Aimathic – Multiply decimals – Free math Worksheets

Multiply decimals

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Difficulty: 1 – 5

5108915

Find each product. a) $0.2 \times 0.4$ b) $0.5 \times 0.6$ c) $0.7 \times 0.8$

Hints

- Each factor is less than $1$, so each product should also be less than $1$. - Think of tenths as fractions with denominator $10$ when deciding the product's place value. - Check whether the product has the expected size.

Solution

1. $0.2 \times 0.4 = 0.08$. 2. $0.5 \times 0.6 = 0.30 = 0.3$. 3. $0.7 \times 0.8 = 0.56$.

Answer

a) $0.08$ b) $0.3$ c) $0.56$

5541645

The written multiplication represents $1.8\times0.4$. A carry is hidden by $*$. Find the missing carry and explain how the whole-number digit work and the place values lead to the displayed product $0.72$.

Hints

- Begin with the rightmost digit multiplication shown by the written algorithm. - Identify what must be regrouped into the next column. - After the digit work, use the place values of the decimal factors to interpret the result.

Solution

1. Ignore the decimal points temporarily and multiply the digits as $18\times4$. 2. The ones digit calculation is $8\times4=32$, so write $2$ and carry $3$. The missing carry is $3$. 3. Then $1\times4+3=7$, giving the digit product $72$. 4. The factors represent tenths, so the product has hundredths value: $1.8\times0.4=0.72$.

Answer

The missing carry is $3$, because the digit calculation $8\times4=32$ writes $2$ and regroups $3$. The digit product is $72$. Since tenths times tenths gives hundredths, the product is $0.72$.

5108865

Use $32 \times 45 = 1440$ to find each product without using the standard written algorithm. a) $3.2 \times 4.5$ b) $0.32 \times 45$ c) $32 \times 0.045$ d) $0.032 \times 0.45$

Hints

- Compare each factor with $32$ or $45$ by powers of ten. - Ask how much smaller each decimal factor is than the corresponding whole-number factor. - Apply the same combined scaling to the known product $1440$.

Solution

1. a) $3.2 = \frac{32}{10}$ and $4.5 = \frac{45}{10}$, so the product is $\frac{1440}{100} = 14.4$. 2. b) $0.32 = \frac{32}{100}$, so $0.32 \times 45 = \frac{1440}{100} = 14.4$. 3. c) $0.045 = \frac{45}{1000}$, so $32 \times 0.045 = \frac{1440}{1000} = 1.44$. 4. d) $0.032 = \frac{32}{1000}$ and $0.45 = \frac{45}{100}$, so the product is $\frac{1440}{100{,}000} = 0.0144$.

Answer

a) $14.4$ b) $14.4$ c) $1.44$ d) $0.0144$

5109185

One notebook weighs $0.35$ lb. A box contains $25$ of these notebooks. a) Find the total weight of all the notebooks in the box. b) A student carries $3$ of these notebooks in a backpack. How much weight do the notebooks add to the backpack?

Hints

- When the same weight occurs several times, multiplication can find the total. - Use place value carefully when multiplying a decimal. - Check whether your product is a reasonable multiple of the weight of one notebook.

Solution

1. For a), multiply the number of notebooks by the weight of one notebook: $25\times0.35=8.75$. The notebooks in the box weigh $8.75$ lb. 2. For b), $3\times0.35=1.05$. The three notebooks weigh $1.05$ lb.

Answer

a) $8.75$ lb b) $1.05$ lb

5116445

A cake recipe uses $2.25$ cups of flour. Tim wants to make $2.4$ times the recipe. How many cups of flour does he need altogether?

Hints

- Think about which operation represents making a multiple of a quantity. - Use place value carefully when multiplying two decimals.

Solution

1. Multiply the original amount of flour by the scale factor: $2.25\times2.4$. 2. The product is $5.4$. 3. Tim needs $5.4$ cups of flour.

Answer

He needs $5.4$ cups of flour.

5122765

Consider $12.4\times5.2$. 1) Which is the best estimate for the product: $6$, $60$, $600$, or $6000$? Explain. 2) Then calculate the exact product.

Hints

- Round each factor to a nearby whole number before choosing the estimate. - For the exact product, connect each decimal factor to an equivalent whole-number factor using powers of ten. - Compare the exact result with the estimate to check its size.

