Last year, Kaitlyn had $30,000 to invest. She invested some of it in an account that paid 5% simple interest per year, and she invested the rest in an account that paid 9% simple interest per year. After one year, she received a total of $2580 in interest. How much did she invest in each account?First account:Second account:

Question

25-10-2022
Mathematics

Answered

Last year, Kaitlyn had $30,000 to invest. She invested some of it in an account that paid 5% simple interest per year, and she invested the rest in an account that paid 9% simple interest per year. After one year, she received a total of $2580 in interest. How much did she invest in each account?First account:Second account:

Answer :

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KeiandreE417590 KeiandreE417590 · Answer 1 · 2022-10-25T18:02:42-04:00

Answer:

She invested an amount of $3,000 in the first accont and $27,000 in the second account

Explanation:

Here, we want to get the amount invested in each of the accounts

To get this,let us write the simple interest formula

Mathematically, we have this as:

[tex]I\text{ = }\frac{PRT}{100}[/tex]

Let the amount invested in the first account be $x, while that of the second account be $y

P represent the amounts invested

R is the rate of investment

T is the time (1 year)

The sum of these two is the $30000 she has to invest

Mathematically, we have this as:

[tex]x\text{ + y = 30000}[/tex]

Let us get the interests on each of the accounts

For the first account, we have the interest as:

[tex]I_1\text{ = }\frac{x\times5\times1}{100}\text{ = }\frac{5x}{100}[/tex]

For the second account, we have the interest as:

[tex]I_2\text{ = }\frac{y\times9\times1}{100}\text{ =}\frac{9y}{100}[/tex]

The sum of these two is the interest value, which is as follows:

[tex]\begin{gathered} I_1+I_2\text{ = }\frac{5x}{100}+\frac{9y}{100}\text{ = 2580} \\ \\ 5x\text{ + 9y = 258000} \end{gathered}[/tex]

So, we have two equations to solve simultaneously

The two equations are:

[tex]\begin{gathered} x\text{ + y = 30000} \\ 5x\text{ + 9y = 258000} \end{gathered}[/tex]

We can solve this by substitution

Let us substitute for y

[tex]y\text{ = 30000-x}[/tex]

Insert this into the second equation, we have this as:

[tex]\begin{gathered} 5x\text{ + 9(30000-x) = 258000} \\ 5x\text{ + 270000-9x = 258000} \\ 9x-5x\text{ = 270000-258000} \\ 4x\text{ = 12000} \\ x\text{ = }\frac{12000}{4} \\ x\text{ = 3,000} \end{gathered}[/tex]

Recall:

[tex]y\text{ = 30000-x = 30000-3000 = 27000}[/tex]

What this mean is that:

She invested an amount of $3,000 in the first accont and $27,000 in the second account