Aimathic – Multiply two-digit numbers

Multiply two-digit numbers

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Difficulty: 1 – 5

5167304

Convert each time to the smaller unit. a) How many minutes are in $12$ hours? b) How many seconds are in $25$ minutes?

Hints

- How many minutes are in $1$ hour? - Can you break the two-digit factor into tens and ones? - How many seconds are in $1$ minute?

Solution

1. a) One hour is $60$ minutes, so calculate $12 \times 60$. Using partial products, $10 \times 60=600$ and $2 \times 60=120$. Then $600+120=720$, so $12$ hours is $720$ minutes. 2. b) One minute is $60$ seconds, so calculate $25 \times 60$. Using partial products, $20 \times 60=1200$ and $5 \times 60=300$. Then $1200+300=1500$, so $25$ minutes is $1500$ seconds.

Answer

a) $720\,\text{minutes}$ b) $1500\,\text{seconds}$

5190914

A watercolor paint set costs $\$14$ at an art supply store. a) How much do $6$ sets cost? b) How much do $20$ sets cost? c) Find the cost of $26$ sets. Explain how you can use your answers from parts a) and b).

Hints

- Break $26$ into tens and ones. - Use the costs for $20$ sets and $6$ sets to find the cost for $26$ sets. - Calculate the easier products first.

Solution

1. For a), $6 \times \$14=\$84$. 2. For b), $20 \times \$14=\$280$. 3. Since $26=20+6$, add the two costs: $\$280+\$84=\$364$.

Answer

a) $6$ sets cost $\$84$. b) $20$ sets cost $\$280$. c) $26$ sets cost $\$364$. Add the answers from parts a) and b) because $26=20+6$.

5191134

Calculate each product. a) $23\times12$ b) $31\times14$ c) $15\times22$

Hints

- Break one factor into tens and ones. - Find the partial products. - Add the partial products using correct place-value alignment.

Solution

1. For a), $23 \times 10=230$ and $23 \times 2=46$. Then $230+46=276$. 2. For b), $31 \times 10=310$ and $31 \times 4=124$. Then $310+124=434$. 3. For c), $15 \times 20=300$ and $15 \times 2=30$. Then $300+30=330$.

Answer

a) $276$ b) $434$ c) $330$

5191714

Two students write a story problem for $15 \times 12$. Lukas writes: “A movie theater has $15$ rows with $12$ seats in each row. How many seats are there?” Marie writes: “I have $\$15$ and buy a toy for $\$12$. How much money remains?” Decide which story matches the multiplication expression, then calculate the product.

Hints

- Multiplication represents equal groups or repeated quantities. - Decide whether each story combines equal groups or finds an amount left. - Break $12$ into $10 + 2$ to calculate.

Solution

1. The expression $15 \times 12$ represents $15$ equal groups of $12$. 2. Lukas’s story describes equal groups, so it matches multiplication. Marie’s story describes subtraction. 3. Use partial products: $15 \times 12 = 15 \times 10 + 15 \times 2 = 150 + 30 = 180$.

Answer

Lukas wrote the matching story. The product is $180$.

5192274

A toy store sells $35$ remote-control cars in June for $\$24$ each. In July, it sells $53$ of the same cars at the same price. Find the revenue for each month.

Hints

- Multiply the number sold by the price of each car. - Break $24$ into tens and ones. - Add the partial products with their place values aligned.

Solution

1. June: $35 \times \$24$. Use partial products: $35 \times \$20=\$700$ and $35 \times \$4=\$140$. The June revenue is $\$840$. 2. July: $53 \times \$24$. Use partial products: $53 \times \$20=\$1060$ and $53 \times \$4=\$212$. The July revenue is $\$1272$.

Answer

The June revenue is $\$840$, and the July revenue is $\$1272$.

5194124

A school library orders $16$ nonfiction books at $\$18$ each. Break $16$ into $10+6$. Give the partial product for $10\times18$, the partial product for $6\times18$, and then the total cost.

Hints

- Use the required decomposition $16=10+6$. - Find the cost for each part of the decomposition separately. - Add the two partial products to get the total.

Solution

1. $10\times18=180$, so ten books cost $\$180$. 2. $6\times18=108$, so six books cost $\$108$. 3. Add the partial products: $180+108=288$. 4. The total cost is $\$288$.

Answer

$10\times18=180$ $6\times18=108$ Total: $\$288$

5202554

Find each product. a) $16\times11$ b) $27\times11$ c) $35\times11$ d) $52\times11$

Hints

- Break $11$ into two numbers that are easy to multiply by. - Start with ten times the number. - Determine what must be added to that product.

Solution

1. Use $n \times 11 = n \times 10 + n$. 2. Part a: $16 \times 11 = 160 + 16 = 176$. 3. Part b: $27 \times 11 = 270 + 27 = 297$. 4. Part c: $35 \times 11 = 350 + 35 = 385$. 5. Part d: $52 \times 11 = 520 + 52 = 572$.

Answer

a) $176$ b) $297$ c) $385$ d) $572$

5203994

Use partial products to calculate each product. Decompose one factor into tens and ones. a) $12 \times 13$ b) $16 \times 12$ c) $14 \times 15$

Hints

- Split one factor into tens and ones. - Multiply the other factor by each part. - Add the partial products.

Solution

1. For a), $12 \times (10 + 3) = 120 + 36 = 156$. 2. For b), $16 \times (10 + 2) = 160 + 32 = 192$. 3. For c), $14 \times (10 + 5) = 140 + 70 = 210$.

