[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