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| Question 1206042:  An actor invests some money at 8%, and $22000 more than twice the amount at 12 %. The total annual interest earned from the investment is $30800. How much did he invest at each amount?
 
 Answer by math_tutor2020(3817)
      (Show Source): 
You can put this solution on YOUR website! x = some amount of money invested at 8% interest rate
 2x+22000 = amount invested at 12%
 
 A = 0.08x = amount of interest earned from the 8% account
 B = 0.12*(2x+22000) = interest earned from the 12% account
 Each duration is for 1 year.
 I'll assume simple interest is used, and not compound interest.
 
 A+B = total interest earned for 1 year
 A+B = 0.08x+0.12*(2x+22000) = 30800
 
 Let's isolate x.
 0.08x+0.12*(2x+22000) = 30800
 0.08x+0.12*(2x)+0.12*(22000) = 30800
 0.08x+0.24x+2640 = 30800
 0.32x+2640 = 30800
 0.32x = 30800-2640
 0.32x = 28160
 x = 28160/0.32
 x = $88,000 was invested at 8%
 
 And,
 2x+22000 = 2*88000+22000 = $198,000 was invested at 12%
 
 Check:
 0.08*88000 + 0.12*198000 = 30800
 The answers are confirmed.
 
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