AMC 8 · 1999 · #9
Easy mode Grade 4Problem
There are three flower beds: A, B, and C. They overlap in some places. (In the figure, bed A overlaps with bed B on one side and with bed C on the other side. Beds B and C do not touch each other.)
Here is how many plants each bed has:
- Bed A has 500 plants.
- Bed B has 450 plants.
- Bed C has 350 plants.
Some plants grow in the overlapping parts, so they get counted twice:
- 50 plants are in both A and B.
- 100 plants are in both A and C.
How many plants are there in total?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Three flower beds $A$, $B$, and $C$ overlap as shown in the picture. Bed $A$ holds $500$ plants, bed $B$ holds $450$, and bed $C$ holds $350$. Beds $A$ and $B$ share $50$ plants, beds $A$ and $C$ share $100$, and from the picture beds $B$ and $C$ do not touch. How many plants are there in all?
Givens: $|A| = 500$, $|B| = 450$, $|C| = 350$; Plants shared by $A$ and $B$: $|A \cap B| = 50$; Plants shared by $A$ and $C$: $|A \cap C| = 100$; From the diagram, beds $B$ and $C$ do not overlap, so $|B \cap C| = 0$ and $|A \cap B \cap C| = 0$; Answer choices: (A) $850$, (B) $1000$, (C) $1150$, (D) $1300$, (E) $1450$
Unknowns: The total number of distinct plants in the three beds, $|A \cup B \cup C|$
Understand
Restated: Three flower beds $A$, $B$, and $C$ overlap as shown in the picture. Bed $A$ holds $500$ plants, bed $B$ holds $450$, and bed $C$ holds $350$. Beds $A$ and $B$ share $50$ plants, beds $A$ and $C$ share $100$, and from the picture beds $B$ and $C$ do not touch. How many plants are there in all?
Givens: $|A| = 500$, $|B| = 450$, $|C| = 350$; Plants shared by $A$ and $B$: $|A \cap B| = 50$; Plants shared by $A$ and $C$: $|A \cap C| = 100$; From the diagram, beds $B$ and $C$ do not overlap, so $|B \cap C| = 0$ and $|A \cap B \cap C| = 0$; Answer choices: (A) $850$, (B) $1000$, (C) $1150$, (D) $1300$, (E) $1450$
Plan
Primary tool: #12 Draw a Venn Diagram
Secondary: #7 Identify Subproblems
The picture is already a Venn diagram: three regions for $A$, $B$, $C$ with $A$ overlapping both $B$ and $C$ but $B$ and $C$ kept apart. Tool #12 (Draw a Venn Diagram) lets us label each region with the right count and read the total straight from the picture, without memorizing a three-set formula. Tool #7 (Identify Subproblems) splits the work into two clean Grade 4 steps: (i) add the three bed totals as if nothing were shared, then (ii) subtract the overlap plants that got counted twice. Because $B$ and $C$ never touch, there is no triple-overlap to add back, which keeps the bookkeeping simple.
Execute — Answer: C
4.OA.A.3 Step 1 - Label the regions of the Venn diagram.
- The picture has bed $A$ (top rectangle) overlapping bed $B$ (right circle) and bed $C$ (left circle), with $B$ and $C$ apart.
- So the diagram has five regions: $A$-only, $B$-only, $C$-only, $A\cap B$, and $A\cap C$.
- There is no $B\cap C$ region and no triple region.
💡 Look at the picture before doing any arithmetic. The shape of the overlaps tells you which regions you need to track and which ones are empty.
4.NBT.B.4 Step 2 - Subproblem 1: pretend nothing is shared and add the three bed counts.
- This over-counts every shared plant exactly once, because each shared plant belongs to two beds and shows up in both totals.
💡 Adding the three sizes treats every plant as if it lived in only one bed. The shared plants get counted twice, so this sum is too big by exactly the amount of overlap.
4.NBT.B.4 Step 3 - Subproblem 2: subtract the overlap that got counted twice.
- The plants on the $A\cap B$ patch were counted in $A$ and again in $B$, so they need to be subtracted once.
- Same for $A\cap C$.
- There is no $B\cap C$ patch, so nothing more to remove.
💡 Each shared plant was counted twice in step 2; subtracting it once leaves it counted exactly once, which is what we want.
4.OA.A.3 Step 4 - Subtract the overlap from the over-count to get the true total.
- Because $B$ and $C$ do not touch, no plant is in all three beds, so there is no triple-overlap to add back.
💡 The Venn-diagram bookkeeping (add the parts, take back the double-count) lands on choice (C).
4.OA.A.3 Label the regions of the Venn diagram. The picture has bed $A$ (top rectangle) o 4.NBT.B.4 Subproblem 1: pretend nothing is shared and add the three bed counts. This over- 4.NBT.B.4 Subproblem 2: subtract the overlap that got counted twice. The plants on the $A\ 4.OA.A.3 Subtract the overlap from the over-count to get the true total. Because $B$ and Review
Reasonableness: Fill in every region of the Venn diagram and add: $A$-only $= 500 - 50 - 100 = 350$, $A\cap B = 50$, $A\cap C = 100$, $B$-only $= 450 - 50 = 400$, $C$-only $= 350 - 100 = 250$. Total $= 350 + 50 + 100 + 400 + 250 = 1150$, matching choice (C). The size also passes a quick estimate: the answer must be less than the naive sum $1300$ (because some plants are shared) and more than the largest single bed $500$, so the only plausible choice is $1150$.
Alternative: Tool #5 (Look for a Pattern / Use a Formula): apply the three-set inclusion-exclusion formula $|A\cup B\cup C| = |A| + |B| + |C| - |A\cap B| - |A\cap C| - |B\cap C| + |A\cap B\cap C|$. Plug in $500 + 450 + 350 - 50 - 100 - 0 + 0 = 1150$, the same answer (C). The Venn-diagram method is really this formula, drawn instead of memorized.
CCSS standards used (min grade 4)
4.OA.A.3Solve multistep word problems with whole numbers using the four operations (Modeling the three flower beds with a Venn diagram and chaining the add-then-subtract steps to land on the total $1150$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers using the standard algorithm (Carrying out $500 + 450 + 350 = 1300$, $50 + 100 + 0 = 150$, and $1300 - 150 = 1150$ with multi-digit accuracy.)
⭐ Add the three bed totals as if no plant were shared, then subtract the shared plants once so they aren't counted twice — $1300 - 150 = 1150$ turns this AMC 8 problem into a Grade 4 Venn-diagram exercise.
⭐ Add the three bed totals as if no plant were shared, then subtract the shared plants once so they aren't counted twice — $1300 - 150 = 1150$ turns this AMC 8 problem into a Grade 4 Venn-diagram exercise.
More like this
Same archetype — closest grade level first.
