college algebra

Question

Roberto invested some money at 2.75%, and then invested $4,000 more than twice this amount at 3%. His total annual income from the two investments was $1,257.50. How much was invested at 3%?

Answer 1

Hello, thank you for taking the time to post your question!

The key on this type of question is finding a way to relate the variables to one another in an algebraic expression. On this one we have

0.0275x + 0.03(2x + 4000) = 1257.50  if you let “x” be the amount invested at the 2.75% rate

From there then you can solve for the value of “x” algebraically

0.0275x + 0.06x + 120 = 1257.50

0.0875x + 120 = 1257.50

0.0875x = 1137.50

x = 1137.50 / 0.0875 = 13000

meaning that $13,000 was invested at the 2.75% rate and then (2*$13,000) + $4000 = $30,000 was invested at the 3% rate

I hope that helps you get moving in a better direction on this type of question! Feel free to reach out if you have any additional questions beyond that :)

Answer 2

Let x = the amount Roberto invested at 2.75%

Let 2x + 4000 = the amount Roberto invested at 3%

 

.0275x + .03(2x + 4000) = 1257.50

.0275x + .06x + 120 = 1257.50

.0875x = 1137.50

x = $13,000

2x + 4000 = 2(13,000) + 4000 = $30,000