Sensim Math Original · sm-10

Easy mode Grade 4

Problem

Imagine a library cart of new books. Every book on the cart is either a paperback or a hardcover.

This morning, the paperbacks and hardcovers on the cart were in the ratio $4 : 3$ (so for every $4$ paperbacks there were $3$ hardcovers).

During the day, two things happened to the cart:

$4$ paperbacks were checked out and taken off the cart.
$6$ donated hardcovers were placed on the cart.

By closing time, the new ratio of paperbacks to hardcovers had changed to $2 : 3$ .

How many books are on the cart at closing time?

$\textbf{(A) } 24\qquad\textbf{(B) } 26\qquad\textbf{(C) } 28\qquad\textbf{(D) } 30\qquad\textbf{(E) } 32$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: A library cart starts the morning with paperbacks and hardcovers in the ratio $4 : 3$. During the day $4$ paperbacks leave the cart (checked out) and $6$ hardcovers arrive (donations). At closing, the new paperback-to-hardcover ratio is $2 : 3$. We need the total number of books on the cart at closing.

Givens: Morning ratio paperbacks : hardcovers $= 4 : 3$ (4 paperbacks for every 3 hardcovers); $4$ paperbacks are removed during the day $\Rightarrow$ paperback count drops by $4$; $6$ hardcovers are added during the day $\Rightarrow$ hardcover count rises by $6$; Closing ratio paperbacks : hardcovers $= 2 : 3$; Answer choices: (A) 24, (B) 26, (C) 28, (D) 30, (E) 32

Unknowns: The total number of books on the cart at closing

Understand

Plan

Primary tool: #6 Guess and Check

Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities

The morning ratio $4 : 3$ means "for every $3$ hardcovers there are $4$ paperbacks," so once we pick the multiplier $k$, both starting counts are forced: paperbacks $= 4k$ and hardcovers $= 3k$. That makes Tool #6 (Guess & Check) the natural attack — try $k = 1, 2, 3, \dots$ in order, apply the $-4 / +6$ change, and check whether the closing ratio is exactly $2 : 3$. Tool #2 (Systematic List) keeps the candidates ordered so nothing is skipped, and Tool #3 (Eliminate Possibilities) finishes by matching the resulting total against the five offered answer choices. The whole path uses only multiplication, addition, and subtraction — no variables or equations needed.

Execute — Answer: D

#2 Make a Systematic List 4.OA.A.2 Step 1

Rewrite the morning ratio in a friendlier form.
"$4 : 3$" means "$3$ hardcovers come with $4$ paperbacks every time," a **multiplicative comparison**.
So pick a positive multiplier $k$ and the morning counts are forced: paperbacks $= 4k$ and hardcovers $= 3k$.
The possible morning pairs are $(4, 3), (8, 6), (12, 9), (16, 12), (20, 15), \dots$ — paperbacks is always $\tfrac{4}{3}$ of hardcovers, or equivalently both sit on the same $k$-bundle.

$$\text{morning } (P, H) \in \{(4,3),(8,6),(12,9),(16,12),(20,15),\dots\}\;\text{(both sides scale by the same } k)$$

💡 Turning a $4 : 3$ ratio into matching multiples of $4$ and $3$ is Grade 4 multiplicative comparison.

#2 Make a Systematic List 2.NBT.B.5 Step 2

Apply the day's change.
Paperbacks lose $4$ and hardcovers gain $6$, so the morning pair $(4k,\, 3k)$ becomes the closing pair $(4k - 4,\, 3k + 6)$ at the end of the day.
Notice both expressions are easy to compute from $k$ alone — no second unknown is needed.

$$\text{closing } (P_{\text{close}},\, H_{\text{close}}) = (4k - 4,\; 3k + 6)$$

💡 Subtracting $4$ and adding $6$ to small two-digit counts is straight Grade 2 add/subtract within 100.

#6 Guess and Check 4.OA.A.2 Step 3

Now guess values of $k$ and check.
For the closing ratio to be $2 : 3$, the hardcovers must be $\tfrac{3}{2}$ of the paperbacks, equivalently $3 \times P_{\text{close}} = 2 \times H_{\text{close}}$.
We will run $k = 1, 2, 3, 4, \dots$ in order and watch the ratio shrink toward $\tfrac{2}{3}$.

$$\text{check: } 3 \times (4k - 4) \;\stackrel{?}{=}\; 2 \times (3k + 6)$$

💡 Asking "is hardcover exactly $1.5$ times paperback?" is again a Grade 4 multiplicative comparison.

