An investor decides to invest some cash in an account paying 11% annual interest, and to put the rest in a stock fund that ends up earning 8% over the course of a year. The investor puts $600 more in the first account than in the stock fund, and at the end of the year finds the total interest from the two investments was $750. How much money was invested at each of the two rates? Round to the nearest integer.
Added by Matthew F.
Step 1
Then, the amount invested in the account paying 11% interest will be \( x + 600 \). Show more…
Show all steps
Close
Your feedback will help us improve your experience
Supreeta N and 65 other Algebra educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
$6,544 is invested, part at 11% and the rest at 8%. If the interest earned from the amount invested at 11% exceeds the interest earned from the amount invested at 8% by $399.50, how much is invested at each rate? (Round to two decimal places if necessary.)
Supreeta N.
Last year, Hong had $30,000 to invest. He invested some of it in an account that paid 7% simple interest per year, and he invested the rest in an account that paid 8% simple interest per year. After one year, he received a total of $2210 in interest. How much did he invest in each account?
Adi S.
You invested 3000 dollars in two accounts paying $6 \%$ and $8 \%$ annual interest. If the total interest earned for the year was 230 dollars, how much was invested at each rate? (Section $1.3,$ Example 5 )
Trigonometric Functions
Applications of Trigonometric Functions
Recommended Textbooks
Elementary and Intermediate Algebra
Algebra and Trigonometry
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD