Question

Emma paid a bill of $2.70 with quarters, dimes, and pennies. She had 5 less quarters than dimes and 4 less quarters than pennies. How many dimes did she have? Please help me answer this ASAP! I hate word problems...

Answer 1

By setting up a system of equations based on the value of coins and the relationships between the number of each type of coin, we find that Emma had 11 dimes.

Step-by-step explanation:

Let's solve this word problem step by step. Emma used quarters (25 cents), dimes (10 cents), and pennies (1 cent) to pay a bill of $2.70. We are given that Emma had 5 fewer quarters than dimes and 4 fewer quarters than pennies.

Let's use algebra to define variables for the different coins:

• Let d represent the number of dimes.
• Since there are 5 fewer quarters than dimes, the number of quarters will be q = d - 5.
• There are also 4 fewer quarters than pennies, so the number of pennies is p = q + 4, which is the same as p = d - 1.

Now, let's translate the total value of the coins into cents and set up an equation:

• 10d + 25q + 1p = 270 cents

Substitute the expressions for q and p in terms of d into the equation:

• 10d + 25(d - 5) + 1(d - 1) = 270
• 10d + 25d - 125 + d - 1 = 270
• 36d - 126 = 270
• 36d = 396
• d = 11

Emma had 11 dimes.