Answer

a) $12 \times (10 + 3) = 120 + 36 = 156$ b) $16 \times (10 + 2) = 160 + 32 = 192$ c) $14 \times (10 + 5) = 140 + 70 = 210$

5363334

Complete this product wall. Each upper brick is the product of the two bricks directly below it.

Hints

- Find the missing factor that makes $15$. - Then multiply the two second-row values.

Solution

1. For the middle bottom value, solve $x \times 3 = 15$, so $x = 15 \div 3 = 5$. 2. The left brick in the second row is $2 \times 5 = 10$. 3. The top is $10 \times 15 = 150$.

Answer

Bottom row: $2$, $5$, $3$ Second row: $10$, $15$ Top: $150$

5544854

Read the written multiplication. a) What product is shown? b) Why does the second partial-product row represent $230$, even though the row is based on $23\times1$?

Hints

- Read the multiplier by place value, not just as separate digits. - The first row comes from the ones digit; the second comes from the tens digit. - Notice how the written layout shifts the tens partial product one place left.

Solution

1. The calculation is $23\times14$. 2. The ones partial product is $23\times4=92$. 3. The tens digit $1$ in $14$ represents $10$, so the second partial product represents $23\times10=230$. 4. $92+230=322$.

Answer

a) $322$ b) The $1$ is in the tens place, so that row has a value ten times $23$.

5544864

Use the written multiplication to write an equivalent distributive equation that uses the place values in the multiplier $21$. Then evaluate the two terms and show how they combine to make the product.

Hints

- Rewrite $21$ by place value before using the rows. - Match each written row to one term of the distributive equation.

Solution

1. The multiplier $21$ is $20+1$. 2. Therefore, $42\times21=(42\times20)+(42\times1)$. 3. The two place-value partial products are $840$ and $42$. 4. $840+42=882$.

Answer

$42\times21=(42\times20)+(42\times1)=840+42=882$.

5544874

The multiplier in the written calculation has a zero in the ones place. a) What does the zero partial-product row mean? b) Find the product.

Hints

- Read both digits of the multiplier, including the zero. - A zero digit contributes a zero partial product. - The $2$ is in the tens place, so its row has tens-place value.

Solution

1. The calculation is $36\times20$. 2. The ones digit is $0$, so $36\times0=0$; that is why a real zero row appears. 3. The tens digit $2$ represents $20$, so the other row represents $36\times20=720$. 4. The product is $720$.

Answer

a) It represents $36\times0=0$. b) $720$

5160534

Fill in the blanks. In each part, use the products in the first two lines to find the product in the third line. a) $10 \times 15 = \dots$ $3 \times 15 = \dots$ $13 \times 15 = \dots$ b) $10 \times 17 = \dots$ $7 \times 17 = \dots$ $17 \times 17 = \dots$

Hints

- Look at how the first factor in the third line is made from the first factors in the two lines above it. - Can you add the two partial products? - For $7 \times 17$, break $17$ into tens and ones.

Solution

1. For a), $10 \times 15 = 150$ and $3 \times 15 = 45$. Since $13 = 10 + 3$, add the partial products: $150 + 45 = 195$. 2. For b), $10 \times 17 = 170$. Also, $7 \times 17 = 7 \times 10 + 7 \times 7 = 70 + 49 = 119$. Since $17 = 10 + 7$, add the partial products: $170 + 119 = 289$.

Answer

a) $10 \times 15 = 150$, $3 \times 15 = 45$, $13 \times 15 = 195$ b) $10 \times 17 = 170$, $7 \times 17 = 119$, $17 \times 17 = 289$

5162014

For a school festival, benches are arranged in $12$ rows with $24$ seats in each row. How many people can sit on the benches?

Hints

- Break $12$ and $24$ into tens and ones. - Use the decomposed parts to label the rows and columns of an area model. - Add all four partial products.

Solution

1. Decompose the factors: $12 = 10 + 2$ and $24 = 20 + 4$. 2. Find the four partial products in the area model: $10 \times 20 = 200$, $10 \times 4 = 40$, $2 \times 20 = 40$, and $2 \times 4 = 8$. 3. Add the partial products: $200 + 40 + 40 + 8 = 288$.

Answer

A total of $288$ people can sit on the benches.

5167394

Calculate each product. a) $24\times56$ b) $37\times42$ c) $58\times19$

Hints

- Reversing the order of the factors uses the commutative property of multiplication. - Break one factor into tens and ones to find partial products. - The original expression and its reversed expression should have the same value.

Solution

1. For a), $24 \times 56 = 24 \times 50 + 24 \times 6 = 1200 + 144 = 1344$. Reversing the factors gives $56 \times 24 = 56 \times 20 + 56 \times 4 = 1120 + 224 = 1344$. 2. For b), $37 \times 42 = 37 \times 40 + 37 \times 2 = 1480 + 74 = 1554$. Reversing the factors gives $42 \times 37 = 42 \times 30 + 42 \times 7 = 1260 + 294 = 1554$. 3. For c), $58 \times 19 = 58 \times 10 + 58 \times 9 = 580 + 522 = 1102$. Reversing the factors gives $19 \times 58 = 19 \times 50 + 19 \times 8 = 950 + 152 = 1102$.

Answer

a) $1344$ b) $1554$ c) $1102$

5167404

Calculate each product. Four expressions have the same value. Which expression does not belong? A: $28\times45$ B: $35\times36$ C: $20\times63$ D: $42\times30$ E: $32\times40$

Hints

- Calculate each product carefully using partial products. - Reverse the order of the factors to check a result. - Compare all five products.

