AMC 10 · 2014 · #1

Grade 3 algebra
linear-equations-one-varmental-arithmetic work-backwards ↑ Prerequisites: linear-equations-one-var
📏 Short solution 💡 1 insight
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Problem
Adding one more of the higher-value coin would even out the two counts. Find the total value.

Pick an answer.

(A)
33
(B)
35
(C)
37
(D)
39
(E)
41
How to solve
Strategy Work Backwards

The clean fact is about a made-up future: after adding one nickel the counts are equal. So Tool #11 (Work Backwards) is natural — first describe that easy equal-split state, then undo the added nickel to recover how many pennies and nickels Leah really has. Tool #8 (Analyze the Units) keeps the money straight at the end: a nickel counts as 5 cents and a penny as 1 cent, so the total must be measured in cents, not in coins.

1STEP 1

Split the pretend total in half

The pretend total splits evenly into 7 and 7.

13+1 = 14, 14 ÷ 2 = 7 nickels and 7 pennies
2STEP 2

Undo the extra nickel

Removing the imagined coin restores the real counts.

7-1 = 6 nickels, 7 pennies, 6+7 = 13 ✓
3STEP 3

Add up the money in cents

Adding the money gives 37, choice (C).

6 × 5 + 7 × 1 = 30 + 7 = 37 cents → (C)
Answer
37
The count 6 nickels and 7 pennies rebuilds 13 coins, and one more nickel would make it 7 and 7 — exactly what the problem demanded, so the split is right. The value 37 cents sits in the middle of the choices (33 to 41), and every choice differs by 2, which is the value gap you get by swapping one penny for one nickel — a sign the problem was built around getting this exact nickel-penny mix.
💡Key takeaway

When a condition describes an imagined future, build that easy picture first, then undo the change to see what is really true now.

  • Split the pretend total in half
  • Undo the extra nickel
  • Add up the money in cents