There are $3$ groups of students, each containing $25$, $50$ and $25$ students respectively. The mean marks obtained by the first two groups are $60$ and $55$. The combined mean of all the three groups is $58$. What is the mean of the marks scored by the third group?

Concept & Strategy This requires calculating an unknown component of a weighted average. The total sum of marks of all three groups combined must equal the sum of the marks of the individual groups. $$ \text{Overall Sum} = \text{Sum}_1 + \text{Sum}_2 + \text{Sum}_3 $$ Step-by-Step Solution * Given: Group sizes are $n_1 = 25$, $n_2 = 50$, $n_3 = 25$. * Total number of students = $25 + 50 + 25 = 100$. * Combined mean = $58$. Therefore, total marks = $100 \times 58 = 5800$. * Calculate marks for Group 1: $25 \times 60 = 1500$. * Calculate marks for Group 2: $50 \times 55 = 2750$. * Let the mean of Group 3 be $x$. Its total marks = $25x$. * Set up the sum equation: $1500 + 2750 + 25x = 5800$. * Combine known sums: $4250 + 25x = 5800$. * Isolate the unknown: $25x = 5800 - 4250 = 1550$. * Solve for $x$: $x = 1550 / 25 = 62$. Exam Strategy & Shortcut Simplify the group sizes by finding their ratio. The ratio of $25 : 50 : 25$ simplifies perfectly to $1 : 2 : 1$. This means total parts = $4$. The combined mean is $58$, so total "ratio sum" = $4 \times 58 = 232$. Sum of parts 1 and 2 = $(1 \times 60) + (2 \times 55) = 60 + 110 = 170$. The missing part for the 3rd group (which is $1$ ratio unit) is $232 - 170 = 62$. Dividing by its ratio weight ($1$), the mean is $62$. This avoids large multiplications entirely. Common Pitfall A common mistake is getting intimidated by large numbers like $5800$ and $4250$, leading to arithmetic errors during subtraction or division under time pressure. Always look for ways to simplify the base populations into smaller ratios first. Final Answer **Therefore, the correct answer is 62.**