Practice · Smallest numerator on top, largest denominator underneath · Sensim Math

Variant 1 easy answer:

\frac{21}{2}

An improper fraction divided by $\frac{3}{8}$ and the same fraction divided by $1\frac{1}{6}$ both give whole numbers. Find the smallest such improper fraction.

Show solution

1 · Understandwhat's really being asked

One improper fraction divides evenly by both 3/8 and 1 and 1/6. We want the smallest one that does.

Givens

Dividing by 3/8 gives a whole number.
Dividing by 1 and 1/6 gives a whole number.
The fraction is improper, so it is at least 1.

Unknowns

The smallest improper fraction meeting both conditions.

Constraints

Both conditions apply to the same fraction at once.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Turn both divisions into multiplications and each one says the same two things: what the numerator must contain and what the denominator may be. Smallest then means smallest on top and largest underneath.

3 · Execute4 carry out the plan

1Write both divisors as fractions

#7 Identify Subproblems 6.NS.A.1

1 and 1/6 is 7 over 6 as one fraction.

\frac{3}{8},\quad 1\frac{1}{6} = \frac{7}{6}

Two divisors, both improper-fraction shaped.

2Say what each condition demands

#13 Convert to Algebra 6.NS.A.1

Dividing by a fraction multiplies by its reciprocal, so the unknown's numerator must be a multiple of 3 and of 7, and its denominator must divide both 8 and 6.

\text{top} \in \{3, 7\}\text{-multiples},\quad \text{bottom} \mid 8,\ 6

One rule for the top, one for the bottom.

3Take the smallest top

#3 Eliminate Possibilities 6.NS.B.4

The smallest number that is a multiple of both 3 and 7.

\text{lcm}(3, 7) = 21

The numerator is 21.

4Take the largest bottom

#3 Eliminate Possibilities 6.NS.B.4

A bigger denominator makes the fraction smaller, so take the largest one allowed: the greatest common factor of 8 and 6.

\text{gcf}(8, 6) = 2

The denominator is 2.

Answer:

\frac{21}{2}

4 · Reviewdoes it hold up?

Both divisions come out whole: $\frac{21}{2} \div \frac{3}{8} = 28$ and $\frac{21}{2} \div \frac{7}{6} = 9$ . And 10 and 1/2 is at least 1, so the fraction really is improper.

Another way: Trying fractions one at a time also works, but there is no way to know when to stop until the two rules are written down.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Turning each division into a condition on numerator and denominator.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using an lcm on top and a gcf underneath.

💡Takeaway. To make a fraction small, make the top as small as the rules allow and the bottom as big.

Variant 2 easy answer:

\frac{111}{2}

An improper fraction divided by $\frac{6}{4}$ and the same fraction divided by $3\frac{7}{10}$ both give whole numbers. Find the smallest such improper fraction.

Show solution

1 · Understandwhat's really being asked

One improper fraction divides evenly by both 6/4 and 3 and 7/10. We want the smallest one that does.

Givens

Dividing by 6/4 gives a whole number.
Dividing by 3 and 7/10 gives a whole number.
The fraction is improper, so it is at least 1.

Unknowns

The smallest improper fraction meeting both conditions.

Constraints

Both conditions apply to the same fraction at once.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Turn both divisions into multiplications and each one says the same two things: what the numerator must contain and what the denominator may be. Smallest then means smallest on top and largest underneath.

3 · Execute4 carry out the plan

1Write both divisors as fractions

#7 Identify Subproblems 6.NS.A.1

3 and 7/10 is 37 over 10 as one fraction.

\frac{6}{4},\quad 3\frac{7}{10} = \frac{37}{10}

Two divisors, both improper-fraction shaped.

2Say what each condition demands

#13 Convert to Algebra 6.NS.A.1

Dividing by a fraction multiplies by its reciprocal, so the unknown's numerator must be a multiple of 3 and of 37, and its denominator must divide both 2 and 10.

\text{top} \in \{3, 37\}\text{-multiples},\quad \text{bottom} \mid 2,\ 10

One rule for the top, one for the bottom.

3Take the smallest top

#3 Eliminate Possibilities 6.NS.B.4

The smallest number that is a multiple of both 3 and 37.

\text{lcm}(3, 37) = 111

The numerator is 111.

