Free math worksheets

- Interpret the final partial dividend as tenths. - Use the divisor to find how many equal groups fit into those tenths.

1. The displayed problem is \(4.2\div2\). 2. After \(4\div2=2\), the final partial dividend is \(2\) tenths. 3. \(2\) tenths divided by \(2\) is \(1\) tenth, so the missing quotient digit is \(1\) and the subtrahend is \(2\).

Missing quotient digit: \(1\); missing subtrahend: \(2\), because \(2\) tenths divided by \(2\) is \(1\) tenth.

- Describe what the total weight and number of bags tell you. - Find the weight of one bag before finding the weight of \(100\) bags. - Think about how multiplying by \(100\) changes place value.

1. For a), divide the total weight by the number of bags: \(12.6\div12=1.05\) lb. 2. For b), \(1.05\times100=105\) lb.

a) \(1.05\) lb b) \(105\) lb

- You may convert each measurement to the requested unit before dividing. - Use place value to decide the quotient's size before calculating. - How many centimeters are in one meter, and how many meters are in one kilometer? - Check by multiplying each quotient by the divisor.

1. For part a, \(4.5\,\text{m} \div 6 = 0.75\,\text{m}\). Since \(0.75\,\text{m} = 75\,\text{cm}\), the result is \(75\,\text{cm}\). 2. For part b, \(0.12\,\text{km} \div 8 = 0.015\,\text{km}\). Since \(0.015\,\text{km} = 15\,\text{m}\), the result is \(15\,\text{m}\).

a) \(75\,\text{cm}\) b) \(15\,\text{m}\)

- Divide the total cost by the number of identical sets. - Keep the decimal point aligned as you divide. - Multiply your quotient by \(8\) to check it.

1. Divide the total cost by the number of sets: \(\$62.00 \div 8 = \$7.75\). 2. Check: \(8 \times \$7.75 = \$62.00\).

One paintbrush set costs \(\$7.75\).

- Divide the total bill by the number of friends. - Keep the decimal point aligned during the division. - Multiply the amount per person by \(9\) to check your answer.

1. Divide the total bill equally among the nine friends: \(\$110.70 \div 9 = \$12.30\). 2. Check: \(9 \times \$12.30 = \$110.70\).

Each friend should pay \(\$12.30\).

- The answer should be a little more than \(2\,\text{m}\). - Use place value to keep the quotient aligned with the dividend. - Check by multiplying your quotient by \(6\).

1. Divide the total length by the number of equal pieces: \(13.5 \div 6\). 2. Using decimal division, \(13.5 \div 6 = 2.25\). 3. Each piece is \(2.25\,\text{m}\) long.

Each piece is \(2.25\,\text{m}\) long.

- Interpret the second displayed partial dividend by place value, not just as the whole number \(15\). - Decide how many groups of \(3\) fit into those tenths. - Match that step to the corresponding place in the quotient.

1. Three goes into \(7\) two times, leaving \(1\) one. 2. Regroup the remaining \(1\) one as \(10\) tenths and combine it with the \(5\) tenths to make the displayed partial dividend \(15\) tenths. 3. Since \(15\div3=5\), the missing quotient digit is \(5\). 4. That digit counts tenths, so it is written in the tenths place. The quotient is \(2.5\).

The missing quotient digit is \(5\). The displayed \(15\) represents \(15\) tenths, and \(15\) tenths divided by \(3\) is \(5\) tenths, so the digit belongs to the right of the decimal point. The quotient is \(2.5\).

- Divide each shipment's total weight by its number of bags. - Compare the two unit weights. - Convert the difference from pounds to ounces.

1. First shipment: \(312.5\div25=12.5\) lb per bag. 2. Second shipment: \(480\div40=12\) lb per bag. 3. The first bag is heavier by \(12.5-12=0.5\) lb. 4. Since \(1\,\text{lb}=16\,\text{oz}\), \(0.5\times16=8\) oz.

a) First shipment: \(12.5\) lb per bag; second shipment: \(12\) lb per bag b) The first bag is heavier by \(8\) oz.

- Divide the total amount equally among the bottles. - For part b), find the total amount needed if every bottle gets the proposed amount.

1. For a), divide the concentrate equally: \(12.5\div50=0.25\) cup per bottle. 2. For b), \(0.3\times50=15\) cups would be needed. 3. Since \(15>12.5\), the club does not have enough concentrate.

a) \(0.25\) cup per bottle b) The plan would require \(15\) cups, but only \(12.5\) cups are available.

- Choose a nearby dollar amount that is easy to divide by \(6\). - You can express the total cost in cents before dividing. - Write the quotient as a dollar amount with two decimal places.

1. Estimate with a nearby compatible number: \(\$120 \div 6 = \$20\). 2. Divide the exact cost by \(6\): \(\$112.50 \div 6 = \$18.75\). 3. The exact amount is close to the estimate, so it is reasonable.

