Percent interest

Question

A man invests his savings in two accounts, one paying 6 percent and the other paying 10 percent simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is 6358 dollars. How much did he invest at each rate?

Answer 1

let x = the amount invested at 10%

then 2x =  the amount invested at 6%

convert the % to decimals; 10% = 0.10 and 6% = 0.06

 

our equation will be 0.10x + 0.06(2x) = $6358

expanding, we get 0.10x + 0.12x = $6358

combining, we get 0.22x = $6358

solving, we get x = $28,900, and 2x = $57,800

 

Proof: $28,900 times 10% = $2,890, and $57,800 times 6% = $3,468; $2,890 + $3,458 = $6358

Answer 2

For this question, you can make a table, with each row representing a different account. Let's call the amount invested in the riskier account x and the amount invested in the safer account 2x (since he invested twice as much in it):

 

INITIAL AMOUNT          INTEREST RATE          INTEREST PER YEAR

2x                                     0.06                               0.06(2x)

x                                       0.1                                  0.1x

 

Now, you can write an equation representing the total interest he earns per year:

 

0.06(2x) + 0.1x = 6358

0.12x + 0.1x = 6358

0.22x = 6358

x = 28900

 

This means the man invested $28,900 in the riskier account and $57,800 (twice that) in the safer account.