Free math worksheets

- The quotient should be smaller than the original unit fraction. - Test a choice by asking whether three equal copies rebuild the starting fraction.

1. Dividing \(\frac{1}{7}\) into \(3\) equal shares makes each share \(\frac{1}{21}\). 2. Check: \(\frac{1}{21}\times3=\frac{3}{21}=\frac{1}{7}\).

\(\frac{1}{21}\)

- Start with the fraction represented by the shaded part. - Splitting that fraction among four equal shares makes each share one fourth as large.

1. The shaded amount is \(\frac{1}{7}\) of the whole. 2. Sharing it among \(4\) people gives \(\frac{1}{7}\div4=\frac{1}{28}\).

Each person gets \(\frac{1}{28}\) of the whole.

- Focus on the shaded unit fraction as the total being shared. - Splitting one part into two equal pieces doubles the total number of equal pieces in the whole.

1. The shaded amount is \(\frac{1}{11}\) of the whole. 2. Divide it into \(2\) equal shares: \(\frac{1}{11}\div2=\frac{1}{22}\).

Each person receives \(\frac{1}{22}\) of the whole.

- Start with the one shaded half in the model. - Split that shaded amount into the two equal shares named in the problem. - Relate one new share to the whole cup.

1. The given amount is one of \(2\) equal parts of a whole cup. 2. Splitting that half into \(2\) equal shares makes \(4\) equal parts in the whole cup. 3. Each cup receives \(\frac{1}{4}\) cup, so \(\frac{1}{2}\div2=\frac{1}{4}\).

Each cup receives \(\frac{1}{4}\) cup.

- Think about what happens to a piece when it is split into several equal smaller pieces. - Check whether putting all the equal pieces back together would recreate the original length.

1. The \(\frac{1}{3}\) yard is shared equally among \(4\) pieces, so the calculation is \(\frac{1}{3}\div4\). 2. Splitting one third into \(4\) equal parts gives \(\frac{1}{12}\).

Each piece is \(\frac{1}{12}\) yard long.

- Focus only on the shaded fraction before thinking about the four shares. - Think about how the shaded part can be split into equal smaller parts.

1. The shaded amount is \(\frac{1}{5}\) of the whole pan. 2. Sharing that amount among \(4\) hikers gives \(\frac{1}{5}\div4=\frac{1}{20}\).

Each hiker receives \(\frac{1}{20}\) of the whole pan.

- Start with the fraction of the entire garden that is available. - Splitting one fraction into equal shares makes each share smaller. - Check that five equal shares rebuild the flower area.

1. The flower area is \(\frac{1}{6}\) of the whole garden. 2. Dividing that fraction into \(5\) equal shares gives \(\frac{1}{6}\div5=\frac{1}{30}\).

Each plant type uses \(\frac{1}{30}\) of the whole garden.

- Both problems start with the same amount. - Think about whether dividing that amount among more groups makes each share larger or smaller.

1. \(\frac{1}{4}\div2=\frac{1}{8}\). 2. \(\frac{1}{4}\div4=\frac{1}{16}\). 3. Splitting the same unit fraction into fewer equal shares makes each share larger, so \(\frac{1}{8}>\frac{1}{16}\).

\(\frac{1}{4}\div2=\frac{1}{8}\) is greater than \(\frac{1}{4}\div4=\frac{1}{16}\).

- Treat the opening time as one quantity being shared equally. - Check that five equal clip lengths add back to the full opening time.

1. The total opening time is \(\frac{1}{3}\) minute and it is split into \(5\) equal parts. 2. \(\frac{1}{3}\div5=\frac{1}{15}\).

Each clip lasts \(\frac{1}{15}\) minute.

- Imagine the whole first divided into ninths. - Then subdivide one of those ninths into six equal parts.

1. One ninth is one of \(9\) equal parts of a whole. 2. Splitting that one ninth into \(6\) equal shares makes \(9\times6=54\) equal shares of the whole. 3. Each share is \(\frac{1}{54}\).

