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Mixture & Money Problems

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Presentation on theme: "Mixture & Money Problems"— Presentation transcript:

1 Mixture & Money Problems
Algebra Honors

2 Let n = number of nickels Counting equation d + n = 22 Value equation
John has $1.70, all in dimes and nickels. He has a total of 22 coins. How many of each kind does he have? Let d = number of dimes Let n = number of nickels Counting equation d + n = 22 Value equation .10d + .05n = 1.70 BACK

3 -10

4 Let A = the number of Adult tickets sold.
Tickets to a movie cost $5.00 for adults and $3.00 for children. If tickets were bought for 50 people for a total of $196 how many adult tickets were sold and how many children tickets were sold? Let A = the number of Adult tickets sold. Let C = the number of children tickets sold. Counting Equation A + C = 50 Value Equation 5A + 3C = 196 BACK

5 23 + C = 50 A + C = 50 5A + 3C = 196 C = 27 -3A – 3C = -150 2A = 46
23 Adults and 27 Children 2A = 46 A = 23 Check 5(23) + 3(27) = 196 = 196

6 Let x = amount of 20% Let y = amount of 50% Counting Equation
How much 20% alcohol solution and 50% alcohol solution must be mixed to get 12 gallons of 30% alcohol solution? Let x = amount of 20% Let y = amount of 50% Counting Equation x + y = 12 Value Equation .20x + .50y = .30(12) .30(12)= 3.6 BACK

7 x + (4) = 12 x + y = 12 -2 .20x + .50y = 3.6 x = 8 10 2x + 5y = 36 8 gallons of 20% 4 gallons of 50% -2x – 2y = -24 3y = 12 y = 4 Check .2(8) + .5(4) = 3.6 = 3.6 BACK

8 Mixture Problems Blue: 10 gal + 2 gal = 12 gal
Green: 4 gal + 3 gal = 7 gal Mixture: 14 gal + 5 gal = 19 gal

9 As a percent of the whole Green Paint:
Look at One Ingredient As a percent of the whole Green Paint: .29(14 gal) + .60(5 gal) = .37(19 gal)

10 Example How many liters of 5% salt solution must I add to 2 liters of 2% salt solution in order to obtain a mixture that is 3.5% salt?

11 Amount of Salt Liters of Salt: .02(2) + .05(x) = .035(2 + x)

12 A typical mixture problem reads like this:
Joe would like to mix 5 lbs of Columbian coffee costing $4.50 per pound with enough flavored coffee costing $3.00 per pound to make a mix worth $4.00 per pound. How many pounds of the flavored coffee should he take? Let x = amount of flavored coffee in pounds Mixture problems can be more easily solved by using a grid.

13 Let’s fill in the chart with what we know.
Joe would like to mix 5 lbs of Columbian coffee costing $4.50 per pound with enough flavored coffee costing $3.00 per pound to make a mix worth $4.00 per pound. How many pounds of the flavored coffee should he take? 5 4.50 3.00 x 4.00 Let’s fill in the chart with what we know. BACK

14 The equation comes from the vertical and horizontal totals.
Now we fill in the remaining cells. Since we are mixing the two types of coffee the mix Amt will be x For total multiply amt times cost and place in total column 5 4.50 22.5 3.00 x 3x x + 5 4.00 Equation The equation comes from the vertical and horizontal totals. BACK

15 He will need 2.5 lbs of the flavored coffee.
BACK

16 An auto mechanic has 300 mL of battery acid solution that is 60% acid
An auto mechanic has 300 mL of battery acid solution that is 60% acid. He must add water to the solution to dilute it so that it is only 45% acid. How much water should he add? 300 60% 180 0% x 300 + x 45% Equation Equation: 180 = 0.45 (300 + x) BACK

17 A health food store sells a mixture of raisins and roasted nuts
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? (Write Equation Only) BACK

18 In a cash register there are 87 bills, $5 and $1 bills only
In a cash register there are 87 bills, $5 and $1 bills only. If their value is $179, how many of each kind of bill are there? Answer: There are 23 five dollar bills and 64 one dollar bills.

19 To raise funds, a school sells two kinds of raffle tickets, some for $6, others for $ Sales for both amounted to $822. If 371 tickets were sold, how many were $1.50 tickets? Answer: 312 were $1.50 tickets.

20 A health food store wishes to blend peanuts that cost $1
A health food store wishes to blend peanuts that cost $1.20/lb with raisins that cost $2.10/lb to make 50 lb of mixture that cost $1.47/lb. How many pounds of peanuts and of raisins are needed. Answer: 35 lb of peanuts and 15 lb of raisins are needed to make a 50 lb mixture.

21 A dairy has milk that is 4% butterfat and cream that is 40% butterfat
A dairy has milk that is 4% butterfat and cream that is 40% butterfat. To make 36 gal of a mixture that is 20% butterfat, how many gallons of milk and cream must be added. Answer: 20 gallons of milk and 16 gallons of cream are added to mixture.

22 A 100 kg mixture of $0. 69/kg pinto beans and $0
A 100 kg mixture of $0.69/kg pinto beans and $0.89/kg kidney beans is valued at $81. How many kilograms of each does it contain? Answer: 40 kilograms of pinto beans and 60 kilograms of kidney beans.

