AlgebraHardCoins of two values in a parking meter
Algebra practice question
A parking meter accepts only 20-cent and 50-cent coins. At the end of a day it holds 26 coins worth $9.10 in total. How many 50-cent coins does it hold?
Answer: 13
With f fifty-cent coins: 20(26 − f) + 50f = 910 gives 30f = 390, so f = 13.
Common wrong answers
- If you answered 8
- 8 solves 20f + 50(26 − f) = 910 with the coin types swapped, giving a non-integer, then rounds.
- If you answered 18
- 18 divides 910 by 50, ignoring the 20-cent coins.
- If you answered 26
- 26 is the total number of coins.
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