Douglas invested $ 1300 altogether
<em><u>Solution:</u></em>
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
<em><u>The simple interest is given by formula:</u></em>
Where,
"p" is the principal
"n" is the number of years
"r" is the rate of interest
<em><u>He earns $152.75 in interest in one year from both accounts combined</u></em>
Therefore,
Combined S.I = 152.75
n = 1 year
<em><u>Considering the account earning 14% interest:</u></em>
<em><u>Considering the account earning 5% interest:</u></em>
<em><u>Since, Combined S.I = 152.75</u></em>
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
<em><u>how much did he invest altogether?</u></em>
Amount invested together = x + y = 975 + 325 = 1300
Thus he invested $ 1300 altogether
