Sensim Math Original · sm-13
SM Original Grade 4 countingProblem
A community library labels every book on its featured shelf in two ways: by genre (fiction or nonfiction) and by binding (hardcover or paperback). The featured shelf holds exactly 24 books. Five eighths of them are fiction, and three fourths of them are hardcover. Some fraction of the shelf consists of books that are both fiction and hardcover. Across all possible ways the shelf could be arranged consistent with these counts, what is the smallest value this fraction can take?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A featured library shelf has $24$ books. Each book is tagged twice: by genre (fiction or nonfiction) and by binding (hardcover or paperback). Exactly $\tfrac{5}{8}$ are fiction and exactly $\tfrac{3}{4}$ are hardcover. We want the **smallest** fraction of the $24$ books that can be both fiction AND hardcover, expressed in lowest terms and matched to one of the choices.
Givens: Total books on the shelf: $24$; Fiction books: $\tfrac{5}{8}$ of the total; the rest are nonfiction; Hardcover books: $\tfrac{3}{4}$ of the total; the rest are paperback; Answer choices: $(A)\ 0,\ (B)\ \tfrac{1}{6},\ (C)\ \tfrac{1}{4},\ (D)\ \tfrac{3}{8},\ (E)\ \tfrac{5}{12}$
Unknowns: The **smallest** possible count of books that are both fiction AND hardcover, divided by $24$
Understand
Restated: A featured library shelf has $24$ books. Each book is tagged twice: by genre (fiction or nonfiction) and by binding (hardcover or paperback). Exactly $\tfrac{5}{8}$ are fiction and exactly $\tfrac{3}{4}$ are hardcover. We want the **smallest** fraction of the $24$ books that can be both fiction AND hardcover, expressed in lowest terms and matched to one of the choices.
Givens: Total books on the shelf: $24$; Fiction books: $\tfrac{5}{8}$ of the total; the rest are nonfiction; Hardcover books: $\tfrac{3}{4}$ of the total; the rest are paperback; Answer choices: $(A)\ 0,\ (B)\ \tfrac{1}{6},\ (C)\ \tfrac{1}{4},\ (D)\ \tfrac{3}{8},\ (E)\ \tfrac{5}{12}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #16 Change Focus / Count the Complement, #6 Guess and Check
Two cross-cutting labels (genre and binding) sit on the same population, which is the signature trigger for a $2 \times 2$ table — Tool #1 lets us see all four cells at once. Asking "make the fiction-hardcover cell as **small** as possible" is awkward to attack directly, so Tool #16 (Change Focus) re-aims the question at "make the fiction-**paperback** cell as **large** as possible," which has obvious ceilings. Tool #6 (Guess and Check) then verifies the candidate value by plugging it back into the table and confirming every cell stays a non-negative integer. We deliberately avoid Tool #13 (Algebra) — the table + complement trick is enough at this grade level.
Execute — Answer: D
4.NF.B.4 Step 1 - Translate the two fractional clues into concrete counts.
- "$\tfrac{5}{8}$ of the $24$ books are fiction" means split the $24$ into $8$ equal groups and take $5$ of them: $\tfrac{5}{8} \times 24 = 15$ fiction.
- Likewise "$\tfrac{3}{4}$ are hardcover" means split into $4$ equal groups and take $3$: $\tfrac{3}{4} \times 24 = 18$ hardcover.
- The leftovers come from subtraction: nonfiction $= 24 - 15 = 9$ and paperback $= 24 - 18 = 6$.
💡 Taking $\tfrac{5}{8}$ or $\tfrac{3}{4}$ of a whole number to find "how many" is exactly the Grade 4 idea of multiplying a fraction by a whole number.
1.MD.C.4 Step 2 - Build a $2 \times 2$ table with genre as the rows and binding as the columns, and write the four edge totals (fiction $15$, nonfiction $9$, hardcover $18$, paperback $6$).
- Let $x$ be the number of books that are both fiction AND hardcover.
- Then the other three cells are forced by the row and column sums: fiction-paperback $= 15 - x$, nonfiction-hardcover $= 18 - x$, and nonfiction-paperback $= x - 9$ (so that the nonfiction row sums to $9$ and the paperback column sums to $6$).
- The picture turns the requirement "every cell is a non-negative integer" into an at-a-glance check.
💡 Sorting items by two labels (genre and binding) into the four cells of a table is the Grade 1 "organize, represent, and interpret data with up to three categories" skill.
2.NBT.B.5 Step 3 - Directly minimizing $x$ is awkward, so **flip the focus**: in the fiction row, making $x$ small is the same as making fiction-paperback $15 - x$ **large**.
- Fiction-paperback sits inside two pools at once, so it cannot exceed either pool: it is at most the $6$ paperbacks that exist, and at most the $15$ fiction books that exist.
- The tighter ceiling is $\min(6, 15) = 6$, so fiction-paperback $\le 6$, i.e.
