Douglas invested $ 1300 altogether
Solution:
Let x represent the amount invested in the account paying 14% interest.
Let y represent the amount invested in the account paying 5% interest
He invests three times as much in an account paying 14% as he does in an account paying 5%
Which means,
x = 3y
The simple interest is given by formula:
[tex]S.I = \frac{p \times n \times r}{100}[/tex]
Where,
"p" is the principal
"n" is the number of years
"r" is the rate of interest
He earns $152.75 in interest in one year from both accounts combined
Therefore,
Combined S.I = 152.75
n = 1 year
Considering the account earning 14% interest:
[tex]S.I = \frac{3y \times 14 \times 1}{100}\\\\S.I = 0.42y[/tex]
Considering the account earning 5% interest:
[tex]S.I = \frac{y \times 5 \times 1}{100}\\\\S.I = 0.05y[/tex]
Since, Combined S.I = 152.75
Therefore,
0.42y + 0.05y = 152.75
0.47y = 152.75
Divide both sides by 0.47
y = 325
Therefore,
x = 3y
x = 3(325)
x = 975
how much did he invest altogether?
Amount invested together = x + y = 975 + 325 = 1300
Thus he invested $ 1300 altogether
