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
New: SSC MTS Tier 1 Answer key 2024 out
Don't Miss: Month-wise Current Affairs | Upcoming Government Exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
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.
Answer Key | Cutoff | Selection Process | Preparation Tips | Eligibility | Application | Exam Pattern
Question : The average of 9 numbers is 47. If a number is removed, then the average becomes 48. Which number was removed?
Question : The average of a certain set of 'b' numbers is 'a', and the average of a certain set of 'a' numbers is 'b'. What is the average of all the numbers taken together?
Question : The average of five consecutive odd numbers is 15. What is the smallest of these five numbers?
Question : The average of nine numbers is 20. If two of these numbers are removed, then the average becomes 18. What is the sum of the two numbers that are removed?
Question : The average of 24 numbers is 43. If we subtract a number ' M ' from each number, then the average becomes 38. What was the value of M?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile