Algebra Tips: How to Solve Word Problems About Ages

These kids will only reveal their ages if you solve a word problem.
These kids will only reveal their ages if you solve a word problem. | Source

Word problems about ages are a popular theme for math puzzles and brain teasers because they make you think analytically; you have to figure out how to apply your math skills to find the solution. Learning how to solve these math problems can help you understand and solve more complicated mathematical problems and riddles. If you always find yourself at a loss for where to begin, the following problem-solving strategies will make these problems a breeze.

Mathematical Notation: Variable Names

A typical age problem names the people whose ages are unknown. The first letters of their names are good choices for variables to represent the unknown ages. For example, if you are trying to determine how old Paula, Quinn, and Robert are right now, you will have an easier time setting up and solving the equations if you represent their present ages with the letters P, Q, and R.

Math word problems about ages deal with current and future ages.
Math word problems about ages deal with current and future ages. | Source

Representing Future Ages

To write equations that describe mathematical relations between future ages, you must add a certain number of years to the variable. For example, if the two current age variables in question are A and B, and you want to describe what happens to their ages 8 years from now, you must replace every instance of "A" with "A + 8," and every instance of "B" with "B + 8."

Consider this simple example problem: The sum of Ben's and Anne's current ages is 48. In 8 years, Anne will be 5/3 times as old as Ben. How old are Anne and Ben now?

The first statement is easy to translate into math terms: A + B = 48. For the second statement, we need to consider the new-ish variables "A + 8" and "B + 8." Using these as placeholders for Anne's and Ben's future ages we get the second equation A + 8 = (5/3)(B + 8).

As two people grow old together, the ratio between their ages approaches 1.
As two people grow old together, the ratio between their ages approaches 1. | Source

The Mathematical Relations Between People's Ages and How They Change Over Time

Given two people's ages, there are several mathematical relations you can calculate using these two numbers. For instance, if the ages are represented by the variables C and D, you could compute any of the following:

  • The ratio of C to D, C/D
  • The ratio of D to C, D/C
  • The difference between C and D, either C - D or D - C
  • The sum of C and D, C + D
  • The product of C and D, CD
  • The arithmetic mean (average) of C and D, (C + D)/2
  • Miscellaneous: Reciprocals 1/C and 1/D, Sum of squares C^2 + D^2, Greatest common factor of C and D, Least common multiple of C and D.

The numerical values of all of these quantities change over time, except for one, the age difference. If the present age difference between Carol and Dan is 5 years, then in 20, 30, or 40 years from now they will still be 5 years apart.

The sum, product, and average of two people's ages will increase as they grow older. If Carol and Dan are currently 12 years old and 16 years old, their current age sum, product, and average are 28, 192, and 14 respectively. In 10 years, their age sum, product, and average will be 48, 572, and 24. Sums of squares and other powers also increase over time.

The ratio between two ages becomes closer to 1 as the years wear on. If Carol and Dan are 12 and 16 right now, their current age ratios are C/D = 0.75 and D/C = 1.3333. But in 10 years these ratios will be (C+10)/(D+10) = 22/26 = 0.8462 and (D+10)/(C+10) = 26/22 = 1.1818.

Reciprocals become closer to 0 as ages increase.

How the greatest common factor and least common multiple change over time depends on the number theoretic properties of the ages. For example, if Carol and Dan are currently 12 and 16, then gcd = 4 and lcm = 48. But if you increase their ages by 1 year to 13 and 17 respectively, you get gcd = 1 and lcm = 221. Increase their ages by 2 years to 14 and 18 respectively and you get gcd = 2 and lcm = 126. The gcd of two people's ages can never exceed the difference between their ages (in this example, 4). The lcm is always equal to the product of the ages divided by the gcd.

Reduce the Number of Variables Whenever Possible

If you know the difference between two people's ages and which one of them is older, you can reduce the number of variables in the problem. For example, suppose you do not know how old Tim and Ursula are, but you do know that Tim is 3 years younger than Ursula. Instead of representing their unknown ages with the pair of variables T and U, you can either replace T with U - 3, or replace U with T + 3. This substitution leaves you with one variable to consider rather than two.

For example, if you are told that the product of Tim's and Ursula's ages is a composite number between 173 and 193, there's not enough information to find a unique solution. But if you know that Ursula is three years older than Tim, you can reduce the number of unknowns and find the unique solution.

The same trick applies to ages that are consecutive numbers (or consecutive evens/odds). If you let the youngest person's age be K, then the older people's ages can be represented by K+1, K+2, K+3, and so on. In the case of consecutive even/odd ages, the sequence is K, K+2, K+4, K+6, and so on, since a pair of consecutive even/odd numbers differ by 2.

You can sometimes reduce the number of variables if you know the sum, product, or ratio of two ages, but you must be mindful that these mathematical relations change over time. For instance, if Liam is currently twice as old as Katie, the substitution L = 2K is only valid for the present. The equation (L+5) = 2(K+5) does not still hold if you are trying to describe the relation between their ages 5 years from now.

Today is the only day you'll ever be exactly four years old. Cheer up girl!
Today is the only day you'll ever be exactly four years old. Cheer up girl! | Source

Assume the Solutions are Whole Numbers

In reality, ages are not whole numbers. If you were born on January 1, 2000 and the current date is October 1, 2009, your true age is 9.75 years. But you will probably tell people you are 9 years old until your next birthday.

