Understanding the Age Word Problem

This question is an age-based word problem that involves setting up and solving linear equations. We are given information about the ages of two individuals, Ravish and Kailash, at different points in time: three years ago and three years from today. We need to find Ravish's current or present age.

Setting Up the Problem with Variables

To solve this type of problem, it's best to represent the unknown present ages using variables.

• Let \(R\) be the present age of Ravish (in years).
• Let \(K\) be the present age of Kailash (in years).

Formulating Equations from the Given Conditions

We are given two conditions that relate the ages of Ravish and Kailash at different times. We can translate these conditions into algebraic equations.

Condition 1: Three years ago

The question states that three years ago, the difference between the age of Ravish and the age of Kailash was 18 years.

• Ravish's age three years ago was \(R - 3\).
• Kailash's age three years ago was \(K - 3\).

The difference in their ages three years ago was 18 years. This can be written as:

\((R - 3) - (K - 3) = 18\)

Let's simplify this equation:

\(R - 3 - K + 3 = 18\)

\(R - K = 18\)     (Equation 1)

This equation tells us that the difference between Ravish's present age and Kailash's present age is 18 years. This makes sense because the age difference between two people remains constant throughout their lives.

Condition 2: Three years from today

The question states that three years from today, Ravish will be three times as old as Kailash.

• Ravish's age three years from today will be \(R + 3\).
• Kailash's age three years from today will be \(K + 3\).

Ravish's age at that time will be three times Kailash's age at that time. This can be written as:

\(R + 3 = 3 \times (K + 3)\)

Let's simplify this equation:

\(R + 3 = 3K + 9\)      (Equation 2)

Solving the System of Equations

Now we have a system of two linear equations with two variables (\(R\) and \(K\)):

1. \(R - K = 18\)
2. \(R + 3 = 3K + 9\)

We can use the substitution method or elimination method to solve this system. Let's use substitution.

From Equation 1, we can express \(R\) in terms of \(K\):

\(R = K + 18\)

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

\((K + 18) + 3 = 3K + 9\)

Simplify and solve for \(K\):

\(K + 21 = 3K + 9\)

Subtract \(K\) from both sides:

\(21 = 2K + 9\)

Subtract 9 from both sides:

\(21 - 9 = 2K\)

\(12 = 2K\)

Divide by 2:

\(K = \frac{12}{2}\)

\(K = 6\)

So, the present age of Kailash is 6 years.

Now that we have the value of \(K\), we can find \(R\) using Equation 1 (\(R = K + 18\)):

\(R = 6 + 18\)

\(R = 24\)

Thus, the present age of Ravish is 24 years.

Verification

Let's check if these present ages satisfy both original conditions:

• Present ages: Ravish = 24, Kailash = 6.
• Three years ago: Ravish was \(24 - 3 = 21\). Kailash was \(6 - 3 = 3\). The difference was \(21 - 3 = 18\). This matches Condition 1.
• Three years from today: Ravish will be \(24 + 3 = 27\). Kailash will be \(6 + 3 = 9\). Is Ravish three times Kailash? \(3 \times 9 = 27\). Yes, it matches Condition 2.

Both conditions are satisfied, so our calculated present age for Ravish is correct.

The present age of Ravish is 24 years.

Revision Table: Age Word Problems

Additional Information: Solving Linear Equations

Age word problems often lead to systems of linear equations. A system of linear equations is a set of two or more linear equations that share the same variables. There are several methods to solve them:

• Substitution Method: Solve one equation for one variable, then substitute that expression into the other equation. This reduces the system to a single equation with one variable.
• Elimination Method: Multiply one or both equations by constants so that the coefficients of one variable are opposites. Then, add the equations together to eliminate that variable.
• Graphical Method: Graph each equation on the same coordinate plane. The point where the lines intersect is the solution to the system. This method is less precise for non-integer solutions.

In this problem, we used the substitution method, which is generally efficient when one variable can be easily isolated in one of the equations.