Section 10.3 Mixture Problems

2. A boy collected 75 coins consisting of nickels and dimes from his allowance. If the coins are worth $5.95, how many of each has he collected?

	Value of coins	# coins	Total value
Nickels	0.05	n	0.05n
Dimes	0.10	75 – n	0.10(75 – n)
Total		75	5.95

0.05n + 0.10(75 – n) = 5.95

0.05n + 7.5 – 0.10n = 5.95

-0.05n + 7.5 = 5.95

-0.05n = -1.55

n = 31 75 – n = 75 – 31 = 44

He had 31 nickels and 44 dimes.

6. Tickets to a school play were sold for $2 for each adult and $0.75 for each child. A total of 350 people paid $450 to see the play. How many adult tickets were sold?

	Value of each ticket	# tickets	Total Value
Adult	$2	n	2n
child	$0.75	350 – n	0.75(350 – n)
Total		350	450

2n + 0.75(350 – n) = 450

2n + 262.5 – 0.75n = 450

1.25n +262.5 = 450

1.25n = 187.5

n = 150 350 – n = 350 – 150 = 200

There were 150 adults and 200 children at the play.

10. A dealer makes a 50-kg mixture of Colombian coffee costing $1.70 per kilogram and Brazilian coffee costing $1.50 per kilogram. How many kilograms of each kind must he use for the mixture to cost $1.56 per kilogram?

	Unit Price	# kg	Total Price
Colombian	$1.70	k	1.70k
Brazilian	$1.50	50 – k	1.50(50 – k)
Mixture	$1.56	50	1.56(50)

1.70k + 1.50(50 – k) = 1.56(50)

1.70k + 75 – 1.50k = 78

0.20k + 75 = 78

0.20k = 3

k = 15 50 – k = 50 – 15 = 35

He should mix 15 kg of the Colombian coffee with 35 kg of the Brazilian coffee to get his mixture.

16. Topsoil sells for $45 a truckload and fill dirt sells for $30 a truckload. A landscape architect estimated that it would take 18 truckloads for a certain job. If he planned to spend $630, how many truckloads of topsoil and how many truckloads of fill dirt would he use?

	Unit Price	# truckloads	Total Price
Top soil	45	T	45t
Fill dirt	30	18 – t	30(18 – t)
Total		18	630

45t + 30(18 – t) = 630

45t + 540 – 30t = 630

15t + 540 = 630

15t = 90

t = 6 18 – t = 18 – 6 = 12

He would need 6 truckloads of top soil and 12 loads of fill dirt.

28. A nurse must administer 45 ml of a 12% solution of medicine. In stock are a 10% solution and a 25% solution of this medicine. How many milliliters of each should be mixed to obtain this 12% solution?

	Percent of Solution	ml of solution	ml of medicine
10% solution	0.10	s	0.10s
25% solution	0.25	45 – s	0.25(45 – s)
Mixture	0.12	45	0.12(45)

0.10s + 0.25(45 – s) = 0.12(45)

0.10s + 11.25 – 0.25s = 5.4

-0.15s = -5.85

s = 39 45 – s = 45 – 39 = 6

She needs to mix 39 ml of the 10% solution with 6 ml of the 25% solution.

30. A candy shop carries a candy that is a favorite of customers and sells for $1.19 per kilogram. The shop has 33 kilograms of a less popular variety that sells for 65¢ per kilogram. In order to reduce his inventory of the less popular variety, the owner decides to mix the two candies. How many kilograms of the $1.19 per kilogram candy would he need to mix with the 33 kilograms of the cheaper candy to get a mixture that would sell for 99¢ per kilogram? Round to the nearest kilogram.

	Unit Price	# kg	Total Price
Favorite	1.19	c	1.19c
Cheaper	0.65	33	0.65(33)
Mixture	0.99	33 + c	0.99(33 + c)

1.19c + 0.65(33) = 0.99(33 + c)

1.19c + 21.45 = 32.67 + 0.99c

1.19c – 0.99c = 32.67 – 21.45

0.20c = 11.22

c = 56.1

He should mix 56 kilograms of the more expensive candy with the 33 kg of the cheaper candy.