4Take the largest bottom

#3 Eliminate Possibilities 6.NS.B.4

A bigger denominator makes the fraction smaller, so take the largest one allowed: the greatest common factor of 2 and 10.

\text{gcf}(2, 10) = 2

The denominator is 2.

Answer:

\frac{111}{2}

4 · Reviewdoes it hold up?

Both divisions come out whole: $\frac{111}{2} \div \frac{3}{2} = 37$ and $\frac{111}{2} \div \frac{37}{10} = 15$ . And 55 and 1/2 is at least 1, so the fraction really is improper.

Another way: Trying fractions one at a time also works, but there is no way to know when to stop until the two rules are written down.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Turning each division into a condition on numerator and denominator.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using an lcm on top and a gcf underneath.

💡Takeaway. To make a fraction small, make the top as small as the rules allow and the bottom as big.

Variant 3 easy answer:

\frac{60}{7}

An improper fraction divided by $\frac{12}{7}$ and the same fraction divided by $1\frac{3}{7}$ both give whole numbers. Find the smallest such improper fraction.

Show solution

1 · Understandwhat's really being asked

One improper fraction divides evenly by both 12/7 and 1 and 3/7. We want the smallest one that does.

Givens

Dividing by 12/7 gives a whole number.
Dividing by 1 and 3/7 gives a whole number.
The fraction is improper, so it is at least 1.

Unknowns

The smallest improper fraction meeting both conditions.

Constraints

Both conditions apply to the same fraction at once.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Turn both divisions into multiplications and each one says the same two things: what the numerator must contain and what the denominator may be. Smallest then means smallest on top and largest underneath.

3 · Execute4 carry out the plan

1Write both divisors as fractions

#7 Identify Subproblems 6.NS.A.1

1 and 3/7 is 10 over 7 as one fraction.

\frac{12}{7},\quad 1\frac{3}{7} = \frac{10}{7}

Two divisors, both improper-fraction shaped.

2Say what each condition demands

#13 Convert to Algebra 6.NS.A.1

Dividing by a fraction multiplies by its reciprocal, so the unknown's numerator must be a multiple of 12 and of 10, and its denominator must divide both 7 and 7.

\text{top} \in \{12, 10\}\text{-multiples},\quad \text{bottom} \mid 7,\ 7

One rule for the top, one for the bottom.

3Take the smallest top

#3 Eliminate Possibilities 6.NS.B.4

The smallest number that is a multiple of both 12 and 10.

\text{lcm}(12, 10) = 60

The numerator is 60.

4Take the largest bottom

#3 Eliminate Possibilities 6.NS.B.4

A bigger denominator makes the fraction smaller, so take the largest one allowed: the greatest common factor of 7 and 7.

\text{gcf}(7, 7) = 7

The denominator is 7.

Answer:

\frac{60}{7}

4 · Reviewdoes it hold up?

Both divisions come out whole: $\frac{60}{7} \div \frac{12}{7} = 5$ and $\frac{60}{7} \div \frac{10}{7} = 6$ . And 8 and 4/7 is at least 1, so the fraction really is improper.

Another way: Trying fractions one at a time also works, but there is no way to know when to stop until the two rules are written down.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Turning each division into a condition on numerator and denominator.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using an lcm on top and a gcf underneath.

💡Takeaway. To make a fraction small, make the top as small as the rules allow and the bottom as big.

Variant 4 easy answer:

\frac{195}{2}

An improper fraction divided by $\frac{10}{12}$ and the same fraction divided by $2\frac{11}{14}$ both give whole numbers. Find the smallest such improper fraction.

Show solution

1 · Understandwhat's really being asked

One improper fraction divides evenly by both 10/12 and 2 and 11/14. We want the smallest one that does.

Givens

Dividing by 10/12 gives a whole number.
Dividing by 2 and 11/14 gives a whole number.
The fraction is improper, so it is at least 1.

Unknowns

The smallest improper fraction meeting both conditions.

Constraints

Both conditions apply to the same fraction at once.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Turn both divisions into multiplications and each one says the same two things: what the numerator must contain and what the denominator may be. Smallest then means smallest on top and largest underneath.

3 · Execute4 carry out the plan

1Write both divisors as fractions

#7 Identify Subproblems 6.NS.A.1

2 and 11/14 is 39 over 14 as one fraction.

\frac{10}{12},\quad 2\frac{11}{14} = \frac{39}{14}

Two divisors, both improper-fraction shaped.

