Question

Kelly has $800, which she divides between two savings accounts. One account earns 2% simple interest and the other earns 5%. If she earns $31 in interest between the two accounts, how much is in each?

Answer 1

Answer:

On the first account, she invests $300, earning 0.02*300 = $6.

On the second acount, she invests 800 - 300 = $500, earning $25.

Step-by-step explanation:

This is a simple interest problem.

The simple interest formula is given by:

$E = P*I*t$

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

In this question:

Two acounts, one with 2% interest(I = 0.02) and the other with 5% interest(I = 0.05).

Each account has two earnings, that i will call $E_{1}$ and $E_{2}$

Two investments adding up to 800. I will call the first investment P and the second is 800 - P.

Time is not given, but for simplicity, i will use 1 year.

First investment:

I at 2%. The earnings are $E_{1}$.

$E_{1} = 0.02P$

Second Investment:

Earnings $E_{2}$, at 5%. So

$E_{2} = 0.05(800 - P)$

The earnings add to 31, so $E_{1} + E_{2} = 31$, then

$E_{2} = 31 - E_{1}$

So

$31 - E_{1} = 0.05(800 - P)$

$31 - 0.02P = 0.05(800 - P)$

$31 - 0.02P = 40 - 0.05P$

$0.03P = 9$

$P = \frac{9}{0.03}$

$P = 300$

So:

On the first account, she invests $300, earning 0.02*300 = $6.

On the second acount, she invests 800 - 300 = $500, earning $25.