Interest problem

Question

Tyler has two savings accounts that his grandparents opened for him. One of the accounts pay 6% annual interest, whereas the other pays 8% annual interest. If you have $600 more invested at 8% than you invested at 6%, how much do you have invested in each account if the total amount of interested you earn in a year is $76?

Answer 1

Hi Lizz.

 

This is a simple interest question ('simple interest' not implied to mean easy, just different from 'compound interest'). The equation for simple interest is:

 

I = P*r*t

 

Where

 

I = interest earned

P = amount invested

r = return rate

t = time

 

You have two separate accounts each earning two separate interest rates with two separate amounts invested.

 

Let's call the interest you return from the 8% account I_e (e for eight), and the interest you return from the 6% interest account I_s.

 

I_e = P_e * r_e * t

 

I_s = P_s * r_s * t

 

Now we need to dissect the information presented in the question. The most obvious are the interest rates "r".

 

r_e = 8% = 0.08

r_s = 6% = 0.06

 

Your question also states that you have $600 more invested in your 8% than you do in your 6%. That means:

 

P_e = P_s + 600

 

You also know that your total interest earned in a year is $76.

 

I_e + I_s = 76

 

The time is the same for both equations. t = 1 year

 

So, let's lay out our equations.

 

I_e = P_e * r_e * t
I_s = P_s * r_s * t

r_e = 8% = 0.08
r_s = 6% = 0.06

P_e = P_s + 600

I_e + I_s = 76

t = 1

 

Now let's substitute what we know into our equations.

 

I_e + I_s = 76

 

P_e * r_e * t + P_s * r_s * t = 76

 

(P_s + 600)*0.08*1 + P_s*0.06*1 = 76

 

0.08*P_s + 48 + 0.06*P_s = 76

 

0.14*P_s = 28

 

P_s = 200

 

P_e = P_s + 600

 

P_e = 200 + 600 = 800

 

Tyler has $200 invested at 6% and $800 invested at 8%.

 

Hope this helps.