Solution

1. Round the factors to nearby whole numbers: $12.4\approx12$ and $5.2\approx5$. Then $12\times5=60$, so $60$ is the best estimate. 2. Write $12.4=\frac{124}{10}$ and $5.2=\frac{52}{10}$. Then $124\times52=6448$, so $12.4\times5.2=\frac{6448}{100}=64.48$. 3. The exact product is close to the estimate of $60$, so its size is reasonable.

Answer

1) $60$, because $12.4\approx12$ and $5.2\approx5$, so $12\times5=60$. 2) $64.48$. The exact product is close to $60$, so its size is reasonable.

5162995

A school garden needs $6$ wooden boards for a border. Each board is exactly $2.30\,\text{m}$ long. How many meters of wood are needed altogether?

Hints

- Imagine placing all $6$ boards end to end. - Multiplication represents adding the same length several times. - Keep the decimal point in the correct place in the product.

Solution

1. Multiply the length of one board by the number of boards: $2.30 \times 6$. 2. Calculate the product: $2.30 \times 6 = 13.80$. 3. The garden needs $13.80\,\text{m}$ of wood.

Answer

A total of $13.80\,\text{m}$ of wood is needed.

5168625

A school-supply store has back-to-school specials. Find how much is saved by buying each set instead of the same items individually: - One gel pen costs $\$1.35$. A box of $10$ gel pens costs $\$11.90$. - One eraser costs $\$0.85$. A set of $5$ erasers costs $\$3.75$.

Hints

- First find what the same number of items would cost individually. - Compare each individual-item total with the set price. - The difference is the amount saved.

Solution

1. Find the individual cost of $10$ gel pens: $10 \times \$1.35 = \$13.50$. Subtract the box price: $\$13.50 - \$11.90 = \$1.60$. 2. Find the individual cost of $5$ erasers: $5 \times \$0.85 = \$4.25$. Subtract the set price: $\$4.25 - \$3.75 = \$0.50$.

Answer

The box of gel pens saves $\$1.60$. The set of erasers saves $\$0.50$.

5168685

One postcard costs $\$0.55$. Find the total cost of: a) $2$ postcards b) $4$ postcards c) $8$ postcards

Hints

- You can think of the price in cents. - Notice how the quantities $2$, $4$, and $8$ are related. - Multiply the single-postcard price by each quantity.

Solution

1. For $2$ postcards: $2 \times \$0.55 = \$1.10$. 2. For $4$ postcards: $4 \times \$0.55 = \$2.20$, or double the cost of $2$ postcards. 3. For $8$ postcards: $8 \times \$0.55 = \$4.40$, or double the cost of $4$ postcards.

Answer

a) $\$1.10$ b) $\$2.20$ c) $\$4.40$

5168695

One small apple costs $\$0.32$. a) How much do $3$ apples cost? b) How much do $6$ apples cost? Use your answer from part a). c) How much do $9$ apples cost?

Hints

- Think about how $3$, $6$, and $9$ are related. - Adding the amounts vertically may help. - You can work in cents and then convert back to dollars.

Solution

1. Find the cost of $3$ apples: $3 \times \$0.32 = \$0.96$. 2. Six is twice $3$, so double the first result: $\$0.96 \times 2 = \$1.92$. 3. Nine is three times $3$, so triple the first result: $\$0.96 \times 3 = \$2.88$.

Answer

a) $\$0.96$ b) $\$1.92$ c) $\$2.88$

5168715

A toy store sells small items. a) One marble costs $\$0.15$. How much do $10$, $20$, $50$, and $100$ marbles cost? b) One small bouncy ball costs $\$0.75$. How much do $2$, $4$, $8$, and $10$ bouncy balls cost?

Hints

- Use doubling when a larger quantity is twice a smaller quantity. - Think about place value when multiplying a decimal by $10$ or $100$. - You may work in cents first and then convert to dollars.

