## 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 payi

Question

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?

in progress
0

Math
2 weeks
2022-01-12T11:31:29+00:00
2022-01-12T11:31:29+00:00 2 Answers
0 views
0
## Answers ( )

Answer:Altogether, he invested $1300.

Step-by-step explanation:This is a

simple interest problem.The simple interest formula is given by:

In which E are the earnings, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.

He invests three times as much in an account paying 14% as he does in an account paying 5%.I am going to call the earnings from the account paying 14% and the earnings from the account paying 5% . The principals are and , in which .

So

He earns $152.75 in interest in one year from both accounts combined.This means that

I am going to write as a function of and replace in the first equation, that of .

So

We also have that

So

In which

So

His earnings are after 1 year, so

His smaller investment is 325.

How much did he invest altogether?This is

Altogether, he invested $1300.

Answer:

Step-by-step explanation:

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%. This means that

x = 3y

The formula for simple interest is expressed as

I = (PRT)/100

Considering the account earning 14% interest,

I = (x × 14 × 1)/100 = 0.14x

Considering the account earning 5% interest,

I = (y × 5 × 1)/100 = 0.05y

If he earns $152.75 in interest in one year from both accounts combined, it means that

0.14x + 0.05y = 152.75 – – – – – – – – – -1

Substituting x = 3y into equation 1, it becomes

0.14(3y) + 0.05y = 152.75

0.42y + 0.05y = 152.75

0.47y = 152.75

y = 152.75!0.47

y = 325

x = 3y = 3 × 325

x = $975

Total amount of money invested is

975 + 325 = $1300