Free math worksheets

- Think about how many eighths make one whole mile. - Then extend that count to all of the miles on the trail.

1. The question asks how many groups of \(\frac{1}{8}\) are in \(4\), so use \(4\div\frac{1}{8}\). 2. Each mile contains \(8\) eighth-mile intervals, so \(4\) miles contain \(4\times8=32\) intervals.

There are \(32\) intervals.

- Read the unit-fraction piece size from the number of equal parts in one whole loaf. - The division question is “How many pieces of that size fit in two wholes?” - Use the image count to check the quotient after writing the division equation.

1. Each loaf is divided into \(6\) equal pieces, so one piece is \(\frac{1}{6}\) of a loaf. 2. The model asks how many \(\frac{1}{6}\)-loaf pieces fit in \(2\) whole loaves, so the equation is \(2\div\frac{1}{6}\). 3. There are \(6\) sixths in each whole and \(12\) sixths in two wholes, so \(2\div\frac{1}{6}=12\).

a) \(\frac{1}{6}\) of a loaf b) \(2\div\frac{1}{6}=12\) c) \(12\) snacks

- Think about how many twelfths make one whole meter. - Repeat that amount for each meter of ribbon.

1. The number of bookmarks is the number of \(\frac{1}{12}\)-meter pieces in \(3\) meters: \(3\div\frac{1}{12}\). 2. Each meter contains \(12\) such pieces, so \(3\times12=36\).

\(36\) bookmarks can be made.

- Compare the sizes of one third and one fifth. - Ask which size piece would fit more times into the same two wholes.

1. \(2\div\frac{1}{3}=6\) because there are \(3\) thirds in each whole. 2. \(2\div\frac{1}{5}=10\) because there are \(5\) fifths in each whole. 3. Smaller pieces fit more times into the same total, so dividing by \(\frac{1}{5}\) gives the greater quotient.

\(2\div\frac{1}{5}=10\) is greater than \(2\div\frac{1}{3}=6\) because fifths are smaller than thirds, so more fifth-sized pieces fit into the same \(2\) wholes.

- Think about how many pieces of the stated size fit in one whole yard. - Extend that count to all of the whole yards. - Consider how using pieces smaller than one whole affects the number of pieces.

1. The number of pieces is \(7\div\frac{1}{4}\). 2. Each whole yard contains \(4\) quarter-yard pieces, so \(7\) yards contain \(7\times4=28\) pieces. 3. The quotient is greater than \(7\) because each whole yard is split into several pieces smaller than one yard.

\(28\) pieces. The count is greater than \(7\) because each yard contains \(4\) quarter-yard pieces.

- Interpret the multiplication as a group count times a group size. - Reverse that relationship to find the number of groups from the total.

1. Eighteen groups of size \(\frac{1}{6}\) make a total of \(3\). 2. Therefore the number of sixth-size groups in \(3\) is \(18\). 3. The division equation is \(3\div\frac{1}{6}=18\).

\(3\div\frac{1}{6}=18\). The multiplication fact says that \(18\) groups of size \(\frac{1}{6}\) make \(3\), so the related division asks how many sixth-size groups fit in \(3\).

- Ask how many pieces of the given size fit in one whole first. - Then extend that count to all four wholes. - Think about why counting pieces smaller than a whole gives more groups than the number of wholes.

1. The number of pieces is represented by \(4\div\frac{1}{7}\). 2. Each whole contains \(7\) one-seventh pieces, so \(4\) wholes contain \(4\times7=28\) pieces. 3. The quotient is larger than \(4\) because the groups being counted are smaller than one whole.

\(4\div\frac{1}{7}=28\), so there are \(28\) pieces. The quotient is larger than \(4\) because each counted piece is only \(\frac{1}{7}\) of a whole, so several pieces fit inside each whole bar.

- First think about how many quarter-hours fit in one hour. - Extend that count to all \(5\) hours. - Keep track of what is being counted by the quotient.

1. The number of sessions is \(5\div\frac{1}{4}\). 2. Each hour contains \(4\) quarter-hour sessions, so \(5\) hours contain \(5\times4=20\) sessions. 3. The quotient counts sessions, not hours.

\(20\) sessions fit in the workshop.

- Start by asking how many copies of the unit fraction make exactly one whole. - Then scale that count to the number of wholes in the dividend. - The quotient counts copies of the unit fraction.

