A man invests his savings in two accounts, one
paying 6% and the other paying 10% simple interest per year.
He puts twice as much in the lower-yielding account because
it is less risky. His annual interest is $3520. How much did
he invest at each rate?
For the given investment problem the man invested $32,000 at a rate of 6% and $16,000 at a rate of 10%.
x = the amount invested in the first account.
y = the amount invested in the second account.
x = 2y
x – 2y = 0 …….(1)
Also the total sum of interest earned would be,
0.06x + 0.1y = 3520 ……….(2)
Solving equation (1) and (2) we have,
20 equation (1) gives 1.2x + 2y = 70400
Now adding this to equation (2), we get
2.2x = 70400
x = 32000
Substituting this in (1)
2y = 32000
Y = 16000
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