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Grade 5 math worksheets

Build your own math worksheets from 35,000+ problems for grades 3 to 12, from fractions to AP Calculus. Every problem comes with step-by-step solutions.

Multiply decimals

Learning goals

  • Finds products of decimal numbers.
  • Explains decimal-point placement in multiplication using place value.
  • Estimates a decimal product to check its magnitude.

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Multiply decimals 1
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) \(0.56\)
Multiply decimals 2
The written multiplication represents \(1.8\times0.4\). A carry is hidden by an empty box. Find the missing carry and explain how the whole-number digit work and the place values lead to the displayed product \(0.72\).
Figure for problem 554164

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 \(0.72\).
Multiply decimals 3
Find \(0.06\times4\). Explain the product using hundredths.

Hints

- Read \(0.06\) as a number of hundredths. - Multiply the number of hundredths by \(4\). - Write the result back as a decimal.

Solution

1. \(0.06\) is \(6\) hundredths. 2. Four groups of \(6\) hundredths make \(24\) hundredths. 3. \(24\) hundredths is \(0.24\).

Answer

\(0.24\); four groups of \(6\) hundredths make \(24\) hundredths.
Multiply decimals 4
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) \(0.0144\)
Multiply decimals 5
One notebook weighs \(0.35\,\text{lb}\). A box contains \(25\) of these notebooks. a) Find the total weight of all the notebooks in the box. b) Rami 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 \(8.75\,\text{lb}\). 2. For b), \(3\times0.35=1.05\). The three notebooks weigh \(1.05\,\text{lb}\).

Answer

a) \(8.75\,\text{lb}\) b) \(1.05\,\text{lb}\)
Multiply decimals 6
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.
Multiply decimals 7
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.
Multiply decimals 8
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 \(\$0.50\).
Multiply decimals 9
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) \(\$4.40\)
Multiply decimals 10
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) \(\$2.88\)
Multiply decimals 11
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) \(\$1.50; \$3.00; \$6.00; \$7.50\)
Multiply decimals 12
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>\(\square\)</td></tr><tr><td>\(12\,\text{oz}\)</td><td>\(\square\)</td></tr><tr><td>\(16\,\text{oz}\)</td><td>\(\square\)</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. \(8\,\text{oz}\) is two \(4\)-ounce portions, so the price is \(2\times\$1.80=\$3.60\). 2. \(12\,\text{oz}\) is three portions, so the price is \(3\times\$1.80=\$5.40\). 3. \(16\,\text{oz}\) is four 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>
Multiply decimals 13
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 \(14 = 10 + 4\).
Multiply decimals 14
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 value represented by that tens row in the whole-number digit work, and explain why the decimal product is \(10.58\).
Figure for problem 554165

Hints

- First use the whole-number digit calculation \(46\times23\). - Identify the place value of the multiplier digit that creates the row with an empty box. - After combining the digit products, use the decimal place values of the original factors.

Solution

1. Ignore the decimal points briefly and use the digit work \(46\times23\). 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 because that \(2\) represents \(20\), the tens row represents \(920\). 4. The whole-number digit product is \(1058\). The original factors have two decimal places altogether, so the product is \(10.58\).

Answer

The missing digit is \(2\). The tens row represents \(920\). The whole-number digit work gives \(46\times23=1058\); because the original factors have two decimal places altogether, the decimal product is \(10.58\).
Multiply decimals 15
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.
Figure for problem 554166

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 \(15.20\).
Multiply decimals 16
Find the rectangle's area and explain the product using tenths.
Figure for problem 554179

Hints

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

Solution

1. The side lengths are \(0.6\) unit and \(0.4\) unit. 2. These are \(6\) tenths and \(4\) tenths. Their product is \(24\) hundredths. 3. 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.
Multiply decimals 17
A roll of ribbon is cut into \(8\) pieces, each \(0.75\,\text{m}\) long. What is the total length of ribbon? Explain one way to check that the product is reasonable.

