Question 532306: An investor has a total $18,000 deposited in three different accounts, which earn annual interest of 9%, 7%, and 5%. The account deposited in the 9% account is twice the amount in the 5% account. If the three accounts earn total annual interest of $1340, how much money is deposited in each account?
Found 2 solutions by stanbon, mananth:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! An investor has a total $18,000 deposited in three different accounts, which earn annual interest of 9%, 7%, and 5%. The account deposited in the 9% account is twice the amount in the 5% account. If the three accounts earn total annual interest of $1340, how much money is deposited in each account?
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Quantity Eq:: n + s + f = 18000
Quantiy Eq: n = 2f
Interest Eq: 9n+7s+5f = 134000
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n + s + f = 18000
n + 0 - 2f = 0
9n +7s + 5f = 134000
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Use any method you know to find:
n = 9% investment = $8000
s = 7% investment = $6000
f = 5% investment = $4000
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Cheers,
Stan H.
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Answer by mananth(16946)  (Show Source):
You can put this solution on YOUR website! An investor has a total $18,000 deposited in three different accounts,
9%,------------ $x
7%,-------------$y
5%.-------------$z
The account deposited in the 9% account is twice the amount in the 5% account.
x=2z
x+y+z=18000
substitute z
2z+y+z=18000
y+3z=18000.................1
the three accounts earn total annual interest of $1340,
9%x+7%y+5%z=1340
x=2z
9%*2z+7%y+5%z=1340
18%z+7%y+5%z=1340
multiply by 100
18z+7y+5z=134000
23z+7y=134000................2
3 z + 1 y = 18000 .............1
23 z + 7 y = 134000 .............2
Eliminate y
multiply (1)by -7
Multiply (2) by 1
-21 z -7 y = -126000
23 z 7 y = 134000
Add the two equations
2 z = 8000
/ 2
z = 4,000.00
plug value of z in (1)
3 z + 1 y = 18000
12000 + 1 y = 18000
1 y = 18000 -12000
1 y = 6000
y = 6000.00
Now that you know z & y you can calculate x
m.ananth@hotmail.ca
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