Solving Age Ratio Problems: Finding A's Present Age

This problem involves the concept of ratios applied to the ages of two individuals, A and B, at different points in time. We are given the ratio of their ages six years ago and the ratio of their ages four years from now. Our goal is to find A's current age.

Setting up Variables for Present Ages

Let's assume:

• A's present age is \(A_p\) years.
• B's present age is \(B_p\) years.

Formulating Equations from Given Ratios

We are given two pieces of information about the ratios of their ages at different times:

Ratio 6 Years Ago

Six years ago, A's age was \(A_p - 6\) and B's age was \(B_p - 6\). The ratio of their ages was 7:5.

This gives us the equation:

\(\qquad \frac{A_p - 6}{B_p - 6} = \frac{7}{5}\)

Cross-multiplying gives:

\(\qquad 5(A_p - 6) = 7(B_p - 6)\)

\(\qquad 5A_p - 30 = 7B_p - 42\)

Rearranging the terms to form a linear equation:

\(\qquad 5A_p - 7B_p = -42 + 30\)

\(\qquad 5A_p - 7B_p = -12 \quad \cdots (1)\)

Ratio 4 Years From Now

Four years from now, A's age will be \(A_p + 4\) and B's age will be \(B_p + 4\). The ratio of their ages will be 11:9.

This gives us the equation:

\(\qquad \frac{A_p + 4}{B_p + 4} = \frac{11}{9}\)

Cross-multiplying gives:

\(\qquad 9(A_p + 4) = 11(B_p + 4)\)

\(\qquad 9A_p + 36 = 11B_p + 44\)

Rearranging the terms to form another linear equation:

\(\qquad 9A_p - 11B_p = 44 - 36\)

\(\qquad 9A_p - 11B_p = 8 \quad \cdots (2)\)

Solving the System of Linear Equations

We now have a system of two linear equations with two variables \(A_p\) and \(B_p\):

• \(5A_p - 7B_p = -12\)
• \(9A_p - 11B_p = 8\)

We can solve this system using methods like substitution or elimination. Let's use the elimination method to find \(A_p\).

To eliminate \(B_p\), we can multiply Equation (1) by 11 and Equation (2) by 7:

Multiplying Equation (1) by 11:

\(\qquad 11 \times (5A_p - 7B_p) = 11 \times (-12)\)

\(\qquad 55A_p - 77B_p = -132 \quad \cdots (3)\)

Multiplying Equation (2) by 7:

\(\qquad 7 \times (9A_p - 11B_p) = 7 \times 8\)

\(\qquad 63A_p - 77B_p = 56 \quad \cdots (4)\)

Now, subtract Equation (3) from Equation (4) to eliminate \(B_p\):

\(\qquad (63A_p - 77B_p) - (55A_p - 77B_p) = 56 - (-132)\)

\(\qquad 63A_p - 77B_p - 55A_p + 77B_p = 56 + 132\)

\(\qquad (63A_p - 55A_p) + (-77B_p + 77B_p) = 188\)

\(\qquad 8A_p + 0 = 188\)

\(\qquad 8A_p = 188\)

Now, solve for \(A_p\):

\(\qquad A_p = \frac{188}{8}\)

Simplify the fraction:

\(\qquad A_p = \frac{94}{4} = \frac{47}{2}\)

Convert the improper fraction to a mixed number:

\(\qquad A_p = 23 \frac{1}{2}\)

So, A's present age is \(23 \frac{1}{2}\) years.

Verifying the Age Calculation

Let's quickly check if this age fits the conditions. If \(A_p = 23.5\), substitute into \(5A_p - 7B_p = -12\) to find \(B_p\):

\(\qquad 5(23.5) - 7B_p = -12\)

\(\qquad 117.5 - 7B_p = -12\)

\(\qquad -7B_p = -12 - 117.5\)

\(\qquad -7B_p = -129.5\)

\(\qquad B_p = \frac{-129.5}{-7} = 18.5\)

So, B's present age is 18.5 years.

• 6 years ago: A was \(23.5 - 6 = 17.5\), B was \(18.5 - 6 = 12.5\). Ratio \(\frac{17.5}{12.5} = \frac{175}{125} = \frac{7}{5}\). Correct.
• 4 years from now: A will be \(23.5 + 4 = 27.5\), B will be \(18.5 + 4 = 22.5\). Ratio \(\frac{27.5}{22.5} = \frac{275}{225} = \frac{11}{9}\). Correct.

The calculated present age of A is consistent with the given information.

The present age of A is \(23\frac{1}{2}\) years.

Revision Table: Age Ratio Problem Steps

Additional Information: Age Problems and Linear Equations

Age problems are a common type of word problem in mathematics that can often be solved using systems of linear equations. The key is to carefully represent the ages of individuals at different points in time (past, present, future) using variables and then translate the given relationships (usually ratios or differences) into algebraic equations.

A system of linear equations consists of two or more linear equations that are considered simultaneously. A solution to a system of linear equations with two variables is a pair of values (\(x, y\)) that satisfies all equations in the system. Common methods for solving systems of linear equations include:

• Substitution Method: Solve one equation for one variable and substitute that expression into the other equation.
• Elimination Method: Multiply equations by constants so that the coefficients of one variable are opposites, then add the equations to eliminate that variable. This is the method used in the solution above.
• Graphical Method: Graph both equations; the intersection point represents the solution. This method is less practical for non-integer solutions.

Understanding how to set up and solve linear equations is fundamental for tackling various word problems, including those involving ages, distances, mixtures, and more.