Answer:
3200
Step-by-step explanation:
0.13x + 0.04*(6100-x) = 532 dollars.
0.13x + 0.04*6100 - 0.04x = 532
(0.13 - 0.04)x = 532 - 0.04*6100
x = 532 - 0.04 - 6100/0.13-0.04 = 3200
The couple invested:
to earn $477 as the total interest for the first year.
The interest over an investment is the additional amount that a person receives on investing his/her money in the cause.
In the question, we are informed that in investing $6,150 of a couple's money, a financial planner put some of it into a savings account paying 6% annual simple interest. The rest was invested in a riskier mini-mall development plan paying 9% annual simple interest. The combined interest earned for the first year was $477.
We are asked to find the amount invested in each rate.
We assume the amount invested in savings account to be $x.
Therefore, the amount invested in development plan is $(6150 - x).
Simple interest on a sum of money (P), at a rate (r) over a time period (t), is given as S.I. = (Prt)/100.
We calculate the interest earned in both investments:-
Investment in savings account:
Amount invested (P) = $x.
Rate of interest (r) = 6%.
Time for investment (t) = 1 year.
Therefore, interest received = (Prt)/100 = $(x*6*1)/100 = $ (6x)/100
Investment in development plan:
Amount invested (P) = $(6150 - x).
Rate of interest (r) = 9%.
Time for investment (t) = 1 year.
Therefore, interest received = (Prt)/100 = $((6150 - x)*9*1)/100 = $ {9(6150 - x)}/100.
We know that the total interest received is $477.
Thus on adding Interests from both the investments, we can write an equation:
(6x)/100 + {9(6150 - x)}/100 = 477
or, 6x + 55350 - 9x = 47700,
or, 55350 - 47700 = 3x,
or, 3x = 7650,
or x = 7650/3 = 2550.
Thus, the investments in:
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