Age and Mixture Problems and Solutions in Algebra

Updated on July 12, 2018
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JR has a Bachelor of Science in Civil Engineering and specializes in Structural Engineering. He loves to write anything about education.

Age and Mixture Problems in Algebra

Age and mixture problems are applications of creating equations from given algebraic problems. It requires good analytical thinking skills and understanding when answering age and mixture problems in algebra. Sometimes, you have to see the word problem twice to understand it fully. Then, write the equations from every phrase or sentence carefully. As much as possible, create a table and categorize the elements of the problem. Write the data in the table in an orderly and organized way. In that way, the formulation of equations will be uncomplicated. Here are some problems in algebra about age and mixtures that you can practice.

Age and Mixture Article Contents:

  • Father and son's ages
  • A person's age
  • Comparison of ages
  • Steel containing nickel mixture problems
  • An alloy containing gold mixture problems
  • The ratio of mixture quantities problems
  • Salt solution mixture problems

Problem 1: Father and Son's Ages

Two times the father's age is eight more than six times the son's age. Ten years ago, the sum of their ages was 36 years. The age of the son is:

Solution

a. Let x be the age of the son and y be the age of the father.

2y = 6x + 8
y = 3x + 8

b. Create a mathematical relation between the father's age and the son's age ten years ago.

(x - 10) + (y - 10) = 36
x + y = 56

c. Substitute the value of y into the equation x + y = 56.

x + y = 56
y = 3x + 8

x + (3x + 8) = 56
4x + 8 = 56
4x = 56 -8
4x = 48
x = 12

Final Answer: The son's age is 12 years old.

Problem 2: A Person's Age

John's age 15 years ago was 1/3 of his age nine years hence. How old is John?

Solution

a. Let x be the age of John now. His age 15 years ago was x- 15 and his age nine years hence is x + 9.

x - 13 = (1/3) (x + 9)
x - 13 = (1/3) x + 3
x - (1/3) x = 3 + 13
(2/3) x = 16
x = 24

Final Answer: Therefore, John's age is 24 years old.

Problem 3: Mother and Daughter's Ages

A mother is 41 years old, and in seven years she will be four times as old as her daughter. How old is her daughter now?

Solution

a. Let x be the age of the daughter and y be the age of the mother.

4 (x + 7) = 41 + 7
4x + 28 = 48
4x = 48 - 28
4x = 20
x = 5

Final Answer: The daughter is five years old.

Problem 4: Father and Son's Ages

A father is four times as old as his son. Six years ago, he was five times as old as his son was at that time. How old is his son?

Solution

a. Let x be the present age of the father and y be the age of the son.

x = 4y

b. Create a mathematical relation between the father's age and the son's age six years ago.

(x - 6) = 5 (y - 6)
x - 6 = 5y - 30
x - 5y = -30 + 6
x - 5y = -24
x = 5y - 24

c. Substitute the value of x = 5 to the first equation.

(5y - 24) = 4y
5y - 4y = 24
y = 24

Final Answer: The son is 24 years old now.

Problem 5: Father and Son's Ages

The ages of the father and son are 50 and 10, respectively. How many years will the father be three times as old as his son?

Solution

a. Let x be the required number of years. Create a mathematical relation between their ages.

50 + x = 3 (10 + x)
50 + x = 30 + 3x
50 + 30 = 3x - x
80 = 2x
x = 40

Final Answer: After 40 years, the father will be three times as old as his son.

Problem 6: Comparison of Ages

Peter is 24 years old. Peter is twice as old as John was when Peter was as old as John is now. How old is John?

Solution

a. Let x be the present age of John. The table shows the relationship between their past and present ages.

 
Past
Present
Peter
x
24
John
24 / 2
x
Past and Present Ages of Peter and John

b. The difference between the ages of two persons is constant.

x - 12 = 24 -x
x + x = 24 + 12
2x = 36
x = 18 years

Final Answer: John is 18 years old now.

Problem 7: Steel Containing Nickel

Mixing steel containing 14% nickel with another steel containing 6% nickel will make two thousand (2000) kg of steel containing 8% nickel. How much of the steel comprising 14% nickel is required?

Mixture Problems in Algebra: Mixture of Steel and Nickel
Mixture Problems in Algebra: Mixture of Steel and Nickel | Source

Solution

a. Create a table representing the equation.

 
Mixture 1
Mixture 2
Final Mixture
Steel
x
y
2000 kg
Nickel
14%
6%
8%

b. Create a mathematical equation for both steel and nickel. Then, create an equation for the summation of mixtures.

Steel:
x + y = 2000
y = 2000 - x

Mixture 1 + Mixture 2 = Final Mixture
14x + 6y = 8 (2000)
7x + 3y = 8000

c. Substitute equation 1 to equation 2.

7x + 3(2000 - x) = 8000
x = 500 kg

Final Answer: 500 kg of steel containing 14% nickel is needed.

Problem 8: Alloy Containing Gold

A 20-gram alloy containing 50% gold melts a 40-gram alloy containing 35% gold. How much percentage of gold is the resulting alloy?

Mixture Problems: Alloy Containing Gold
Mixture Problems: Alloy Containing Gold | Source

Solution

a. Solve for the total number of grams of the alloy.

Total alloy = 20 + 40
Total alloy = 60 grams

b. Create a table representing the mixtures.

 
Mixture 1
Mixture 2
Final Mixture
Alloy
40 g
20 g
60 g
Gold
35%
50%
x

c. Create an equation for the mixtures.

35% (40) + 50% (20) = x (60)
x = 40%

Final Answer: The resulting alloy contains 40% gold.

Problem 9: Ratio of Mixtures

In what ratio must a peanut costing $240 per kilogram be mixed with a peanut costing $340 per kilogram so that a profit of 20% is made by selling the mixture at $360 per kilogram?

Solution

a. Let x be the quantity of $240 per kilogram and y be the quantity of $340 per kilogram of peanuts. Write an equation for the capital and total sales.

Capital = 240x + 340y
Total sales = 360 (x + y)
Total sales = 360x + 360y

b. The formula for profit is:

Profit = Total Sales - Capital
Profit = (360x + 360y) - (240x + 340y)
Profit = 120x + 20y

c. Since profit is 20% of the capital, the equation would be:

120x + 20y = 0.20 (240x + 340y)
120x + 20y = 48x + 68y
72x = 48y

d. Write the ratio of x and y variables.

(x) / (y) = 48 / 72
(x) / (y) = 2 / 3

Final Answer: The final ratio is 2/3.

Problem 10: Salt Solution

A 100-kg salt solution initially 4% by weight. Salt in water is boiled to reduce water content until the concentration is 5% by weight. How much water evaporated?

Mixture Problems: Salt Solution
Mixture Problems: Salt Solution | Source

Solution

a. Create a mathematical equation for the mixtures.

4% (100) - 0 = 5% (100 - x)
400 = 500 - 5x
x = 20 kg

b. Check the water.

96% (100) - 100% (x) = 95% (100 - x)
1920 - 20x = 1900 - 19x
1920 - 1900 = -19x + 20x
x = 20 kg

Final Answer: 20 kg of water evaporated.

Problem 11: Sum of Ages

A boy is one-third as old as his brother and eight years younger than his sister. The sum of their ages is 38 years. How old is his sister?

Solution

a. Let x be the age of the boy. Create a mathematical equation for the ages.

3x = age of the brother
x + 8 = age of sister

x + 3x + (x + 8) = 38
5x = 30
x =  6 years (age of boy)

x + 8 = 14 years

Final Answer: The age of the sister is 14 years old.

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    © 2018 John Ray

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