Question

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.

Answer 1

Answer:

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

Answer 2

Answer:

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