1. Eleven copies of \(\frac{1}{11}\) make one whole because \(11\times\frac{1}{11}=1\). 2. Two wholes therefore contain \(2\times11=22\) copies of \(\frac{1}{11}\). 3. Thus \(2\div\frac{1}{11}=22\).

There are \(22\) copies of \(\frac{1}{11}\) in \(2\) wholes.

- Estimate whether the quotient should be larger or smaller than the whole number. - Think about how many ninths fit in one whole before considering all five wholes.

1. Dividing \(5\) wholes into pieces of size \(\frac{1}{9}\) should produce more than \(5\) pieces, not a value less than \(1\). 2. Each whole contains \(9\) ninths, so \(5\) wholes contain \(5\times9=45\) ninths. 3. Therefore \(5\div\frac{1}{9}=45\).

Nora's answer is not reasonable because \(\frac{5}{9}<1\), but five wholes contain many pieces of size \(\frac{1}{9}\). The correct quotient is \(45\).

- Think about how many \(\frac{1}{n}\) pieces fit in one whole. - Relate the total number of pieces in three wholes to the given quotient.

1. If one whole is divided into \(n\) equal unit-fraction pieces, then \(3\) wholes contain \(3n\) pieces. 2. Set \(3n=24\), giving \(n=8\). 3. The divisor is \(\frac{1}{8}\), so there are \(24\) eighths in \(3\) wholes.

\(n=8\). It means each whole is partitioned into \(8\) equal parts, so the divisor is \(\frac{1}{8}\) and there are \(24\) eighths in \(3\) wholes.

- For each expression, think about how many copies of the unit fraction make one whole. - Scale that count by the whole-number dividend.

1. A gives \(2\times9=18\). 2. B gives \(3\times6=18\). 3. C gives \(6\times3=18\). 4. D gives \(4\times4=16\), so it does not match.

A, B, and C.

- Think about how many packets make one whole pound. - Group the packet count into complete pounds. - Use your total to write the related division statement.

1. Seven packets of \(\frac{1}{7}\) pound make \(1\) pound. 2. Group \(35\) packets into groups of \(7\): \(35\div7=5\) groups, so the total weight is \(5\) pounds. 3. This means \(5\div\frac{1}{7}=35\): there are \(35\) one-seventh-pound packets in \(5\) pounds.

\(5\) pounds, and \(5\div\frac{1}{7}=35\).

- The unit-fraction group size stays the same in all three expressions. - Think about what happens to the number of equal groups when the total amount is halved or doubled.

1. Halving the whole-number dividend from \(6\) to \(3\) halves the group count: \(30\div2=15\). 2. Doubling the dividend from \(6\) to \(12\) doubles the group count: \(30\times2=60\). 3. Therefore \(3\div\frac{1}{5}=15\) and \(12\div\frac{1}{5}=60\).

\(3\div\frac{1}{5}=15\) and \(12\div\frac{1}{5}=60\). With the divisor fixed, halving the dividend halves the quotient and doubling the dividend doubles the quotient.

- For each expression, count how many divisor-sized pieces fit in one whole. - Then scale by the number of wholes.

1. Four wholes contain \(4\times6=24\) sixths. 2. Six wholes contain \(6\times4=24\) fourths. 3. Both quotients are \(24\), so they are equal.

The quotients are equal; both are \(24\). Four wholes contain \(4\times6=24\) sixths, while six wholes contain \(6\times4=24\) fourths.

- Determine how many quarter-mile pieces are in one whole mile. - Test each allowed whole-number length and compare its piece count with \(10\).

1. One whole mile contains \(4\) quarter-mile pieces. 2. For lengths of \(1\), \(2\), \(3\), and \(4\) miles, the piece counts are \(4\), \(8\), \(12\), and \(16\). 3. The counts greater than \(10\) come from lengths of \(3\) miles and \(4\) miles.

The possible lengths are \(3\) miles and \(4\) miles. Each whole mile contains \(4\) fourths, so \(3\) miles make \(12\) pieces and \(4\) miles make \(16\) pieces; these are the only allowed counts greater than \(10\).

- Count how many divisor-sized pieces fit in one whole for each expression. - Multiply that count by the number of wholes.

1. Two wholes contain \(2\times7=14\) sevenths. 2. Seven wholes contain \(7\times2=14\) halves. 3. Both divisions count \(14\) unit-fraction groups.