Solution

1. For the marbles: - $10 \times \$0.15 = \$1.50$ - $20 \times \$0.15 = \$3.00$ - $50 \times \$0.15 = \$7.50$ - $100 \times \$0.15 = \$15.00$ 2. For the bouncy balls: - $2 \times \$0.75 = \$1.50$ - $4 \times \$0.75 = \$3.00$ - $8 \times \$0.75 = \$6.00$ - $10 \times \$0.75 = \$7.50$

Answer

a) $\$1.50; \$3.00; \$7.50; \$15.00$ b) $\$1.50; \$3.00; \$6.00; \$7.50$

5168805

Salami costs $\$1.80$ per $4\,\text{oz}$. Complete the table. <table><tr><th>Weight</th><th>Price</th></tr><tr><td>$4\,\text{oz}$</td><td>$\$1.80$</td></tr><tr><td>$8\,\text{oz}$</td><td>?</td></tr><tr><td>$12\,\text{oz}$</td><td>?</td></tr><tr><td>$16\,\text{oz}$</td><td>?</td></tr></table>

Hints

- Express each target weight as a whole-number multiple of $4\,\text{oz}$. - Apply the same scaling factor to the price. - Check that larger multiples of the weight have proportionally larger prices.

Solution

1. Eight ounces is two $4$-ounce portions, so the price is $2 \times \$1.80=\$3.60$. 2. Twelve ounces is three $4$-ounce portions, so the price is $3 \times \$1.80=\$5.40$. 3. Sixteen ounces is four $4$-ounce portions, so the price is $4 \times \$1.80=\$7.20$.

Answer

<table><tr><th>Weight</th><th>Price</th></tr><tr><td>$4\,\text{oz}$</td><td>$\$1.80$</td></tr><tr><td>$8\,\text{oz}$</td><td>$\$3.60$</td></tr><tr><td>$12\,\text{oz}$</td><td>$\$5.40$</td></tr><tr><td>$16\,\text{oz}$</td><td>$\$7.20$</td></tr></table>

5190935

A granola bar costs $\$0.60$ in the school cafeteria. a) How much do $10$ granola bars cost? b) How much do $4$ granola bars cost? c) How much do $14$ granola bars cost? d) Explain how to use your answers to parts a and b to find the answer to part c.

Hints

- Use place value to multiply the price by $10$. - Break $14$ into $10 + 4$. - Add the two partial costs.

Solution

1. Ten granola bars cost $10 \times \$0.60 = \$6.00$. 2. Four granola bars cost $4 \times \$0.60 = \$2.40$. 3. Fourteen granola bars cost $14 \times \$0.60 = \$8.40$. 4. Since $14 = 10 + 4$, add the two partial costs: $\$6.00 + \$2.40 = \$8.40$.

Answer

a) $\$6.00$ b) $\$2.40$ c) $\$8.40$ d) Add the cost of $10$ bars and the cost of $4$ bars because $14 = 10 + 4$.

5209275

Pine Street Produce prepares $12$ bags of oranges. Each bag weighs $3.2\,\text{lb}$. Give a reasonable estimate of the total weight, calculate the exact total, and compare the two.

Hints

- Round one bag's weight to a nearby whole number for the estimate. - Multiply the decimal weight by $12$. - You can split $12$ into $10 + 2$ to check the product.

Solution

1. Estimate by rounding $3.2\,\text{lb}$ to $3\,\text{lb}$: $3 \times 12 = 36$, so the total is about $36\,\text{lb}$. 2. Multiply exactly: $3.2 \times 12 = 38.4$. 3. The exact total weight is $38.4\,\text{lb}$.

Answer

Estimate: about $36\,\text{lb}$ Exact: $38.4\,\text{lb}$

5222115

A class has $25$ students. During morning break, each student drinks an average of $1.5$ cups of water. How many cups of water does the class drink during a $5$-day school week?

Hints

- First find how much water all the students drink in one day. - Then use the number of school days to find the weekly total. - Check the place value in each product.

Solution

1. Find the amount the class drinks in one day: $25\times1.5=37.5$ cups. 2. Multiply by $5$ school days: $37.5\times5=187.5$ cups.

Answer

$187.5$ cups

5541655

In the written multiplication for $4.6\times2.3$, one digit in the tens partial-product entry is hidden. Find the missing digit, state the number displayed by that shifted row in the whole-number work, and explain why the final decimal product is $10.58$.

Hints

- Read the partial-product rows as the whole-number digit calculation first. - Identify the place value of the multiplier digit that creates the starred row. - Only after combining the digit products should you interpret the result using the decimal place values of the factors.

Solution

1. The tool performs the digit work as $46\times23$. 2. The units partial product is $46\times3=138$. 3. The tens digit is $2$, so $46\times2=92$. The missing digit is $2$, and the English layout shifts that row to display $920$. 4. The whole-number digit product is $1058$. Since the original factors have two decimal places altogether, the product is $10.58$.

