Question

In a jar there are 73 coins consisting of quarters, dimes, and pennies. There is $9.19. The number of quarters is 2 more than twice the number of dimes. How many quarters, nickels, and dimes are there?

Answer 1

• 30 quarters, 14 dimes and 29 pennies

Explanation:

• Note: there are no nickels, just pennies. With quarters, dimes and nickels there is no solution since all three coins are multiples of 5 but the sum ($9.19) is not.

Coins:

• Quarters | q = 25 cents
• Dimes | d = 10 cents
• Pennies | p = 1 cent
• Total coins: 73
• Total amount: $9.19 = 919 c

Equations:

• q + d + p= 73
• q = 2d + 2
• 25q + 10d + p = 919

Substitute q in the first equation:

• 2d+2 +d +p = 73 ⇒ 3d + p = 71 ⇒ p = 71 -3d

Substitute p and q in the third equation:

• 25q + 10d + p = 919
• 25(2d+2) + 10d + 71 - 3d = 919
• 50d + 50 + 7d = 848
• 57d = 798
• d= 798/57
• d= 14

Then, finding p and q:

• q = 2d + 2 = 2*14 + 2 = 30
• p = 71 - 3d = 71 - 3*14 = 29

So there are 30 quarters, 14 dimes and 29 pennies

Proof:

• 30+14+29 = 73
• 73 = 73
• 30*25 + 14*10 + 29 = 919
• 919= 919