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Bunuel
Average of 75 students is 82, out of which average of 42 students is 79. What is the average of the remaining 33 students rounded to the nearest integer?

A. 84
B. 85
C. 86
D. 87
E. 88

The sum of the 75 students is 75 x 82 = 6,150.

The sum of the 42 students is 42 x 79 = 3,318.

Thus, the sum of the remaining 33 students is 6,150 - 3,3318 = 2,832, and their average is 2,832/33 = 85.8, rounded to 86.

Answer: C
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Is this the right procedure to solve such questions cause looks very easy!
Archit3110


sum for 75 students ; 75*82 = 6150
sum of 42 students ; 42 * 79 = 3138
so for 33 students sum = 6150-3138 ; 2832
avg ; 2832/33; 85.8 ~ 86 IMO C
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nandini14
Is this the right procedure to solve such questions cause looks very easy!

nandini14 Yes, it is. You can use it for weighted average problems like this.

Why This Works:
The fundamental relationship is: \(\text{Average} = \frac{\text{Sum}}{\text{Count}}\), which means \(\text{Sum} = \text{Average} \times \text{Count}\). When you know the overall average and one subgroup's average, you can isolate the remaining subgroup's sum through subtraction. This is the most efficient approach for this problem type.

What to watch for:
  • Some problems give you individual values instead of subgroup averages—then you'd add those values directly to your sum calculations
  • Watch for problems asking for the average of combined groups (different formula needed)
  • Always verify that subgroup counts add up to the total count
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