Age Relationship Between Meera and Heena

Understanding age relationships often involves setting up algebraic equations based on given conditions. Let’s explore the age relationship between Meera and Heena through an example to see how we can solve such problems.

Example Problem

Suppose we know the following about Meera and Heena:

  1. Meera is 4 years older than Heena.
  2. The sum of their ages is 24 years.

We need to find their individual ages.

Step-by-Step Solution

  1. Define Variables
    First, let’s define variables for their ages. Let:

    • $M$ be Meera’s age.
    • $H$ be Heena’s age.

  1. Set Up Equations
    Based on the problem, we can set up the following equations:

    1. Meera is 4 years older than Heena:
      $M = H + 4$
    2. The sum of their ages is 24 years:
      $M + H = 24$

  1. Substitute and Solve
    We can substitute the first equation into the second equation to solve for $H$
    Substitute $M$ in the second equation:
    $(H + 4) + H = 24$

    Combine like terms:
    $2H + 4 = 24$

    Subtract 4 from both sides:
    $2H = 20$

    Divide by 2:
    $H = 10$

    So, Heena is 10 years old.

  1. Find Meera’s Age
    Now, substitute $H = 10$ back into the first equation to find Meera’s age:
    $M = 10 + 4$
    $M = 14$

    So, Meera is 14 years old.

Verification

Let’s verify our solution by checking if the sum of their ages is 24:
$M + H = 14 + 10 = 24$
The solution is correct.

Conclusion

Through this example, we see that setting up algebraic equations based on given conditions helps us solve age-related problems efficiently. Meera is 14 years old, and Heena is 10 years old.

Additional Examples

Let’s explore a few more examples to solidify our understanding.

Example 2

Suppose Meera is twice as old as Heena, and the sum of their ages is 36 years. Find their ages.

  1. Define Variables
    Let:

    • $M$ be Meera’s age.
    • $H$ be Heena’s age.

  1. Set Up Equations
    1. Meera is twice as old as Heena:
      $M = 2H$
    2. The sum of their ages is 36 years:
      $M + H = 36$

  1. Substitute and Solve
    Substitute the first equation into the second equation:
    $2H + H = 36$

    Combine like terms:
    $3H = 36$

    Divide by 3:
    $H = 12$

    So, Heena is 12 years old.

  1. Find Meera’s Age
    Substitute $H = 12$ back into the first equation:
    $M = 2 times 12$
    $M = 24$

    So, Meera is 24 years old.

Verification

Check the sum of their ages:
$M + H = 24 + 12 = 36$
The solution is correct.

Example 3

Suppose Meera is 3 years younger than twice Heena’s age, and the sum of their ages is 27 years. Find their ages.

  1. Define Variables
    Let:

    • $M$ be Meera’s age.
    • $H$ be Heena’s age.

  1. Set Up Equations
    1. Meera is 3 years younger than twice Heena’s age:
      $M = 2H – 3$
    2. The sum of their ages is 27 years:
      $M + H = 27$

  1. Substitute and Solve
    Substitute the first equation into the second equation:
    $(2H – 3) + H = 27$

    Combine like terms:
    $3H – 3 = 27$

    Add 3 to both sides:
    $3H = 30$

    Divide by 3:
    $H = 10$

    So, Heena is 10 years old.

  1. Find Meera’s Age
    Substitute $H = 10$ back into the first equation:
    $M = 2 times 10 – 3$
    $M = 20 – 3$
    $M = 17$

    So, Meera is 17 years old.

Verification

Check the sum of their ages:
$M + H = 17 + 10 = 27$
The solution is correct.

Conclusion

By setting up and solving algebraic equations, we can determine the age relationship between Meera and Heena in various scenarios. This method is useful for solving a wide range of age-related problems.

Citations

  1. 1. Khan Academy – Solving Age Problems
  2. 2. Purplemath – Age Word Problems
  3. 3. Math is Fun – 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