Moe and Joe are twins. Moe invested $1,000, earned 9% annually, and now has $1,992.56. Joe invested $1,000, earned 6.47%, and now has $1,992.97. Joe invested his money _____ years before Moe.

A) 2.5 years
B) 2.8 years
C) 3.0 years
D) 3.2 years
E) 3.5 years

Moe's investment grew at 9% annually to double in value, which can be calculated using the formula for compound interest: $A=P(1+r{)}^t$ , where $A$ is the amount after time $t$ , $P$ is the principal amount, $r$ is the annual interest rate, and $t$ is the time in years. For Moe, solving $1992.56=1000(1+0.09{)}^t$ gives $t$ close to 8 years. For Joe, using the same formula with his interest rate, $1992.97=1000(1+0.0647{)}^t$ , gives $t$ close to 11 years. The difference in time, therefore, is about 3 years, indicating Joe invested his money 3 years before Moe.