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15 July, 18:58

a person invested $6300 for 1 year part at 8% part at 10% and the remainder at 15%. The total annual income from these investments was $766. The amount of money invested at 15% was $100 more than the amounts invested at 8% and 10% combined. Find the amount invested at each rate.

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  1. 15 July, 19:01
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    Step-by-step explanation:

    Let the part at 8% be x; at 10% be y; at 12% be z.

    Total investment is $6300:

    x + y + z = 6300

    Total annual income is $766:

    0.08x + 0.10y + 0.12z = 766

    The amount of money invested at 15% was $100 more than the amounts invested at 8% and 10% combined:

    z - x - y = 100

    Solving the three equations with x, y and z:

    x = $1200

    y = $1900

    z = $3200
  2. 15 July, 19:22
    0
    Step-by-step explanation:

    let x be part at 8%, y be part at 10% and z be part at 15%

    (1) x+y+z=6300

    the amount of money invested at 15% was $100 more than the amounts invested at 8% and 10% combined. so

    (2) z=x+y+100

    The total annual income from these investments was $766. so

    0.08x+0.1y+0.15z=766 or

    (3) 8x+10y+15z=76600

    substitute (2) into (1)

    x+y+x+y+100=6300

    (4) 2x+2y=6200

    substitute (2) inot (3)

    8x+10y+15 (x+y+100) = 76600

    (5) 23x+25y=75100

    solving (4) n (5)

    x=1200

    y=1900

    z=1200+1900+100=3200
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