Both quotients are \(14\). Two wholes contain \(14\) sevenths, and seven wholes contain \(14\) halves, so each division counts the same number of divisor-sized groups.

- First determine how many pieces of the given fractional length fit in all \(3\) yards. - Only after finding the total number of pieces should you account for the pieces set aside.

1. Find the total number of pieces: \(3\div\frac{1}{6}=18\). 2. Subtract the \(4\) pieces set aside: \(18-4=14\). 3. Therefore \(14\) pieces remain for decorating.

\(14\) pieces remain.

- Count how many twelfths make one whole before considering both circles. - Ask whether the proposed answer accounts for every whole shown in the model.

1. One whole contains \(12\) pieces of size \(\frac{1}{12}\). 2. There are \(2\) wholes, so there are \(2\times12=24\) such pieces. 3. Therefore \(2\div\frac{1}{12}=24\). Mateo counted the twelfths in only one whole.

The correct quotient is \(24\). Mateo counted only one whole instead of both wholes shown in the model.

- Compare both the number of wholes and the size of the unit-fraction pieces being counted. - Smaller pieces fit more times into a whole. - After predicting, count how many of each unit fraction make one whole and scale up.

1. The second expression starts with more wholes and counts smaller pieces, so it should have the larger quotient. 2. \(3\div\frac{1}{8}=3\times8=24\). 3. \(4\div\frac{1}{10}=4\times10=40\). 4. Since \(40>24\), the prediction is confirmed.

\(4\div\frac{1}{10}\) should be greater because it starts with more wholes and counts smaller pieces. The exact quotients are \(24\) and \(40\), confirming the prediction.

- First determine how many fractional-size bottles can be filled from all the drink mix. - After you know the bottle count, group those bottles by the crate capacity. - Keep the two division steps separate because they answer different questions.

1. Find the number of bottles: \(5\div\frac{1}{8}=40\). 2. Pack the \(40\) bottles in groups of \(6\): \(40=6\times6+4\). 3. Therefore there are \(6\) full crates and \(4\) bottles left over.

\(6\) full crates can be packed, with \(4\) bottles left over.

- First find how many of the equal pieces come from one foot. - A foot divided into that many equal pieces gives the unit-fraction length. - Use the unit fraction in a division equation to check the total piece count.

1. Divide the total piece count by the number of feet: \(63\div7=9\). So each foot contains \(9\) equal pieces. 2. If one foot is split into \(9\) equal pieces, each piece is \(\frac{1}{9}\) foot long. 3. Check with the required division equation: \(7\div\frac{1}{9}=63\).

Each piece is \(\frac{1}{9}\) foot long. The corresponding whole-number ÷ unit-fraction equation is \(7\div\frac{1}{9}=63\), because \(63\) ninth-foot pieces fit in \(7\) feet.

- For each divisor, ask how many of those unit-fraction pieces make one whole. - Multiply that count by the number of wholes in the dividend. - Look for different factor pairs that produce the same total count.

1. a) Each whole contains \(10\) tenths, so \(3\) wholes contain \(3\times10=30\) tenths. 2. b) Each whole contains \(6\) sixths, so \(5\) wholes contain \(5\times6=30\) sixths. 3. c) Each whole contains \(5\) fifths, so \(6\) wholes contain \(6\times5=30\) fifths. 4. All three quotients equal \(30\).

a), b), and c) all equal \(30\).

- The pieces are all the same size, so the rope lengths can be combined before dividing. - Think about how many sixths fit in one foot, then in the total number of feet.

1. Combine the rope lengths: \(1+2+3=6\) feet. 2. Count one-sixth-foot pieces in the total length: \(6\div\frac{1}{6}=36\). 3. Therefore \(36\) pieces are made.

\(36\) pieces are made altogether.

- Find how many fractional-size pieces fit in each whole-number length. - Compare the two resulting piece counts only after solving both divisions. - The final question asks for the difference between the counts.

1. The first cord produces \(8\div\frac{1}{4}=32\) pieces. 2. The second cord produces \(6\div\frac{1}{3}=18\) pieces. 3. The first cord produces \(32-18=14\) more pieces.

The \(8\)-foot cord produces more pieces, by \(14\) pieces.

- First determine how many unit-fraction pieces fit in the full roll. - The second step uses the stated fraction of that piece count. - Keep length units separate from the final count of pieces.

1. Find the total number of pieces: \(9\div\frac{1}{12}=108\). 2. Find one third of the pieces: \(\frac{1}{3}\times108=36\). 3. Therefore \(36\) pieces are used.

