Free math worksheets

- First follow the written whole-number multiplication as if the decimal point were absent. - Use the carry marks to track regrouping from one place to the next. - After multiplying, use the number of decimal places in the factors to place the decimal point.

1. Ignore the decimal point temporarily and multiply \(234\times3\) using the standard algorithm. 2. The whole-number product is \(702\). 3. The factor \(23.4\) has one decimal place, so place the decimal one digit from the right in the product. 4. Therefore, \(23.4\times3=70.2\).

\(70.2\)

- First multiply as though the decimal points were absent. - Count the decimal places in both factors together. - Estimate using \(3.4\times2.6\) to check the product's size.

1. Multiply the whole-number digits: \(3407\times26=88{,}582\). 2. The factors have a total of four decimal places: three in \(3.407\) and one in \(2.6\). 3. Place the decimal four places from the right to get \(8.8582\).

\(8.8582\)

- Temporarily ignore the decimal points and identify the multiplier digit used for each row. - Remember that a tens-place multiplier digit shifts its partial product one place. - Count the decimal places in both factors before placing the decimal point in the final product.

1. Ignoring decimal points, the written work multiplies \(408\) by the digits of \(25\). 2. The first partial product, \(2040\), comes from \(408\times5\). The second, \(816\), comes from \(408\times2\) in the tens place, so that row is shifted one place in the written algorithm. 3. The whole-number rows combine to \(10{,}200\). 4. The factors have three decimal places altogether, so the product is \(10.200\), which equals \(10.2\).

a) The rows represent \(408\times5\) and \(408\times2\), with the second row shifted one place. b) \(10.2\)

- Temporarily ignore the decimal points and look at the digits of \(205\). - Identify which multiplier digit creates each partial-product row. - Count the decimal places in both factors before placing the decimal point in the result.

1. In the original factor \(2.05\), the \(0\) is the tenths digit. When the decimal point is temporarily ignored, the multiplier is written as \(205\), where that same \(0\) occupies the tens place of the whole-number calculation and produces a zero partial-product row. 2. The nonzero partial products come from \(314\times5=1570\) and \(314\times2=628\), with the hundreds-place row shifted appropriately in the written algorithm. 3. Combining the rows gives \(64{,}370\) before replacing the decimal point. 4. The factors have four decimal places altogether, so \(3.14\times2.05=6.4370=6.437\).

The zero row comes from the \(0\) tenths digit in \(2.05\), which becomes the tens-place zero when the decimal point is temporarily ignored. The product is \(6.437\).

- Follow the one-digit written multiplication before placing the decimal point. - Count the decimal places in the two factors. - Express the final money amount to the nearest cent.

1. Ignore the decimal points temporarily and multiply \(1260\times4=5040\), following the displayed carries. 2. The factors \(12.60\) and \(0.4\) have three decimal places altogether, so place the decimal three digits from the right: \(5.040\). 3. As a money amount, \(5.040\) dollars is \(\$5.04\).

\(\$5.04\)

- Follow the partial dividends in the written work from left to right. - Keep place values aligned as you bring down each digit. - With a whole-number divisor, the decimal point in the quotient aligns with the dividend's decimal point.

1. Divide \(15\) by \(3\) to get \(5\). 2. Continue through the decimal places: \(7\div3\) gives \(2\) with a remainder, and bringing down the \(5\) completes the final division step. 3. For division by a whole number, place the quotient's decimal point directly above the dividend's decimal point. 4. The quotient is \(5.25\).

\(5.25\)

- Make the divisor a whole number by moving its decimal point. - Apply the same place-value change to the dividend so the quotient stays unchanged. - Then follow the whole-number long-division steps shown.

1. Multiply both the dividend and divisor by \(10\), giving the equivalent division \(75.6\div12\). 2. This keeps the quotient unchanged while making the divisor a whole number. 3. The written steps give \(75\div12=6\) with remainder \(3\), then bring down the next digit to get \(36\div12=3\). 4. Therefore, \(7.56\div1.2=6.3\).

\(6.3\)

- Find how much Anton earns from the small mugs. - Subtract that amount from the total revenue. - Determine how many large-mug prices fit into the remaining revenue.

1. Find the revenue from the small mugs: \(8 \times \$1.30 = \$10.40\). 2. Find the revenue from the large mugs: \(\$24.40 - \$10.40 = \$14.00\). 3. Find the number of large mugs: \(\$14.00 \div \$2.80 = 5\).

Anton sells \(5\) large mugs.

- Reverse a division equation with the related multiplication when useful. - For a decimal divisor, scale the dividend and divisor by the same power of \(10\). - Compare the numerical quotients only after evaluating both expressions.

1. For a), reverse the division by multiplying: \(x=0.075\times40=3\). 2. For b), \(0.6\div x=0.12\) means \(0.12\times x=0.6\), so \(x=0.6\div0.12=5\). 3. For c), \(5\div0.2=25\). Also, multiply both numbers in \(30\div1.2\) by \(10\): \(300\div12=25\). 4. The two quotients are equal.

a) \(x=3\) b) \(x=5\) c) Both quotients are \(25\).

