[mixture problems][section 30]

(1.) A baker mixes cookies worth $0.95 per pound with cookies

worth $1.70 per pound. How many of each kind must be used

to produce a 45-pound mixture that sells for $1.25 per

pound?

x + y = 45

0.95x + 1.7y = 1.25(45) here is the problem

0.95x + 1.7y = 56.25 just multiply

-0.95x - 0.95y = -42.75 multiply eq 1 thru by -0.95
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0.75y = 13.50 subtract equations
_____ _______
0.75 0.75 divide each side by 0.75

y = 18 divide and cancel

x + 18 = 45 replace y with 18

-18 -18 subtract 18 from each side
__________________
x = 27 subtract

result: 27 pounds of the first kind of cookie.

18 pounds of the second kind of cookie.

(2.) A 4 % salt solution is mixed with an 8 % salt solution.

How many grams of each solution are needed to obtain a

400 grams of a 5 % solution?

x + y = 400

.04x + .08y = .05(400) here is the problem

.04x + .08y = 20 just multiply

-.04x - .04y = -16 multiply eq 1 thru by -.04
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.04y = 4 subtract equations
____ ____
.04 .04 divide each side by .04

y = 100 divide and cancel

x + 100 = 400 replace y with 100

-100 -100 subtract 100 from each side
_________________
x = 300 subtract

result: 300 grams of the 4 % solution.

100 grams of the 8 % solution.

(3.) An alloy that is 35 % copper is melted with a second

alloy that is 65 % copper. How many kilograms of each

alloy must be melted to obtain 20 kilograms of an alloy

that is 41 % copper?

x + y = 20

.35x + .65y = .41(20) here is the problem

.35x + .65y = 8.2 just multiply

-.35x - .35y = -7 multiply eq 1 thru by -.35
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.3y = 1.2 subtract equations
____ _____
.3 .3 divide each side by .3

y = 4 divide and cancel

x + 4 = 20 replace y with 4

-4 -4 subtract 4 from each side
________________
x = 16 subtract

result: 16 kg of the 35 % solution.

4 kg of the 65 % solution.

(4.) Solution A, which is 10 % iodine, is mixed with solution

B, which is 18 % iodine, to obtain 320 grams of a

solution that is 15 % iodine. How many gram of each

solution are needed?

A + B = 320

.10A + .18B = .15(320) here is the problem

.10A + .18B = 48 just multiply

-.10A - .10B = -32 multiply eq 1 thru by -.10
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.08B = 16 subtract equations
_____ ____
.08 .08 divide each side by .08

B = 200 divide and cancel

A + 200 = 320 replace B with 200

-200 -200 subtract 200 from each side
___________________
A = 120 subtract

result: 120 grams of solution A, and 200 grams of solution B

(5.) A milk distributer has cream that is 24 % butterfat and

cream that is 18 % butterfat. How many quarts of each

must be used to obtain 90 quarts of cream that is

22 % butterfat?

x + y = 90

.24x + .18y = .22(90) here is the problem

.24x + .18y = 19.8 just multiply

- .18x - .18y = -16.2 multiply eq 1 thru by -.18
____________________
.06x = 3.6 subtract equations
_____ _____
.06 .06 divide each side by .06

x = 60 divide and cancel

60 + y = 90 replace x with 60

-60 -60 subtract 60 from each side
_________________

y = 30 subtract

result: 60 quarts of the first cream.

30 quarts of the second cream.

(6.) A tank contains 100 liters of a solution of acid and

water. The solution is 20 % acid. How much water

must be evaporated to obtain a solution that is 50 %

acid?

.80(100) - w = .5(100 - w) here is the problem

80 - w = 50 - .5w multiply thru parentheses

-80 + w = -50 + .5w multiply thru by -1

-.5w - .5w subtract .5w from each side
_______________________

-80 + .5w = -50 subtract

+ 80 + 80 add 80 to each side
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.5w = 30 add
____ ___
.5 .5 divide each side by .5

w = 60 divide and cancel

result: 60 liters of water must be evaporated

(7.) An 8 % plant food solution is mixed with a 12 % solution.

How many liter of each solutions are needed to obtain

10 liters of a 9 % solution?

x + y = 10

.08x + .12y = .09(10) here is the problem

.08x + .12y = .9 just multiply

- .08x - .08y = - .8 multiply eq 1 thru by - .08
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.04y = .1 subtract equations
____ ____
.04 .04 divide each side by .04

y = 2.5 divide and cancel

x + 2.5 = 10 replace y with 2.5

-2.5 -2.5 subtract 2.5 from each side
_________________
x = 7.5 subtract

result: 7.5 liters of the 8 % solution

2.5 liters of the 12 % solution

(8.) Wild rice worth $9.50 per pound is mixed with brown

rice worth $2.25 per pound. How many pounds of each

kind of rice must be used to produce a 50-pound mixture

selling for $3.70 per pound?

w + b = 50

9.5w + 2.25b = 3.7(50) here is the problem

9.5w + 2.25b = 185 just multiply

-2.25w - 2.25b = -112.5 multiply eq 1 thru by -2.25
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7.25w = 72.5 subtract equations
______ _____
7.25 7.25 divide each side by 7.25

w = 10 divide and cancel

10 + b = 50 replace w with 10

-10 -10 subtract 10 from each side
_________________

b = 40 subtract

result: 10 pounds of wild rice and 40 pounds of brown rice.

(9.) A solution contains 8 grams of acid and 32 grams of

water. How many grams of water must be evaporated to

obtain a solution that is 40 % acid?

.80(40) - w = .60(40 - w) here is the problem

32 - w = 24 - .6w multiply thru parentheses

+ w + w add w to each side
_____________________
32 = 24 + .4w add

-24 -24 subtract 24 from each side
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8 = .4w subtract
____ ___
.4 .4 divide each side by .4

20 = w divide and cancel

result: 20 grams of water must be evaporated

(10.) A chemist has one solution that is 40 % acid and a

second solution that is 15 % acid. How many grams of

each should be used to obtain 40 grams of a solution

that is 25 % acid?

x + y = 40

.40x + .15y = .25(40) here is the problem

.4x + .15y = 10 just multiply

-.15x - .15y = -6 multiply eq 1 thru by -.15
___________________
.25x = 4 subtract equations
____ ___
.25 .25 divide each side by .25

x = 16 divide and cancel

16 + y = 40 replace x with 16

- 16 -16 subtract 16 from each side
___________________
y = 24 subtract

result: 16 grams of the 1st solution. 24 grams of the 2nd

solution.