Solution

1. For A, $28 \times 45 = 28 \times 40 + 28 \times 5 = 1120 + 140 = 1260$. Reversing the factors gives $45 \times 28 = 45 \times 20 + 45 \times 8 = 900 + 360 = 1260$. 2. For B, $35 \times 36 = 35 \times 30 + 35 \times 6 = 1050 + 210 = 1260$. Reversing the factors gives $36 \times 35 = 36 \times 30 + 36 \times 5 = 1080 + 180 = 1260$. 3. For C, $20 \times 63 = 20 \times 60 + 20 \times 3 = 1200 + 60 = 1260$. Reversing the factors gives $63 \times 20 = 1260$. 4. For D, $42 \times 30 = 1260$. Reversing the factors gives $30 \times 42 = 30 \times 40 + 30 \times 2 = 1200 + 60 = 1260$. 5. For E, $32 \times 40 = 1280$. Reversing the factors gives $40 \times 32 = 40 \times 30 + 40 \times 2 = 1200 + 80 = 1280$. 6. Expressions A, B, C, and D equal $1260$, but expression E equals $1280$.

Answer

E: $32 \times 40 = 1280$ does not belong because the other four products equal $1260$.

5167414

Two produce companies are packing apples. Green Orchard Produce packs $18$ boxes with $24$ apples in each box. Market Lane Produce packs $24$ boxes with $18$ apples in each box. a) Find the total number of apples packed by each company. b) What do you notice about the totals? Explain why this happens.

Hints

- Write a multiplication expression for each company. - Compare the factors in the two expressions. - What property says that reversing the factors does not change a product?

Solution

1. Green Orchard Produce: $18\times24=18\times20+18\times4=360+72=432$. 2. Market Lane Produce: $24\times18=24\times10+24\times8=240+192=432$. 3. The totals are equal because multiplication is commutative: $18\times24=24\times18$.

Answer

a) Each company packs $432$ apples. b) The totals are equal because the factors are reversed, and multiplication has the commutative property.

5167534

Use efficient strategies for these problems. a) Calculate $24 \times 2$ and $24 \times 20$. b) Explain how you can use those two products to find $24 \times 22$. c) Calculate $24 \times 44$. How is this product related to your answer in part b)?

Hints

- How can you write $22$ and $44$ as sums of tens and ones? - Use the distributive property to split a product into easier partial products. - What happens to a product when one factor is doubled?

Solution

1. For a), $24 \times 2 = 48$ and $24 \times 20 = 480$. 2. For b), use the distributive property: $24 \times 22 = 24 \times (20+2)=480+48=528$. 3. For c), $24 \times 44 = 1056$. Because $44$ is twice $22$, the product $1056$ is twice $528$.

Answer

a) $48$ and $480$ b) Add the two products: $480+48=528$, so $24 \times 22=528$. c) $24 \times 44=1056$. This is twice $528$ because $44$ is twice $22$.

5167554

Use the digit cards $2,3,5,6$ exactly once to form two two-digit factors. Which multiplication equation has the greatest product? Explain how you chose the tens digits.

Hints

- Decide which place value has the greatest effect on a two-digit number. - Put the two greatest digits in those places. - Calculate both possible arrangements of the remaining ones digits. - Compare the two products.

Solution

1. To make the factors large, place the two greatest digits, $6$ and $5$, in the tens places. 2. The remaining arrangements are $62\times 53$ and $63\times 52$. 3. Their products are $3286$ and $3276$, respectively. 4. Since $3286>3276$, the greatest product is $62\times 53=3286$.

Answer

$62\times 53=3286$ gives the greatest product. The digits $6$ and $5$ belong in the tens places because those places have the greatest effect on the values of the factors.

5167584

Use the digit cards $1,2,4,6$ exactly once to form two two-digit factors. Which arrangement gives the smallest product? Calculate the product.

Hints

- Decide whether the tens digits should be large or small. - Place the two smallest digits in the tens places. - Test both arrangements of the remaining digits. - Compare the products.

Solution

1. To make the product small, place the two smallest digits, $1$ and $2$, in the tens places. 2. The remaining arrangements are $14\times 26$ and $16\times 24$. 3. Their products are $364$ and $384$. 4. Therefore, the smallest product is $14\times 26=364$.

Answer

$14\times 26=364$ gives the smallest product. The reversed equation $26\times 14=364$ is equivalent.

5170184

The product $12\times 40$ is $480$. Write three other multiplication equations with a product of $480$. Change the factors in ways that keep the product unchanged.

Hints

- Try doubling one factor and halving the other. - Try multiplying one factor by $4$ and dividing the other by $4$. - Check each new product.

Solution

1. If one factor is divided by a number while the other factor is multiplied by the same number, the product stays unchanged. 2. Halving $12$ and doubling $40$ gives $6\times 80=480$. 3. Doubling $12$ and halving $40$ gives $24\times 20=480$. 4. Multiplying $12$ by $4$ and dividing $40$ by $4$ gives $48\times 10=480$.

Answer

One possible answer is: $6\times 80=480$ $24\times 20=480$ $48\times 10=480$

5170204

Consider $28\times 15$. Which expressions have the same value? a) $14\times 30$ b) $56\times 30$ c) $7\times 60$ d) $140\times 3$ Justify your choices without fully calculating every product.