- For a decimal divisor, create an equivalent division with a whole-number divisor. - Recall that percent means “per hundred.” - Multiplying both dividend and divisor by the same power of \(10\) does not change the quotient.

1. For a), multiply both dividend and divisor by \(10\): \(48\div6=8\). 2. For b), multiply both dividend and divisor by \(10\): \(12.5\div5=2.5\). 3. For c), \(12\%=0.12\), and \(0.12\div5=0.024\).

a) \(8\) b) \(2.5\) c) \(0.024\)

- Which operations must be completed before the subtraction? - For the multiplication, use place value to place the decimal point in the product. - For the division, make an equivalent division problem with a whole-number divisor.

1. Multiply: \(4.5\times1.2=5.4\). 2. Divide: \(1.2\div0.3=4\). 3. Subtract: \(5.4-4=1.4\).

\(1.4\)

- Find the cost of all six bags. - Subtract the amount Lucas is short. - Divide his money by the price per bag.

1. Six bags cost \(6 \times \$1.10 = \$6.60\). 2. Lucas has \(\$6.60 - \$1.10 = \$5.50\). 3. The greatest number he can buy is \(\$5.50 \div \$1.10 = 5\) bags.

a) Lucas has \(\$5.50\). b) He can buy at most \(5\) bags.

- Compare the second factor with \(1\) before calculating. - Multiply the digits without decimal points. - Use both decimal-place count and the size prediction to check the result.

1. Since \(0.35<1\), multiplying \(6.4\) by \(0.35\) must give a product less than \(6.4\). 2. Multiply \(64\times35=2240\). 3. The factors have three decimal places in total, so the product is \(2.240=2.24\). 4. The result \(2.24<6.4\), so it agrees with the prediction.

The product is less than \(6.4\), and \(6.4\times0.35=2.24\).

- Multiply the digits first without the decimal point. - Use the decimal places in the factor to scale the product. - Bound the decimal factor with familiar benchmark decimals.

1. Multiply \(625\times48=30{,}000\). 2. The decimal factor has three decimal places, so \(0.625\times48=30.000=30\). 3. Since \(0.625\) is between \(0.5\) and \(0.75\), the product must be between \(0.5\times48=24\) and \(0.75\times48=36\). The exact product \(30\) is in that interval.

\(30\)

- Use the inverse operation to isolate the missing factor. - Make the divisor a whole number by shifting both decimal points equally. - Substitute the result to check the product.

1. Divide both sides by \(3.25\): \(x=24.375\div3.25\). 2. Rewrite the division as \(2437.5\div325\) and calculate \(x=7.5\). 3. Check: \(3.25\times7.5=24.375\).

\(x=7.5\)

- Rewrite the division statement as a multiplication fact. - Use the quotient and dividend to recover the divisor. - Verify by multiplying the divisor and quotient.

1. The related multiplication equation is \(4.2d=18.9\). 2. Divide: \(d=18.9\div4.2=189\div42=4.5\). 3. Check: \(4.2\times4.5=18.9\).

\(d=4.5\)

- Compute the two operations separately. - Compare \(0.8\) with one. - Interpret multiplication as scaling and division as counting groups.

1. The product is \(7.2\times0.8=5.76\). 2. The quotient is \(7.2\div0.8=72\div8=9\). 3. Multiplying by a positive number less than \(1\) scales a value down, while dividing by a positive number less than \(1\) counts more than one small group and gives a larger quotient.

a) \(5.76\) and \(9\) b) The factor and divisor \(0.8\) are both less than \(1\), causing multiplication to decrease the value and division to increase it.

- Make the divisor a whole number by shifting both decimal points equally. - Perform the resulting whole-number divisor calculation. - Compare the original divisor with one to reason about quotient size.

1. Move both decimal points two places right: \(6.804\div0.06=680.4\div6\). 2. Divide to get \(113.4\). 3. The divisor \(0.06\) is less than \(1\), so many small groups fit into \(6.804\), making the quotient greater than the dividend.

\(113.4\)

- Calculate the product and quotient separately. - Compare the common factor or divisor with one. - Use scaling and grouping ideas to explain the directions of change.

1. Multiply: \(6.3\times1.4=8.82\). 2. Divide: \(6.3\div1.4=63\div14=4.5\). 3. Since \(1.4>1\), multiplying by it increases \(6.3\), while dividing by it produces fewer than \(6.3\) groups.

\(6.3\times1.4=8.82\) and \(6.3\div1.4=4.5\).

- Relate the area of a rectangle to its two side lengths. - Decide which operation recovers a missing factor. - Verify that multiplying the two side lengths returns the given area.

1. Use \(\text{width}=\text{area}\div\text{height}\). 2. Compute \(9.5625\div1.125=8.5\). 3. Check: \(1.125\times8.5=9.5625\), so the dimensions give the stated area.

\(8.5\,\text{m}\)

- Compare each factor with one before calculating. - Multiply the digits without decimal points. - Check that the decimal placement agrees with your size prediction.