An estimate is \(\$20\) per student. Each student pays exactly \(\$18.75\).

- Find the total cost of the notebooks first. - Subtract that amount from the full bill. - Divide the remaining amount equally among the four pens.

1. The notebooks cost \(5 \times \$2.40 = \$12.00\). 2. The four pens cost \(\$22.00 - \$12.00 = \$10.00\). 3. One pen costs \(\$10.00 \div 4 = \$2.50\).

One pen costs \(\$2.50\).

- First find the total amount Lucas spends. - How much do the three highlighters cost altogether? - Subtract the highlighter cost to find the cost of all five notebooks. - Then divide to find the cost of one notebook.

1. Find the total amount spent: \(\$10.00 - \$2.50 = \$7.50\). 2. Find the cost of the three highlighters: \(3 \times \$1.50 = \$4.50\). 3. Find the cost of all five notebooks: \(\$7.50 - \$4.50 = \$3.00\). 4. Divide by the number of notebooks: \(\$3.00 \div 5 = \$0.60\).

One notebook costs \(\$0.60\).

- Find the price of one pencil. - Multiply the unit price by \(15\). - Check that six times the unit price equals \(\$4.20\).

1. One pencil costs \(\$4.20 \div 6 = \$0.70\). 2. Fifteen pencils cost \(15 \times \$0.70 = \$10.50\).

The \(15\) pencils cost \(\$10.50\).

- Divide Offer A’s total price by the number of bottles. - Compare the two unit prices. - Subtract the smaller unit price from the larger one.

1. Offer A costs \(\$3.90 \div 3 = \$1.30\) per bottle. 2. Since \(\$1.30 < \$1.45\), Offer A costs less per bottle. 3. The difference is \(\$1.45 - \$1.30 = \$0.15\).

a) Offer A costs \(\$1.30\) per bottle. b) Offer A costs less by \(\$0.15\) per bottle.

- How much does the entire purchase cost after accounting for the change? - How much do the two loaves cost? - What is the total cost of the rolls? - Divide that amount by ten to find the price of one roll.

1. Find the total cost of the purchase: \(\$20 - \$7 = \$13\). 2. Find the cost of the two loaves: \(2 \times \$4 = \$8\). 3. Find the cost of all ten rolls: \(\$13 - \$8 = \$5.00\). 4. Divide by the number of rolls: \(\$5.00 \div 10 = \$0.50\).

One roll costs \(\$0.50\).

- Find the price of one binder at each store. - You can change each dollar amount to cents before dividing. - Compare the two unit prices.

1. Find the unit price at the first store: \(840\text{ cents} \div 4 = 210\text{ cents}\), or \(\$2.10\). 2. Find the unit price at the second store: \(660\text{ cents} \div 3 = 220\text{ cents}\), or \(\$2.20\). 3. Compare the unit prices: \(\$2.10 < \$2.20\).

One binder is less expensive at the first store because it costs \(\$2.10\), compared with \(\$2.20\) at the second store.

- Interpret the remainder after the tenths step in terms of tenths. - An equivalent decimal can include a trailing zero without changing its value. - Use the displayed final partial dividend to determine the last quotient digit and subtraction row.

1. Divide \(12\) by \(4\) to get \(3\). The next partial dividend is \(6\) tenths; subtract \(4\) tenths, leaving \(2\) tenths. 2. Writing \(12.6\) as \(12.60\) does not change its value. The remaining \(2\) tenths become \(20\) hundredths, which is the displayed final partial dividend. 3. Since \(20\div4=5\), the missing hundredths digit is \(5\) and the missing subtrahend is \(20\). 4. The quotient is \(3.15\).

The missing quotient digit is \(5\) and the missing subtrahend is \(20\). Writing \(12.6\) as \(12.60\) does not change its value; the remaining \(2\) tenths become \(20\) hundredths, and \(20\div4=5\). The quotient is \(3.15\).

- First compare the ones in the dividend with the divisor. - Use the displayed partial dividend \(33\) to determine the tenths quotient digit. - The same multiplier determines the missing first subtraction row.

1. Since \(3\) ones are fewer than \(8\), the quotient has \(0\) ones. 2. Regroup to consider \(33\) tenths. The greatest multiple of \(8\) not exceeding \(33\) is \(32=8\times4\). Therefore the missing tenths quotient digit is \(4\) and the missing subtrahend is \(32\). 3. The remainder is \(1\) tenth; bringing down \(6\) gives \(16\) hundredths, and \(16\div8=2\). 4. The quotient is \(0.42\).

The quotient begins with \(0\) ones because \(3<8\). The displayed \(33\) tenths gives \(8\times4=32\), so the missing tenths digit is \(4\) and the missing subtrahend is \(32\). Continuing with \(16\div8=2\) gives \(0.42\).