A company invested $29,000, putting part of it into a savings account that earned 3.2% annual simple interest and the remainder in a stock fund that earned 12.4% annual simple interest. If the investments earned $1572 annually, how much was in each account?
Do you know the simple interest formula?
No :/
Interest = principle*rate*time
oh how would i set it up?? lost about that
\[I = P_{0}*r_{1}*t_{1}\] \[I_{2} = P_{2}*r_{2}*t_{2}\] annually means 1 year so t2 = t1 which is just 1 \[I_{1} = P_{0}(r_{1})\] \[I_{2} = P_{1}(r_{2})\] we know that the initial values invested are different P0 in one account and P1 in the other but their sums are \[P_{0}+P_{1} = (29,000)\] and we know that the sum of the interest is 1,572. that includes the two here. \[P_{1}r_{2}+P_{0}r_{1} = $1,572\] because we've got two different variables we need to equations.
ok how do i solve for that ti looks confusing :(
alright so we know that the sum of the deposits of both accounts must equal 29,000 right?
@candycanegone
Like we know that we deposited two different amounts into two separate investment accounts. well call those P0 and P1
Here are the two equations you must solve. \[P_{0}+P_{1} = 29,000\] \[P_{0}(0.032)+P_{1}(0.12) = 1,572\]
The second equation is just the simple interest formula. note I took the decimal form of both percentages. now the key here, is to add these up because Interest = P*r*t so the total interest earned would be P*r*t1 + P*r*t1 since t is just a year we can just disregard it and now it becomes: \[P_{0}(0.032)+P_{1}(0.124) = 1,574\] \[p_{0}+p_{1} = 29,000\] \[P_{0} = 29,000-P_{1}\] Substitute this into the first equation. \[(29,000)*(0.032)-(P_{1})0.032+P_{1}(0.124) = 1,572\] We clean this up a bit \[928+0.092P_{1} = $1,572\] \[P_{1} =\frac{ 644 }{ 0.092 } = $7,000\] so one account 7,000 was placed into it. can you find the amount of money placed into the other account? it's more straight forward.
Join our real-time social learning platform and learn together with your friends!