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svetoff [14.1K]
3 years ago
5

Douglas invests money in two simple interest accounts. He invests three times as much in an account paying 14% as he does in an

account paying 5%. If he earns $152.75 in interest in one year from both accounts combined, how much did he invest altogether?
Mathematics
1 answer:
Rainbow [258]3 years ago
3 0

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>

S.I = \frac{p \times n \times r}{100}

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>

S.I = \frac{3y \times 14 \times 1}{100}\\\\S.I = 0.42y

<em><u>Considering the account earning 5% interest:</u></em>

S.I = \frac{y \times 5 \times 1}{100}\\\\S.I = 0.05y

<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

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