Hints

- Compare each new first factor with $28$. - Compare each new second factor with $15$. - Decide whether the two factor changes compensate for each other. - Remember that changing both factors in the same direction does not keep the product fixed.

Solution

1. In part a, one factor is halved and the other is doubled, so the product stays unchanged. 2. In part b, both factors are doubled, so the product becomes four times as large. 3. In part c, one factor is divided by $4$ and the other is multiplied by $4$, so the product stays unchanged. 4. In part d, one factor is multiplied by $5$ and the other is divided by $5$, so the product stays unchanged. 5. Therefore, parts a, c, and d have the same value as $28\times 15$.

Answer

a), c), and d) have the same value as $28\times 15$.

5183284

In a school garden, the Green Team plants $12$ rows with $25$ carrots in each row. The Yellow Team plants $14$ rows with $22$ carrots in each row. Which team plants more carrots, and how many more?

Hints

- Find each team's total separately. - Break apart the two-digit factors if helpful. - Compare the products and find their difference.

Solution

1. Find the Green Team's total: $12 \times 25 = 300$. 2. Find the Yellow Team's total: $14 \times 22 = 308$. 3. Since $308 > 300$, the Yellow Team plants more. 4. Find the difference: $308 - 300 = 8$.

Answer

The Yellow Team plants $8$ more carrots than the Green Team.

5183524

A store owner buys $15$ boxes of frozen treats for $\$12$ per box. She calculates mentally by finding $15 \times 10$ and $15 \times 2$, then adding the partial products. How much does she pay in all? Write the calculation she uses.

Hints

- Break $12$ into $10 + 2$. - Find each partial product. - Add the partial products to find the total.

Solution

1. Decompose $12$ as $10 + 2$: $15 \times 12 = 15 \times (10 + 2)$. 2. Find the partial products: $15 \times 10 = 150$ and $15 \times 2 = 30$. 3. Add them: $150 + 30 = 180$. The total cost is $\$180$.

Answer

$15 \times 12 = 15 \times 10 + 15 \times 2 = 150 + 30 = 180$. She pays $\$180$.

5184394

A school festival committee buys prizes for a game booth. It orders $15$ packages with $24$ small stuffed animals in each package. Each stuffed animal costs $\$3$. It also orders $5$ boxes with $120$ colored pencils in each box. Each pencil costs $\$1$. What is the total cost of all the prizes?

Hints

- Find the total number of each kind of prize. - Find the cost of each kind separately. - Add the two costs.

Solution

1. The committee buys $15 \times 24 = 360$ stuffed animals. 2. The stuffed animals cost $360 \times \$3 = \$1080$. 3. The committee buys $5 \times 120 = 600$ colored pencils. 4. The colored pencils cost $600 \times \$1 = \$600$. 5. The total cost is $\$1080 + \$600 = \$1680$.

Answer

All the prizes cost $\$1680$.

5187274

A school library receives $12$ boxes with $15$ nonfiction books in each box and $8$ boxes with $24$ fiction books in each box. Which type has more books, and what is the difference?

Hints

- Find the total number of each type of book. - Break apart the factors if helpful. - Compare the products and subtract to find the difference.

Solution

1. Find the number of nonfiction books: $12 \times 15 = 180$. 2. Find the number of fiction books: $8 \times 24 = 192$. 3. Find the difference: $192 - 180 = 12$.

Answer

The library receives $12$ more fiction books than nonfiction books.

5187434

Two classes prepare drinks for a school fair. Class A has $12$ cases with $12$ bottles of apple juice in each case. Class B has $9$ cases with $15$ bottles of orange juice in each case. Which class prepares more bottles, and what is the difference?

Hints

- Find each class's total separately. - Break apart the factors if helpful. - Compare the products and subtract.

Solution

1. Find Class A's total: $12 \times 12 = 144$. 2. Find Class B's total: $9 \times 15 = 135$. 3. Since $144 > 135$, Class A prepares more bottles. 4. Find the difference: $144 - 135 = 9$.

Answer

Class A prepares $9$ more bottles than Class B.

5188344

A school orders $15$ packages of notebooks. Each package contains $10$ notebooks. a) How many notebooks does the school receive altogether? b) Each package costs $\$12$. What is the total cost of the order?

Hints

- Multiply the number of packages by the number of notebooks in each package. - For the total cost, multiply the number of packages by the price of one package. - Break $12$ into $10 + 2$ if that helps.

Solution

1. Find the number of notebooks: $15 \times 10 = 150$. 2. Find the total cost: $15 \times \$12 = 15 \times (\$10 + \$2) = \$150 + \$30 = \$180$.

Answer

a) $150$ notebooks b) $\$180$

5190494

A bicycle shop has a $\$5000$ budget for new accessories. It orders $45$ helmets at $\$38$ each and $62$ bike locks at $\$24$ each. Is the budget enough? Find the amount left over or still needed.

Hints

- Find the cost of each product group. - Add the two costs. - Compare the total with the budget and subtract to find the difference.

Solution

1. The helmets cost $45 \times \$38 = \$1710$. 2. The bike locks cost $62 \times \$24 = \$1488$. 3. The total order costs $\$1710 + \$1488 = \$3198$. 4. Since $\$3198 < \$5000$, the budget is enough. 5. The amount left is $\$5000 - \$3198 = \$1802$.

Answer

Yes. The order costs $\$3198$, and $\$1802$ remains.