- $15 - x \le 6$, which gives $x \ge 15 - 6 = 9$.
💡 Swapping a hard "minimize" question for the easier "maximize the leftover" question, then doing $15 - 6 = 9$, is a Grade 2 subtraction-within-$100$ move.
2.NBT.B.5 Step 4 - Guess and check the small candidates $x = 0, 1, \ldots, 9$ by reading off the cell formulas.
- For any $x < 9$, the nonfiction-paperback cell $x - 9$ would be negative — already a contradiction.
- Equivalently, fiction-paperback $15 - x$ would exceed the $6$ paperbacks available.
- So every candidate below $9$ fails.
- For $x = 9$: fiction-paperback $= 6$, nonfiction-hardcover $= 9$, nonfiction-paperback $= 0$, all non-negative — the table closes.
💡 Plugging candidate values in and checking each subtraction for non-negativity is Grade 2 subtraction-within-$100$ work.
4.NF.A.1 Step 5 - Divide the minimum count $9$ by the total $24$ to form the fraction: $\tfrac{9}{24}$.
- Both numerator and denominator share a factor of $3$, so reduce: $\tfrac{9}{24} = \tfrac{3}{8}$.
- Compare to the choices: $\tfrac{3}{8}$ is exactly (D).
- The other options each fail.
- (A) $0$ would need $x = 0$, but we showed $x \ge 9$.
- (B) $\tfrac{1}{6} = \tfrac{4}{24}$ needs $x = 4 < 9$.
- (C) $\tfrac{1}{4} = \tfrac{6}{24}$ needs $x = 6 < 9$.
- (E) $\tfrac{5}{12} = \tfrac{10}{24}$ corresponds to $x = 10$, which is feasible but larger than the minimum, so it is not the **smallest** value.
💡 Reducing $\tfrac{9}{24}$ to $\tfrac{3}{8}$ by dividing top and bottom by the common factor $3$ is the standard Grade 4 equivalent-fraction move.
4.NF.B.4 Translate the two fractional clues into concrete counts. "$\tfrac{5}{8}$ of the 1.MD.C.4 Build a $2 \times 2$ table with genre as the rows and binding as the columns, an 2.NBT.B.5 Directly minimizing $x$ is awkward, so **flip the focus**: in the fiction row, m 2.NBT.B.5 Guess and check the small candidates $x = 0, 1, \ldots, 9$ by reading off the ce 4.NF.A.1 Divide the minimum count $9$ by the total $24$ to form the fraction: $\tfrac{9}{ Review
Reasonableness: Re-add the $x = 9$ table by rows and columns: row sums give $9 + 6 = 15$ (fiction) and $9 + 0 = 9$ (nonfiction); column sums give $9 + 9 = 18$ (hardcover) and $6 + 0 = 6$ (paperback). All four edge totals match exactly, so the configuration is valid. The minimum being nonzero also matches a pigeonhole feel: $15$ fiction books cannot all hide among the $6$ paperbacks, so at least $15 - 6 = 9$ fiction books are forced to be hardcover. As a fraction, $\tfrac{3}{8} = 0.375$ sits sensibly between $\tfrac{1}{4} = 0.25$ and $\tfrac{5}{12} \approx 0.417$, which is the expected neighborhood for "smallest possible" given the tight $\tfrac{3}{4}$-hardcover constraint.
Alternative: Tool #12 (Venn diagram) gives the same answer. Draw circles "Fiction" and "Hardcover" inside a universe of $24$ books. If the intersection is $x$, then fiction-only $= 15 - x$, hardcover-only $= 18 - x$, and the outside region (nonfiction paperback) $= 24 - (15 + 18 - x) = x - 9$. Requiring that outside region $\ge 0$ immediately gives $x \ge 9$, so the minimum fraction is $\tfrac{9}{24} = \tfrac{3}{8}$. Same conclusion via inclusion-exclusion instead of the $2 \times 2$ table.
CCSS standards used (min grade 4)
1.MD.C.4Organize, represent, and interpret data with up to three categories (Sorting the books by the two labels (genre and binding) into the four cells of a $2 \times 2$ table.)2.NBT.B.5Fluently add and subtract within 100 (Filling the table with $24 - 15 = 9,\ 24 - 18 = 6,\ 15 - 6 = 9$ and checking each candidate $x$ for non-negative cells.)4.NF.A.1Explain why a fraction is equivalent to another fraction (Reducing $\tfrac{9}{24}$ to its lowest-terms form $\tfrac{3}{8}$ by dividing numerator and denominator by $3$ to match the answer choice.)4.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a whole number (Computing $\tfrac{5}{8} \times 24 = 15$ fiction books and $\tfrac{3}{4} \times 24 = 18$ hardcover books.)
⭐ This problem only needs Grade 4 "fraction of a whole number" plus a simple 2-by-2 table you already know!
⭐ This problem only needs Grade 4 "fraction of a whole number" plus a simple 2-by-2 table you already know!
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