an actor invest some money at 5 and 30000 more than twice th

an actor invest some money at 5% and $30,000 more than twice the amount at 11% . The total annual interest earned from the investment is $20,850. How much did he invest at each amount? Use the 6 step method

Solution

Let first amount be $ x invested at 5%.

Second amount is $30,000 more than twice of x. That is twice of x is 2x and 30,000 more than twice = (2x+30000)

Therefore, $ (2x+30000) was invested at 11%.

Total interest earned from both amounts = $20,850.

So, Interest earned from first amount + Interest earned from second amount = $20,850.

Interest = Principal * Rate * Time.

Interest earned from first amount = x * 0.05 * 1 = 0.05x.

Interest earned from second amount = (2x+30000)* 0.11* 1 = 0.11(2x +30000) = 0.22x + 3300.

Therefore,

0.05x + 0.22x + 3300 = 20850

0.27x +3300 = 20850

Subtracting 3300 from both sides.

0.27x +3300-3300 = 20850-3300

0.27x = 17550

Dividing both sides by 0.27.

0.27x/ 0.27= 17550/0.27

x= 65000.

Therefore, amount invested at 5% = $65,000.

Amount invested at 11% = 2x+30,000 = 2* 65000 +30000 = 1,60,000.