23 You have a total of 25 coins, all nickels and quarters
You have a total of 25 coins, all nickels and quarters. The total value is $3.85. Find the number of nickels and quarters you have.? Answer:12 nickels, 13 quarters.

24 A farmer raises wheat and corn on 430 acres of land
A farmer raises wheat and corn on 430 acres of land. He wants to plant 62 more acres of wheat than corn. How many acres of each should he plant? 246 acres of wheat and 184 acres of corn.

25 The math club and the science club had fundraisers to buy supplies for a hospice. The math club spent $135 buying six cases juice and one case of bottled water. The science club spent $110 buying four cases of juice and two cases of bottled water. How much did each cases of juice and each case of water cost? Answer: $20 cases of juice and $15 cases of water.

26 Suppose you bought supplies for a party
Suppose you bought supplies for a party. Three rolls of streamers and 15 party hats that cost $30. Later, you bought 2 rolls of streamers and 4 party hats for $11. How much did each roll of streamers and each party hat cost? Answer:$2.50 Streamers, $1.50 Party Hats

27 The Problem Chocolate sells for $8 per pound. How many pounds of mocha selling for $6 per pound should be mixed with 10 pounds of chocolate in order to obtain a mixture that sells for $7.50 per.

28 Table Rate x time = distance Chocolate Cost $ Pounds Total Mocha Mix
- Cost of Chocolate per pound: $ 8 - Cost of Mocha per pound: - $ 6 - Cost of the Mix: $ 7.50 Chocolate Cost $ Pounds Total Mocha Mix - Pounds used in mixing for Chocolate: 10 $ 8 10 80 m 6m - Pounds used in mixing for Mocha: m $ 6 - Total pounds used in mixing: m + 10 $ 7.50 m + 10 7.50 (m +10) - Total of pounds and cost for chocolate: 80 - Total of pounds and cost for mocha: 6m Rate x time = distance - Total of pounds and cost $ for mix: 7.50 (m +10)

29 This is our equation 80 + 6m = 7.50 (m + 10) 80 + 6m = 7.50m + 75
- Add the totals of Mocha and Chocolate to the total of the mixture This is our equation 80 + 6m = 7.50 (m + 10) - Distribute the total of the mixture 80 + 6m = 7.50m + 75 - Minus the numbers on one side and the family of m’s on the other side 80 – 75 = 7.50m – 6m 5 = 1.50 m - The total 5 = 1.50 - Divide both sides by 1.50 m = 3 1/3 pounds - The total is 3 1/3 pounds

30 Our Answer You need 3 1/3 pounds of mocha to mix with the 10 pounds of chocolate in order to create a mix of $7.50

31 A health food store sells a mixture of raisins and roasted nuts
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? (Write Equation Only) x 3.5x 3.50 20 - x 4.75 4.75(20 - x) 20 4.00 80 3.5x (20 – x) = 80 BACK

32 Money Mohammed has a box of coins containing only dimes and half-dollars. There are 26 coins, and the total value is $8.60. How many of each type does he have?

33 Money Mohammed has a box of coins containing only dimes and half-dollars. There are 26 coins, and the total value is $8.60. How many of each type does he have? D = dimes, H = half-dollars

34 D = dimes, H = half-dollars D + H = 26
Money Mohammed has a box of coins containing only dimes and half-dollars. There are 26 coins, and the total value is $8.60. How many of each type does he have? D = dimes, H = half-dollars D + H = 26

35 D = dimes, H = half-dollars D + H = 26 .10D + .50H = 8.60
Money Mohammed has a box of coins containing only dimes and half-dollars. There are 26 coins, and the total value is $8.60. How many of each type does he have? D = dimes, H = half-dollars D + H = 26 .10D + .50H = 8.60

36 Money Mohammed has a box of coins containing only dimes and half-dollars. There are 26 coins, and the total value is $8.60. How many of each type does he have? D = dimes, H = half-dollars D + H = D + H = 26 .10D + .50H = D – 5H = -86

37 Money Mohammed has a box of coins containing only dimes and half-dollars. There are 26 coins, and the total value is $8.60. How many of each type does he have? D = dimes, H = half-dollars D + H = D + H = 26 .10D + .50H = D – 5H = -86 - 4H = -60 H = 15

38 Money Mohammed has a box of coins containing only dimes and half-dollars. There are 26 coins, and the total value is $8.60. How many of each type does he have? D = dimes, H = half-dollars D + H = D + H = 26 .10D + .50H = D – 5H = -86 - 4H = -60 H = 15 D + 15 = 26, D = 11

39 There are 15 half-dollars and 11 dimes.
Check Mohammed has a box of coins containing only dimes and half-dollars. There are 26 coins, and the total value is $8.60. How many of each type does he have? D = dimes, H = half-dollars CHECK H = 15 D = 11 = 26 ?? OK 15 (.50) + 11 (.10) = ?? OK There are 15 half-dollars and 11 dimes.

40 Mixture & Money Problems
Algebra Honors


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