SOLUTION: Last year, Carlos had $30,000 to invest. He invested some of it in account that paid 10% simple interest per year, and he invested the rest in an account that paid 5% simple intere

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Last year, Carlos had $30,000 to invest. He invested some of it in account that paid 10% simple interest per year, and he invested the rest in an account that paid 5% simple intere      Log On


   

Question 1200225: Last year, Carlos had $30,000 to invest. He invested some of it in account that paid 10% simple interest per year, and he invested the rest in an account that paid 5% simple interest per year. After one year, he received a total of $2500 in interest. How much did he invest in each account?
Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


With formal algebra....

x = amount invested at 10%
30000-x = amount at 5%

The total interest was $2500:

.10(x)+.05(30000-x) = 2500

Solve using basic algebra... I leave that to you.

Informally, if a formal algebraic solution is not required....

All $30000 invested at 5% would have returned $1500 interest; all at 10% would have returned $3000 interest.
The actual interest $2500 is 2/3 of the way from $1500 to $3000.
That means 2/3 of the total was invested at the higher rate.

ANSWER: 2/3 of the $30000, or $20000, at 10%; the other $10000 at 5%

CHECK: .10($20000)+.05($10000) = $2000+$500 = $2500