2Say what each condition demands

#13 Convert to Algebra 6.NS.A.1

Dividing by a fraction multiplies by its reciprocal, so the unknown's numerator must be a multiple of 5 and of 39, and its denominator must divide both 6 and 14.

\text{top} \in \{5, 39\}\text{-multiples},\quad \text{bottom} \mid 6,\ 14

One rule for the top, one for the bottom.

3Take the smallest top

#3 Eliminate Possibilities 6.NS.B.4

The smallest number that is a multiple of both 5 and 39.

\text{lcm}(5, 39) = 195

The numerator is 195.

4Take the largest bottom

#3 Eliminate Possibilities 6.NS.B.4

A bigger denominator makes the fraction smaller, so take the largest one allowed: the greatest common factor of 6 and 14.

\text{gcf}(6, 14) = 2

The denominator is 2.

Answer:

\frac{195}{2}

4 · Reviewdoes it hold up?

Both divisions come out whole: $\frac{195}{2} \div \frac{5}{6} = 117$ and $\frac{195}{2} \div \frac{39}{14} = 35$ . And 97 and 1/2 is at least 1, so the fraction really is improper.

Another way: Trying fractions one at a time also works, but there is no way to know when to stop until the two rules are written down.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Turning each division into a condition on numerator and denominator.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using an lcm on top and a gcf underneath.

💡Takeaway. To make a fraction small, make the top as small as the rules allow and the bottom as big.

Variant 5 medium answer:

\frac{175}{4}

An improper fraction divided by $\frac{7}{12}$ and the same fraction divided by $1\frac{9}{16}$ both give whole numbers. Find the smallest such improper fraction.

Show solution

1 · Understandwhat's really being asked

One improper fraction divides evenly by both 7/12 and 1 and 9/16. We want the smallest one that does.

Givens

Dividing by 7/12 gives a whole number.
Dividing by 1 and 9/16 gives a whole number.
The fraction is improper, so it is at least 1.

Unknowns

The smallest improper fraction meeting both conditions.

Constraints

Both conditions apply to the same fraction at once.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Turn both divisions into multiplications and each one says the same two things: what the numerator must contain and what the denominator may be. Smallest then means smallest on top and largest underneath.

3 · Execute4 carry out the plan

1Write both divisors as fractions

#7 Identify Subproblems 6.NS.A.1

1 and 9/16 is 25 over 16 as one fraction.

\frac{7}{12},\quad 1\frac{9}{16} = \frac{25}{16}

Two divisors, both improper-fraction shaped.

2Say what each condition demands

#13 Convert to Algebra 6.NS.A.1

Dividing by a fraction multiplies by its reciprocal, so the unknown's numerator must be a multiple of 7 and of 25, and its denominator must divide both 12 and 16.

\text{top} \in \{7, 25\}\text{-multiples},\quad \text{bottom} \mid 12,\ 16

One rule for the top, one for the bottom.

3Take the smallest top

#3 Eliminate Possibilities 6.NS.B.4

The smallest number that is a multiple of both 7 and 25.

\text{lcm}(7, 25) = 175

The numerator is 175.

4Take the largest bottom

#3 Eliminate Possibilities 6.NS.B.4

A bigger denominator makes the fraction smaller, so take the largest one allowed: the greatest common factor of 12 and 16.

\text{gcf}(12, 16) = 4

The denominator is 4.

Answer:

\frac{175}{4}

4 · Reviewdoes it hold up?

Both divisions come out whole: $\frac{175}{4} \div \frac{7}{12} = 75$ and $\frac{175}{4} \div \frac{25}{16} = 28$ . And 43 and 3/4 is at least 1, so the fraction really is improper.

Another way: Trying fractions one at a time also works, but there is no way to know when to stop until the two rules are written down.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Turning each division into a condition on numerator and denominator.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using an lcm on top and a gcf underneath.

💡Takeaway. To make a fraction small, make the top as small as the rules allow and the bottom as big.

Variant 6 medium answer:

\frac{715}{3}

An improper fraction divided by $\frac{11}{15}$ and the same fraction divided by $3\frac{2}{21}$ both give whole numbers. Find the smallest such improper fraction.

Show solution

1 · Understandwhat's really being asked

One improper fraction divides evenly by both 11/15 and 3 and 2/21. We want the smallest one that does.