5191144

Calculate and compare the products. Replace each blank with $<$, $>$, or $=$. a) $36 \times 25\;\square\;45 \times 20$ b) $58 \times 14\;\square\;42 \times 19$

Hints

- Calculate each product before comparing. - Compare the exact values. - Recall the meanings of $<$, $>$, and $=$.

Solution

1. For a), $36 \times 25=900$ and $45 \times 20=900$, so the products are equal. 2. For b), $58 \times 14=812$ and $42 \times 19=798$, so the left product is greater.

Answer

a) $=$ b) $>$

5191224

Calculate each product. Then add the three products. a) $27 \times 43$ b) $52 \times 19$ c) $36 \times 36$ What is the total?

Hints

- Use partial products for each multiplication. - Keep the place values aligned. - Add all three products after calculating them. - Estimate to check whether the total is reasonable.

Solution

1. For a), use partial products: $27 \times 40=1080$ and $27 \times 3=81$. Then $1080+81=1161$. 2. For b), use partial products: $52 \times 10=520$ and $52 \times 9=468$. Then $520+468=988$. 3. For c), use partial products: $36 \times 30=1080$ and $36 \times 6=216$. Then $1080+216=1296$. 4. Add the products: $1161+988+1296=3445$.

Answer

a) $1161$ b) $988$ c) $1296$ The total is $3445$.

5191764

The first factor in a multiplication expression is $14$. The second factor is $6$ times the first factor. Find the product of the two factors.

Hints

- First use the given relationship to find the second factor. - Then multiply the two factors. - This problem requires two multiplication steps.

Solution

1. Find the second factor: $14 \times 6=84$. 2. Multiply the two factors: $14 \times 84=1176$.

Answer

The product is $1176$.

5192144

A school auditorium has $24$ rows with $16$ seats in each row. Are there enough seats for all $400$ students? Justify your answer with calculations.

Hints

- Multiply the number of rows by the number of seats in each row. - Compare the total number of seats with $400$. - If there are not enough seats, subtract to find how many are missing.

Solution

1. Find the number of seats: $24 \times 16$. 2. Use partial products: $24 \times 10=240$ and $24 \times 6=144$. 3. Add: $240+144=384$ seats. 4. Compare: $384<400$. There are $400-384=16$ fewer seats than students.

Answer

No. There are $384$ seats, so $16$ more seats are needed for $400$ students.

5192464

A landscaper buys $40$ bags of sand for $\$160$. One bag of specialty soil costs $12$ times as much as one bag of sand. The landscaper needs $35$ bags of specialty soil and pays a $\$45$ delivery fee. What is the total cost of the specialty soil and delivery?

Hints

- Use a multiplication fact to find the price of one bag of sand. - Use the multiplicative comparison to find the soil price per bag. - Find the cost of all soil bags, then add delivery.

Solution

1. Since $40 \times \$4 = \$160$, one bag of sand costs $\$4$. 2. One bag of specialty soil costs $12 \times \$4 = \$48$. 3. The specialty soil costs $35 \times \$48 = \$1680$. 4. With delivery, the total is $\$1680 + \$45 = \$1725$.

Answer

The specialty soil and delivery cost $\$1725$ altogether.

5199064

A school receives $20$ cases with $12$ bottles of apple juice in each case and $20$ cases with $15$ bottles of orange juice in each case. Which kind of juice has more bottles, and what is the difference?

Hints

- Find the total number of bottles of each kind. - You can also find the difference per case and multiply it by $20$. - Compare the two totals.

Solution

1. Find the number of apple-juice bottles: $20 \times 12 = 240$. 2. Find the number of orange-juice bottles: $20 \times 15 = 300$. 3. Find the difference: $300 - 240 = 60$.

Answer

The school receives $60$ more bottles of orange juice.

5202364

A garden center packs $14$ flowers in each box. a) Find the number of flowers in $2$ boxes, $10$ boxes, and $20$ boxes. b) How many flowers are in $12$ boxes? c) How many flowers are in $22$ boxes?

Hints

- Break $12$ into $10 + 2$. - Break $22$ into $20 + 2$. - Add the useful results from part a.

Solution

1. For part a, $2 \times 14 = 28$, $10 \times 14 = 140$, and $20 \times 14 = 280$. 2. For part b, combine the amounts for $10$ boxes and $2$ boxes: $140 + 28 = 168$. 3. For part c, combine the amounts for $20$ boxes and $2$ boxes: $280 + 28 = 308$.

Answer

a) $28$, $140$, and $280$ flowers b) $168$ flowers c) $308$ flowers

5202834

A community garden has $24$ rows with $14$ seedlings in each row. Break $24$ into $20+4$. Give the partial product for $14\times20$, the partial product for $14\times4$, and then the total number of seedlings.

Hints

- Use the required decomposition $24=20+4$. - Find the product for each part of the decomposition. - Add the two partial products.

Solution

1. $14\times20=280$. 2. $14\times4=56$. 3. Add the partial products: $280+56=336$. 4. The garden has $336$ seedlings.

Answer

$14\times20=280$ $14\times4=56$ Total: $336$ seedlings

5204004

A school auditorium has $17$ rows with $14$ seats in each row. Use partial products to find the total number of seats.

Hints

- Multiply the number of rows by the seats in each row. - Split $14$ into tens and ones. - Add the two partial products.

Solution

1. Write the product $17 \times 14$. 2. Decompose $14$ as $10 + 4$: $17 \times 10 + 17 \times 4$. 3. Add the partial products: $170 + 68 = 238$.