\(\frac{1}{54}\)

- The whole-number divisor tells how many equal shares the fractional area is split into. - Your answer should be smaller than the original unit fraction. - Multiply one share by the number of varieties to check that you recover the original area.

1. Divide the unit fraction by the number of equal shares: \(\frac{1}{8}\div6=\frac{1}{48}\). 2. Each variety receives \(\frac{1}{48}\) acre. 3. Check: \(6\times\frac{1}{48}=\frac{6}{48}=\frac{1}{8}\).

Each variety receives \(\frac{1}{48}\) acre.

- Six equal container amounts must combine to the original \(\frac{1}{5}\) cup. - Use the relationship between one share and the total to judge the quotient's size. - A valid multiplication check should rebuild the starting amount exactly.

1. Divide the unit fraction by \(6\): \(\frac{1}{5}\div6=\frac{1}{30}\) cup. 2. The reverse multiplication check is \(6\times\frac{1}{30}=\frac{6}{30}=\frac{1}{5}\). 3. The check confirms that six equal shares rebuild the original amount.

Each container gets \(\frac{1}{30}\) cup, and \(6\times\frac{1}{30}=\frac{1}{5}\).

- Begin with the meaning of \(\frac{1}{4}\) as one of four equal parts. - Subdivide that one part into the seven equal shares named in the problem. - Decide how many equal small parts the whole liter would contain.

1. The sample is one of \(4\) equal parts of a liter. 2. Dividing that one-fourth-liter part into \(7\) equal shares makes \(4\times7=28\) equal parts in one liter. 3. Each container receives \(\frac{1}{28}\) liter, so \(\frac{1}{4}\div7=\frac{1}{28}\).

Each container receives \(\frac{1}{28}\) liter.

- Rewrite the starting unit fraction using twentieths. - Ask how many copies of the quotient make the original fraction.

1. The quotient \(\frac{1}{20}\) must be repeated \(n\) times to rebuild \(\frac{1}{4}\). 2. Since \(\frac{1}{4}=\frac{5}{20}\), five copies of \(\frac{1}{20}\) make \(\frac{1}{4}\). 3. Therefore \(n=5\).

\(n=5\).

- Make the unit fraction the total amount being shared. - Make the whole number the number of equal shares. - Interpret the quotient as the size of one share.

1. One valid story is: \(\frac{1}{5}\) pound of clay is shared equally among \(6\) students. 2. \(\frac{1}{5}\div6=\frac{1}{30}\). 3. In this story, each student receives \(\frac{1}{30}\) pound of clay.

Answers will vary. A valid story must use \(\frac{1}{5}\) as the total amount, \(6\) as the number of equal shares, and \(\frac{1}{30}\) as the size of each share.

- All three expressions start with the same amount. - Think about how the size of one share changes when the number of shares increases.

1. The same amount \(\frac{1}{3}\) is split into different numbers of equal shares, so fewer shares give a larger share. 2. \(\frac{1}{3}\div2=\frac{1}{6}\), \(\frac{1}{3}\div5=\frac{1}{15}\), and \(\frac{1}{3}\div8=\frac{1}{24}\). 3. Therefore \(\frac{1}{6}>\frac{1}{15}>\frac{1}{24}\).

\(\frac{1}{3}\div2>\frac{1}{3}\div5>\frac{1}{3}\div8\), with quotients \(\frac{1}{6},\frac{1}{15},\frac{1}{24}\).

- Think about subdividing one equal part into three smaller equal parts. - Compare each original denominator with the denominator of its quotient.

1. \(\frac{1}{2}\div3=\frac{1}{6}\). 2. \(\frac{1}{5}\div3=\frac{1}{15}\). 3. \(\frac{1}{8}\div3=\frac{1}{24}\). 4. Each original unit fraction is split into \(3\) equal parts, so the whole is partitioned into three times as many equal parts.

