[mixture problems][section 30]

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
_________________________
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
___________________
.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
________________________
.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
_____________________
.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
__________________________
____  ___
.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
______________________
.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
___________________________
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
_________________________
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.