Ray is a Licensed Civil Engineer 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:
a. Let x be the age of the son and y be the age of the father.
b. Create a mathematical relation between the father's age and the son's age ten years ago.
c. Substitute the value of y into the equation x + y = 56.
Final Answer: The son's age is 13 years old.
Problem 2: A Person's Age
John's age 13 years ago was 1/3 of his age nine years hence. How old is John?
a. Let x be the age of John now. His age 13 years ago was x- 13 and his age nine years hence is x + 9.
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?
a. Let x be the age of the daughter and y be the age of the mother.
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?
a. Let x be the present age of the father and y be the age of the son.
b. Create a mathematical relation between the father's age and the son's age six years ago.
c. Substitute the value of x = 5 to the first equation.
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?
a. Let x be the required number of years. Create a mathematical relation between their ages.
Final Answer: After 10 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?
a. Let x be the present age of John. The table shows the relationship between their past and present ages.
24 / 2
b. The difference between the ages of two persons is constant.
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?
a. Create a table representing the equation.
|Mixture 1||Mixture 2||Final Mixture|
b. Create a mathematical equation for both steel and nickel. Then, create an equation for the summation of mixtures.
c. Substitute equation 1 to equation 2.
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?
a. Solve for the total number of grams of the alloy.
b. Create a table representing the mixtures.
|Mixture 1||Mixture 2||Final Mixture|
c. Create an equation for the mixtures.
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?
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.
b. The formula for profit is:
c. Since profit is 20% of the capital, the equation would be:
d. Write the ratio of x and y variables.
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?
a. Create a mathematical equation for the mixtures.
b. Check the water.
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?
a. Let x be the age of the boy. Create a mathematical equation for the ages.
Final Answer: The age of the sister is 14 years old.
Questions & Answers
Question: Kit is twice as old as Sam. Sam is 5 yrs older than Cara. In 5 years, Kit will be three times as old as Cara. How old is Sam?
Answer: Let age of Carla: x
Sam's age: x + 5
Kit's age: 2(x+5) or 2x+10
Their ages in 5 years (future):
Carla: X + 5
Sam: x+5 + 5 or x +10
Kit: 2x+10+5 or 2x+15
Condition in 5 years:
Kit's age will be three times as old as Carla
2x + 15 = 3(x+5)
2x+15 = 3x + 15
3x-2x = 15-15
x = 0
Carla: x = 0 (she's maybe a newborn or infant)
Sam: X + 5
0 + 5 = 5 years old
Kit: 2x + 10
2(0) + 10 = 10 years old
Sam is 5 years old
Question: What is the age of Jeremy and Rain after 3 years if Jeremy is 5 years older than Rain?
Answer: I believe this is unsolvable. The problem might be lacking some more given. To show you,
Let x be Jeremy's age and y be Rain's age.
x = y + 5
Their ages after 3 years will be x + 3 and y + 3. There must be one more provision or relationship in order to compute for their ages. We need two equations to solve two unknowns.
Question: In 8 years, Mane will be three times her current age. In how many years will she be 20 years old?
Answer: Let x be the present age of Mane.
x + 8 = 3x
8 = 3x - x
8 = 2x
x = 4 years old
Mane's current age is 4. In 16 years, she will be 20 years old.
Therefore, the answer is 16 years.
Question: What do you mean by sum of ages?
Answer: Basically, the sum of ages is when you add the ages of two persons. Either it is their present ages, previous ages or their future ages depending on what is stated in the problem. Solving age problems really requires a lot of critical thinking and analyzation skills. Just practice more problems so you can master solving age problems.
Question: The present age of Hina's mother is four times that of her daughter. After 15 years the sum of their ages will be 75 years. Find the present age of Hina and her mother?
Answer: First you have to set-up variables. Let x be the present age of Hina's and y be her mother's present age.
From the first sentence, we can create an equation like this.
y = 4x (eq.1)
After 15 years, Hina's age will be x + 15 and her mother's age will be y + 15. Since the sum of their ages is 75, the equation will be:
x + 15 + y + 15 = 75
x + y = 75 - 30
x + y = 45 (eq. 2)
Substitute equation 1 in equation 2
x + 4x = 45
5x = 45
x = 9 years old
y = 4 x 9
y = 36 years old
Therefore, Hina's present age is 9 and her mother's present age is 36.
© 2018 Ray
Agaba amos on October 21, 2019:
MissEngr on September 30, 2018:
I believe the answer on #1 should be 13 for the son's age.
You forgot to divide all terms by 2.
2y = 6x + 8
y = 3x + 8 (should be y = 3x + 4)
For #2, it should be 27 since it's not 13 years ago but 15.
For #5, you forgot to change the sign. Hence, it should be 10 years.
50 + x = 30 + 3x
50 + 30 = 3x - x (should be 50 - 30 = 3x - x)
I did not review the rest but hope you change the incorrect solutions. :)