[mixture problems 2][section 51]

(1.)   How many grams of salt must be added to 600 grams of

a 20% salt solution to obtain a 25% salt solution?

.20(600) + x = .25(600 + x)     here is the problem

120 + x = 150 + .25x        multiply thru parentheses

-120      -120            subtract 120 from each side
_______________________
x = 30 + .25x          subtract

-.25x    -  .25x       subtract .25x from each side
_____________________
.75x = 30             subtract
_____  ___
.75    .75             divide each side by .75

x = 40             divide and cancel

result:  40 grams of salt

(2.)  How many liters of sulfuric acid must be added to 63

liters of a 25 % sulfuric acid solution to obtain a

60 % sulfuric acid solution?

.25(63) + x = .60(63 + x)       here is the problem

15.75 + x = 37.8 + .60x      multiply thru parentheses

-15.75     -15.75             subtract 15.75 fr ea side
______________________________
x = 22.05 + .60x       subtract

-.60x         - .60x         subtract .60x fr ea side
_____________________________
.40x = 22.05                 subtract
____  _______
.40   .40            divide each side by .40

x = 55.125             divide and cancel

result:  55.125 liters of sulfuric acid

(3.)  How many oz. of salt must be added to 60 oz. of a 10 %

sal solution to obtain a 46 % salt solution?

.10(60) + x = .46(60 + x)      here is the problem

6 + x = 27.6 + .46x      multiply thru parentheses

-6       -6              subtract 6 from each side
__________________________
x = 21.6 + .46x           subtract

-  .46x      - .46x        subtract .46x from each side
__________________________
.54x = 21.6             subtract
____  _____
.54   .54            divide each side by .54

x = 40              divide and cancel

result:  40 oz of salt

(4.)  How many quarts of antifreeze must be added to 30 quarts

of a 10 % antifreeze solution to obtain a 30 % antifreeze

solution?

.10(30) + x = .30(30 + x)      here is the problem

3 + x = 9 + .3x           multiply thru parentheses

-3      -3             subtract 3 from each side
_________________________

x = 6 + .3x            subtract

-.3x    -  .3x        subtract .3x from ea side
__________________________
.7x = 6                subtract
____ ___
.7   .7           divide each side by .7

x = 8 4/7             divide and cancel

result:  8 4/7 qt of antifreeze

(5.)  A pharmacist has 10 liters of a 20 % peroxide solution.

How much peroxide must be added to obtain a 50 % peroxide

solution?

.20(10) + x = .50(10 + x)       here is the problem

2 + x = 5 + .5x            multiply thru

-2      -2               subtract 2 from each side
_______________________________
x = 3 + .5x              subtract

-  .5x    - .5x          subtract .5x fr ea side
____________________
.5x = 3                 subtract
___ __
.5  .5          divide each side by .5

x = 6                divide and cancel

result:  6 liters of peroxide

(6.)  A 100-gram solution is 4 % salt.  How much water must

be evaporated to obtain a 12 % salt solution?

.96(100) - x = .88(100 - x)    here is the problem

96 - x = 88 - .88x         multiply thru

+ x         + x     add x to each side
_____________________________
96     =  88 + .12x     add

- 88      - 88            subtract 88 from each side
___________________________
8 =          .12x        subtract
____           _____
.12            .12        divide each side by .12

66 2/3 = x              divide and cancel

result:  66 2/3 grams of water

(7.)  How many grams of a 30 % gold alloy must be mixed with

how many grams of a 5 % gold alloy to obtain 25 grams

of a 12 % gold alloy?

x + y = 25

.30x + .05y = .12(25)       here is the problem

.3x + .05y = 3                    just multiply

- .05x - .05y = -1.25         multiply eq 1 thru by -.05
________________________
.25x      =  1.75          subtract equations
____        ______
.25          .25        divide each side by .25

x = 7                  divide and cancel

7 + y = 25              replace x with 7

-  7     -  7           subtract 7 from each side
______________________
y = 18           subtract

result:

mix 7 grams of the 30 % alloy with 18 grams of the 5 % alloy

(8.)  A car radiator is filled with 18 quarts of a 20 %

antifreeze solution.  How much of this solution must

be drained and replaced by pure antifreeze to obtain a

50 % antifreeze solution?

.20(18) - .20x + x = .50(18)      here is the problem

3.6 + .8x = 9    multiply and combine like terms

-3.6        -  3.6   subtract 3.6 from each side
________________________
.8x = 5.4       subtract

x = 6 3/4      divide and cancel

result:  6 3/4 quarts