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

Hint:

In word problems with headcount and revenue constraints, first set up two equations: one for the total number of people and another for total money. Then eliminate one variable to get a simple linear relation and apply given inequalities to find extrema.

Answer 1

Correct Answer: 700

Approach: Use the two given totals (head count and fee) to collapse three unknowns into a single relation between Science and Arts, then push Science as high as the constraint \(S \le A\) allows.

Step 1: Let \(S, A, C\) be the number of science, arts and commerce students. Head count: \[ S + A + C = 1500. \] Fee: \[ 1100S + 1000A + 800C = 1550000. \]

Step 2: Eliminate \(C\) by writing \(C = 1500 - S - A\) and substituting into the fee equation: \[ 1100S + 1000A + 800(1500 - S - A) = 1550000. \]

Step 3: Expand and simplify: \[ 1100S + 1000A + 1200000 - 800S - 800A = 1550000 \] \[ 300S + 200A = 350000 \implies 3S + 2A = 3500. \]

Step 4: Apply the condition \(S \le A\). From \(2A = 3500 - 3S\) we get \(A = \dfrac{3500 - 3S}{2}\), so \[ S \le \frac{3500 - 3S}{2} \implies 2S \le 3500 - 3S \implies 5S \le 3500 \implies S \le 700. \]

Step 5: Check \(S = 700\) is feasible: then \(A = \dfrac{3500 - 2100}{2} = 700\) and \(C = 1500 - 700 - 700 = 100\) — all non-negative whole numbers. Valid.

Final answer: Maximum number of science students \(= \boxed{700}\).

Why this works: Two equations on three unknowns leave one degree of freedom; the inequality \(S \le A\) is what pins the maximum, so the answer always sits at the boundary \(S = A\).

Answer 2

Alternate approach — using a weighted-average / replacement idea:
If all 1500 students paid the lowest fee (Rs 800, commerce), total collection would be \( 1500\times800=1{,}200{,}000 \), which is Rs \( 350{,}000 \) less than the actual Rs \( 1{,}550{,}000 \).
Switching one student from commerce to arts adds Rs 200 to the total; switching one from commerce to science adds Rs 300. If \( a \) students are in arts and \( s \) in science (instead of commerce), then \( 200a+300s=350000 \), i.e. \( 2a+3s=3500 \).
We want the largest \( s \) with \( s\le a \). From \( a=\frac{3500-3s}{2} \), the constraint \( s\le a \) gives \( s\le700 \); since \( a \) must be a whole number, \( s \) must be even, so the maximum possible is \( s=700 \) (giving \( a=700 \), \( c=100 \)), which satisfies all conditions.
So the maximum number of science students is 700.