SOLUTION: A person invested $4800, part of it in bank bonds bearing 3.10% interest and the remainder in Canada Savings Bonds bearing 4.85% interest
and she received the same income from ea
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-> SOLUTION: A person invested $4800, part of it in bank bonds bearing 3.10% interest and the remainder in Canada Savings Bonds bearing 4.85% interest
and she received the same income from ea
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Question 909711: A person invested $4800, part of it in bank bonds bearing 3.10% interest and the remainder in Canada Savings Bonds bearing 4.85% interest
and she received the same income from each. How much, to the nearest dollar, was invested in each?
Do not round intermediate results. Round only final answers. Round to the nearest integer.
You can put this solution on YOUR website! We know the total amount of money invested. $4800
x+y=4800,
We know that the difference in interest earned by the two accounts is $0
0.0485*x-0.031*y=0
x=4800-y
We substitute for x
0.0485*(4800-y)-0.031*y=0
We multiply out
232.8-0.0485y-0.031*y=0
We combine like terms.
232.8=0.0795*y
Isolate y
y=232.8/0.0795
y=2928.30189 at 3.1% 2928 rounded
Calculate x
x=4800-2928.30189
x=1871.69811 at 4.85% 1872 rounded
Interest earned at 4.85% is 90.7773585 rounded 91
Interest earned at 3.1% is 90.7773585 rounded 91
We check
0.0485*1871.69811-0.031*2928.30189=0
90.7773585-90.7773585=0
-0.56843419e-13=0
Since this statement is TRUE and neither x nor y is negative then all is well
