SOLUTION: 70. Nickels, dimes, and quarters. Bernard has 41 coins consisting of nickels, dimes, and quarters, and they are worth a total of $4.00. If the number of dimes plus the number of qu

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Question 955348: 70. Nickels, dimes, and quarters. Bernard has 41 coins consisting of nickels, dimes, and quarters, and they are worth a total of $4.00. If the number of dimes plus the number of quarters is one more than the number of nickels,
then how many of each does he have?
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
N=nickels; D=dimes; Q=quarters
D+Q=N+1
N+N+1=41
2N=40
N=20 ANSWER 1 there are 20 nickels worth $1.00 so there 21 quarters and dimes worth $3.00
D+Q=21
D=21-Q
$0.10D+$0.25Q=$3.00 Substitute for D.
$0.10(21-Q)+$0.25Q=$3.00
$2.10-$0.10q+$0.25Q=$3.00 Subtract $2.10 from each side.
$0.15Q=$0.90 Divide each side by $0.15.
Q=6 ANSWER 2: There are 6 quarters.
D=21-Q=21-6=15 ANSWER 3: There are 15 dimes.
CHECK:
$0.05N+$0.10D+$0.25Q=$4.00
$0.05(20)+$0.10(15)+$0.25(6)=$4.00
$1.00+$1.50+$1.50=$4.00
$4.00=$4.00