Answer

$17 \times 14 = 17 \times 10 + 17 \times 4 = 170 + 68 = 238$ seats

5204014

Two students calculate $13 \times 16$ using different partial products. Lukas uses $13 \times 10 + 13 \times 6$. Marie uses $10 \times 16 + 3 \times 16$. Evaluate both methods. Do they give the same product? Explain why.

Hints

- Evaluate Lukas’s two partial products. - Evaluate Marie’s two partial products. - Compare how each method decomposes one of the factors.

Solution

1. Lukas calculates $130 + 78 = 208$. 2. Marie calculates $160 + 48 = 208$. 3. Both methods give the same product because each decomposes one factor and applies the distributive property to $13 \times 16$.

Answer

Yes. Both methods give $208$ because each method decomposes one factor and applies the distributive property.

5209414

Compare the products. Write $<$, $>$, or $=$ in each box. a) $15 \times 40 \square 12 \times 50$ b) $21 \times 30 \square 32 \times 20$ c) $18 \times 50 \square 44 \times 20$

Hints

- Use place value to multiply by multiples of $10$. - Find both products in each part. - Compare the final values.

Solution

1. For part a, $15 \times 40 = 600$ and $12 \times 50 = 600$, so the products are equal. 2. For part b, $21 \times 30 = 630$ and $32 \times 20 = 640$, so $630 < 640$. 3. For part c, $18 \times 50 = 900$ and $44 \times 20 = 880$, so $900 > 880$.

Answer

a) $15 \times 40 = 12 \times 50$ b) $21 \times 30 < 32 \times 20$ c) $18 \times 50 > 44 \times 20$

5210574

Compare the products. Write $<$, $>$, or $=$ in each box. a) $12 \times 40 \square 14 \times 30$ b) $25 \times 20 \square 50 \times 10$ c) $18 \times 30 \square 20 \times 27$ d) $16 \times 50 \square 15 \times 60$

Hints

- Estimate each product before calculating. - Use place value to multiply by multiples of $10$. - Compare the final products.

Solution

1. For part a, $12 \times 40 = 480$ and $14 \times 30 = 420$, so $480 > 420$. 2. For part b, $25 \times 20 = 500$ and $50 \times 10 = 500$, so the products are equal. 3. For part c, $18 \times 30 = 540$ and $20 \times 27 = 540$, so the products are equal. 4. For part d, $16 \times 50 = 800$ and $15 \times 60 = 900$, so $800 < 900$.

Answer

a) $>$ b) $=$ c) $=$ d) $<$

5210774

Compare the products. Replace each blank with $<$, $>$, or $=$. a) $16\times25\;\square\;8\times50$ b) $20\times25\;\square\;12\times40$ c) $32\times25\;\square\;16\times60$ d) $44\times25\;\square\;22\times50$

Hints

- Calculate each product or compare how the factors change from one expression to the other. - Ask whether a change in one factor is balanced by the change in the other factor. - Compare exact products if the factor relationship is not enough to decide.

Solution

1. a) $16\times25=400$ and $8\times50=400$, so the products are equal. 2. b) $20\times25=500$ and $12\times40=480$, so the left product is greater. 3. c) $32\times25=800$ and $16\times60=960$, so the left product is less. 4. d) $44\times25=1100$ and $22\times50=1100$, so the products are equal.

Answer

a) $=$ b) $>$ c) $<$ d) $=$

5210804

Find each product. a) $63\times9$ b) $24\times19$ c) $15\times49$ d) $32\times99$

Hints

- Each factor is one less than a multiple of $10$ or $100$. - Multiply by the nearby round number first. - Subtract one group of the other factor.

Solution

1. $63 \times 9=63 \times 10-63=630-63=567$. 2. $24 \times 19=24 \times 20-24=480-24=456$. 3. $15 \times 49=15 \times 50-15=750-15=735$. 4. $32 \times 99=32 \times 100-32=3200-32=3168$.

Answer

a) $567$ b) $456$ c) $735$ d) $3168$

5213074

Maya wants to calculate $63 \times 14$. She plans to find $63 \times 10$ and $63 \times 4$, then add. 1. Calculate the two partial products. 2. Add them to find $63 \times 14$. 3. How would the product change if Maya calculated $63 \times 15$ instead? Explain how to find the new product without starting over.

Hints

- Use the product for $63 \times 14$ as your starting point. - Increasing a factor by $1$ adds one group of the other factor.

Solution

1. The partial products are $63 \times 10=630$ and $63 \times 4=252$. 2. Add: $630+252=882$. 3. Increasing the second factor from $14$ to $15$ adds one more group of $63$. Therefore, $882+63=945$.

Answer

1. $630$ and $252$ 2. $882$ 3. The product increases by $63$ to $945$ because there is one additional group of $63$.

5374214

There are $48$ packs with $17$ trading cards in each pack. Use $48\times10+48\times7$ to find the total number of cards, and explain why the decomposition works.

Hints

- Calculate the two partial products separately. - Add them because the two decomposed parts combine to make the original factor.

Solution

1. Decompose $17$ as $10+7$. 2. $48\times10=480$ and $48\times7=336$. 3. Add the partial products: $480+336=816$. 4. The decomposition works because $10+7=17$.

Answer

$48\times17=48\times10+48\times7=480+336=816$. The decomposition works because $17=10+7$.

5374224

A student calculates the number of bottles in $36$ crates that each hold $14$ bottles: $36\times14=36\times10+36\times4=360+44=404$. Find the error and determine the correct total.

