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Gordon invested $3600 for one year in three different accounts. The accounts paid simple annual interest of 6%, 8% and 7% respectively. The total interest at the end of one year was $260. If Gordon invested twice as much at 8% than 6%, find the amount he invested in each account.

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sqdancefan

Answer:

  • 6%: $800
  • 7%: $1200
  • 8%: $1600

Step-by-step explanation:

You want the amounts invested in each of three accounts, earning 6%, 7%, and 8%. Given is the total interest for one year ($260), the total amount invested ($3600), and the fact that twice as much was invested at 8% as at 6%.

Setup

Let x represent the amount invested at 6%. Then 2x is the amount invested at 8%, and (3600 -x -2x) is the amount invested at 7%. The total interest earned is ...

  0.06x +0.07(3600 -3x) +0.08(2x) = 260

Solution

Simplifying the equation, we find ...

  0.01x +252 = 260

  x = (260 -252)/0.01 = 800 . . . . . . subtract 252 and divide by 0.01

The amounts invested in the various accounts are ...

6%: $800

7%: $1200

8%: $1600

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