To find the number of two-digit positive numbers that satisfy the given condition, we start by formulating the problem mathematically.

Let the original two-digit number be denoted as \(10x + y\), where \(x\) and \(y\) represent the tens and units digits, respectively, and \(x, y \in \{0, 1, 2, \ldots, 9\}\) with \(x \neq 0\) since it's a two-digit number.

The reversed number will then be \(10y + x\).

According to the problem, seven times the original number equals four times the reversed number:

\(7(10x + y) = 4(10y + x)\)

Expanding both sides, we get:

\(70x + 7y = 40y + 4x\)

Rearranging the equation to isolate terms with \(x\) and \(y\) on opposite sides, we obtain:

\(70x - 4x = 40y - 7y\)

\(66x = 33y\)

Simplifying this equation by dividing both sides by 33 yields:

\(2x = y\)

This shows that for any two-digit number, the unit digit \(y\) must be exactly twice the tens digit \(x\).

Now, we find possible values for \(x\) such that \(y\) is a valid single-digit number:

• If \(x = 1\), then \(y = 2\). The number is 12.
• If \(x = 2\), then \(y = 4\). The number is 24.
• If \(x = 3\), then \(y = 6\). The number is 36.
• If \(x = 4\), then \(y = 8\). The number is 48.

Since \(y\) must be a single-digit number, \(x = 5\) would lead to \(y = 10\), which is not valid. Thus, no more valid digits exist for \(x\) beyond 4.

Therefore, there are 4 such two-digit numbers: 12, 24, 36, and 48.

The correct answer is 4.