How to Determine Someone’s Age from a Problem?

Determining someone’s age from a problem often involves setting up and solving algebraic equations. Let’s break down the process step by step with examples to make it easier to understand.

Step-by-Step Process

  1. Understand the Problem
    First, read the problem carefully to understand what is being asked. Identify the information given and what you need to find. For example:

    Example Problem:
    Sarah is twice as old as her brother John. In 5 years, the sum of their ages will be 40. How old are they now?

  1. Define Variables
    Assign variables to the unknowns. In our example, let:

    • $S$ be Sarah’s current age
    • $J$ be John’s current age

  1. Set Up Equations
    Use the information given to set up equations. According to the problem:

    1. Sarah is twice as old as John: $S = 2J$
    2. In 5 years, the sum of their ages will be 40: $(S + 5) + (J + 5) = 40$

  1. Simplify and Solve the Equations
    Simplify the equations to solve for the variables. Start with the second equation:

    $(S + 5) + (J + 5) = 40$

    $S + J + 10 = 40$

    $S + J = 30 tag{1}$

    Now, substitute $S = 2J$ from the first equation into equation (1):

    $2J + J = 30$

    $3J = 30$

    $J = 10$

    Now, substitute $J = 10$ back into $S = 2J$:

    $S = 2 times 10$

    $S = 20$

    So, John is 10 years old, and Sarah is 20 years old.

  1. Verify the Solution
    Always verify your solution by plugging the values back into the original problem to ensure they satisfy all conditions.

    In 5 years:

    • Sarah’s age: $20 + 5 = 25$
    • John’s age: $10 + 5 = 15$
    • Sum of their ages: $25 + 15 = 40$

    The solution is correct.

Additional Examples

Example 2: Age Difference

Problem:
Emma is 4 years older than Mia. In 3 years, Emma will be twice as old as Mia. How old are they now?

Solution:

  1. Let $E$ be Emma’s current age and $M$ be Mia’s current age.
  2. Set up the equations based on the problem:
    • $E = M + 4$
    • $E + 3 = 2(M + 3)$
  3. Simplify and solve the equations:
    • Substitute $E = M + 4$ into the second equation:

      $M + 4 + 3 = 2(M + 3)$

      $M + 7 = 2M + 6$

      $7 – 6 = 2M – M$

      $1 = M$

    • So, $M = 1$
    • Substitute $M = 1$ back into $E = M + 4$:

      $E = 1 + 4$

      $E = 5$

  4. Verify the solution:
    • In 3 years, Emma’s age: $5 + 3 = 8$
    • In 3 years, Mia’s age: $1 + 3 = 4$
    • Emma will be twice as old as Mia: $8 = 2 times 4$

The solution is correct. Emma is 5 years old, and Mia is 1 year old.

Example 3: Age Sum

Problem:
The sum of the ages of a father and his son is 50 years. Five years ago, the father’s age was four times the son’s age. How old are they now?

Solution:

  1. Let $F$ be the father’s current age and $S$ be the son’s current age.
  2. Set up the equations based on the problem:
    • $F + S = 50$
    • $F – 5 = 4(S – 5)$
  3. Simplify and solve the equations:
    • From the first equation, express $F$ in terms of $S$:

      $F = 50 – S$

    • Substitute $F = 50 – S$ into the second equation:

      $(50 – S) – 5 = 4(S – 5)$

      $45 – S = 4S – 20$

      $45 + 20 = 4S + S$

      $65 = 5S$

      $S = 13$

    • So, $S = 13$
    • Substitute $S = 13$ back into $F = 50 – S$:

      $F = 50 – 13$

      $F = 37$

  4. Verify the solution:
    • Five years ago, father’s age: $37 – 5 = 32$
    • Five years ago, son’s age: $13 – 5 = 8$
    • Father’s age was four times the son’s age: $32 = 4 times 8$

The solution is correct. The father is 37 years old, and the son is 13 years old.

Conclusion

Determining someone’s age from a problem involves understanding the problem, defining variables, setting up equations, solving the equations, and verifying the solution. By following these steps, you can solve various age-related problems efficiently. Practice with different problems to become proficient in this method.

Citations

  1. 1. Khan Academy – Solving Age Problems
  2. 2. Purplemath – Age Word Problems
  3. 3. Mathway – Age Problems

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