SOLUTION: The annual interest on a
$6000 investment exceeds the interest earned on a
$3000 investment by
$93.
The
$6000 is invested at a
0.3% higher rate of interest than the
$3
Algebra ->
Finance
-> SOLUTION: The annual interest on a
$6000 investment exceeds the interest earned on a
$3000 investment by
$93.
The
$6000 is invested at a
0.3% higher rate of interest than the
$3
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Question 1026830: The annual interest on a
$6000 investment exceeds the interest earned on a
$3000 investment by
$93.
The
$6000 is invested at a
0.3% higher rate of interest than the
$3000.
What is the interest rate of each investment?
You can put this solution on YOUR website! The annual interest rate on$6000 investment be x%
and on $3000 investment be y%
x%*6000-y%*3000 = 93 ............(1) (exceeds the interest earned on a
$93.)
x%-y% =0.3%.....................(2)
Multiply (1) by 100
6000x-3000y = 9300
/100
60x-30y= 93............(3)
x-y =0.3
multiply by 30
30x-30y=9..............(4)
add 3&4
30x=84
x= 84/30 = 2.8%
y = 2.5%
The
$6000 is invested at a 2.8%
the $3000 is invested at 2.5%
You can put this solution on YOUR website! 6000x+3000y+93
x = y+0.003 use this value for x in the formula above:
6000(y+0.003) = 3000y+93
6000y+18 = 3000y+93
3000y = 75 = 0.025(100)= 2.5
The 3000 are invested at 2.5 and the 6000 at 2.5+0.3 = 2.8
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Check:
6000*0.028 = 168
3000*0.025 = 75
. . . . . . .-----
Subtract . . : 93 The investment of 6000 makes 93 more/year than the 3000, just like the problem says
