Let the two numbers be \( x \) and \( y \). According to the problem:
- \( x = 5y \) (A positive number is 5 times another number).
Next, when 21 is added to both numbers, one of the new numbers becomes twice the other. This gives us:
- \( x + 21 = 2(y + 21) \) or \( y + 21 = 2(x + 21) \).
Let's first use the form \( x + 21 = 2(y + 21) \):
Substituting \( x = 5y \) into the equation:
\[ 5y + 21 = 2(y + 21) \]
Expanding the right side:
\[ 5y + 21 = 2y + 42 \]
Now, we will isolate \( y \):
\[ 5y - 2y = 42 - 21 \] \[ 3y = 21 \] \[ y = 7 \]
Now, substitute \( y \) back to find \( x \):
\[ x = 5y = 5 \times 7 = 35 \]
Thus, the numbers are:
\[ x = 35 \quad \text{and} \quad y = 7 \]
To verify, when we add 21:
- \( x + 21 = 35 + 21 = 56 \)
- \( y + 21 = 7 + 21 = 28 \)
Checking the condition \( 56 = 2 \times 28 \):
\[ 56 = 2 \times 28 \quad \text{(True)} \]
Thus, the numbers are indeed \( \boxed{35} \) and \( \boxed{7} \).