CAT 2025Slot 3QAQuestion & SolutionQues & Sol
Question
In a school with 1500 students, each student chooses any one of the streams out of science, arts, and commerce, by paying a fee of Rs 1100, Rs 1000, and Rs 800, respectively. The total fee paid by all the students is Rs 15,50,000. If the number of science students is not more than the number of arts students, then the maximum possible number of science students in the school is
Solution
1. Concept Used
- Topic: Linear Equations in Three Variables with Optimization Constraint
- Formula: $$S + A + C = 1500 \quad \text{and} \quad 1100S + 1000A + 800C = 15{,}50{,}000 \quad \text{with} \quad S \leq A$$
2. Calculation
Let the number of students who chose Science, Arts, and Commerce be $S$, $A$, and $C$ respectively.
From the total number of students: $$S + A + C = 1500 \implies C = 1500 - S - A \quad \cdots (1)$$
From the total fee collected: $$1100S + 1000A + 800C = 15{,}50{,}000$$
Dividing the entire equation by $100$ to simplify: $$11S + 10A + 8C = 15500 \quad \cdots (2)$$
Substituting the value of $C$ from equation $(1)$ into equation $(2)$: $$11S + 10A + 8(1500 - S - A) = 15500$$
$$11S + 10A + 12000 - 8S - 8A = 15500$$
$$3S + 2A = 3500 \quad \cdots (3)$$
Now, the constraint is $S \leq A$. We want to maximize $S$.
From equation $(3)$: $$A = \frac{3500 - 3S}{2}$$
For $A$ to be a valid (non-negative integer), we need $3500 - 3S$ to be even and non-negative, which means $S$ must be even.
Applying the constraint $S \leq A$: $$S \leq \frac{3500 - 3S}{2}$$
$$2S \leq 3500 - 3S$$
$$5S \leq 3500$$
$$S \leq 700$$
So the maximum value of $S$ is $700$.
Let's verify: if $S = 700$, then: $$A = \frac{3500 - 3(700)}{2} = \frac{3500 - 2100}{2} = \frac{1400}{2} = 700$$
$$C = 1500 - 700 - 700 = 100$$
Check: $S = 700 \leq A = 700$ ✅, and all values are non-negative integers ✅.
Fee check: $$1100(700) + 1000(700) + 800(100) = 7{,}70{,}000 + 7{,}00{,}000 + 80{,}000 = 15{,}50{,}000$$
3. Solution
Answer = 700 ✅
The maximum possible number of science students in the school is 700.