SOLUTION: Arnold had $1.70 in dimes and quarters. He had 3 more dimes than quarters. How man of each coin did he have?
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Question 1010764: Arnold had $1.70 in dimes and quarters. He had 3 more dimes than quarters. How man of each coin did he have?
Thank you. Found 2 solutions by macston, addingup:Answer by macston(5194) (Show Source):
You can put this solution on YOUR website! .
D=number of dimes; Q=number of quarters=D-3
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$0.10(D)+$0.25(Q)=$1.70
$0.10(D)+$0.25(D-3)=$1.70
$0.10(D)+$0.25D-$0.75=$1.70
$0.35D=$2.45
D=7
ANSWER 1: There were 7 dimes.
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Q=D-3-7-3=4
ANSWER 2: There were 4 quarters
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CHECK:
$0.10(D)+$0.25(Q)=$1.70
$0.10(7)+$0.25(4)=$1.70
$0.70+$1.00=$1.70
$1.70=$1.70
You can put this solution on YOUR website! dimes: d
quarters: q
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d= q+3 we'll use this value for d next
.10d+.25q = 1.70 now substitute for d:
.10(q+3)+.25q= 1.70
.10q+.30+.25q= 1.70 subtract .30 on both sides and add q on left
.35q= 1.40 divide both sides by .35
q= 4 He has 4 quarters and
d= q+3= 4+3= 7 he has 7 dimes
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Check:
4*.25= 1
7*.10= .70
1+.70= 1.70 We have the right answer
J