Answer

The missing digit is $2$, because $46\times2=92$. Since that $2$ is the tens digit of $23$, the rendered shifted row is $920$. The digit product is $1058$, and the original tenths factors make the product $10.58$.

5541665

In the written multiplication for $3.04\times5$, the carry into the tenths column is hidden. Find the missing carry and explain how the internal zero in $3.04$ affects the next step.

Hints

- Start with the rightmost decimal digit. - Determine what is written in that place and what is regrouped into the next place. - Do not skip the zero digit; include the incoming regrouped amount when you process it.

Solution

1. Multiply the hundredths digit: $4\times5=20$. Write $0$ in the hundredths place and carry $2$ into the tenths column. 2. The missing carry is $2$. 3. The tenths digit is $0$, so $0\times5+2=2$ tenths. 4. Continuing gives $15.20$.

Answer

The missing carry is $2$, because $4\times5=20$. The next digit is $0$, so $0\times5+2=2$ tenths; the zero cannot be skipped. Continuing gives $15.20$.

5541795

The rectangle's side lengths are shown only in the diagram. Find its area and explain the product using tenths.

Hints

- Read both side lengths from the diagram. - Rewrite each side length in tenths. - Think about what unit results when tenths are multiplied by tenths.

Solution

1. The diagram shows side lengths $0.6$ unit and $0.4$ unit. 2. These are $6$ tenths and $4$ tenths. Their product is $24$ hundredths. 3. Therefore the area is $0.24$ square unit.

Answer

The area is $0.24$ square unit. Since $0.6$ is $6$ tenths and $0.4$ is $4$ tenths, $6\times4=24$ gives $24$ hundredths, or $0.24$.

5109195

A market sells two varieties of apples. Variety A costs $\$2.40$ per pound. Variety B is sold in $1.5$-lb bags for $\$3.45$ per bag. First find the cost of $1.5$ lb of Variety A. Which variety is less expensive for the same amount of apples?

Hints

- To compare prices fairly, compare the cost of the same amount of apples. - Find what $1.5$ lb of Variety A costs before comparing the two prices.

Solution

1. Find the cost of $1.5$ lb of Variety A: $1.5\times\$2.40=\$3.60$. 2. Compare equal amounts: Variety A costs $\$3.60$ for $1.5$ lb, while Variety B costs $\$3.45$. 3. Since $\$3.45<\$3.60$, Variety B is less expensive.

Answer

$1.5$ lb of Variety A costs $\$3.60$. Variety B is less expensive.

5109315

Start with $4.5\times100=450$. Find each new product without recomputing from scratch. Briefly explain how changing the factors changes the product. a) Divide the factor $4.5$ by $10$, while $100$ stays the same. b) Keep $4.5$ the same, but multiply $100$ by $10$. c) Divide both factors, $4.5$ and $100$, by $10$.

Hints

- Think about how a product changes when one factor is scaled by a power of $10$. - Use the original product, $450$, as your starting point. - Before calculating, decide whether each new product should be greater or less than $450$.

Solution

1. For a), dividing one factor by $10$ divides the product by $10$: $450\div10=45$. 2. For b), multiplying one factor by $10$ multiplies the product by $10$: $450\times10=4500$. 3. For c), dividing both factors by $10$ divides the product by $100$: $450\div100=4.5$.

Answer

a) $45$; the product is divided by $10$. b) $4500$; the product is multiplied by $10$. c) $4.5$; the product is divided by $100$.

5168135

At a grocery store, one gallon of milk costs $\$1.20$, and one carton of eggs costs $\$3.50$. Ms. Schmidt buys $3$ gallons of milk and $2$ cartons of eggs. How much does she pay in all?

Hints

- First find the total cost of the milk. - Then find the total cost of the eggs. - Add the two amounts.

Solution

1. Find the cost of the milk: $3 \times \$1.20 = \$3.60$. 2. Find the cost of the eggs: $2 \times \$3.50 = \$7.00$. 3. Add the costs: $\$3.60 + \$7.00 = \$10.60$.

Answer

Ms. Schmidt pays $\$10.60$ in all.

