1. Correct Answer: Option B: 15
 2. Explanation:
     To solve this problem, we need to determine the pattern in which the ring is passed and find when it returns to child-1.
     Step-by-step Analysis:
         ○ The ring starts with child-1.
     ● Child-1 passes the ring to child-2.  
     ● Child-2 passes the ring to child-4.  
     ● Child-4 passes the ring to child-7.  
     Observing the pattern, we see that the number of the child receiving the ring increases as follows: 1, 2, 4, 7, ...
     This sequence can be described by the formula for the nth term:
     \[
     a_n = a_{n-1} + (n-1)
     \]
     where \( a_1 = 1 \).
     Let's calculate the sequence until it returns to child-1:
         ○ \( a_1 = 1 \)
         ○ \( a_2 = 1 + 1 = 2 \)
         ○ \( a_3 = 2 + 2 = 4 \)
         ○ \( a_4 = 4 + 3 = 7 \)
         ○ \( a_5 = 7 + 4 = 11 \)
         ○ \( a_6 = 11 + 5 = 16 \)
         ○ \( a_7 = 16 + 6 = 22 \)
         ○ \( a_8 = 22 + 7 = 29 \)
         ○ \( a_9 = 29 + 8 = 37 \)
         ○ \( a_{10} = 37 + 9 = 46 \)
     Since there are 40 children, we take the modulo 40 of each term to find the position in the circle:
         ○ \( a_1 \equiv 1 \mod 40 \)
         ○ \( a_2 \equiv 2 \mod 40 \)
         ○ \( a_3 \equiv 4 \mod 40 \)
         ○ \( a_4 \equiv 7 \mod 40 \)
         ○ \( a_5 \equiv 11 \mod 40 \)
         ○ \( a_6 \equiv 16 \mod 40 \)
         ○ \( a_7 \equiv 22 \mod 40 \)
         ○ \( a_8 \equiv 29 \mod 40 \)
         ○ \( a_9 \equiv 37 \mod 40 \)
         ○ \( a_{10} \equiv 46 \equiv 6 \mod 40 \)
         ○ \( a_{11} \equiv 6 + 10 = 16 \mod 40 \)
         ○ \( a_{12} \equiv 16 + 11 = 27 \mod 40 \)
         ○ \( a_{13} \equiv 27 + 12 = 39 \mod 40 \)
         ○ \( a_{14} \equiv 39 + 13 = 52 \equiv 12 \mod 40 \)
         ○ \( a_{15} \equiv 12 + 14 = 26 \equiv 1 \mod 40 \)
     After 15 changes, the ring returns to child-1.
     Therefore, the correct answer is Option B: 15.