The quotients are \(\frac{1}{6},\frac{1}{15},\frac{1}{24}\). Each denominator is \(3\) times the original denominator.

- Read the multiplication as a statement about equal copies. - Reverse the relationship to describe one share of the total.

1. The multiplication says seven equal copies of \(\frac{1}{56}\) make \(\frac{1}{8}\). 2. Therefore splitting \(\frac{1}{8}\) into \(7\) equal shares gives \(\frac{1}{56}\). 3. The division equation is \(\frac{1}{8}\div7=\frac{1}{56}\).

\(\frac{1}{8}\div7=\frac{1}{56}\).

- Think about how many equal pieces the whole has after both stages of partitioning. - Compare the total number of final equal pieces in the two expressions.

1. Splitting \(\frac{1}{5}\) into \(7\) equal shares gives \(\frac{1}{35}\). 2. Splitting \(\frac{1}{7}\) into \(5\) equal shares also gives \(\frac{1}{35}\). 3. In both cases, the whole ends up partitioned into \(5\times7=35\) equal parts.

The quotients are equal; both are \(\frac{1}{35}\).

- Think about what happens to the size of a piece when a unit fraction is shared among several cups. - The quotient should be smaller than the original \(\frac{1}{6}\) liter. - Look for a model whose single shaded part has the correct size.

1. The division is \(\frac{1}{6}\div4\). 2. Splitting one sixth into \(4\) equal pieces creates \(24\) equal pieces in the whole liter, so one share is \(\frac{1}{24}\) liter. 3. Model a) shows one of \(24\) equal parts, so it represents one cup. Model b) shows \(\frac{4}{6}\), not the quotient.

a); each cup contains \(\frac{1}{24}\) liter.

- Find the size of one equal share in each situation. - Both answers are unit fractions, so compare their denominators carefully. - Remember that a larger denominator means smaller equal parts when the numerator is \(1\).

1. In a), \(\frac{1}{4}\div2=\frac{1}{8}\). 2. In b), \(\frac{1}{6}\div3=\frac{1}{18}\). 3. Since \(\frac{1}{8}>\frac{1}{18}\), a person in a) receives the larger share.

a) gives the larger share: \(\frac{1}{8}\) compared with \(\frac{1}{18}\).

- First find the size of one equal share of the original fraction of an hour. - After you have the fraction of an hour, use the number of minutes in one hour to interpret it.

1. Divide the unit fraction by the number of clips: \(\frac{1}{12}\div5=\frac{1}{60}\) hour. 2. One hour is \(60\) minutes, so \(\frac{1}{60}\) hour is \(1\) minute.

Each clip is \(\frac{1}{60}\) hour, or \(1\) minute, long.

- Focus on what it means to share one fixed amount between two people. - Compare the size of one new share with the shaded amount you started with. - Use the model to think about how many equal pieces the whole would have after the split.

1. The shaded fifth is the amount being split into \(2\) equal shares. 2. Each share must be smaller than the original \(\frac{1}{5}\). 3. Splitting one fifth into \(2\) equal parts gives \(\frac{1}{10}\), so \(\frac{1}{5}\div2=\frac{1}{10}\).

Tara’s claim is false. \(\frac{1}{5}\div2=\frac{1}{10}\), which is smaller than \(\frac{1}{5}\).

- Rewrite the original amount using twelfths so it can be compared directly with one cup’s share. - Count how many equal shares of the stated size make the original fraction. - Check by dividing the original unit fraction by your number of cups.

1. The question asks how many equal shares of size \(\frac{1}{12}\) make the original \(\frac{1}{2}\) liter. 2. Rewrite \(\frac{1}{2}=\frac{6}{12}\). 3. Therefore \(6\) shares of \(\frac{1}{12}\) make \(\frac{1}{2}\), so there are \(6\) cups. 4. This matches \(\frac{1}{2}\div6=\frac{1}{12}\).

There are \(6\) cups.

