SOLUTION: A man invests his savings in 2 accounts, one paying 6% per year
and the other 7%. He has twice as much invested in the 7% account
as in the 6% account, and his annual interest in
Algebra.Com
Question 677102: A man invests his savings in 2 accounts, one paying 6% per year
and the other 7%. He has twice as much invested in the 7% account
as in the 6% account, and his annual interest income is $600. How
much is invested in each account? I=PRT
Answer by partha_ban(41) (Show Source): You can put this solution on YOUR website!
Let he put $d in 6% interest account (1st account)
So, he put $2d in 7% interest account (2nd account)
Interest earned from 1st account = d * (6/100) * 1 = 6d/100
Interest earned from 2nd account = 2d * (7/100) * 1 = 14d/100
Total interest earned = 6d/100 + 14d/100 = (6d+14d)/100 = 20d/100 = d/5
By condition, d/5 = 600
Therefore, d = 600 * 5 = 3,000
Therefore, the man invested $3,000 in 6% interest account and $6,000 in 7% interest account.
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