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5108915
Find each product. a) \(0.2\times0.4\) b) \(0.5\times0.6\) c) \(\frac{1}{2}\times0.8\)

Hints

- Think about the place value of the factors before placing the decimal point in the product. - You can first multiply the digits as whole numbers. - For c), write \(\frac{1}{2}\) as a decimal.

Solution

1. \(0.2\times0.4=0.08\). 2. \(0.5\times0.6=0.30=0.3\). 3. Since \(\frac{1}{2}=0.5\), \(\frac{1}{2}\times0.8=0.5\times0.8=0.4\).

Answer

a) \(0.08\) b) \(0.3\) c) \(0.4\)
5108865
Use \(32\times45=1440\) to find each product without using the standard written algorithm: a) \(3.2\times4.5\) b) \(0.32\times45\) c) \(32\times0.045\) d) \(0.032\times0.45\)

Hints

- Compare the place value of each factor with \(32\) and \(45\). - Use the total number of decimal places to place the decimal point in the product. - Add leading zeros when the product is less than \(1\).

Solution

1. For a), the factors have two decimal places altogether, so \(3.2\times4.5=14.4\). 2. For b), the factors have two decimal places altogether, so \(0.32\times45=14.4\). 3. For c), the factors have three decimal places altogether, so \(32\times0.045=1.44\). 4. For d), the factors have five decimal places altogether, so \(0.032\times0.45=0.0144\).

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) Use an estimate to choose the most reasonable product. Explain your choice. A: \(6.448\); B: \(64.48\); C: \(644.8\); D: \(6448\) 2) Then calculate the exact product.

Hints

- Round the factors to whole numbers for a quick estimate. - Use the estimate to judge the size of the product before calculating exactly. - Compare your exact product with your estimate.

Solution

1. Estimate with whole numbers: \(12\times5=60\). Of the choices, \(64.48\) is the only value close to \(60\), so B is reasonable. 2. Multiply the digits: \(124\times52=6448\). 3. The factors have two decimal places altogether, so the exact product is \(64.48\).

Answer

1) B, because \(12\times5=60\) is close to \(64.48\). 2) \(64.48\)
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) \(\$1.50; \$3.00; \$6.00; \$7.50\)
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
A produce seller prepares \(12\) bags of oranges. Each bag weighs \(3.2\,\text{lb}\). First estimate the total weight, then calculate the exact total.

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
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

The apples cost \(\$5.85\), the pears cost \(\$4.98\), and the potatoes cost \(\$3.75\). The total is \(\$14.58\).
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 \(\$0.75\). A student 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 \(\$5.25 > \$5.00\), the student does not have enough money. 3. Find the amount short: \(\$5.25 - \$5.00 = \$0.25\).

Answer

No. The \(7\) erasers cost \(\$5.25\), so the student is \(\$0.25\) short.
5168735
An art teacher buys watercolor sets that cost \(\$4.25\) each. a) How much do \(2\), \(4\), and \(8\) sets cost? b) How much do \(10\) sets cost? c) The teacher 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. Double repeatedly: - \(2 \times \$4.25 = \$8.50\) - \(4\) sets cost \(2 \times \$8.50 = \$17.00\) - \(8\) sets cost \(2 \times \$17.00 = \$34.00\) 2. Find the cost of \(10\) sets: \(10 \times \$4.25 = \$42.50\). 3. Find the cost of \(12\) sets by combining \(10\) and \(2\) sets: \(\$42.50 + \$8.50 = \$51.00\). 4. Since \(\$51.00 > \$50.00\), the teacher does not have enough money.

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
A produce seller buys \(40\,\text{lb}\) of apples for \(\$32\). Transportation and storage cost another \(\$8\). The seller 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 seller’s total profit?

Hints

- First find the seller’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 seller’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

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