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
Show Hint
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.
Top Questions on Linear Equations
- Let $f(x)$ be a differentiable function satisfying \[ f(x)=e^x+\int_0^1 (y+xe^x)f(y)\,dy \] Find $f(0)+e$, where $e$ is Napier's constant.
- JEE Main - 2026
- Mathematics
- Linear Equations
- Let $5000<N<9000$ and $N$ has digits from the set $\{0,1,2,5,9\}$. If digits can be repeated, then find the number of such numbers $N$ which are divisible by $3$.
- JEE Main - 2026
- Mathematics
- Linear Equations
- Evaluate the series \[ \frac{1}{25!}+\frac{1}{3!\,23!}+\frac{1}{5!\,21!}+\cdots \text{ up to 13 terms.} \]
- JEE Main - 2026
- Mathematics
- Linear Equations
- Find the number of real solutions of \[ x|x-3|+|x-1|+3=0 \]
- JEE Main - 2026
- Mathematics
- Linear Equations
- Consider a geometric sequence $729,\,81,\,9,\,1,\ldots$ If $P_n$ denotes the product of first $n$ terms of the G.P. such that \[ \sum_{n=1}^{40} (P_n)^{\frac{1}{n}}=\frac{3^{\alpha}-1}{2\times 3^{\beta}}, \] then find the value of $(\alpha+\beta)$.
- JEE Main - 2026
- Mathematics
- Linear Equations
Questions Asked in CAT exam
- The passage given below is followed by four summaries. Choose the option that best captures the essence of the passage.
In the dynamic realm of creativity, artists often find themselves at the crossroads between drawing inspiration from diverse cultures and inadvertently crossing into the territory of cultural appropriation. Inspiration is the lifeblood of creativity, driving artists to create works that resonate across borders. In a globalized era of the modern world, artists draw from a vast array of cultural influences. When approached respectfully, inspiration becomes a bridge, fostering understanding and appreciation of cultural diversity. However, the line between inspiration and cultural appropriation can be thin and easily blurred.
Cultural appropriation occurs when elements from a particular culture are borrowed without proper understanding, respect, or acknowledgment. This leads to the commodification of sacred symbols, the reinforcement of stereotypes, and the erasure of the cultural context from which these elements originated. It is essential to recognize that the impact of cultural appropriation extends beyond the realm of artistic expression, influencing societal perceptions and perpetuating power imbalances.- CAT - 2025
- Para Summary
- The number of distinct integers $n$ for which $\log_{\left(\frac14\right)}(n^2 - 7n + 11)>0$ is:
- CAT - 2025
- Linear Inequalities
- In the set of consecutive odd numbers $\{1, 3, 5, \ldots, 57\}$, there is a number $k$ such that the sum of all the elements less than $k$ is equal to the sum of all the elements greater than $k$. Then, $k$ equals?
- CAT - 2025
- Number Systems
- The number of distinct pairs of integers $(x, y)$ satisfying the inequalities $x>y \ge 3$ and $x + y<14$ is:
- CAT - 2025
- Number Systems
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is:
- CAT - 2025
- Linear Inequalities