\(36\) pieces are used.

- First find how many one-eighth-mile lengths fit in \(6\) miles. - Distinguish spaces between markers from the marker locations themselves. - A row of intervals has one more endpoint than interval.

1. The number of intervals is \(6\div\frac{1}{8}=48\). 2. The first marker is at the start before any interval is traveled. 3. Each of the \(48\) intervals ends at another marker, so there are \(48+1=49\) markers. 4. The marker count is one greater than the interval count because markers include both endpoints.

There are \(48\) intervals and \(49\) checkpoint markers. The markers include both the starting and ending points.

- Count the whole bars and the equal parts within one whole in each model. - Use the size of one small part as the divisor in each expression. - Compare the two divisor sizes before comparing the two piece counts.

1. Both models show \(2\) whole yards. In panel a), each yard is split into \(12\) equal parts, so each piece is \(\frac{1}{12}\) yard. In panel b), each yard is split into \(6\) equal parts, so each piece is \(\frac{1}{6}\) yard. 2. Panel a) represents \(2\div\frac{1}{12}=24\) pieces. Panel b) represents \(2\div\frac{1}{6}=12\) pieces. 3. The second piece is twice as long as the first, so only half as many pieces fit in the same total length.

a) Both rolls are \(2\) yards long. The piece sizes are \(\frac{1}{12}\) yard and \(\frac{1}{6}\) yard. b) \(2\div\frac{1}{12}=24\) and \(2\div\frac{1}{6}=12\). c) Doubling the piece length halves the piece count.

- First count how many unit-fraction pieces fit in the full strip. - Subtract the number of pieces used. - Convert the remaining piece count back into a length using the size of one piece.

1. The strip makes \(2\div\frac{1}{7}=14\) pieces. 2. After \(5\) pieces are used, \(14-5=9\) pieces remain. 3. Nine pieces of length \(\frac{1}{7}\) yard total \(9\times\frac{1}{7}=\frac{9}{7}=1\frac{2}{7}\) yards.

\(1\frac{2}{7}\) yards remain.

- For each divisor, determine how many of those unit fractions make one whole. - Multiply that count by the whole-number dividend. - Look for the common total created by different factor pairs.

1. a) One whole contains \(12\) twelfths, so the quotient is \(12\). 2. b) Each whole contains \(6\) sixths, so \(2\) wholes contain \(2\times6=12\) sixths. 3. c) Each whole contains \(4\) fourths, so \(3\) wholes contain \(3\times4=12\) fourths. 4. In each expression, the number of wholes times the number of unit-fraction pieces per whole equals \(12\).

a) \(12\) b) \(12\) c) \(12\)

- Multiply the reported count by the length of one piece to test the claim. - Determine how many eighth-foot pieces make one whole foot. - Rebuild the original \(10\)-foot length to check the corrected count.

1. If there were \(18\) pieces, their total length would be \(18\times\frac{1}{8}=\frac{18}{8}=2\frac{1}{4}\) feet, not \(10\) feet. 2. Each whole foot contains \(8\) pieces of length \(\frac{1}{8}\) foot. 3. Ten feet contain \(10\times8=80\) pieces, so \(10\div\frac{1}{8}=80\). 4. Check: \(80\times\frac{1}{8}=10\) feet.

The report is impossible because \(18\) pieces total only \(2\frac{1}{4}\) feet. The correct count is \(80\) pieces.

- Use estimation only to recognize that the two answers should be close. - For each expression, count the unit-fraction pieces in one whole and then in all the wholes. - Compare the exact counts and explain why their small difference matters.

1. Both divisors are unit fractions with denominators close to \(10\), and the dividends differ by only \(1\). A rough estimate would place both quotients near the mid-forties, so it would not settle which is greater. 2. \(5\div\frac{1}{9}=45\), because \(5\) wholes contain \(5\times9\) ninths. 3. \(4\div\frac{1}{11}=44\), because \(4\) wholes contain \(4\times11\) elevenths. 4. Therefore the first quotient is greater by \(1\). Exact calculation is needed for this near-tie.

\(5\div\frac{1}{9}=45\) and \(4\div\frac{1}{11}=44\). The first is greater by \(1\); rough estimation alone does not resolve the near-tie.

- Count the equal parts in one whole bar in each panel. - Keep the total ribbon length the same in both division expressions. - Compare how many divisor-sized pieces fit in one yard, then extend that relationship to \(L\) yards.

