SajjadAhmad wrote:

Daniel invested a total of $10,000 for a period of one year. Part of the money he put into an investment that earned 6 percent simple interest, and the rest of the money into an investment that earned 8 percent simple interest. How much money did he put into the investment that earned 6 percent?

(1) The total interest earned on the $10,000 for the year was $640.

(2) The dollar value of the investment that earned 6 percent was four times the dollar value of the investment that earned 8 percent.

Let the investment that earned 6% be \(x\), hence investment that earned 8%\(=10000-x\)

Statement 1: implies \(\frac{x*6*1}{100}+\frac{(1000-x)*8*1}{100}=640\). One variable and one equation can be solved. Hence

SufficientStatement 2: implies \(x=4*(10000-x)\). Again One variable and one equation can be solved. Hence

SufficientOption

D