lb bag Of peanuts $58.30 $80.85 + = 35 lb+55lb =90 lb $ 80. 85 + $58.30 =$139.15 Unit Price = $2.31/lb Unit Price = $1.06/lb Unit Price = $1.56/lb
7
Minilesson (cont’d): Unit Price X Total Price = # of units Red Fescue Total Combined Blueseed 0.85r r 0.85 (63 —r) 63 0.90 1.20(63 —r) 56.70 1.20 Let r = Pounds of Red seed 63 — r = lbs Blueseed Handout Q 1, first page This column provides the equation to solve
8
Minilesson (cont’d): Let r = Pounds of Red seed 63 — r = lbs Blueseed Answer: 54 lbs if red fescue 9 lbs of blue seed.
9
Minilesson (cont’d): Let r = Pounds of Red seed 63 — r = lbs Blueseed YOU MUST CHECK YOUR WORK!!
10
Minilesson (cont’d): Unit Price X Total Price = # of units 62 ¢ candy Total Combined 18.60 30 0.62 x (30 + x) 0.70 0.82x 0.70(30 + x) 0.82 Let x = pounds of 82 ¢ candy Handout Q 5, first page 82 ¢ candy This column provides the equation to solve
11
Minilesson (cont’d): Let x = pounds of 82 ¢ candy Answer: 20 lbs of the 82-cent candy.
12
Minilesson (cont’d): YOU MUST CHECK YOUR WORK!!
13
Minilesson (cont’d):
14
Minilesson: Solution problems concern the concentration of a certain substance within H 2 O. The substance as a % of the solution You see this especially at the pharmacy when you need to buy hydrocortisone, rubbing alcohol, hydrogen peroxide, etc. The packaging tells you the percent of concentration of the active ingredient. The amt of solution The amt of substance IN the solution × =
15
Minilesson (cont’d): + = 25% Salt15% Salt19.4% Salt
16
Guided Practice Handout, Side 1, qq. 1, 5 Handout, Side 2, qq. 19, 23
17
Independent Practice Handout, Side 1, qq. 2, 9 Handout, Side 2, qq. 20, 24
18
Our Textbook, p. 324, qq. 9--14 Answers on next page
19
Our Textbook, p. 324, ANSWERS 9) 25 L 10) 30 mL 11) 3 g 12) 15 kg of the $5; 25 kg of the $5.80 13) 8 kg dried apples; 12 kg dried apricots 14) 80 ₵ per Liter
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.