Question

A man invests $3100 in three accounts that pay 6 %, 8% and 10% in annual interest, respectively. He has 3 times as much invested at 10% than be does at 6 %. If his total interest for the year is $268, how much is invested at each rate?

Answer 1

Let x be the amount he invest in the 6% account.

We know that he invested 3 times more in the 10% account, then he invested 3x in that account.

We know that in total he invested $3100, then he invested:

`3100-x-3x=3100-4x`

in the 8% account.

Now, the simple interest formula is given as:

`A=Prt`

where A is the interest generated in the year, P is the principal, r is the interest rate (in decimal form) and t is the time.

We know that the total interest in a year (that is t=1) is $268, then we apply the formula for each principal and each rate, and equate it to $268, that is:

`0.06x+0.1(3x)+0.08(3100-4x)=268`

Solving for x we have:

`\begin{gathered} 0.06x+0.1(3x)+0.08(3100-4x)=268 \\ \text{0}.06x+0.3x+248-0.32x=268 \\ 0.04x=268-248 \\ 0.04x=20 \\ x=(20)/(0.04) \\ x=500 \end{gathered}`

Once we know the value of x we can find how much he invested in each account:

He invested $500 in the 6% account.

He invested $1100 in the 8% account.

He invested $1500 in the 10% account.