The present ages of A and B are in the ratio of 6:4. 5 years ago their ratio were in the 5:3. How old is A now? 

Let the present ages of A and B be \(6x\) and \(4x\) respectively, where \(x\) is a constant multiplier for the ratio.

### Step 1: Form equation based on the information
We know that 5 years ago, the ratio of their ages was 5:3. Therefore:

\[
\frac{6x - 5}{4x - 5} = \frac{5}{3}
\]

### Step 2: Solve the equation
Cross-multiply to eliminate the fractions:

\[
3(6x - 5) = 5(4x - 5)
\]

Expanding both sides:

\[
18x - 15 = 20x - 25
\]

### Step 3: Isolate \(x\)
Simplify and move the terms involving \(x\) to one side:

\[
18x - 20x = -25 + 15
\]

\[
-2x = -10
\]

\[
x = 5
\]

### Step 4: Calculate A's age
Since \(x = 5\), the present age of A is:

\[
6x = 6 \times 5 = 30
\]

So, A is 30 years old.

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Do you want further details or have any questions?

Here are 5 related questions:
1. What is B's current age?
2. How old was A five years ago?
3. How will the ratio of their ages change in 10 years?
4. Can we find the total of their ages using the given information?
5. How would the solution change if the ratios were different?

**Tip:** When solving problems involving ratios and ages, always express ages in terms of a single variable for easier manipulation.

The present ages of A and B are in the ratio of 6:4. 5 years ago their ratio was 5:3. How old is A now?

Learn how to solve an age problem involving ratios and linear equations. This example covers the present ages of A and B, using the ratio 6:4 and the ratio 5 years ago of 5:3. A step-by-step solution is provided, suitable for students in Grades 6-8.

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