AMC 8 · 2022 · #5

Grade 2 arithmeticalgebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

linear-equations-one-varmulti-digit-arithmeticformula-substitution convert-to-algebraidentify-subproblems ↑ Prerequisites: multi-digit-arithmeticlinear-equations-one-var

📏 Short solution 💡 2 insights

📘 View easy version →

Problem

Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned $6$ years old, she received a newborn kitten as a birthday present. Today the sum of the ages of the two children and the kitten is $30$ years. How many years older than Bella is Anna?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Anna and Bella share a birthday. Five years ago today, Bella turned $6$ and was given a newborn kitten as a present. Right now, the three ages — Anna's, Bella's, and the kitten's — add up to $30$. How many years older than Bella is Anna?

Givens: Five years ago, Bella's age was $6$; Five years ago, the kitten's age was $0$ (newborn); Today the three ages sum to $30$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$

Unknowns: The difference $\text{Anna's age} - \text{Bella's age}$ today

Understand

Givens: Five years ago, Bella's age was $6$; Five years ago, the kitten's age was $0$ (newborn); Today the three ages sum to $30$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$

Plan

Primary tool: #6 Guess and Check

Secondary: #11 Work Backwards

Two of the three current ages are already fixed once we work backwards five years: Bella is $6 + 5 = 11$ and the kitten is $0 + 5 = 5$. That leaves Anna as the only unknown. Tool #11 (Work Backwards) carries the two known ages forward five years; Tool #6 (Guess and Check) then lets us try the five answer-choice age differences on top of Bella's $11$ to see which makes Anna $+ 11 + 5 = 30$. This keeps the solution arithmetic and avoids reaching for Tool #13 (Algebra), which is overkill for a Grade 2 word problem.

Execute — Answer: C

#11 Work Backwards 1.OA.A.1 Step 1

Carry Bella's and the kitten's ages forward five years to today.
Bella was $6$ five years ago, so she is $6 + 5 = 11$ today.
The kitten was $0$ (newborn), so it is $0 + 5 = 5$ today.

$$\text{Bella} = 6 + 5 = 11, \quad \text{kitten} = 0 + 5 = 5$$

💡 Adding $5$ to small ages to roll the clock forward is a Grade 1 add-within-$20$ word-problem move.

#11 Work Backwards 2.OA.A.1 Step 2

Add Bella's and the kitten's current ages so we know what Anna's age must complete to reach the total of $30$.
Together Bella and the kitten account for $11 + 5 = 16$ of the $30$ years.

$$11 + 5 = 16$$

💡 Combining the two known ages into one running total is a one-step "put together" word problem within $100$.

#6 Guess and Check 2.OA.A.1 Step 3

Try each answer choice as the age gap between Anna and Bella, and check which one makes the three ages sum to $30$.
Bella is $11$, so a gap of $d$ means Anna is $11 + d$, and the total is $(11 + d) + 11 + 5 = 27 + d$.
We need $27 + d = 30$, so the right gap is the choice that gives $d = 3$ — i.e., $\textbf{(C)}$.
Checking: $14 + 11 + 5 = 30$.
✓ The other choices give totals $28, 29, 31, 32$, all off the target $30$.

$$d = 1 \Rightarrow 12 + 11 + 5 = 28; \; d = 2 \Rightarrow 13 + 11 + 5 = 29; \; \boxed{d = 3 \Rightarrow 14 + 11 + 5 = 30}; \; d = 4 \Rightarrow 31; \; d = 5 \Rightarrow 32$$

💡 Plugging each candidate gap into a one-step addition check is exactly the Grade 2 "add/subtract within $100$" word-problem skill.

#6 Guess and Check 1.OA.A.1 Step 4

Read off the answer.
The gap that makes the sum hit $30$ is $d = 3$, so Anna is $3$ years older than Bella.

$$\text{Anna} - \text{Bella} = 14 - 11 = 3 \;\Rightarrow\; \textbf{(C)}$$

💡 Finding the difference of two small whole numbers is a Grade 1 subtract-within-$20$ skill.

[1] #11 1.OA.A.1 Carry Bella's and the kitten's ages forward five years to today. Bella was $6$ f

[2] #11 2.OA.A.1 Add Bella's and the kitten's current ages so we know what Anna's age must comple

[3] #6 2.OA.A.1 Try each answer choice as the age gap between Anna and Bella, and check which on

[4] #6 1.OA.A.1 Read off the answer. The gap that makes the sum hit $30$ is $d = 3$, so Anna is

Review

Reasonableness: All three current ages are whole, positive, and reasonable: Bella $11$, kitten $5$, Anna $14$. Their sum is $14 + 11 + 5 = 30$, matching the problem exactly. The gap $3$ is also one of the listed choices, so the arithmetic and the multiple-choice structure agree.

Alternative: Tool #11 (Work Backwards) on its own: from the total $30$, peel off the two ages we already know — $30 - 11 - 5 = 14$ is Anna's age. Then $14 - 11 = 3$. Same answer with no guessing, just inverse arithmetic.

CCSS standards used (min grade 2)

1.OA.A.1 Solve addition and subtraction word problems within 20 (Rolling Bella's and the kitten's ages forward five years ($6 + 5 = 11$, $0 + 5 = 5$) and computing the final age difference $14 - 11 = 3$.)
2.OA.A.1 Solve one- and two-step word problems using addition and subtraction within 100 (Adding the two known ages ($11 + 5 = 16$) and checking each answer-choice gap against the target sum of $30$ — both are one-step add/subtract steps within $100$.)

⭐ This AMC 8 problem only needs Grade 2 add-and-subtract-within-100 word-problem skills you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 2)

More like this