Hints

- Check each partial product separately. - When multiplying $36$ by $4$, account for both the tens and ones.

Solution

1. Decomposing $14$ as $10+4$ is correct. 2. The second partial product is incorrect: $36\times4=144$, not $44$. 3. Add the correct partial products: $360+144=504$. 4. Therefore, the $36$ crates hold $504$ bottles.

Answer

The error is $36\times4=44$. The correct calculation is $36\times14=360+144=504$, so there are $504$ bottles.

5374234

A shipment has $63$ cartons with $18$ notebooks in each carton. A calculation gives a total of $1134$ notebooks. Check the result first with an estimate and then with an exact calculation.

Hints

- Round both factors to convenient multiples of ten for the estimate. - For the exact product, use a decomposition of $18$ that makes multiplication efficient.

Solution

1. Estimate: $63\approx60$ and $18\approx20$, so $60\times20=1200$. A result of $1134$ is reasonably close to $1200$. 2. Calculate exactly using $18=20-2$: $63\times18=63\times20-63\times2=1260-126=1134$. 3. The exact calculation confirms the stated total.

Answer

The estimate is $1200$, so $1134$ is reasonable. The exact calculation $63\times18=1134$ confirms the result.

5544884

A student looks at the second partial-product row in the written calculation and says, “That row is worth only $81$ because the digits $81$ come from $27\times3$.” Explain the error, state the actual value of that row, and use both partial-product values to verify the final product.

Hints

- Ask what place value the $3$ has in $34$. - Distinguish the unshifted basic product from the value represented by its position in the written layout.

Solution

1. The digit $3$ in $34$ is in the tens place, so it represents $30$, not $3$. 2. The basic product $27\times3=81$ is shifted one place left in the written algorithm, so the row represents $810$. 3. The ones partial product is $27\times4=108$. 4. $810+108=918$, which verifies the product.

Answer

The statement is wrong because the $3$ represents $30$. The second row represents $810$, not $81$. The row values are $810$ and $108$, and $810+108=918$.

5544894

One digit is hidden in the tens partial product of the written multiplication. Find the missing digit and state the value of that full partial-product row.

Hints

- Focus on the multiplier's tens digit. - Compute the basic product before applying the place-value shift. - Distinguish the digits shown in the row from the row's full place-value amount.

Solution

1. The calculation is $46\times23$. 2. The tens digit of the multiplier is $2$, so the unshifted basic product is $46\times2=92$. The hidden digit is $2$. 3. Because this is the tens row, its value in the full multiplication is $920$. 4. The other row is $46\times3=138$, and $138+920=1058$.

Answer

The missing digit is $2$, and the full tens partial product is $920$.

5544904

One digit in the ones partial-product row and one digit in the final product are hidden. Find the missing digit in the ones row. Then state the full place value represented by the second row and write the addition of the two shifted rows that determines the completed product.

Hints

- The first row comes from the ones digit of the multiplier. - The second row comes from a tens digit, so its rendered place value is ten times its basic product. - Add the two row values, not just the unshifted strings.

Solution

1. The ones multiplier digit is $6$, so the first row is $58\times6=348$. Its missing digit is $4$. 2. The second basic row is $58$, but it comes from the tens digit $1$, so the rendered shift makes its value $580$. 3. Add the written row values: $348+580=928$. 4. Therefore the missing final-product digit is $2$.

Answer

Missing ones-row digit: $4$, giving $348$. Second-row value: $580$. Row addition: $348+580=928$. Missing final-product digit: $2$.

5544914

A student made an error in the written multiplication. Identify the first incorrect partial product and give the correct product.

Hints

- Check each partial product separately before checking the final sum. - Begin with the multiplier's ones digit. - After correcting that row, combine it with the shifted tens row.

Solution

1. The calculation is intended to be $32\times24$. 2. The ones partial product should be $32\times4=128$, not $118$. 3. The tens partial product is based on $32\times2=64$, representing $640$ after the shift. 4. The correct total is $128+640=768$.

Answer

The first partial product is wrong. It should be $128$, and the correct product is $768$.

5544924

Explain why the tens partial-product row in the written multiplication has a value of $1000$, even though the basic product shown for that row is based on $50\times2=100$. Then give the final product.

Hints

- Identify the value of the $2$ in $27$. - The tens row is not multiplying by $2$ ones; it is multiplying by $2$ tens. - Add the place-value amounts represented by both rows.

Solution

1. The calculation is $50\times27$. 2. The ones partial product is $50\times7=350$. 3. The $2$ in $27$ represents $20$, so $50\times20=1000$. The written layout shifts the basic product $100$ one place left. 4. $350+1000=1350$.

Answer

The tens row represents $1000$, and the product is $1350$.

5544934

The final product row is hidden. Identify what each visible partial-product row represents in place value, then add the shifted rows to find the exact product. Finally compare it with the estimate $70\times20$.

Hints

- Do not treat the second visible row as only $67$; use its tens-place shift. - Add the represented values of the two rows before using the estimate.

Solution

1. The units multiplier digit $8$ gives the first partial product $536$. 2. The tens multiplier digit $1$ gives a basic partial product $67$, which the English layout shifts to represent $670$. 3. Add the represented row values: $536+670=1206$. 4. The estimate $70\times20=1400$ is in the same general range.

Answer

The first row represents $67\times8=536$. The second row $67$ is shifted and represents $670$. Exact product: $536+670=1206$. Estimate: $70\times20=1400$, so $1206$ is reasonable.

