CAT 2020 Quant Question [Slot 2] with Solution 17

Given \(x + 9 = z = y + 1\) and \(x + y < z + 5\)

\( \Rightarrow (z - 9) + (z - 1) < z + 5\)

\( \Rightarrow z < 15\)

Hence the maximum value of \(z = 14,\max \) of \(x = 5\) and \(\max \) of \(y = 13\)

Required answer, \(2x + y = 2 \times 5 + 13 = 23\)