# 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.

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

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

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

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

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

The first solution percent = 64%

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$(60-n) and mixture of x and y =60$\times3.78

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

cashew x = 30kg and y = 30kg were added

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

Pure water =30 ounces

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
70% solution = 100 gallons
40% solution = 50 gallons

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

Brand A contain 45% of fruit juice

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

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%