- Find the equal share produced in each situation. - Both answers will be unit fractions, which makes comparing them easier. - For unit fractions, think about how the denominator affects the size of one part.

1. In a), \(\frac{1}{8}\div2=\frac{1}{16}\). 2. In b), \(\frac{1}{4}\div8=\frac{1}{32}\). 3. Since \(\frac{1}{16}>\frac{1}{32}\), the share in a) is larger.

a) \(\frac{1}{16}\) b) \(\frac{1}{32}\) The share in a) is larger.

- Keep separate the piece's length in meters and its size relative to the original strip. - One of \(7\) equal pieces is \(\frac{1}{7}\) of the strip, but the strip itself is only \(\frac{1}{9}\) meter. - Use multiplication to test whether seven copies of your length rebuild the strip.

1. Divide the original unit fraction by \(7\): \(\frac{1}{9}\div7=\frac{1}{63}\) meter. 2. Because the strip is divided into \(7\) equal pieces, one piece is \(\frac{1}{7}\) of the original strip. 3. Thus each piece is \(\frac{1}{63}\) meter and \(\frac{1}{7}\) of the starting strip.

Each piece is \(\frac{1}{63}\) meter, which is \(\frac{1}{7}\) of the original strip.

- First divide the original unit fraction into the stated number of equal zones. - Then compare the two unit fractions by considering the size of equal parts with different denominators.

1. One zone has length \(\frac{1}{3}\div4=\frac{1}{12}\) mile. 2. Compare \(\frac{1}{12}\) and \(\frac{1}{10}\). With the same numerator \(1\), the fraction with denominator \(12\) is smaller. 3. Therefore each zone is shorter than \(\frac{1}{10}\) mile.

Each zone is \(\frac{1}{12}\) mile long, which is shorter than \(\frac{1}{10}\) mile.

- Start with one of the equal parts represented by the original unit fraction. - Think about how splitting that single part again changes the total number of equal pieces in the whole. - The quotient should be smaller than \(\frac{1}{11}\).

1. Divide the unit fraction by \(9\): \(\frac{1}{11}\div9=\frac{1}{99}\). 2. One eleventh of the mile is one of \(11\) equal parts. Splitting that part into \(9\) equal pieces makes \(11\times9=99\) equal pieces in the whole mile. 3. Therefore one zone is \(\frac{1}{99}\) mile.

One zone is \(\frac{1}{99}\) mile.

- Divide the original unit fraction into the stated number of equal shares. - The result will still be a unit fraction. - Compare unit fractions by thinking about the size of equal parts with different denominators.

1. One share is \(\frac{1}{12}\div4=\frac{1}{48}\) mile. 2. Compare \(\frac{1}{48}\) and \(\frac{1}{50}\). For unit fractions, the smaller denominator gives the larger fraction. 3. Therefore \(\frac{1}{48}>\frac{1}{50}\).

One part is \(\frac{1}{48}\) mile, which is greater than \(\frac{1}{50}\) mile.

- Work backward by combining all of the equal shares. - Multiplication reverses the equal-sharing division. - Simplify the rebuilt fraction before stating the original unit fraction.

1. Four equal shares of \(\frac{1}{44}\) rebuild the original fraction. 2. Multiply: \(4\times\frac{1}{44}=\frac{4}{44}=\frac{1}{11}\). 3. Therefore the original division was \(\frac{1}{11}\div4=\frac{1}{44}\).

The original unit fraction was \(\frac{1}{11}\).

- First find the size of one equal share of the original unit fraction. - Then combine the stated number of equal shares. - Simplify the resulting fraction of the whole container.

1. One cup share is \(\frac{1}{3}\div4=\frac{1}{12}\) of the whole container. 2. Three such shares contain \(3\times\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\). 3. Therefore the combined amount is \(\frac{1}{4}\) of the whole container.

The \(3\) shares together make \(\frac{1}{4}\) of the whole container.

