Let Ashok be born on the \(x^{th}\) day of the \(y^{th}\) month in \(1995\).
Given that,
\(24x+60y = 1284\)
\(2x+5y = 107\)
\(y\) = \(\frac{107-2x}{5}\) = \(21+\frac{2(1-x)}{5}\)
\((1 – x)\) must be a multiple of \(5\).
As \(1≤x≤31\), \(1≤y≤12\)
Then, the only values of \((x, y)\) satisfying the equation and other conditions are \((26, 11)\) and \((31, 9)\).
Thus, the birthday can be on \(26\) November or \(31\) September.
As September does not have \(31\) days, this case is rejected.
Thus, the required sum = \(26 + 11 + 1995 = 2032\)
Hence, option A is the correct answer.
A | B | C | D | Average |
---|---|---|---|---|
3 | 4 | 4 | ? | 4 |
3 | ? | 5 | ? | 4 |
? | 3 | 3 | ? | 4 |
? | ? | ? | ? | 4.25 |
4 | 4 | 4 | 4.25 |