Mr. Brown has $12,000 invested...

Question

Mr. Brown has $12,000 invested. Part is invested at 6% and the remainder at 8%. If the interest from the 6% investment is $230 more than the interest from the 8% investment, how much is invested at each rate?

Answer 1

Hi Jonelle!

This problem basically gives us two different equations with two unknowns (the amount invested at each percentage). The first equation is that both parts combined add up to $12,000:

P_6% + P_8% = 12,000

where P_6% is the amount invested at 6% and P_8% is the amount invested at 8%. The second equation is that the interest from the 6% investment is $230 more than the interest from the 8% investment. Recall that interest is calculated as:

I = P*(p/100)

where I is the interest, P is the principal invested, and p is the interest percentage. The information from the problem statement gives us the equation:

I_6% = I_8% + 230

or

P_6%*(6/100) = P_8%*(8/100) + 230

We can easily solve for P_8% in the first equation:

P_8% = 12,000 - P_6%

Then we can plug the value for P_8% into the second equation:

P_6%*(6/100) = (12,000 - P_6%)*(8/100) + 230

and move all P_6% values to the left side:

P_6%*(6/100 + 8/100) = (12,000*8/100) + 230

Simplify:

P_6%*(14/100) = 1,190

Multiply both sides by 100/14:

P_6% = 8,500

This tells us that $8,500 was invested at 6% and then plugging this value into the equation for P_8%:

P_8% = 12,000 - P_6% = 12,000 - 8,500

P_8% = 3,500

In total $8,500 was invested at 6% interest and $3,500 was invested at 8% interest.

I hope this helps!

Answer 2

.06S = .08(12,000-S)+ 230

.06S = 960 -.08S + 230

.14S = 1190

S = 1190/.14 =8500

S= amount invested at 6% = $8,500

12000-8500= 3,500

12000-S = amount invested at 8% = $3,500

check the answer

.06(8,500)= .08(3,500) + 230

510 = 280 +230

interest on 6% = $510

interest on 8% = $280