SOLUTION: Lane invested $22,000 part at 8% and part at 7%. If the actual interest at the end of the year is $1710. How much did she invest at 8%?
(a)$18,000
(b)$16,000
(c)$5,000
(d)$17
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Question 82580: Lane invested $22,000 part at 8% and part at 7%. If the actual interest at the end of the year is $1710. How much did she invest at 8%?
(a)$18,000
(b)$16,000
(c)$5,000
(d)$17,000
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=amount invested at 8%
Then $22000-x=amount invested at 7%
Interest(I)=Principal(P) times Rate(R) times time(T) or I=PRT
Interest on amount invested at 8%= 0.08x
Interest on amount invested at 7%= 0.07($22000-x)
Now we are told that the actual interest at the end of the year is $1710. So:
0.08x+0.07($22000-x)=$1710 get rid of parens
0.08x+$1540-0.07x=$1710 subtract $1540 from both sides
0.08x+$1540-$1540-0.07x=$1710-$1540 collect like terms
0.01x=$170 divide both sides by 0.01
x=$17000-------------------------amount invested at 8%
$22000-x=$22000-$17000=$5000 -----------amount invested at 7%
CK
$5000(0.07)+$17000(0.08)=$1710
$350+$1360=$1710
$1710=$1710
Hope this helps----ptaylor
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