5168595

A family buys produce at a farmers market. Complete the table by finding each item’s cost and the total. Then estimate to check whether the total is reasonable. <table> <tr> <th>Amount</th> <th>Item</th> <th>Price per pound</th> <th>Cost</th> </tr> <tr> <td>$3\,\text{lb}$</td> <td>Apples</td> <td>$\$1.95$</td> <td></td> </tr> <tr> <td>$2\,\text{lb}$</td> <td>Pears</td> <td>$\$2.49$</td> <td></td> </tr> <tr> <td>$5\,\text{lb}$</td> <td>Potatoes</td> <td>$\$0.75$</td> <td></td> </tr> <tr> <td></td> <td><strong>Total</strong></td> <td></td> <td></td> </tr> </table>

Hints

- Multiply each amount by its price per pound. - Add the three item costs to find the total. - Round the unit prices to convenient amounts for an estimate.

Solution

1. Find each item’s cost: $3 \times \$1.95 = \$5.85$, $2 \times \$2.49 = \$4.98$, and $5 \times \$0.75 = \$3.75$. 2. Add the costs: $\$5.85 + \$4.98 + \$3.75 = \$14.58$. 3. Estimate using $\$2.00$, $\$2.50$, and $\$0.80$ per pound: $3 \times \$2.00 + 2 \times \$2.50 + 5 \times \$0.80 = \$6.00 + \$5.00 + \$4.00 = \$15.00$. The exact total of $\$14.58$ is reasonable.

Answer

Apples: $\$5.85$ Pears: $\$4.98$ Potatoes: $\$3.75$ Exact total: $\$14.58$ One reasonable estimate is $\$15.00$, so the exact total is reasonable.

5168605

Leon buys breakfast at a bakery: - $5$ multigrain rolls at $\$0.45$ each - $2$ butter croissants at $\$1.30$ each - $1$ slice of strawberry cake for $\$2.75$ How much does he pay in all? If he pays with a $\$10$ bill, how much change does he receive?

Hints

- Find the combined cost of the rolls and croissants. - Add the price of the cake. - Subtract the total cost from the amount paid.

Solution

1. Find the cost of the rolls: $5 \times \$0.45 = \$2.25$. 2. Find the cost of the croissants: $2 \times \$1.30 = \$2.60$. 3. Add all the costs: $\$2.25 + \$2.60 + \$2.75 = \$7.60$. 4. Find the change: $\$10.00 - \$7.60 = \$2.40$.

Answer

Leon pays $\$7.60$ and receives $\$2.40$ in change.

5168615

Two students buy school supplies. Lukas buys $4$ individual markers at $\$1.25$ each and $3$ notebooks at $\$0.95$ each. Mia buys a set of $4$ markers for $\$4.80$ and a three-pack of notebooks for $\$2.80$. Who pays more, and what is the difference?

Hints

- Find Lukas’s total cost first. - Mia’s two package prices only need to be added. - Compare the totals and subtract to find the difference.

Solution

1. Find Lukas’s marker cost: $4 \times \$1.25 = \$5.00$. 2. Find Lukas’s notebook cost: $3 \times \$0.95 = \$2.85$. 3. Find Lukas’s total: $\$5.00 + \$2.85 = \$7.85$. 4. Find Mia’s total: $\$4.80 + \$2.80 = \$7.60$. 5. Lukas pays more. The difference is $\$7.85 - \$7.60 = \$0.25$.

Answer

Lukas pays $\$7.85$, and Mia pays $\$7.60$. Lukas pays $\$0.25$ more.

5168705

One eraser costs $\$0.75$. Sofia wants to buy $7$ erasers and has a $\$5$ bill. Is that enough money? Justify your answer with a calculation.

Hints

- First find the total cost of the $7$ erasers. - Compare the cost with the $\$5$ bill. - Determine how much is missing or left over.

Solution

1. Find the total cost: $7 \times \$0.75 = \$5.25$. 2. Since $\$5.25 > \$5.00$, Sofia does not have enough money. 3. The amount short is $\$5.25 - \$5.00 = \$0.25$.

Answer

No. The $7$ erasers cost $\$5.25$, so Sofia is $\$0.25$ short.

5168735

Ms. Nakamura buys watercolor sets that cost $\$4.25$ each for an art club. a) How much do $2$, $4$, and $8$ sets cost? b) How much do $10$ sets cost? c) Ms. Nakamura has a $\$50$ bill. Is that enough for $12$ sets? Justify your answer.