Givens

Dividing by 11/15 gives a whole number.
Dividing by 3 and 2/21 gives a whole number.
The fraction is improper, so it is at least 1.

Unknowns

The smallest improper fraction meeting both conditions.

Constraints

Both conditions apply to the same fraction at once.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Turn both divisions into multiplications and each one says the same two things: what the numerator must contain and what the denominator may be. Smallest then means smallest on top and largest underneath.

3 · Execute4 carry out the plan

1Write both divisors as fractions

#7 Identify Subproblems 6.NS.A.1

3 and 2/21 is 65 over 21 as one fraction.

\frac{11}{15},\quad 3\frac{2}{21} = \frac{65}{21}

Two divisors, both improper-fraction shaped.

2Say what each condition demands

#13 Convert to Algebra 6.NS.A.1

Dividing by a fraction multiplies by its reciprocal, so the unknown's numerator must be a multiple of 11 and of 65, and its denominator must divide both 15 and 21.

\text{top} \in \{11, 65\}\text{-multiples},\quad \text{bottom} \mid 15,\ 21

One rule for the top, one for the bottom.

3Take the smallest top

#3 Eliminate Possibilities 6.NS.B.4

The smallest number that is a multiple of both 11 and 65.

\text{lcm}(11, 65) = 715

The numerator is 715.

4Take the largest bottom

#3 Eliminate Possibilities 6.NS.B.4

A bigger denominator makes the fraction smaller, so take the largest one allowed: the greatest common factor of 15 and 21.

\text{gcf}(15, 21) = 3

The denominator is 3.

Answer:

\frac{715}{3}

4 · Reviewdoes it hold up?

Both divisions come out whole: $\frac{715}{3} \div \frac{11}{15} = 325$ and $\frac{715}{3} \div \frac{65}{21} = 77$ . And 238 and 1/3 is at least 1, so the fraction really is improper.

Another way: Trying fractions one at a time also works, but there is no way to know when to stop until the two rules are written down.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Turning each division into a condition on numerator and denominator.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using an lcm on top and a gcf underneath.

💡Takeaway. To make a fraction small, make the top as small as the rules allow and the bottom as big.

Variant 7 medium answer:

\frac{60}{7}

An improper fraction divided by $\frac{4}{21}$ and the same fraction divided by $2\frac{1}{7}$ both give whole numbers. Find the smallest such improper fraction.

Show solution

1 · Understandwhat's really being asked

One improper fraction divides evenly by both 4/21 and 2 and 1/7. We want the smallest one that does.

Givens

Dividing by 4/21 gives a whole number.
Dividing by 2 and 1/7 gives a whole number.
The fraction is improper, so it is at least 1.

Unknowns

The smallest improper fraction meeting both conditions.

Constraints

Both conditions apply to the same fraction at once.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Turn both divisions into multiplications and each one says the same two things: what the numerator must contain and what the denominator may be. Smallest then means smallest on top and largest underneath.

3 · Execute4 carry out the plan

1Write both divisors as fractions

#7 Identify Subproblems 6.NS.A.1

2 and 1/7 is 15 over 7 as one fraction.

\frac{4}{21},\quad 2\frac{1}{7} = \frac{15}{7}

Two divisors, both improper-fraction shaped.

2Say what each condition demands

#13 Convert to Algebra 6.NS.A.1

Dividing by a fraction multiplies by its reciprocal, so the unknown's numerator must be a multiple of 4 and of 15, and its denominator must divide both 21 and 7.

\text{top} \in \{4, 15\}\text{-multiples},\quad \text{bottom} \mid 21,\ 7

One rule for the top, one for the bottom.

3Take the smallest top

#3 Eliminate Possibilities 6.NS.B.4

The smallest number that is a multiple of both 4 and 15.

\text{lcm}(4, 15) = 60

The numerator is 60.

4Take the largest bottom

#3 Eliminate Possibilities 6.NS.B.4

A bigger denominator makes the fraction smaller, so take the largest one allowed: the greatest common factor of 21 and 7.

\text{gcf}(21, 7) = 7

The denominator is 7.

Answer:

\frac{60}{7}

4 · Reviewdoes it hold up?

Both divisions come out whole: $\frac{60}{7} \div \frac{4}{21} = 45$ and $\frac{60}{7} \div \frac{15}{7} = 4$ . And 8 and 4/7 is at least 1, so the fraction really is improper.

