A coaching institute has students in 3 batches - X, Y and Z. In a certain examination, the average marks obtained by these batches are 72, 60 and 50 respectively. The average marks of batches X and Y taken together is 69. If the ratio of the number of students in batches Y and Z is 6 : 7, what is the average score of all the three batches put together?

### Concept & Logic This problem requires the application of **Alligation** or weighted averages in two stages to find the true ratio of students in all three batches. Once the combined ratio $n_x : n_y : n_z$ is unified, we apply the standard weighted average formula to find the final aggregate average. ### Step-by-Step Solution * **Step 1: Find the ratio of students in Batches X and Y.** Average of X = $72$ Average of Y = $60$ Combined Average of X and Y = $69$ Using Alligation (Cross difference method): Ratio of X : Y = $(69 - 60) : (72 - 69) = 9 : 3 = 3 : 1$. * **Step 2: Combine the ratios.** We found $X : Y = 3 : 1$ We are given $Y : Z = 6 : 7$ To merge these, make the value of Y consistent in both ratios. Multiply the first ratio by $6$: $X : Y = 18 : 6$ Therefore, the cleanly combined ratio of students is $X : Y : Z = 18 : 6 : 7$. * **Step 3: Calculate the overall average.** Let the number of students be $18k$, $6k$, and $7k$. (Total parts = $31k$). Total marks = $(18 \times 72) + (6 \times 60) + (7 \times 50)$ Total marks = $1296 + 360 + 350 = 2006$. Overall Average = $2006 / 31 \approx 64.709$. ### Exam Strategy & Shortcut Using Alligation to find the $X : Y$ ratio visually is the fastest start. Instead of calculating massive numbers like $18 \times 72$, you can shift the base to $60$ (the middle value). Deviations from $60$: X has a deviation of $+12$ for $18$ parts = $+216$. Y has a deviation of $0$ for $6$ parts = $0$. Z has a deviation of $-10$ for $7$ parts = $-70$. Net deviation = $216 - 70 = 146$. Distribute this net deviation evenly over all $31$ parts ($18+6+7$). $146 / 31 \approx 4.7$. Add this to the assumed base of $60$: $60 + 4.7 = 64.7$. This requires much smaller calculations and slashes the chance of multiplication errors. ### Common Pitfall Failing to align the ratios correctly before combining them. If you blindly add the parts or use unaligned variables (e.g., mixing up the $Y$ value of $1$ and $6$), the final weighted average will be entirely warped and lead to a trap answer. ### Final Answer **Therefore, the correct answer is 64.7.**