A person invested $50,000 into two accounts that pay 4 and 6% simple interest annually, respectively. How do you find the amount invested into the two accounts if the total interest earned after one year is $2400?

Question

Avery Alexander · Accepted Answer

The person invested #$30000# at #4%# SI and #$20000# at #6%# SI

Let the person invested #$x# at #4%# SI. Then the person invested #$(50000-x)# at #6%# SI. The total interest earned after one year is # x*0.04+(50000-x)*0.06# which is equal to #$2400# So #0.04x-0.06x+3000=2400 or -0.02x = -600 or x=600/0.02=$30000 :. 50000-x=50000-30000=$20000# Therefore, the person invested #$30000# at #4%# SI. and invested #$20000# at #6%# SI.[Ans]

Savannah Anderson · Answer

undefined Let x be the amount invested at 4% and y be the amount invested at 6%. Then, the total interest earned can be represented as 0.04x + 0.06y = 2400. Since the total amount invested is $50,000, we have x + y = 50000. Solving these two equations simultaneously will give the amounts invested in each account.