Another way: Trying fractions one at a time also works, but there is no way to know when to stop until the two rules are written down.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Turning each division into a condition on numerator and denominator.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using an lcm on top and a gcf underneath.

💡Takeaway. To make a fraction small, make the top as small as the rules allow and the bottom as big.

Variant 8 medium answer:

\frac{150}{11}

An improper fraction divided by $\frac{10}{11}$ and the same fraction divided by $3\frac{9}{22}$ both give whole numbers. Find the smallest such improper fraction.

Show solution

1 · Understandwhat's really being asked

One improper fraction divides evenly by both 10/11 and 3 and 9/22. We want the smallest one that does.

Givens

Dividing by 10/11 gives a whole number.
Dividing by 3 and 9/22 gives a whole number.
The fraction is improper, so it is at least 1.

Unknowns

The smallest improper fraction meeting both conditions.

Constraints

Both conditions apply to the same fraction at once.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Turn both divisions into multiplications and each one says the same two things: what the numerator must contain and what the denominator may be. Smallest then means smallest on top and largest underneath.

3 · Execute4 carry out the plan

1Write both divisors as fractions

#7 Identify Subproblems 6.NS.A.1

3 and 9/22 is 75 over 22 as one fraction.

\frac{10}{11},\quad 3\frac{9}{22} = \frac{75}{22}

Two divisors, both improper-fraction shaped.

2Say what each condition demands

#13 Convert to Algebra 6.NS.A.1

Dividing by a fraction multiplies by its reciprocal, so the unknown's numerator must be a multiple of 10 and of 75, and its denominator must divide both 11 and 22.

\text{top} \in \{10, 75\}\text{-multiples},\quad \text{bottom} \mid 11,\ 22

One rule for the top, one for the bottom.

3Take the smallest top

#3 Eliminate Possibilities 6.NS.B.4

The smallest number that is a multiple of both 10 and 75.

\text{lcm}(10, 75) = 150

The numerator is 150.

4Take the largest bottom

#3 Eliminate Possibilities 6.NS.B.4

A bigger denominator makes the fraction smaller, so take the largest one allowed: the greatest common factor of 11 and 22.

\text{gcf}(11, 22) = 11

The denominator is 11.

Answer:

\frac{150}{11}

4 · Reviewdoes it hold up?

Both divisions come out whole: $\frac{150}{11} \div \frac{10}{11} = 15$ and $\frac{150}{11} \div \frac{75}{22} = 4$ . And 13 and 7/11 is at least 1, so the fraction really is improper.

Another way: Trying fractions one at a time also works, but there is no way to know when to stop until the two rules are written down.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Turning each division into a condition on numerator and denominator.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using an lcm on top and a gcf underneath.

💡Takeaway. To make a fraction small, make the top as small as the rules allow and the bottom as big.

Variant 9 hard answer:

\frac{649}{2}

An improper fraction divided by $\frac{11}{10}$ and the same fraction divided by $2\frac{11}{24}$ both give whole numbers. Find the smallest such improper fraction.

Show solution

1 · Understandwhat's really being asked

One improper fraction divides evenly by both 11/10 and 2 and 11/24. We want the smallest one that does.

Givens

Dividing by 11/10 gives a whole number.
Dividing by 2 and 11/24 gives a whole number.
The fraction is improper, so it is at least 1.

Unknowns

The smallest improper fraction meeting both conditions.

Constraints

Both conditions apply to the same fraction at once.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Turn both divisions into multiplications and each one says the same two things: what the numerator must contain and what the denominator may be. Smallest then means smallest on top and largest underneath.

3 · Execute4 carry out the plan

1Write both divisors as fractions

#7 Identify Subproblems 6.NS.A.1

2 and 11/24 is 59 over 24 as one fraction.

\frac{11}{10},\quad 2\frac{11}{24} = \frac{59}{24}

Two divisors, both improper-fraction shaped.

2Say what each condition demands

#13 Convert to Algebra 6.NS.A.1

Dividing by a fraction multiplies by its reciprocal, so the unknown's numerator must be a multiple of 11 and of 59, and its denominator must divide both 10 and 24.

\text{top} \in \{11, 59\}\text{-multiples},\quad \text{bottom} \mid 10,\ 24

One rule for the top, one for the bottom.