1. Both factors are positive and less than \(1\), so the product will be less than each factor. 2. Multiply the digits: \(48\times375=18{,}000\). 3. The factors have five decimal places in total, so the product is \(0.18000=0.18\), which is less than \(0.48\) and \(0.375\).

The product is less than each factor, and \(0.48\times0.375=0.18\).

- Determine how many places are needed to make the divisor a whole number. - Apply the same place-value change to the dividend. - Explain the rewrite as multiplying both numbers by the same amount.

1. Multiplying both the dividend and divisor by \(1000\) gives an equivalent division expression: \(2.184\div0.024=2184\div24\). 2. Divide: \(2184\div24=91\). 3. Scaling both numbers by the same nonzero factor leaves the quotient unchanged.

\(91\)

- Identify which multiplier digit produces the row containing the missing digit. - Compute that one-digit multiplication before thinking about the row's place-value shift. - Use both partial-product rows to check the displayed final product.

1. The tens partial-product row comes from multiplying \(427\) by \(3\): \(427\times3=1281\). 2. Because the \(3\) represents \(30\), the English renderer appends one place-value zero, so that row is displayed as \(12{,}810\). Therefore, the missing digit in \(1*810\) is \(2\). 3. The units partial-product row is \(427\times6=2562\). 4. Adding the displayed place-value rows gives \(2562+12{,}810=15{,}372\).

\(2\); the final product is \(15{,}372\).

- Check the whole-number multiplication separately from the decimal placement. - Count the decimal places in both factors. - Estimate: multiplying \(2.4\) by a number less than \(1\) should make the result smaller than \(2.4\).

1. Ignoring decimal points, \(24\times35=840\), so the displayed partial products are consistent. 2. The factor \(2.4\) has one decimal place and \(0.35\) has two, for three decimal places altogether. 3. Therefore, the digits \(840\) must represent \(0.840\), not \(8.40\). 4. The correct product is \(0.84\).

The student placed the decimal point one place too far right. The correct product is \(0.84\).

- Recall why \(4.2=4.20=4.200\). - Follow each new partial dividend created after bringing down a zero. - Keep the quotient digits aligned with the decimal place values they represent.

1. Writing \(4.2\) as \(4.20\) or \(4.200\) does not change its value. 2. The first useful partial dividend is \(42\), giving \(5\) tenths with remainder \(2\). 3. Bring down an appended zero to make \(20\), then another to make \(40\), completing the division. 4. The quotient is \(0.525\).

Appending zeros to the right of a decimal does not change its value. The quotient is \(0.525\).

- Look at the partial dividend that corresponds to the missing quotient place. - Find the largest multiple of \(7\) that does not exceed that partial dividend. - Check the next step to make sure the remainder and brought-down digit agree.

1. The first step gives \(8\div7=1\) with remainder \(1\). 2. Bringing down the next digit makes \(15\). Since \(14=7\times2\), the missing quotient digit is \(2\). 3. The last step uses \(14\div7=2\), so the complete quotient is \(12.2\).

\(2\); the quotient is \(12.2\).

- First compare the divisor with \(1\) and predict how division by that positive number should affect the size of the dividend. - Check whether shifting both decimal points creates an equivalent division. - Use the displayed subtraction step to determine how many groups of \(9\) fit into \(63\).

1. Because the positive divisor \(0.9\) is less than \(1\), dividing \(6.3\) by \(0.9\) should give a quotient greater than \(6.3\). Thus, \(0.7\) is not reasonable. 2. Shifting both decimal points one place gives the equivalent division \(63\div9\), so that transformation is correct. 3. The long-division step shows \(63\) is exactly \(7\times9\). 4. Therefore, the correct quotient is \(7\). The student's error was moving the quotient decimal point after making the equivalent division.

The quotient should be greater than \(6.3\). The decimal-divisor transformation is correct, but the quotient decimal was placed incorrectly; the correct quotient is \(7\).

- Divide total cost by the number of pounds to get a unit price. - Make the decimal divisor a whole number while keeping the division equivalent. - Interpret the decimal quotient as a money amount to the nearest cent.

1. The unit price is \(18.90\div4.5\). 2. Shift both decimal points one place to form the equivalent division \(189.0\div45\). 3. The written steps give \(189\div45=4\) with remainder \(9\), then \(90\div45=2\). 4. The quotient is \(4.2\), so the apples cost \(\$4.20\) per pound.

\(\$4.20\) per pound

- Express the colony mass and one bee’s mass in the same unit. - Divide the total mass by the mass of one bee to find a count. - Use each previous answer in the next part.

1. Convert the colony mass: \(2.1\,\text{kg} = 2100\,\text{g}\). 2. Find the number of bees: \(2100\,\text{g} \div 0.1\,\text{g} = 21{,}000\). 3. Find the number of foragers: \(\frac{1}{3} \times 21{,}000 = 7000\). 4. Find the nectar mass: \(7000 \times 0.03\,\text{g} = 210\,\text{g}\). 5. Convert: \(210\,\text{g} = 0.21\,\text{kg}\).

a) About \(21{,}000\) bees b) About \(7000\) forager bees c) \(0.21\,\text{kg}\) of nectar