The answer to the question regarding the distribution of ₹16 among A, B, C, and D is 3. Explanation: To distribute the coins, we need to assign different amounts to each boy under the following constraints: 1. Each boy receives at least 1 rupee. 2. D receives 4 more rupees than B. 3. B receives more than C but less than A. Let’s denote: - C receives \( c \) rupees. - B receives \( b \) rupees (where \( b > c \)). - A receives \( a \) rupees (where \( a > b \)). - D receives \( d \) rupees (where \( d = b + 4 \)). The equation governing the total distribution can be set up as: \[ a + b + c + d = 16 \] Substituting \( d \) with \( b + 4 \): \[ a + b + c + (b + 4) = 16 \] \[ a + 2b + c + 4 = 16 \] \[ a + 2b + c = 12 \] To explore different distributions that satisfy all conditions: 1. The minimum value for \( b \) can be 2 (since B should get more than C). If \( b \) is 2, \( d \) becomes 6. Hence, we can explore further: - If \( b = 2 \), then: - \( d = 6 \) - Substitute back to find \( c \) and \( a \): \[ a + 2 + c = 12 \] \[ a + c = 10 \] The distribution could be: - If \( c = 1 \), then \( a = 9 \) (values: A=9, B=2, C=1, D=6). - If \( c = 2 \), then \( a = 8 \) (values: A=8, B=2, C=2, D=6 - not valid since C and B must be different). - If \( c = 3 \), then \( a = 7 \) (values: A=7, B=3, C=3, D=7 - not valid). Continuing this way, valid distributions yield \( A \) as either 7 or 5 while keeping \( b \) as high as 3 or 2. Evaluating the difference for \( A \): - Maximum value of \( A \): 7 - Minimum value of \( A \): 4 Thus, the difference between the maximum and minimum rupees that A can receive is: \[ 7 - 4 = 3 \] This concludes the reasoning, confirming that the difference between the maximum and minimum amounts that A can have is 3【4:12†source】.