Practice · The overlap is what double-counting reveals · Sensim Math

Variant 1 easy

In a class, $\dfrac{5}{8}$ of all the students like soccer, $\dfrac{5}{8}$ of all the students like baseball, and $\dfrac{1}{4}$ of all the students like neither soccer nor baseball. What fraction of all the students like both soccer and baseball?

Your answer

Show solution

1 · Understandwhat's really being asked

A class is split by two sports. $\dfrac{5}{8}$ like soccer, $\dfrac{5}{8}$ like baseball, and $\dfrac{1}{4}$ like neither. We need the fraction who like both.

Givens

Soccer: $\dfrac{5}{8}$ of the class.
Baseball: $\dfrac{5}{8}$ of the class.
Neither sport: $\dfrac{1}{4}$ of the class.

Unknowns

The fraction of the class who like both sports.

Constraints

Every fraction is of the same whole -- the class -- and the whole class is 1.

2 · Planchoose the strategy

#12 Draw a Venn Diagram · also uses: #7 Identify Subproblems

Two overlapping circles inside a box. The part outside both circles is given, so the part inside at least one circle is what is left. Adding the two circles counts the overlap twice, and the difference is the overlap.

3 · Execute3 carry out the plan

1Find the 'soccer or baseball' part

#12 Draw a Venn Diagram 5.NF.A.2

The whole class is 1. The students who like neither sport are

\dfrac{1}{4}

, so everyone else likes soccer, baseball, or both. Subtract the 'neither' fraction from the whole.

1 - \dfrac{1}{4} = \dfrac{8}{8} - \dfrac{2}{8} = \dfrac{3}{4}

\dfrac{3}{4}

of the class is inside at least one circle.

2Add the two circles with a common denominator

#7 Identify Subproblems 5.NF.A.1

If I add the soccer fraction and the baseball fraction, the overlap (students who like both) gets counted twice. First write both over 8, then add.

\dfrac{5}{8} + \dfrac{5}{8} = \dfrac{5}{8} + \dfrac{5}{8} = \dfrac{5}{4}

\dfrac{5}{4}

-- bigger than the class allows, because of the double count.

3The double-counted part is 'both'

#7 Identify Subproblems 5.NF.A.2

The true 'soccer or baseball' part is

\dfrac{3}{4}

, but adding the two circles gave

\dfrac{5}{4}

. The extra is exactly the overlap that was counted twice, so it is the fraction who like both.

\dfrac{5}{4} - \dfrac{3}{4} = \dfrac{10}{8} - \dfrac{6}{8} = \dfrac{1}{2}

\dfrac{1}{2}

of the class likes both sports.

Answer:

\frac{1}{2}

4 · Reviewdoes it hold up?

Add the four parts of the picture back up: soccer only $\dfrac{1}{8}$ , baseball only $\dfrac{1}{8}$ , both $\dfrac{1}{2}$ , neither $\dfrac{1}{4}$ . Together they make 1, the whole class.

Another way: Or start from the overlap: if $\dfrac{1}{2}$ like both, then soccer-only is $\dfrac{1}{8}$ and baseball-only is $\dfrac{1}{8}$ , and adding those two to the overlap gives $\dfrac{3}{4}$ -- matching the part that is not 'neither'.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two sport fractions over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Taking 'neither' off the whole and reading the leftover as the overlap.

💡Takeaway. Adding two groups counts whoever is in both of them twice, so the extra you get is exactly the overlap.

Variant 2 easy

In a class, $\dfrac{7}{10}$ of all the students like soccer, $\dfrac{3}{5}$ of all the students like baseball, and $\dfrac{1}{10}$ of all the students like neither soccer nor baseball. What fraction of all the students like both soccer and baseball?

Your answer

Show solution

1 · Understandwhat's really being asked

A class is split by two sports. $\dfrac{7}{10}$ like soccer, $\dfrac{3}{5}$ like baseball, and $\dfrac{1}{10}$ like neither. We need the fraction who like both.

Givens

Soccer: $\dfrac{7}{10}$ of the class.
Baseball: $\dfrac{3}{5}$ of the class.
Neither sport: $\dfrac{1}{10}$ of the class.

Unknowns

The fraction of the class who like both sports.

Constraints

Every fraction is of the same whole -- the class -- and the whole class is 1.

2 · Planchoose the strategy

#12 Draw a Venn Diagram · also uses: #7 Identify Subproblems

Two overlapping circles inside a box. The part outside both circles is given, so the part inside at least one circle is what is left. Adding the two circles counts the overlap twice, and the difference is the overlap.

