Bunuel
What is the average (arithmetic mean) of 10 consecutive integers?
(1) The average (arithmetic mean) of the first 7 integers is 7.
(2) The average (arithmetic mean) of the last 7 integers is 10.
Solution: Pre Analysis: - We are asked the average of 10 consecutive integers
- Let the 10 consecutive integers be \(a, a+1, a+2, a+3....., a+8, a+9\)
- So, average of the same \(=\frac{a+a+1+a+2+a+3+...+a+8+a+9}{10}=\frac{10a+45}{10}=a+4.5\)
- Thus to get the answer we need the value of \(a\)
Statement 1: The average (arithmetic mean) of the first 7 integers is 7
- According to this statement, \(\frac{a+a+1+a+2+a+3....+a+6}{7}=7\)
- Using the above equation we can easily get the value of \(a\)
- Thus, statement 1 alone is sufficient and we can eliminate options B, C and E
Statement 2: The average (arithmetic mean) of the last 7 integers is 10
- According to this statement, \(\frac{a+3+a+4+a+5+a+6....+a+9}{7}=10\)
- Using the above equation we can easily get the value of \(a\)
- Thus, statement 2 alone is also sufficient
Hence the right answer is
Option D