Question

Use the three steps to solve the problem. Betty has 10 more dimes than quarters. If she has $3.45, how many coins does she have?

{
quarters,
dimes }

Answer 1

1. Represent the problem: Let's use the variables q for quarters and d for dimes.
Given that Betty has 10 more dimes than quarters, we can express this as: d = q + 10.
Additionally, we know that the total value of the coins is $3.45.

2. Translate the problem into an equation: The value of each quarter is $0.25, so the total value of the quarters is 0.25q. Similarly, the value of each dime is $0.10, so the total value of the dimes is 0.10d. The total value of all the coins can be represented by the equation: 0.25q + 0.10d = 3.45.

3. Solve the equation: Let's substitute the value of d from step 1 into the equation in step 2:
0.25q + 0.10(q + 10) = 3.45.
Simplifying the equation:
0.25q + 0.10q + 1 = 3.45.
Combining like terms:
0.35q + 1 = 3.45.
Subtracting 1 from both sides:
0.35q = 2.45.
Finally, dividing both sides by 0.35 gives:
q ≈ 7.
So, Betty has approximately 7 quarters.

Substituting this value back into the equation in step 2 gives:
0.25(7) + 0.10d = 3.45.
Simplifying:
1.75 + 0.10d = 3.45.
Subtracting 1.75 from both sides:
0.10d = 1.70.
Dividing both sides by 0.10 gives:
d ≈ 17.
Therefore, Betty has approximately 17 dimes.

In conclusion, Betty has 7 quarters and 17 dimes, totaling 7 + 17 = 24 coins.