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

Variant 1 easy

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.

Your answer

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

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.

Your answer

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

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.

Your answer

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

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.

Your answer

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

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.

Your answer

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

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.

Your answer

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

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.

Your answer

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

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.

Your answer

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

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.

Your answer

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

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.

Your answer

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

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.

Your answer

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

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.

Your answer

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

Generated variants — 12

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Where next?