Hints

- Use the cost of $2$ sets to find the cost of $4$, then $8$. - Find the cost of $10$ sets and combine it with the cost of $2$ sets. - Compare the cost of $12$ sets with $\$50$.

Solution

1. a) Double repeatedly: $2$ sets cost $\$8.50$, $4$ sets cost $\$17.00$, and $8$ sets cost $\$34.00$. 2. b) $10 \times \$4.25 = \$42.50$. 3. c) Combine the costs of $10$ and $2$ sets: $\$42.50 + \$8.50 = \$51.00$. 4. Since $\$51.00 > \$50.00$, the bill is not enough.

Answer

a) $\$8.50; \$17.00; \$34.00$ b) $\$42.50$ c) No. The $12$ sets cost $\$51.00$.

5185435

A print shop charges a fixed setup fee of $\$8.50$ for each order. It also charges: - $\$0.10$ for each black-and-white flyer - $\$0.25$ for each color flyer Find the total cost of each order. a) A sports club orders $150$ black-and-white flyers. b) A school orders $80$ color flyers.

Hints

- Multiply the number of flyers by the per-flyer price. - Add the setup fee once to each order. - Keep the decimal points aligned when adding money amounts.

Solution

1. The printing cost for $150$ black-and-white flyers is $150 \times \$0.10 = \$15.00$. 2. With the setup fee, order a costs $\$15.00 + \$8.50 = \$23.50$. 3. The printing cost for $80$ color flyers is $80 \times \$0.25 = \$20.00$. 4. With the setup fee, order b costs $\$20.00 + \$8.50 = \$28.50$.

Answer

a) The black-and-white flyer order costs $\$23.50$. b) The color flyer order costs $\$28.50$.

5200885

A cheese pizza costs $\$6.50$. A pepperoni pizza costs $\$1.25$ more. How much do three pepperoni pizzas cost?

Hints

- First find the price of one pepperoni pizza. - Then multiply that price by $3$.

Solution

1. One pepperoni pizza costs $\$6.50 + \$1.25 = \$7.75$. 2. Three pepperoni pizzas cost $3 \times \$7.75 = \$23.25$.

Answer

Three pepperoni pizzas cost $\$23.25$.

5209425

Pine Street Produce buys $40\,\text{lb}$ of apples for $\$32$. Transportation and storage cost another $\$8$. The store normally charges $\$1.50$ per pound. At the end of the day, $5\,\text{lb}$ of bruised apples remain and are sold for $\$0.80$ per pound. What is the store's total profit?

Hints

- First find the store's total cost. - Separate the apples sold at the regular price from those sold at the reduced price. - Find the revenue from each group and add. - Profit equals total revenue minus total cost.

Solution

1. Find the total cost: $\$32 + \$8 = \$40$. 2. Find the amount sold at the regular price: $40\,\text{lb} - 5\,\text{lb} = 35\,\text{lb}$. 3. Find the regular-price revenue: $35 \times \$1.50 = \$52.50$. 4. Find the reduced-price revenue: $5 \times \$0.80 = \$4.00$. 5. Find the total revenue: $\$52.50 + \$4.00 = \$56.50$. 6. Subtract the total cost from the revenue: $\$56.50 - \$40 = \$16.50$.

Answer

The store's total profit is $\$16.50$.

5209485

An elevator can carry at most $1320\,\text{lb}$. Three people enter: Mr. Smith weighs $194\,\text{lb}$, Ms. Weber weighs $141\,\text{lb}$, and Lucas weighs $99\,\text{lb}$. They also have $5$ identical packages that each weigh $27.5\,\text{lb}$. How many more pounds can the elevator carry?

Hints

- Add the three people’s weights. - Multiply to find the total weight of the five packages. - Subtract the current load from the elevator’s capacity.

Solution

1. Add the people’s weights: $194 + 141 + 99 = 434$, so the people weigh $434\,\text{lb}$. 2. Find the packages’ total weight: $5 \times 27.5\,\text{lb} = 137.5\,\text{lb}$. 3. Find the current load: $434\,\text{lb} + 137.5\,\text{lb} = 571.5\,\text{lb}$. 4. Subtract from capacity: $1320\,\text{lb} - 571.5\,\text{lb} = 748.5\,\text{lb}$.

Answer

The elevator can carry $748.5\,\text{lb}$ more.

