AMC 10 · 2014 · #1
Easy mode Grade 3Leah has 13 coins. Each coin is either a penny (worth 1 cent) or a nickel (worth 5 cents). If you add just one more nickel, she would then have the same number of pennies as nickels. How much are Leah's 13 coins worth, in cents?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Leah has 13 coins made of only pennies and nickels. Adding one more nickel would make her pennies and nickels equal in number. Find the total value of her coins in cents.
Givens: Leah has $13$ coins total, all pennies or nickels; If she had one more nickel, she would have the same number of pennies and nickels; A penny is worth $1$ cent and a nickel is worth $5$ cents; Answer choices: (A) $33$, (B) $35$, (C) $37$, (D) $39$, (E) $41$
Unknowns: The total value of Leah's 13 coins, in cents
Understand
Restated: Leah has 13 coins made of only pennies and nickels. Adding one more nickel would make her pennies and nickels equal in number. Find the total value of her coins in cents.
Givens: Leah has $13$ coins total, all pennies or nickels; If she had one more nickel, she would have the same number of pennies and nickels; A penny is worth $1$ cent and a nickel is worth $5$ cents; Answer choices: (A) $33$, (B) $35$, (C) $37$, (D) $39$, (E) $41$
Plan
Primary tool: #11 Work Backwards
Secondary: #8 Analyze the Units
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.
Execute — Answer: C
3.OA.A.3 Step 1 Split the pretend total in half
- Imagine Leah already added the extra nickel.
- Then she has $13+1=14$ coins, and the problem says the pennies and nickels are now equal in number.
- Fourteen coins split into two equal groups gives $14\div 2=7$, so in this pretend picture there are $7$ nickels and $7$ pennies.
💡 "Same number of each" out of a known total just means split the total into two equal groups.
2.OA.A.1 Step 2 Undo the extra nickel
- That equal split counted the nickel that was only added in the story.
- Take it back off: the pennies were never touched, so there are still $7$ pennies, but the real number of nickels is one fewer, $7-1=6$.
- So right now Leah has $6$ nickels and $7$ pennies, which correctly adds back to $6+7=13$ coins.
💡 To reverse a step, remove exactly what was added and leave everything else alone.
2.MD.C.8 Step 3 Add up the money in cents
- Now count value, not coins.
- The $6$ nickels are worth $6\times 5 = 30$ cents and the $7$ pennies are worth $7\times 1 = 7$ cents.
- Together that is $30+7 = 37$ cents, which is choice (C).
💡 Coins of different kinds only add up correctly once you turn each into its cent value first.
3.OA.A.3 Imagine Leah already added the extra nickel. Then she has $13+1=14$ coins, and t 2.OA.A.1 That equal split counted the nickel that was only added in the story. Take it ba 2.MD.C.8 Now count value, not coins. The $6$ nickels are worth $6\times 5 = 30$ cents and Review
Reasonableness: 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.
Alternative: Use a variable instead. Let $n$ be the current number of nickels; pennies are $13-n$. One more nickel makes nickels $n+1$, set equal to pennies: $n+1 = 13-n$, so $2n = 12$ and $n = 6$. Then pennies $=7$, and the value is $6\times 5 + 7\times 1 = 37$ cents, matching (C).
CCSS standards used (min grade 3)
3.OA.A.3Solve multiplication and division word problems within 100 (Splitting the pretend total of $14$ coins into two equal groups, $14\div 2 = 7$.)2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Undoing the added nickel with $7-1=6$ and checking $6+7=13$.)2.MD.C.8Solve word problems involving dollar bills, quarters, dimes, nickels, and pennies (Converting $6$ nickels and $7$ pennies to their cent values and totaling $30+7=37$ cents.)
⭐ When a condition describes an imagined future, build that easy picture first, then undo the change to see what is really true now.
⭐ When a condition describes an imagined future, build that easy picture first, then undo the change to see what is really true now.
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