For age and birthday math problems, you should assume all the ages to be whole numbers (positive integers) unless otherwise noted. If a problem is posed in such a way that there may exist mathematically consistent solutions that are not integers, the problem may specify that you should discard solutions that are not whole numbers.

Example: Ellie is older than Fred, who in turn is older than Gerta. The sum of their three ages is 43 and the product is 2640. How old are these three people?

As this problem is stated, it actually has infinitely many correct solutions, a few of which are

  • E = 18.832, F = 14.5, G = 9.9668
  • E = 18.444, F = 15.037, G = 9.519
  • E = 19.227, F = 13.882, G = 9.891
  • E = 20.05, F = 11.544, G = 11.406

However, there is only one integer solution to this word problem.

This math problem would be clearer if it explicitly stated that you should discard non-integral solutions, or if it mentioned that today was their common birthday, which would imply that their current ages are whole numbers.

Verbal Clues: Older, Oldest, Younger, Youngest

If a problem states that certain people are older or younger than others, this information can help you reduce the number of possibilities when solving for people's ages. For example, if Marc's and Nadine's ages add up to 44 and Nadine is older, then Nadine must be at least 23 and Marc is no older than 21.

Similary, if you are trying to find a set of ages and you are given a clue about the oldest individual member, then you can rule out the possibility that a set of twins or triplets is older than everyone else. This is a key clue in the classic age problem "The Proof Is in the Pudding."

Remember Basic Algebra and Other Math Concepts

Most birthday and age word problems are set up to be systems of linear equations. You can solve a system of n linear equations in n variables using the techniques of substitution, elimination, matrices, or by using an algebra solver.

Age puzzles that give you sums and products of ages can be solved by finding the roots of polynomials. For example, a math puzzle that gives you the sum and product of two ages can be solved with a quadratic equation.

For math puzzles about the divisibility and factors of an age, you can apply basic number theory concepts. For example, if the product of two ages is an even number, the at least one of the ages is an even number (assuming whole number ages). If the sum of two ages is even, then either both ages are even or both are odd.

Example Age Word Problem

Bob and Billy are twins. Together with their older cousins Romeo and Tara, the sum of all their ages at present is 36. Six years ago, the sum of the twins' ages was half the sum of their cousins' ages. If the sum of the twins' and Romeo's current ages is a square number, how old is everyone?

For this problem, we can represent Bob's and Billy's current age with B, Romeo's age with R, and Tara's age with T. The first equation we can get out of the problem is 2B + R + T = 36, or equivalently 2B = 36 - (R + T).

The second statement of the problem can be translated into mathematical notation as

(B-6) + (B-6) = 0.5[(R-6) + (T-6)]

This equation can be simplified to 2B = 0.5(R + T) + 6. Before we analyze the last piece of information in the problem, let's use these two equations to find out more about the people's ages. Since we know that 2B = 36 - (R + T) and 2B = 0.5(R + T) + 6, we can eliminate the quantity 2B to make the equation

36 - (R + T) = 0.5(R + T) + 6

After some algebraic simplification, we can see that R + T = 20. While this doesn't tell us the individual ages of Romeo and Tara, it does tell us that B = 8.

Since the problem stated that both Romeo and Tara are older than Bob and Billy, there are only a few possibilities for the values of R and T:

  • R = 9 and T = 11
  • R = 10 and T = 10
  • R = 11 and T = 9

Only the first possibility satisfies the last condition of the problem, that 2B + R equal a square number. Therefore the complete solution is

  • Bob = 8
  • Billy = 8
  • Romeo = 9
  • Tara = 11

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Comments 5 comments

vickie 4 years ago

Thanks, this is really useful. I usually get stuck when they say "in 7 years so-and-so will be half as old as such-and-such was 8 years ago..."

ok so how do you solve... 4 years ago

the sum of Herp and Derp's ages is a square number, and in 7 years Herp will be half as old as Derp was 8 years ago?

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calculus-geometry 4 years ago from Germany Author

For the second condition, if you simplify the expression H + 7 = (1/2)(D - 8) you get D - 2H = 22. And the first condition is D + H = N^2, where N is an integer. Eliminating D from both equations gives you 3H = N^2 - 22, or H = (N^2 - 22)/3. Since N^2 can be any square number, this problem actually has infinitely many (whole number) solutions if you don't put any upper limits on Herp's and Derp's ages.

But if neither of them can be older than the world-record human age (currently 122 years), then you get the following solutions:

D = 24, H = 1

D = 40, H = 9

D = 50, H = 14

D = 74, H = 26

D = 88, H = 33

D = 120, H = 49

Tim 2 years ago

Thanks, I solved this one with trial and error, but am curious to know how you can do it with equations.

Ben is currently 2/3 of Steven's age. When Ben is as old as Steve is now, Ben will be 3/4 of Steven's age.

calculus-geometry profile image

calculus-geometry 2 years ago from Germany Author

Hi Tim, as stated, your age problem has infinitely many solutions. For example their current ages could be (B, S) = (2, 3), (4, 6), (6, 9), (8, 12), (10, 15), etc. The first equation you get out of the problem is

B = (2/3)S

The second equation you get is when you add the difference in their ages, S-B, to get the new relation when Ben is as old as Steven. This gives you

B + (S-B) = (3/4)[S + (S-B)]

But this equation reduces to the first:

B + (S-B) = (3/4)[S + (S-B)]

S = (3/4)[2S - B]

S - (6/4)S = -B

(-2/3)S = -B

(2/3)S = B

Without two distinct equations in two variables you can't get a unique solution. So there may be some information missing in the problem.

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