5222125

An orchard ships apples in wooden crates. An empty crate weighs $2.5$ lb. Each crate holds $48$ apples, and one apple weighs an average of $0.4$ lb. What is the total weight of $15$ filled crates?

Hints

- Find the weight of the apples in one crate first. - Remember to include the weight of the empty crate. - Once you know the weight of one filled crate, find the total for all $15$ crates.

Solution

1. Find the weight of the apples in one crate: $48\times0.4=19.2$ lb. 2. Add the empty crate weight: $19.2+2.5=21.7$ lb for one filled crate. 3. Find the weight of $15$ filled crates: $15\times21.7=325.5$ lb.

Answer

$325.5$ lb

5541675

In the written multiplication for $0.48\times6.5$, one digit in the first partial-product row is hidden. a) Find the missing digit and identify which multiplier digit creates that row. b) Explain how the two whole-number partial-product rows combine to support the decimal product $3.12$.

Hints

- First interpret the renderer's rows as the digit multiplication $48\times65$. - Match the first row to the rightmost multiplier digit. - After combining the rows, use the place values in the original decimal factors to interpret the product.

Solution

1. The tool uses the digit work $48\times65$. The rightmost multiplier digit is $5$, so the first partial product is $48\times5=240$. The missing digit is $4$. 2. The other digit product is $48\times6=288$, shifted one place in the written layout because the $6$ is the tens digit in the digit work. 3. These rows give the digit product $3120$. The original factors have three decimal places altogether, so the product is $3.120=3.12$.

Answer

a) The missing digit is $4$; the row is generated by the digit $5$. b) The partial products combine to the digit product $3120$, which represents $3.12$ for the original decimal factors.

5541685

The written multiplication shows $2.4 \times 1.7 = 40.8$. The whole-number digit multiplication is otherwise correct. Explain the place-value error and give the correct product.

Hints

- Compare each decimal factor with its corresponding whole-number form. - Combine the two powers-of-ten changes before scaling the whole-number product back. - Estimate $2.4 \times 1.7$ to check whether a result near $40$ is plausible.

Solution

1. The corresponding whole-number multiplication is $24 \times 17 = 408$. 2. But $2.4 = \frac{24}{10}$ and $1.7 = \frac{17}{10}$, so their product is $\frac{408}{100} = 4.08$. 3. The displayed result $40.8$ is ten times too large because the combined scaling was not fully accounted for.

Answer

The correct product is $4.08$. The whole-number digit work gives $24\times17=408$, but $2.4=24/10$ and $1.7=17/10$, so the product is $408/100=4.08$. The displayed $40.8$ is ten times too large.

5541695

The written multiplication has a missing tenths digit in the multiplicand and a missing tenths digit in the result. The shown carry is $2$. Use the displayed written work to find both missing digits and complete the multiplication.

Hints

- Use the visible final digit and the shown carry together; either one alone is not enough. - Ask which one-digit product ends in the displayed hundredths digit and regroups $2$. - Then use the carry in the next column to determine the other starred digit.

Solution

1. The digit work multiplies $1*$ by $4$. The rightmost product digit is $8$, and the shown carry is $2$. 2. The only missing digit whose product with $4$ ends in $8$ and produces carry $2$ is $7$, because $7\times4=28$. 3. Then $1\times4+2=6$, so the missing tenths digit of the result is $6$. 4. The completed multiplication is $1.7\times0.4=0.68$.

Answer

Missing multiplicand digit: $7$ Missing result digit: $6$ Completed multiplication: $1.7\times0.4=0.68$

5541805

The rectangle's side lengths are shown only in the diagram. The whole-number digit product corresponding to the side lengths is $12\times3=36$. Find the rectangle's area and explain why the answer is not $3.6$ or $36$.

Hints

- Read the two factors from the diagram. - Express each decimal factor in tenths before using the given digit product. - Determine the place-value unit of tenths times tenths.

Solution

1. The diagram shows side lengths $1.2$ units and $0.3$ unit. 2. Write them as $12$ tenths and $3$ tenths. Their product is $36$ hundredths. 3. $36$ hundredths is $0.36$, so the area is $0.36$ square unit.

Answer

The area is $0.36$ square unit. The diagram gives $1.2=12$ tenths and $0.3=3$ tenths, so $12\times3=36$ counts hundredths: $36/100=0.36$.

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