AlgebraHardCoin problem as a system
Algebra practice question
A jar contains 25 coins, all dimes and quarters, worth $4.00 in total. How many quarters are in the jar?
Answer: 10
Let q be quarters and d dimes: q + d = 25 and 25q + 10d = 400 (in cents). Substitute d = 25 − q: 25q + 250 − 10q = 400, so 15q = 150 and q = 10. Check: 10 quarters ($2.50) + 15 dimes ($1.50) = $4.00.
Common wrong answers
- If you answered 15
- 15 is the number of DIMES. The question asks for quarters: 25 − 15 = 10.
- If you answered 16
- 16 quarters would be $4.00 by themselves, leaving no room for the 9 other coins to add value. Both coin types contribute.
- If you answered 25
- 25 is the total number of coins, not the quarters alone.
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