To get the best deal on Tutoring, call 1-855-666-7440 (Toll Free)

Mixture Problems


Mixture Problems
  • Page 1
 1.  
Suppose in a pitcher you have 20% of grape juice in 10L of water. Find how many litter of 80% grape juice should be added to make the mixture of 75% grape juice? 
a.
80%=0.166 L,20%= 9.834 L
b.
80%=9.166 L,20%= 0.834 L
c.
80%=9.00 L,20%= 0.75 L
d.
80%=20 L ,20%= 0.40 L


Solution:

Let  x = number of L of water with 20% grape juice
and y= number of L of water with 80% grape juice

as given 
x + y = 10....(1)
0.2x + 0.80y = 0.75 $\times$ 10....(2)

use elimination method,
from equation (1)
x = 10 - y
put in equation (2)
0.2(10 - y) + 0.80y = 0.75 $\times$ 10
y = 9.166
substitute y value in equation (1)
x + 9.166 = 10
x = 0.834

Hence, To make juice which contain 75% of grape juice. Mix 9.166 L water in 80% grape juice and mix 0.834 L water in 20% grape juice.  




Correct answer : (2)
 2.  
Drink X contain 20% calories and drink Y contain 40% of calories.Find how much X and Y needed to make 40L of drink that contain 35% calories.  
a.
X=20L and Y=20L
b.
X=30L and Y=10L
c.
X=25L and Y=15L
d.
X=10L and Y=30L


Solution:

Given X = 20% calories
Y = 40% calories
So, 
X + Y = 40...(1)
0.20X + 0.40Y = 0.35 $\times$ 40.....(2)

Use elimination,
X = 40 - Y

substitute in equation (1)
0.20(40 - Y) + 0.40Y=0.35 $\time$40
Y = 30L

substitute Y value in equation (1)
X + 30 = 40
X = 10L

Answer is:
X = 10L and Y = 30L needed to make 40L drink that contain 35% calories 




Correct answer : (4)
 3.  
Solution A contain 70% alcohol and solution B contain 40% of alcohol.Find how many liters of A and B should be added to make 50L of solution that contain 50% alcohol. 
a.
A=17.67 L and B=32.33L
b.
A=33.33 L and B=16.67L
c.
A=16.67 L and B=33.33L
d.
A=25 L and B=25L


Solution:

Let x = number of liters of solution A
y = number of liters of solution B

x + y=50.....(1)
0.7x + 0.4y = 50$\times$0.5....(1)

Use elimination method,
from equation (1)
x = 50 - y

Substitute in equation (2)
0.7(50 - y) + 0.4y = 50$\times$0.5
y=33.33 

Substitute y value in equation (1)
x + 33.33=50
x = 16.67

Answer is:
Solution A=16.67 L and B=33.33L needed to make 50L of solution that contain 50% of alcohol. 



Correct answer : (3)
 4.  
5L of salt solution was mixed with 3L of a 40% salt solution to make 55% salt solution. How much percent of concentration the first solution have? 
a.
64%
b.
74%
c.
60%
d.
55%


Solution:

Let the percent  = x
By given, 
5L solution with x = 5x
3L solution with 40% = 3$\times$0.40=1.2
Total 8L solution with 55% = 8$\times$0.55 = 4.4

Now to get the percent of the first solution,
5x + 1.2 = 4.4
x = 0.64

Answer is: 
The first solution percent = 64%



Correct answer : (1)
 5.  
Suppose you have 60 Kg of cashew mixture using x cashew and y cashew. x cashew cost $4.08/kg and y cashew cost $3.48/kg.If the cashew mixture cost $3.78/kg.Find how many kg of each cashew were added?
a.
x =20kg and y=40kg
b.
x =30kg and y=30kg
c.
x =10kg and y=50kg
d.
Non of them


Solution:

Let cashew x = n kg = $4.08n
then cashew y = 60 - n = $3.48$\times$(60-n)
and mixture of x and y =60$\times$ $3.78

