Xavier Aptitude Test 2025 Paper

Concept:The goal is to choose the best combination of single, double, and triple rooms so that 55 people occupy 25 rooms while generating the highest revenue.Explanation:Let $x$, $y$, and $z$ be the numbers of single, double, and triple rooms used.Each room is used exactly once, so $x+y+z=25$.Each single room has 1 person, each double has 2, and each triple has 3, so $x+2y+3z=55$.From $x+y+z=25$, substitute $x=25-y-z$ into the people equation:$25-y-z+2y+3z=55$$y+2z=30$$y=30-2z$.Also, $x=25-y-z=z-5$.Revenue is $R=2000x+3000y+3500z$.Putting $x=z-5$ and $y=30-2z$:$R=2000(z-5)+3000(30-2z)+3500z$$R=80000-500z$.Since $x\ge0$, we need $z\ge5$.Because revenue decreases as $z$ increases, use the smallest possible value, $z=5$.Then $y=30-2(5)=20$ and $x=5-5=0$.Maximum revenue:$20\times3000+5\times3500=60000+17500=77500$Answer:Maximum possible revenue is $\mathrm{Rs.\ }77{,}500$, which is Option C.