#6 Guess and Check 3.OA.C.7 Step 4

Step through the candidates in order, computing the closing pair and its ratio.
$k=1: (4,3)\to(0,9)$ — paperbacks vanish, ratio undefined, skip.
$k=2: (8,6)\to(4,12)$, ratio $4 : 12 = 1 : 3$ — hardcovers still too dominant.
$k=3: (12,9)\to(8,15)$, ratio $8 : 15$ — not a clean small ratio.
$k=4: (16,12)\to(12,18)$, ratio $12 : 18 = 2 : 3$ — **hit!** The paperback-to-hardcover ratio drops $1{:}3 \to 8{:}15 \to 2{:}3$, so $k=4$ is the first (and only positive-integer) match.

$$k=4:\; (16,12) \to (12,18),\;\; 12 : 18 = 2 : 3\;\checkmark$$

💡 Computing $4 \times 4 = 16$, $3 \times 4 = 12$, and reducing $12 : 18$ by dividing both by $6$ are all Grade 3 multiplication/division within 100.

#3 Eliminate Possibilities 2.NBT.B.5 Step 5

Answer the actual question: total books on the cart at closing.
Add the closing paperback and hardcover counts: $12 + 18 = 30$.
Among the choices $24, 26, 28, 30, 32$, this matches (D) exactly.
A quick sweep of the traps — the morning total is $16 + 12 = 28$ and the day's net change is $-4 + 6 = +2$, so the correct closing total is $28 + 2 = 30$.
(A) $24 = 28 - 4$ applies the sales but forgets the donations.
(B) $26 = 28 - 2$ flips the direction of the net change.
(C) $28$ is the morning total — the day's swap was ignored entirely.
(E) $32 = 28 + 4$ adds where it should have subtracted (and still forgets the donations).

$$12 + 18 = 30 \;\Rightarrow\; \textbf{(D)}$$

💡 Adding two two-digit numbers ($12 + 18$) is core Grade 2 addition within 100.

[1] #2 4.OA.A.2 Rewrite the morning ratio in a friendlier form. "$4 : 3$" means "$3$ hardcovers

[2] #2 2.NBT.B.5 Apply the day's change. Paperbacks lose $4$ and hardcovers gain $6$, so the morn

[3] #6 4.OA.A.2 Now guess values of $k$ and check. For the closing ratio to be $2 : 3$, the hard

[4] #6 3.OA.C.7 Step through the candidates in order, computing the closing pair and its ratio.

[5] #3 2.NBT.B.5 Answer the actual question: total books on the cart at closing. Add the closing

Review

Reasonableness: Sanity-check by re-reading the closing ratio $2 : 3$. Treat the closing cart as $5$ equal bundles: $2$ bundles paperbacks and $3$ bundles hardcovers. Each bundle has $12 \div 2 = 6$ books, so total $= 5 \times 6 = 30$, agreeing with the direct sum $12 + 18 = 30$. A second check: paperbacks went $16 \to 12$ (a drop of $4$ ✓) and hardcovers went $12 \to 18$ (a gain of $6$ ✓), so both color counts respect the day's swap. Net change in total is $-4 + 6 = +2$, taking the morning total $28$ to a closing total $30$ — consistent.

Alternative: Tool #13 (Convert to Algebra) is the standard alternative. Let the morning multiplier be $k$, so $(P, H) = (4k,\, 3k)$ and the closing pair is $(4k - 4,\, 3k + 6)$. Set $\frac{4k-4}{3k+6} = \frac{2}{3}$, cross-multiply to get $3(4k-4) = 2(3k+6)$, i.e. $12k - 12 = 6k + 12$, so $6k = 24$ and $k = 4$. The total is $7k + 2 = 30$. Same answer, but for an elementary student the guess-and-check + systematic list combo above introduces no variables and stays inside familiar multiplication tables.

CCSS standards used (min grade 4)

2.NBT.B.5 Fluently add and subtract within 100 (Applying the day's $-4 / +6$ change to the morning pair and computing the final total $12 + 18 = 30$.)
3.OA.C.7 Fluently multiply and divide within 100 (Computing the morning $(4k, 3k)$ pairs for $k = 1, 2, 3, 4$ and reducing $12 : 18$ to $2 : 3$ by dividing both terms by $6$.)
4.OA.A.2 Multiply or divide to solve word problems involving multiplicative comparison (Reading "$4 : 3$" and "$2 : 3$" as multiplicative comparisons so each candidate $k$ can be checked by multiplication instead of by setting up equations.)

⭐ This problem only needs Grade 4 multiplicative comparison ("$\sim$ times as many") you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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