Question

a person invested $60000 into two accounts that pays 8% and 6% simple interest annually, respectively. Find the amount invested into the two accounts if the total interest earned after one year is $ 3600

Answer 1

\[.08\times x + .06\times (60000-x)=3600\] solve for x

Answer 2

that is, let x = amount invested at 8% and so 6000-x is amount invested at 6%. do you know how to solve this for x?

Answer 3

turns out it was all invested at 6% because 6% of $60,000 = $3600 exactly

Answer 4

let sum invested at 8% = A

interest1 = PRT/100 = (A x 8 x 1)/100 = 8A/100

sum invested at 6% = 60000 - A

interest2 = PRT/100 = (60000 - A) x 6 x 1/100 = (360000 - 6A)/100

total interest = interest1 + interest 2
= 8A/100 + (36000 - 6A)/100
= (8A - 6A + 36000)/100
= (2A + 36000)/100
ATQ total interest is 3600

so 2A + 360000
------------ = 3600
100
2A + 360000 = 3600 x 100
2A = 360000 - 360000
2A = 0 so A = 0/2 = 0

This means that sum invested at 8% = 0
and sum invested at 6% = 60000