1. Panel a) divides one yard into \(4\) equal parts, so each piece is \(\frac{1}{4}\) yard. Panel b) divides one yard into \(8\) equal parts, so each piece is \(\frac{1}{8}\) yard. 2. For a ribbon of \(L\) whole yards, the counts are \(L\div\frac{1}{4}\) and \(L\div\frac{1}{8}\). 3. Each yard contains \(4\) fourth-yard pieces but \(8\) eighth-yard pieces. Every yard therefore contributes twice as many pieces in panel b), so the total count is always doubled.

a) Panel a): \(\frac{1}{4}\) yard; panel b): \(\frac{1}{8}\) yard. b) \(L\div\frac{1}{4}\) and \(L\div\frac{1}{8}\). c) Each yard contains \(4\) fourths but \(8\) eighths, so the second piece count is always twice the first.

- Compare the new dividend with the dividend in the known fact. - The divisor stays the same, so the number of unit-fraction groups scales with the number of wholes. - Apply the same scale factor to the known quotient.

1. The dividend \(12\) is \(4\) times the dividend \(3\). 2. With the same unit-fraction divisor, multiplying the number of wholes by \(4\) multiplies the group count by \(4\). 3. Therefore \(12\div\frac{1}{16}=4\times48=192\).

\(12\div\frac{1}{16}=192\).

- Count the whole bars and the equal parts in each whole directly from the model. - The divisor is the size of one small part, not the number of parts. - After finding the available piece count, convert any leftover pieces back into a ribbon length.

1. The model shows \(3\) whole bars, so there are \(3\) yards in all. Each whole is divided into \(4\) equal parts, so one small piece is \(\frac{1}{4}\) yard. 2. The number of pieces is \(3\div\frac{1}{4}=12\). 3. Since \(12\ge10\), there is enough ribbon. Two quarter-yard pieces remain, and \(2\times\frac{1}{4}=\frac{1}{2}\) yard.

a) \(3\) yards total; each piece is \(\frac{1}{4}\) yard. b) \(3\div\frac{1}{4}=12\). c) Yes; \(\frac{1}{2}\) yard remains.

- Read both the number of wholes and the number of equal parts in each whole from each model. - Smaller unit-fraction pieces fit more times into one whole. - Make a size prediction from the models before multiplying the pieces-per-whole count by the number of wholes.

1. Model a) shows \(2\) wholes divided into eighths, so it represents \(2\div\frac{1}{8}=16\). 2. Model b) shows \(3\) wholes divided into fifths, so it represents \(3\div\frac{1}{5}=15\). 3. Model a) has fewer wholes but smaller divisor-sized pieces. Eight eighths fit in each whole, compared with five fifths in each whole. 4. Therefore model a) has the greater quotient: \(16>15\).

Model a) is greater. It represents \(2\div\frac{1}{8}=16\), while model b) represents \(3\div\frac{1}{5}=15\). The smaller eighth-size pieces more than offset the smaller number of wholes.

- Think separately about the effect of doubling the total amount and making each piece half as large. - Each change affects how many groups fit. - Compare the final counts after considering both effects.

1. \(2\div\frac{1}{4}=8\), because each whole contains \(4\) fourths. 2. \(4\div\frac{1}{8}=32\), because each whole contains \(8\) eighths. 3. Doubling the total amount doubles the group count, and halving the piece size doubles the count again. 4. Therefore the second quotient is \(2\times2=4\) times the first: \(32=4\times8\).

The quotients are \(8\) and \(32\). The second is four times the first.

- Let the denominator of the unit-fraction piece tell how many pieces fit in one yard. - List the multiples of \(4\) that are strictly between \(20\) and \(30\). - Use each possible total piece count to determine the piece length.

1. If each piece is \(\frac{1}{d}\) yard, then one yard contains \(d\) pieces and \(4\) yards contain \(4d\) pieces. 2. The piece count must be strictly between \(20\) and \(30\). 3. The multiples of \(4\) in that interval are \(24\) and \(28\). 4. A count of \(24\) means \(d=6\), and a count of \(28\) means \(d=7\). 5. Therefore the possible piece lengths are \(\frac{1}{6}\) yard and \(\frac{1}{7}\) yard.

The possible piece lengths are \(\frac{1}{6}\) yard and \(\frac{1}{7}\) yard.