Question 1204205: Thomas has $6000 invested among a checking account paying 2% annual interest, a savings account paying 5% annual interest, and a bond paying 7% annual interest. He earns $355 in annual interest and he has $2300 less invested in his savings account than in his bond. How much does he have invested in each account?
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! Thomas has $6000 invested among a checking account paying 2% annual interest, a savings account paying 5% annual interest, and a bond paying 7% annual interest. He earns $355 in annual interest and he has $2300 less invested in his savings account than in his bond. How much does he have invested in each account?
Thomas has invested in a checking account paying 2% annual interest, Let amount be $x
A savings account paying 5% annual interest, let investment be $y
a bond paying 7% annual interest. Let investment be $z
x+y+z=6000 ----------------------(1)
He earns $355 in annual interest
0.02x+0.05y+0.07z = 355-----------------(2)
he has $2300 less invested in his savings account than in his bond.
z-2300 =y
0.02x+0.05y+0.07z= 355 From (2)
0.02x+0.05(z-2300)+0.07z =355 (substitute y)
0.02x+ 0.05z -115+0.07z=355
0.02x+0.12z = 470-----------------------(3)
multiply by 100
2x +12z = 47000-------------------(4)
x+y+z=6000
substitute z-2300=y
x+z-2300 +z = 6000
x+2z = 8300 -------------------(5)
Solve (4) &(5)
2x +12z = 47000
x+2z = 8300 multiply by 2 and subtract
2x+4z=16600
8z = 30400
z= 3800
y= z-2300
3800-2300 = 1500
x= 6000-3800-1500 =700
Thomas has $700 invested in the checking account, $1500 in the savings account, and $3800 in the bond.
CHECK
700*2%+1500*5%+3800*7%= 355
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