We get the equation is,
4.08n + 3.48(60 - n) = 60$\times$3.78
n=30kg  
cashew y =  60 - 30 = 30kg

Answer is: 
cashew x = 30kg and y = 30kg were added 



Correct answer : (2)
 6.  
Suppose you have 60 ounces of a 15% saline solution. The solution contain 10% of salt. Find the ounces of pure water. 
a.
20 ounces
b.
25 ounces
c.
30 ounces
d.
40 ounces


Solution:


Let the ounces of pure water = x
As given
15% saline solution have 60 ounces = 60$\times$0.15 = 9
10% salt  = (60 + x) $\times$0.10
We have the equation,

9 = (60 + x)$\times$0.10 
x = 30 

Answer is: 
Pure water =30 ounces



Correct answer : (3)
 7.  
A mixture of a solution have 70% of solution A and 40% of solution B to obtain 150 gallons of 60% solution. Find how many of gallons of each solution are there in the mixture.
a.
70% solution=50 gallons,40% solution=100 gallons
b.
70% solution=100 gallons, 40% solution=50 gallons
c.
70% solution=75 gallons, 40% solution=75 gallons
d.
70% solution=130 gallons, 40% solution=20 gallons


Solution:

Let gallons of 70% solution = x
and gallons of 40% solution = y

70% solution = 0.7x
40% solution = 0.4y
Mixture solution = 150$\times$0.60

BY given,
x + y=150....(1)
0.70x + 0.4y = 0.60(150).....(2)

Use elimination method,
from equation (1)
x = 150 - y
Substitute in equation (2)
0.70(150 - y) + 0.4y = 0.60(150)
y = 50 
Substitute y value in equation (1)
x + 50 = 150
x = 100
Answer is:
70% solution = 100 gallons
40% solution = 50 gallons



Correct answer : (2)
 8.  
Suppose a mixture of 2 gallons brand A juice and 6 gallons brand B juice which contain 55% fruit juice. If the mixture contain 35% fruit juice.Find the percentage of fruit juice in brand A?
a.
50%
b.
30%
c.
45%
d.
55%


Solution:

Let the percentage of juice in brand A = x

2 gallons brand A = 2x
6 gallons brand B = 6$\times$0.55    
12 gallons brand mixture = 12$\times$0.35

We get the equation,
2x + 6$\times$0.55 = 12$\times$0.35
x = 0.45

Answer is:
Brand A contain 45% of fruit juice



Correct answer : (3)
 9.  
Solution X is 40% silicon and solution Y is 75% silicon. Suppose you want to make 75 liters of a solution which contain 60% silicon. Find how many liters of each solution used?
a.
X=45 L,Y=30 L
b.
X=30 L,Y=45 L
c.
X=42.857 L,Y=32.1429 L
d.
X=32.1429 L,Y=42.8571 L


Solution:

Let 40% used solution X = 0.40n
75% used solution Y = 0.75(75-n)  
60% new solution = 75$\times$0.60 

We get equation 
0.40n + 0.75(75 - n) = 0.60$\times$75
n = 32.1429

Hence, 32.1429 liters of solution X and 42.8571 liters of solution Y



Correct answer : (4)
 10.  
Gloria invested money at 7% and $8000 more than this at 8%. Find the amount invested at each rate if the total annual interest is $1540.
a.
$3000 at 7% and $14000 at 8%
b.
$6000 at 7% and $14000 at 8%
c.
$3000 at 7% and $10000 at 8%
d.
$6000 at 7% and $10000 at 8%


Solution:

Let x= the $ invested at 7%= 0.07x
and y= the $ invested at 8%=0.08y 

Total is y = x + 8000....(1)
0.07x + 0.08y = 1540......(2)

substitute y value from equation (1)to equation (2)
0.07(x + 8000) + 0.08y = 1540
x = 6000

Substitute x value in equation (1)
y = 6000 + 8000
y = 14000

Hence, $6000 invested at 7%
and $14000 invested at 8%



Correct answer : (2)

*AP and SAT are registered trademarks of the College Board.