- Think about how the original denominator combines with the number of equal shares. - Compare the products formed by each denominator and whole-number divisor. - Equal final denominators produce equal unit-fraction quotients.

1. a) \(\frac{1}{6}\div4=\frac{1}{24}\). 2. b) \(\frac{1}{8}\div3=\frac{1}{24}\). 3. c) \(\frac{1}{12}\div2=\frac{1}{24}\). 4. In each case, splitting one unit-fraction part into the stated number of equal shares creates \(24\) equal parts in the whole because \(6\times4=8\times3=12\times2=24\).

a) \(\frac{1}{24}\) b) \(\frac{1}{24}\) c) \(\frac{1}{24}\)

- Think about what it means to divide one amount into two equal shares. - Compare that idea with selecting one of two equal parts of the same amount. - Check that both descriptions lead to the same unit fraction.

1. Split one fifteenth into \(2\) equal shares: \(\frac{1}{15}\div2=\frac{1}{30}\). 2. Taking one half of \(\frac{1}{15}\) gives \(\frac{1}{2}\times\frac{1}{15}=\frac{1}{30}\). 3. Both operations find one of two equal parts of the original amount.

\(\frac{1}{15}\div2=\frac{1}{30}\), which is also one half of \(\frac{1}{15}\).

- Interpret \(\frac{1}{9}\) as one equal part of a whole. - Subdivide that one part into the eight equal shares named by the divisor. - Check that the quotient is smaller than the amount being shared.

1. The original amount is one of \(9\) equal parts of a whole. 2. Splitting that one part into \(8\) equal shares makes \(9\times8=72\) equal shares in the whole, not \(9+8=17\). 3. Therefore \(\frac{1}{9}\div8=\frac{1}{72}\). 4. The quotient is smaller than \(\frac{1}{9}\), as an equal share must be.

The student’s rule is incorrect. \(\frac{1}{9}\div8=\frac{1}{72}\).

- Follow the sharing process one stage at a time. - Count how many final equal packets result from the two stages together. - Compare the two-stage quotient with dividing by that total number of packets.

1. First share: \(\frac{1}{5}\div3=\frac{1}{15}\). 2. Split each share into \(4\) packets: \(\frac{1}{15}\div4=\frac{1}{60}\). 3. There are \(3\times4=12\) equal packets altogether, and \(\frac{1}{5}\div12=\frac{1}{60}\). 4. Both procedures give the same packet size.

One packet is \(\frac{1}{60}\) of the whole strip, and \(\frac{1}{5}\div12=\frac{1}{60}\).

- Find each equal share before comparing the situations. - Both answers are unit fractions, so their denominators can be compared directly. - Consider both the starting fraction and the number of people sharing it.

1. In a), \(\frac{1}{7}\div4=\frac{1}{28}\). 2. In b), \(\frac{1}{8}\div3=\frac{1}{24}\). 3. Since \(\frac{1}{24}>\frac{1}{28}\), b) gives the larger share. 4. Although \(\frac{1}{8}<\frac{1}{7}\), it is split among fewer people, which more than offsets the smaller starting amount.

b) gives the larger share: \(\frac{1}{24}\) compared with \(\frac{1}{28}\).

- Describe the denominator of one share using \(8\) and the unknown number of students. - For unit fractions, a larger denominator means a smaller value. - Find the multiple of \(8\) that lies strictly between the two benchmark denominators.

1. If the ribbon is shared among a whole-number count of students, each share has the form \(\frac{1}{8}\div\text{(student count)}\). 2. The final denominator is \(8\) times the student count. 3. For the share to lie strictly between \(\frac{1}{56}\) and \(\frac{1}{40}\), that denominator must lie strictly between \(40\) and \(56\). 4. The only multiple of \(8\) strictly between \(40\) and \(56\) is \(48\), and \(48=8\times6\). 5. Therefore \(6\) students share the ribbon, and each receives \(\frac{1}{48}\) yard.

\(6\) students; each share is \(\frac{1}{48}\) yard.