Multiple choice

A portion of $3,360 was invested at 6% simple interest per year, while the remaining was invested at 4% simple interest per year. If the annual income earned from the amount invested at 6% was double the amount invested at 4%, then what would be the total income from both the investments after 3 years?

  1. $115.20
  2. $172.80
  3. $345.60
  4. $518.40
  5. $718.20
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let x be the amount at 6% and y be the amount at 4%. x+y=3360. Income 0.06x = 2 * 0.04y. 0.06x = 0.08y. x = 4/3 y. (4/3)y + y = 3360. 7/3 y = 3360. y = 1440. x = 1920. Annual income = 0.06(1920) + 0.04(1440) = 115.2 + 57.6 = 172.8. Total income for 3 years = 172.8 * 3 = 518.4.

AI explanation

Let x be the portion invested at 6 percent, so the interest from it is 0.06x, and the interest from the remaining portion of (3360 minus x) at 4 percent is 0.04 multiplied by (3360 minus x). Since the interest at 6 percent is double the interest at 4 percent, the equation is 0.06x equals 2 multiplied by 0.04 multiplied by (3360 minus x). Solving this gives 0.06x equals 268.8 minus 0.08x, so 0.14x equals 268.8, and x equals 1920, meaning 1440 was invested at 4 percent. The total interest for one year is (0.06 multiplied by 1920) plus (0.04 multiplied by 1440), which equals 115.2 plus 57.6, totaling 172.8, and multiplying by 3 years gives $518.40.

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