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
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Correct Answer: 700
Approach Solution - 1
To solve the problem, let's define the variables: S represents the number of science students, A represents the number of arts students, and C represents the number of commerce students. The total number of students is given by:
S + A + C = 1500
The total fee paid by all students is:
1100S + 1000A + 800C = 1550000
Given that the number of science students is not more than the number of arts students, we have:
S ≤ A
We are required to find the maximum number of science students, S. First, express C in terms of S and A:
C = 1500 - S - A
Substitute C into the fee equation:
1100S + 1000A + 800(1500 - S - A) = 1550000
Simplify this equation:
1100S + 1000A + 1200000 - 800S - 800A = 1550000
300S + 200A = 350000
Divide the entire equation by 100:
3S + 2A = 3500
Next, solve for A in terms of S:
2A = 3500 - 3S
A = (3500 - 3S) / 2
Since S ≤ A, we have:
S ≤ (3500 - 3S) / 2
Multiplying through by 2 to eliminate the fraction gives:
2S ≤ 3500 - 3S
Add 3S to both sides:
5S ≤ 3500
Divide by 5:
S ≤ 700
Therefore, the maximum possible number of science students is 700. This value fits within the given range of 700,700.
Approach Solution -2
Let the number of science, arts, and commerce students be S, A, and C respectively. Given: S + A + C = 1500 and the total fees paid is Rs 15,50,000, hence 1100S + 1000A + 800C = 15,50,000. We need to maximize S under the condition S ≤ A.
We substitute C = 1500 - S - A into the fee equation:
1100S + 1000A + 800(1500 - S - A) = 15,50,000.
Simplifying:
1100S + 1000A + 12,00,000 - 800S - 800A = 15,50,000.
300S + 200A = 3,50,000.
Divide by 100:
3S + 2A = 3500.
Rewriting: A = (3500 - 3S) / 2.
Since S ≤ A, it follows from A=(3500-3S)/2 that S ≤ (3500-3S)/2. Solving: 2S ≤ 3500 - 3S.
5S ≤ 3500.
S ≤ 700.
Check S = 700:
If S = 700, then A = (3500 - 3*700) / 2 = 700; C = 1500 - 700 - 700 = 100.
Condition is satisfied: S = 700, A = 700, S ≤ A.
Fees check: 1100*700 + 1000*700 + 800*100 = 770000 + 700000 + 80000 = 1550000.
All conditions fit, and S = 700 is valid within the range.
The maximum possible number of science students in the school is 700.
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