5544944

The written multiplication shows $35\times42$. State the value represented by each partial-product row, then explain how the rows combine to make the final product.

Hints

- Separate the multiplier into ones and tens. - Translate the second row from its basic product into its shifted place-value amount. - Add the two row values.

Solution

1. The ones row is $35\times2=70$, so its value is $70$. 2. The tens row is based on $35\times4=140$, but the $4$ represents $40$, so that row has value $1400$. 3. $70+1400=1470$.

Answer

The partial-product values are $70$ and $1400$, and the product is $1470$.

5544954

The tens digit of the multiplier and one digit of the final product are hidden. Use the second partial-product row to find the multiplier digit, state the shifted value that row represents, and complete the product.

Hints

- Use the value of the second basic partial-product row to identify the hidden multiplier digit. - Then apply the tens-place shift before adding the two rows.

Solution

1. The second basic partial product is $96$, so the hidden tens digit satisfies $24\times d=96$. Thus $d=4$. 2. Because that digit is in the tens place, the row represents $960$, not $96$. 3. The ones row represents $72$. Add $960+72=1032$.

Answer

The hidden multiplier digit is $4$ because $24\times4=96$. The shifted row represents $960$. The complete product is $1032$, so the hidden result digit is $3$.

5191774

The factors are $25$ and $12$. a) Find their product. b) What happens to the product if you double the first factor and divide the second factor by $2$? c) What happens to the original product if you double both factors?

Hints

- Calculate the original product first. - Write the changed factors for each part. - Compare each new product with the original product. - Think about how multiplying or dividing a factor changes a product.

Solution

1. For a), $25 \times 12=300$. 2. For b), the new factors are $25 \times 2=50$ and $12 \div 2=6$. Their product is $50 \times 6=300$, so the product is unchanged. 3. For c), the new factors are $50$ and $24$. Their product is $50 \times 24=1200$, which is $4$ times the original product. Doubling both factors multiplies the product by $2 \times 2=4$.

Answer

a) The product is $300$. b) The product stays the same: $300$. c) The product becomes $4$ times as great: $1200$.

5193194

Find the missing digits so that the multiplication equation is true: $2\square \times 14 = \square22$

Hints

- Start with the ones digit of the product. - Find digits whose product with $4$ ends in $2$. - Test each possible two-digit factor completely.

Solution

1. The ones digit of the product is $2$. The missing ones digit in $2\square$, when multiplied by $4$, must therefore produce a product ending in $2$. The possible digits are $3$ and $8$. 2. Test $3$: $23 \times 14 = 322$, which has the required form. 3. Test $8$: $28 \times 14 = 392$, which does not have the required form. 4. Both missing digits are $3$.

Answer

$23 \times 14 = 322$

5544964

Several digits are hidden in the written multiplication. Determine every hidden digit and give the completed calculation.

Hints

- Start with the ones partial product because it is not shifted. - Use the final digit $6$ and the multiplier digit $4$ to constrain the hidden multiplicand digit. - After recovering the multiplicand, recompute both partial products and the sum.

Solution

1. The multiplicand is $6\Box$, and the multiplier is $24$. 2. The ones partial product comes from multiplying by $4$. The pattern $2\Box6$ must be $256$, so the multiplicand is $64$. 3. Then $64\times2=128$, so the second basic partial product is $128$, representing $1280$ in the tens row. 4. $256+1280=1536$.

Answer

The completed calculation is $64\times24=1536$, with basic partial products $256$ and $128$.

5544974

A student made an error in the tens partial product of the displayed multiplication. Identify the error and give the correct product.

Hints

- Check the ones and tens basic products separately. - The tens row should use the multiplier digit $3$. - Correct the row before applying its tens-place shift and adding.

Solution

1. The calculation is $47\times32$. 2. The ones partial product is $47\times2=94$, which is correct. 3. The tens partial product should be based on $47\times3=141$, not $131$. Because it is the tens row, it represents $1410$. 4. $94+1410=1504$.

Answer

The tens basic partial product should be $141$, and the correct product is $1504$.

5544984

One digit in each factor is hidden. Use the two partial products to determine both missing digits.

Hints

- Use the tens partial product first because its multiplier digit $2$ is visible. - Recover the multiplicand before solving for the hidden ones digit of the multiplier. - Check both partial products against the completed factors.

Solution

1. The tens basic partial product is $96$, and it comes from multiplying the multiplicand by $2$. Therefore, the multiplicand is $48$. 2. The ones basic partial product is $240$. Since $48\times5=240$, the hidden ones digit of the multiplier is $5$. 3. The completed factors are $48$ and $25$, and $48\times25=1200$.

Answer

The missing digits are $4$ and $5$. The completed calculation is $48\times25=1200$.

5193204

Find the missing digits in the multiplication equation: $\square7\times2\square=851$

Hints

- Use the ones digit of the product to constrain the missing ones digit in the second factor. - Then separate the contribution from the visible $7$ in the first factor. - Ask what multiple of $10$ times the completed second factor supplies the rest of the product.

Solution

1. The ones digit of the product is $1$. Since the first factor ends in $7$, the second factor must end in $3$, because $7\times3$ ends in $1$. So the second factor is $23$. 2. The contribution from the ones digit of the first factor is $7\times23=161$. 3. The remaining contribution is $851-161=690$. Since $30\times23=690$, the tens part of the first factor is $30$. 4. Therefore the first factor is $37$, and $37\times23=851$.

Answer

$37\times23=851$

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