10.8 Mixture Problems
Goal: To solve problems involving the
mixture of substances
Mixture Problems
One solution is 80% acid and another
is 30% acid. How much of each is
required to make 200 L of solution
that is 62% acid?
Steps to Solve Mixture Problems
• Set up a chart (4x4)
Amount of Percent of
Solution
_______
Solution 1
Solution 2
Final
Solution
Amount
of _____
Steps to Solve Mixture Problems
• Convert the percentages to decimals and fill
out the chart
• Multiply going across the chart
• Add going down the chart
• Set up 2 equations with 2 variables (system)
• Solve the system by substitution or addition
One solution is 80% acid and another is
30% acid. How much of each is required to
make 200 L of solution that is 62% acid?
Let x =
y=
Amount of • Percent
solution
Acid
= Amount of
pure Acid
1st Solution
x
0.80
.80(x)
2nd Solution
y
0.30
.30(y)
3rd Solution
200
0.62
.62(200)
124
One solution is 80% acid and another is
30% acid. How much of each is required to
make 200 L of solution that is 62% acid?
.8x  .3 y  124
x  y  200
Y= 200-x
Y = 200 -128
1st
8x + 3y =1240
8x + 3 (200-x) =1240
Amount of • Percent
8x +600 -3x =1240
solution
Acid
x
0.80
Y = 72 L
Solution
y
5x +600 =1240
5x = 640
2nd Solution
200
3rd Solution
0.30
X= 128
L
0.62
= Amount of
Acid
.80(x)
.30(y)
.62(200)
124
x chemist
y  750
A
has one solution.6that
x is.360%
y  375
and 
another
30%
.6is750
 acid.
y   .How
3 y  375
xacid
 750
y that
much of each solution is needed to make a
x

500


450

.
6
y

.
3
y

375
750ml solution that is 50% acid?
Amount of
solution
 .3 y  75
• Percent = Amount of
y

250
Acid
Acid
1st Solution
x
0.60
.60(x)
2nd Solution
y
0.30
.30(y)
3rd Solution
750
0.50
.50(750)
375
x chemist
y  300
A
has one solution
28%
.28that
x is.40
y oil108
another
40%
oil.How
y  much
.40 y of
 108
xand
 300
 ythat is.28
300
each solution is needed to make a 300 L
84

.28
y

.40
y

108
x

100
solution that is 36% oil?
Amount of
solution
.12 y  24
• Percent = Amount of
y

200
Acid
Acid
1st Solution
x
0.28
.28(x)
2nd Solution
y
0.40
.4(y)
3rd Solution
300
0.36
.36(300)
108
Try to make your own chart
• How many gallons of a 50% salt solution
must be mixed with 60 gallons of a 15%
solution to obtain a solution that is 40%
salt?
How many gallons of a 50% salt
solution must be mixed with 60
gallons of a 15% solution to obtain a
solution that is 40% salt?
Amount of
Solution
(gallons)
Percent of
Salt
Amount of
Salt
Solution 1
x
.50
.5x
Solution 2
60
.15
9
Final Solution
y
.40
.4y
System
x + 60 =y
150 + 60 = y
210 gallons =y
0 .5x + 9 = 0.4y
5x +90 = 4y
5x + 90 = 4 (x +60)
5x + 90 = 4x + 240
x + 90 =240
x =150 gallons
Coffee Beans
• How many pounds of coffee beans selling
for $2.20 per pound should be mixed with 2
pounds of coffee beans selling for $1.40 per
pound to obtain a mixture selling for $2.04
per pound?
How many pounds of coffee beans selling for $2.20
per pound should be mixed with 2 pounds of coffee
beans selling for $1.40 per pound to obtain a mixture
selling for $2.04 per pound?
Pounds of
Coffee
$ per
pound
total cost
Coffee mix 1
X
$2.20
2.20 x
Coffee mix 2
2
$1.40
2.80
Final Coffee
mix
Y
$2.04
2.04 y
System
X + 2 =y
2.20x +2.80 = 2.04 y
Your Turn
• Come up with your own mixture word
problem. Make it interesting!
• Remember to include:
Amount of Solution
Wieght of Object
% of (acid /water / oil/salt/etc)
Cost per weight
Amount of (acid / water
/oil/salt/etc)
Total cost
Solution 1 and 2
2 objects
Final Solution
Mixture of 2 objects
Assignment:
Page 462
(1 -9) odd
1963
0.06x  1  0.20
Amount
of •Percent
.06
x  .06
0.20
lake
salt
.06 x  0.14
1
.20
x  2.33 liters
2nd Solution
1984
= Amount of
salt
.20
x
.00
.00(x)
x+1
.06
.20
Vocabulary
• Mixture- two substances combined
• Concentrate or Solution- how much nonwater is mixed (juice)
• 10% solution -10% concentration and 90%
water