Problem 7 · 1989 AJHSME
Medium
Algebra & Patterns
coin-value
If the value of 20 quarters and 10 dimes equals the value of 10 quarters and n dimes, then n =
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Answer: D — 35.
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Hint 1 of 3
Both sides already have 10 quarters and 10 dimes in common. What's left over once you mentally cross out the shared coins?
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Hint 2 of 3
Cancel anything identical on both sides of an equation first — you only need to balance the difference, not the whole pile.
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Hint 3 of 3
After cancelling, you're left with 10 extra quarters on one side that the extra dimes must match in value.
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Approach: cancel the shared coins, balance the rest
- Both sides carry 10 quarters and 10 dimes — cross those out, since identical amounts on each side don't affect the balance. The left keeps 10 extra quarters; the right keeps (n − 10) extra dimes.
- So 10 quarters must equal (n − 10) dimes in value: 10 × 25¢ = 250¢, and 250 ÷ 10 = 25 extra dimes. Then n = 10 + 25 = 35.
- Why this transfers: in any 'this equals that' setup, deleting whatever is common to both sides shrinks the problem to its real difference — here, 'trade 10 quarters for dimes' instead of juggling 600¢ totals.
Another way — compute both totals in cents:
- Left side: 20×25 + 10×10 = 500 + 100 = 600¢.
- Right side: 10×25 + n×10 = 250 + 10n. Set 250 + 10n = 600, so 10n = 350 and n = 35.
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