Hints

- Think of the problem as \(8\) equal groups of \(0.75\). - You can use place value or a familiar fraction equivalent to check the size of the product. - The product should be less than \(8\) because each piece is shorter than \(1\,\text{m}\).

Solution

1. Multiply: \(0.75\times8=6.00\), so the total length is \(6\,\text{m}\). 2. A reasonableness check is to note that \(0.75\) is \(\frac{3}{4}\), and \(\frac{3}{4}\) of \(8\) is \(6\).

Answer

\(6\,\text{m}\). A check is \(0.75=\frac{3}{4}\), and \(\frac{3}{4}\) of \(8\) is \(6\).
Multiply decimals 18
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.
Multiply decimals 19
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 \(100\).
Multiply decimals 20
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.
Multiply decimals 21
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.
Multiply decimals 22
A family buys produce at a farmers market. Complete the cost cells and 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>\(\square\)</td></tr><tr><td>\(2\,\text{lb}\)</td><td>Pears</td><td>\(\$2.49\)</td><td>\(\square\)</td></tr><tr><td>\(5\,\text{lb}\)</td><td>Potatoes</td><td>\(\$0.75\)</td><td>\(\square\)</td></tr><tr><td>—</td><td><strong>Total</strong></td><td>—</td><td>\(\square\)</td></tr></table>

Hints

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

Solution

1. \(3\times\$1.95=\$5.85\), \(2\times\$2.49=\$4.98\), and \(5\times\$0.75=\$3.75\). 2. \(\$5.85+\$4.98+\$3.75=\$14.58\). 3. One estimate is \(3\times\$2.00+2\times\$2.50+5\times\$0.80=\$15.00\), which is close to the exact total.

Answer

Apples: \(\$5.85\) Pears: \(\$4.98\) Potatoes: \(\$3.75\) Exact total: \(\$14.58\) One reasonable estimate: \(\$15.00\) The exact total \(\$14.58\) is close to the estimate, so the total is reasonable.
Multiply decimals 23
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.
Multiply decimals 24
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.
Multiply decimals 25
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.
Multiply decimals 26
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 \(\$51.00\).
Multiply decimals 27
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 \(\$28.50\).
Multiply decimals 28
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 \(\$23.25\).
Multiply decimals 29
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 exact decimal weight by \(12\). - Subtract the estimate from the exact total to compare them.

Solution

1. Round \(3.2\,\text{lb}\) to \(3\,\text{lb}\): \(3\times12=36\), so a reasonable estimate is \(36\,\text{lb}\). 2. Exactly, \(3.2\times12=38.4\), so the total is \(38.4\,\text{lb}\). 3. The exact total is \(38.4-36=2.4\,\text{lb}\) greater than the estimate.

Answer

Estimate: about \(36\,\text{lb}\) Exact: \(38.4\,\text{lb}\) The exact total is \(2.4\,\text{lb}\) greater than the estimate.
Multiply decimals 30
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 \(\$16.50\).
Multiply decimals 31
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.
Multiply decimals 32
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
Multiply decimals 33
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
Multiply decimals 34
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 \(3.12\).
Figure for problem 554167

Hints

- First use 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.

Solution

1. Use 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 \(48\times6=288\). Because that \(6\) is in the tens place of \(65\), its row represents \(2880\). 3. The rows combine to \(3120\). 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 comes from the multiplier digit \(5\). b) The partial products represent \(240\) and \(2880\), which total \(3120\). Interpreting the three decimal places in the original factors gives \(3.12\).
Multiply decimals 35
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.
Figure for problem 554168

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.
Multiply decimals 36
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.
Figure for problem 554169

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 digit in a box.

Solution

1. The digit work multiplies \(1\square\) 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: \(1.7\times0.4=0.68\)
Multiply decimals 37
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 \(36\).
Figure for problem 554180

Hints

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

Solution

1. The side lengths are \(1.2\) units and \(0.3\) unit. 2. These are \(12\) tenths and \(3\) tenths, so the 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 because \(12\) tenths times \(3\) tenths is \(36\) hundredths.

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