Solving the Population Increase Problem

The question asks us to find the initial number of males in a town given the total initial population, the percentage increase in males and females, and the new total population.

Let's break down the information given:

• Initial total population = 50,000
• New total population = 56,000
• Percentage increase in males = 10%
• Percentage increase in females = 15%

We need to find the number of males in the town initially.

Setting Up Equations for Population

Let \(M\) be the initial number of males and \(F\) be the initial number of females.

Based on the initial total population, we can write the first equation:

\(M + F = 50,000\) (Equation 1)

Now, let's consider the increase in population. The number of males increases by 10%, and the number of females increases by 15%. The new population is the sum of the increased number of males and the increased number of females.

New number of males = \(M + 10\%\) of \(M = M + 0.10M = 1.10M\)

New number of females = \(F + 15\%\) of \(F = F + 0.15F = 1.15F\)

The new total population is 56,000. So, we can write the second equation:

\(1.10M + 1.15F = 56,000\) (Equation 2)

Solving the System of Equations

We have a system of two linear equations:

1. \(M + F = 50,000\)
2. \(1.10M + 1.15F = 56,000\)

We can solve this system using substitution. From Equation 1, we can express \(F\) in terms of \(M\):

\(F = 50,000 - M\)

Now, substitute this expression for \(F\) into Equation 2:

\(1.10M + 1.15(50,000 - M) = 56,000\)

Distribute 1.15 on the left side:

\(1.10M + (1.15 \times 50,000) - (1.15 \times M) = 56,000\)

\(1.10M + 57,500 - 1.15M = 56,000\)

Combine the terms with \(M\):

\((1.10M - 1.15M) + 57,500 = 56,000\)

\(-0.05M + 57,500 = 56,000\)

Subtract 57,500 from both sides:

\(-0.05M = 56,000 - 57,500\)

\(-0.05M = -1,500\)

Divide both sides by -0.05 to find \(M\):

\(M = \frac{-1,500}{-0.05}\)

\(M = \frac{1,500}{0.05}\)

To remove the decimal from the denominator, multiply the numerator and denominator by 100:

\(M = \frac{1,500 \times 100}{0.05 \times 100}\)

\(M = \frac{150,000}{5}\)

\(M = 30,000\)

So, the initial number of males in the town was 30,000.

Verifying the Answer

If the initial number of males (\(M\)) is 30,000, then from Equation 1:

\(30,000 + F = 50,000\)

\(F = 50,000 - 30,000\)

\(F = 20,000\)

Initial number of females was 20,000.

Let's check if the new population matches 56,000:

• Increased males = \(1.10 \times 30,000 = 33,000\)
• Increased females = \(1.15 \times 20,000 = 23,000\)
• New total population = \(33,000 + 23,000 = 56,000\)

This matches the given new population, confirming our answer.

The number of males in the town initially was 30,000.

Revision Table: Population Increase Problem

Additional Information: Solving Word Problems

Solving word problems often involves translating the words into mathematical expressions and equations. Here are some tips:

• Read Carefully: Understand what the problem is asking and what information is given.
• Define Variables: Assign letters to the unknown quantities you need to find.
• Translate to Equations: Write down mathematical equations that represent the relationships described in the problem.
• Solve the Equations: Use algebraic techniques to solve for the variables.
• Check Your Answer: Plug your solution back into the original problem or equations to make sure it makes sense and satisfies all conditions.

In this problem, we used two variables (\(M\) and \(F\)) and formed two equations. Problems involving percentages and population changes are common applications of linear equations.