3Take the smallest top

#3 Eliminate Possibilities 6.NS.B.4

The smallest number that is a multiple of both 11 and 59.

\text{lcm}(11, 59) = 649

The numerator is 649.

4Take the largest bottom

#3 Eliminate Possibilities 6.NS.B.4

A bigger denominator makes the fraction smaller, so take the largest one allowed: the greatest common factor of 10 and 24.

\text{gcf}(10, 24) = 2

The denominator is 2.

Answer:

\frac{649}{2}

4 · Reviewdoes it hold up?

Both divisions come out whole: $\frac{649}{2} \div \frac{11}{10} = 295$ and $\frac{649}{2} \div \frac{59}{24} = 132$ . And 324 and 1/2 is at least 1, so the fraction really is improper.

Another way: Trying fractions one at a time also works, but there is no way to know when to stop until the two rules are written down.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Turning each division into a condition on numerator and denominator.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using an lcm on top and a gcf underneath.

💡Takeaway. To make a fraction small, make the top as small as the rules allow and the bottom as big.

Variant 10 hard answer:

\frac{63}{4}

An improper fraction divided by $\frac{7}{20}$ and the same fraction divided by $2\frac{6}{24}$ both give whole numbers. Find the smallest such improper fraction.

Show solution

1 · Understandwhat's really being asked

One improper fraction divides evenly by both 7/20 and 2 and 6/24. We want the smallest one that does.

Givens

Dividing by 7/20 gives a whole number.
Dividing by 2 and 6/24 gives a whole number.
The fraction is improper, so it is at least 1.

Unknowns

The smallest improper fraction meeting both conditions.

Constraints

Both conditions apply to the same fraction at once.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Turn both divisions into multiplications and each one says the same two things: what the numerator must contain and what the denominator may be. Smallest then means smallest on top and largest underneath.

3 · Execute4 carry out the plan

1Write both divisors as fractions

#7 Identify Subproblems 6.NS.A.1

2 and 6/24 is 9 over 4 as one fraction.

\frac{7}{20},\quad 2\frac{6}{24} = \frac{9}{4}

Two divisors, both improper-fraction shaped.

2Say what each condition demands

#13 Convert to Algebra 6.NS.A.1

Dividing by a fraction multiplies by its reciprocal, so the unknown's numerator must be a multiple of 7 and of 9, and its denominator must divide both 20 and 4.

\text{top} \in \{7, 9\}\text{-multiples},\quad \text{bottom} \mid 20,\ 4

One rule for the top, one for the bottom.

3Take the smallest top

#3 Eliminate Possibilities 6.NS.B.4

The smallest number that is a multiple of both 7 and 9.

\text{lcm}(7, 9) = 63

The numerator is 63.

4Take the largest bottom

#3 Eliminate Possibilities 6.NS.B.4

A bigger denominator makes the fraction smaller, so take the largest one allowed: the greatest common factor of 20 and 4.

\text{gcf}(20, 4) = 4

The denominator is 4.

Answer:

\frac{63}{4}

4 · Reviewdoes it hold up?

Both divisions come out whole: $\frac{63}{4} \div \frac{7}{20} = 45$ and $\frac{63}{4} \div \frac{9}{4} = 7$ . And 15 and 3/4 is at least 1, so the fraction really is improper.

Another way: Trying fractions one at a time also works, but there is no way to know when to stop until the two rules are written down.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Turning each division into a condition on numerator and denominator.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using an lcm on top and a gcf underneath.

💡Takeaway. To make a fraction small, make the top as small as the rules allow and the bottom as big.

Variant 11 hard answer:

\frac{76}{3}

An improper fraction divided by $\frac{12}{9}$ and the same fraction divided by $3\frac{4}{24}$ both give whole numbers. Find the smallest such improper fraction.

Show solution

1 · Understandwhat's really being asked

One improper fraction divides evenly by both 12/9 and 3 and 4/24. We want the smallest one that does.

Givens

Dividing by 12/9 gives a whole number.
Dividing by 3 and 4/24 gives a whole number.
The fraction is improper, so it is at least 1.

Unknowns

The smallest improper fraction meeting both conditions.

Constraints

Both conditions apply to the same fraction at once.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Turn both divisions into multiplications and each one says the same two things: what the numerator must contain and what the denominator may be. Smallest then means smallest on top and largest underneath.

3 · Execute4 carry out the plan

1Write both divisors as fractions

#7 Identify Subproblems 6.NS.A.1

3 and 4/24 is 19 over 6 as one fraction.

\frac{12}{9},\quad 3\frac{4}{24} = \frac{19}{6}

Two divisors, both improper-fraction shaped.

2Say what each condition demands

#13 Convert to Algebra 6.NS.A.1

Dividing by a fraction multiplies by its reciprocal, so the unknown's numerator must be a multiple of 4 and of 19, and its denominator must divide both 3 and 6.

\text{top} \in \{4, 19\}\text{-multiples},\quad \text{bottom} \mid 3,\ 6

One rule for the top, one for the bottom.

3Take the smallest top

#3 Eliminate Possibilities 6.NS.B.4

The smallest number that is a multiple of both 4 and 19.

\text{lcm}(4, 19) = 76

The numerator is 76.

4Take the largest bottom

#3 Eliminate Possibilities 6.NS.B.4

A bigger denominator makes the fraction smaller, so take the largest one allowed: the greatest common factor of 3 and 6.

\text{gcf}(3, 6) = 3

The denominator is 3.

Answer:

\frac{76}{3}

4 · Reviewdoes it hold up?

Both divisions come out whole: $\frac{76}{3} \div \frac{4}{3} = 19$ and $\frac{76}{3} \div \frac{19}{6} = 8$ . And 25 and 1/3 is at least 1, so the fraction really is improper.

Another way: Trying fractions one at a time also works, but there is no way to know when to stop until the two rules are written down.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Turning each division into a condition on numerator and denominator.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using an lcm on top and a gcf underneath.

💡Takeaway. To make a fraction small, make the top as small as the rules allow and the bottom as big.

Variant 12 hard answer:

\frac{70}{3}

An improper fraction divided by $\frac{7}{15}$ and the same fraction divided by $3\frac{9}{27}$ both give whole numbers. Find the smallest such improper fraction.

Show solution

1 · Understandwhat's really being asked

One improper fraction divides evenly by both 7/15 and 3 and 9/27. We want the smallest one that does.

Givens

Dividing by 7/15 gives a whole number.
Dividing by 3 and 9/27 gives a whole number.
The fraction is improper, so it is at least 1.

Unknowns

The smallest improper fraction meeting both conditions.

Constraints

Both conditions apply to the same fraction at once.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Turn both divisions into multiplications and each one says the same two things: what the numerator must contain and what the denominator may be. Smallest then means smallest on top and largest underneath.

3 · Execute4 carry out the plan

1Write both divisors as fractions

#7 Identify Subproblems 6.NS.A.1

3 and 9/27 is 10 over 3 as one fraction.

\frac{7}{15},\quad 3\frac{9}{27} = \frac{10}{3}

Two divisors, both improper-fraction shaped.

2Say what each condition demands

#13 Convert to Algebra 6.NS.A.1

Dividing by a fraction multiplies by its reciprocal, so the unknown's numerator must be a multiple of 7 and of 10, and its denominator must divide both 15 and 3.

\text{top} \in \{7, 10\}\text{-multiples},\quad \text{bottom} \mid 15,\ 3

One rule for the top, one for the bottom.

3Take the smallest top

#3 Eliminate Possibilities 6.NS.B.4

The smallest number that is a multiple of both 7 and 10.

\text{lcm}(7, 10) = 70

The numerator is 70.

4Take the largest bottom

#3 Eliminate Possibilities 6.NS.B.4

A bigger denominator makes the fraction smaller, so take the largest one allowed: the greatest common factor of 15 and 3.

\text{gcf}(15, 3) = 3

The denominator is 3.

Answer:

\frac{70}{3}

4 · Reviewdoes it hold up?

Both divisions come out whole: $\frac{70}{3} \div \frac{7}{15} = 50$ and $\frac{70}{3} \div \frac{10}{3} = 7$ . And 23 and 1/3 is at least 1, so the fraction really is improper.

Another way: Trying fractions one at a time also works, but there is no way to know when to stop until the two rules are written down.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Turning each division into a condition on numerator and denominator.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using an lcm on top and a gcf underneath.

💡Takeaway. To make a fraction small, make the top as small as the rules allow and the bottom as big.

Smallest numerator on top, largest denominator underneath · 12 practice problems

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