Question

Katie has a collection of nickels dimes and quarters with a total value of $4.20 there are 7 more dimes than nickels and six more quarters and nickels how many of each coin does she have

Answer 1

Katie, Let the number of nickles = N Since there are 7 more dimes than nickles, the number of dimes D = N + 7 Since there are 4 more quarters than nickles, the number of quarters Q = N + 4 Since quarters = $0.25, dimes = $0.10, and nickles = $0.05 and all added together = $4.90 ⇒  0.25·Q + 0.10·D + 0.05·N = 4.90 Now multiply through by 100 to eliminate the decimal: 25Q + 10D + 5N = 490 Substitute for Q and D 25(N+4) + 10(N+7) + 5N = 490 25N + 10N + 5N +100 +70 = 490 40N + 170 = 490 ⇒ 40N = 320 ⇒ N = 320/40 N = 8, D = N+7 = 15 Q = N+4 = 12 Check: 25(12) + 10(15) + 5(8) = 300 + 150 + 40 = 490