Question : What is the average of the squares of the first 19 natural numbers?
Option 1: 124
Option 2: 127.5
Option 3: 130
Option 4: 133.5
Correct Answer: 130
Solution : Given: 19 natural numbers. Sum of first n natural numbers = $\frac{\text{n(n+1)(2n+1)}}{6}$ Sum of first 19 natural numbers = $\frac{\text{19(19+1)(2×19+1)}}{6}$ = $\frac{\text{14820}}{6}$ = 2470 Average of these numbers = $\frac{\text{Sum of first 19 natural numbers}}{19}$ = $\frac{\text{2470}}{19}$ = 130 Hence, the correct answer is 130.
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Question : The average of 19 numbers is 22.8. The average of the first ten numbers is 18.4 and that of the last ten numbers is 28.6. If the $10^{\text {th}}$ number is excluded from the given numbers, then what is the average of the remaining numbers? (Your answer should be nearest to an integer.)
Option 1: 21
Option 2: 23
Option 3: 22
Option 4: 20
Question : What is the average of the first 29 even numbers?
Option 1: 30
Option 2: 31
Option 3: 32
Option 4: 33
Question : What is the average of the first six prime numbers?
Option 1: $6 \frac{1}{3}$
Option 2: $6 \frac{5}{6}$
Option 3: $6 \frac{2}{3}$
Option 4: $6 \frac{1}{6}$
Question : Which one of the following is a factor of the sum of the first twenty-five natural numbers?
Option 1: 26
Option 2: 24
Option 3: 13
Option 4: 12
Question : The average of n numbers is 42. If 75% of the numbers are increased by 4 each and the remaining numbers are decreased by 8 each, then what is the average of the numbers, so obtained?
Option 1: 44
Option 2: 42.5
Option 3: 43
Option 4: 43.8
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