Question : The average of a set of 15 numbers is 41. Five numbers 21, 24, 39, 47, and 97 are removed from the above set. What will the average of the remaining 10 numbers be?
Option 1: 39.2
Option 2: 37.6
Option 3: 38.7
Option 4: 36.4
Correct Answer: 38.7
Solution :
According to the question,
Original average = 41
Original sum of 15 numbers = 41 × 15 = 615
⇒ Sum of removed numbers = 21 + 24 + 39 + 47 + 97 = 228
⇒ Updated sum of remaining 10 numbers = 615 – 228 = 387
$\therefore$ Average of the remaining 10 numbers $=\frac{387}{10} =38.7$
Hence, the correct answer is 38.7.
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