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 |

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 |

Given X = 20% calories

Y = 40% calories

So,

X + Y = 40...(1)

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 |

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% |

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 |

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 |

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 |

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% |

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

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 |

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% |

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)