Alan has twice as many dimes as he has quarters and 10 more nickels than dimes. The remaining 35 coins are pennies. How many of quarters, dimes, and nickels does he have if he has $2.50?
The Correct Answer and Explanation is:
To solve this problem, we will use algebra to set up and solve a system of equations based on the information provided.
Let:
- qq = number of quarters
- dd = number of dimes
- nn = number of nickels
- pp = number of pennies
Given relationships:
- Alan has twice as many dimes as quarters:
d=2qd = 2q - Alan has 10 more nickels than dimes:
n=d+10n = d + 10 - Alan has 35 pennies:
p=35p = 35 - The total value of all coins is $2.50 or 250 cents.
Now let’s write the value equation using cents:
- Quarters are worth 25 cents: 25q25q
- Dimes are worth 10 cents: 10d10d
- Nickels are worth 5 cents: 5n5n
- Pennies are worth 1 cent: 1p=351p = 35
So the total value equation is: 25q+10d+5n+35=25025q + 10d + 5n + 35 = 250
Now substitute the expressions for dd and nn:
- d=2qd = 2q
- n=d+10=2q+10n = d + 10 = 2q + 10
Substitute into the total value equation: 25q+10(2q)+5(2q+10)+35=25025q + 10(2q) + 5(2q + 10) + 35 = 250
Now simplify: 25q+20q+10q+50+35=25025q + 20q + 10q + 50 + 35 = 250
Combine like terms: (25q+20q+10q)+85=250⇒55q+85=250(25q + 20q + 10q) + 85 = 250 \Rightarrow 55q + 85 = 250
Subtract 85 from both sides: 55q=16555q = 165
Divide by 55: q=3q = 3
Now solve for the other coins:
- d=2q=2(3)=6d = 2q = 2(3) = 6
- n=d+10=6+10=16n = d + 10 = 6 + 10 = 16
Final Answer:
- Quarters: 3
- Dimes: 6
- Nickels: 16
- Pennies: 35
Check Total Value:
- Quarters: 3 × 25 = 75
- Dimes: 6 × 10 = 60
- Nickels: 16 × 5 = 80
- Pennies: 35 × 1 = 35
Total: 75 + 60 + 80 + 35 = 250 cents = $2.50
✅ The answer is correct.