To solve the problem, we can summarize the information given and set up equations based on the requirements: 1. Let the amounts received by A, B, C, and D be represented as \(a\), \(b\), \(c\), and \(d\) respectively. 2. According to the problem: - Each boy receives at least one rupee and receives a different amount. - D receives 4 more rupees than B: \(d = b + 4\) - B receives more rupees than C but less than A: \(c < b < a\) 3. The total amount distributed is ₹16: \(a + b + c + d = 16\). From these conditions, substituting \(d\) in the total amount: \[ a + b + c + (b + 4) = 16 \] This can be simplified to: \[ a + 2b + c + 4 = 16 \quad \Rightarrow \quad a + 2b + c = 12 \] Now, we know: - Each amount is different, and \(a\), \(b\), \(c\), and \(d\) must be at least 1 rupee each. To explore the possible amounts: 1. The smallest values for \(c\) and \(b\) could be: - For \(c = 1\): Then \(b\) would need to be at least 2, - Thus \(a\) needs to satisfy: \(a + 2(2) + 1 = 12 \rightarrow a = 8\) (which is valid, \(d = 6\)). So one configuration is: - \(a = 8\), \(b = 2\), \(c = 1\), \(d = 6\). From this setup, we can explore the maximum and minimum amounts that A can get. - Exploring higher values for \(b\): - \(b = 3\) -> \(c = 2\) -> \(d = 7\) -> \(a + 2(3) + 2 = 12 \rightarrow a = 4\). - . . . - Adjusting lower values will find more configurations while keeping distinct values. Finally: - Minimum possible \(a = 4\) - Maximum possible \(a = 8\) Calculating the difference: \[ \text{Difference} = \text{Maximum} - \text{Minimum} = 8 - 4 = 4 \] Hence, the answer is 4. The answer choice is (a) 4, and the difference between the maximum and minimum rupees that A can receive is indeed 4【4:0†source】.