3 · Execute3 carry out the plan

1Find the 'soccer or baseball' part

#12 Draw a Venn Diagram 5.NF.A.2

The whole class is 1. The students who like neither sport are

\dfrac{1}{10}

, so everyone else likes soccer, baseball, or both. Subtract the 'neither' fraction from the whole.

1 - \dfrac{1}{10} = \dfrac{10}{10} - \dfrac{1}{10} = \dfrac{9}{10}

\dfrac{9}{10}

of the class is inside at least one circle.

2Add the two circles with a common denominator

#7 Identify Subproblems 5.NF.A.1

If I add the soccer fraction and the baseball fraction, the overlap (students who like both) gets counted twice. First write both over 10, then add.

\dfrac{7}{10} + \dfrac{3}{5} = \dfrac{7}{10} + \dfrac{6}{10} = \dfrac{13}{10}

\dfrac{13}{10}

-- bigger than the class allows, because of the double count.

3The double-counted part is 'both'

#7 Identify Subproblems 5.NF.A.2

The true 'soccer or baseball' part is

\dfrac{9}{10}

, but adding the two circles gave

\dfrac{13}{10}

. The extra is exactly the overlap that was counted twice, so it is the fraction who like both.

\dfrac{13}{10} - \dfrac{9}{10} = \dfrac{13}{10} - \dfrac{9}{10} = \dfrac{2}{5}

\dfrac{2}{5}

of the class likes both sports.

Answer:

\frac{2}{5}

4 · Reviewdoes it hold up?

Add the four parts of the picture back up: soccer only $\dfrac{3}{10}$ , baseball only $\dfrac{1}{5}$ , both $\dfrac{2}{5}$ , neither $\dfrac{1}{10}$ . Together they make 1, the whole class.

Another way: Or start from the overlap: if $\dfrac{2}{5}$ like both, then soccer-only is $\dfrac{3}{10}$ and baseball-only is $\dfrac{1}{5}$ , and adding those two to the overlap gives $\dfrac{9}{10}$ -- matching the part that is not 'neither'.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two sport fractions over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Taking 'neither' off the whole and reading the leftover as the overlap.

💡Takeaway. Adding two groups counts whoever is in both of them twice, so the extra you get is exactly the overlap.

Variant 3 easy

In a class, $\dfrac{7}{12}$ of all the students like soccer, $\dfrac{1}{2}$ of all the students like baseball, and $\dfrac{1}{6}$ of all the students like neither soccer nor baseball. What fraction of all the students like both soccer and baseball?

Your answer

Show solution

1 · Understandwhat's really being asked

A class is split by two sports. $\dfrac{7}{12}$ like soccer, $\dfrac{1}{2}$ like baseball, and $\dfrac{1}{6}$ like neither. We need the fraction who like both.

Givens

Soccer: $\dfrac{7}{12}$ of the class.
Baseball: $\dfrac{1}{2}$ of the class.
Neither sport: $\dfrac{1}{6}$ of the class.

Unknowns

The fraction of the class who like both sports.

Constraints

Every fraction is of the same whole -- the class -- and the whole class is 1.

2 · Planchoose the strategy

#12 Draw a Venn Diagram · also uses: #7 Identify Subproblems

Two overlapping circles inside a box. The part outside both circles is given, so the part inside at least one circle is what is left. Adding the two circles counts the overlap twice, and the difference is the overlap.

3 · Execute3 carry out the plan

1Find the 'soccer or baseball' part

#12 Draw a Venn Diagram 5.NF.A.2

The whole class is 1. The students who like neither sport are

\dfrac{1}{6}

, so everyone else likes soccer, baseball, or both. Subtract the 'neither' fraction from the whole.

1 - \dfrac{1}{6} = \dfrac{12}{12} - \dfrac{2}{12} = \dfrac{5}{6}

\dfrac{5}{6}

of the class is inside at least one circle.

2Add the two circles with a common denominator

#7 Identify Subproblems 5.NF.A.1

If I add the soccer fraction and the baseball fraction, the overlap (students who like both) gets counted twice. First write both over 12, then add.

\dfrac{7}{12} + \dfrac{1}{2} = \dfrac{7}{12} + \dfrac{6}{12} = \dfrac{13}{12}

\dfrac{13}{12}

-- bigger than the class allows, because of the double count.

3The double-counted part is 'both'

#7 Identify Subproblems 5.NF.A.2

The true 'soccer or baseball' part is

\dfrac{5}{6}

, but adding the two circles gave

\dfrac{13}{12}

. The extra is exactly the overlap that was counted twice, so it is the fraction who like both.

\dfrac{13}{12} - \dfrac{5}{6} = \dfrac{13}{12} - \dfrac{10}{12} = \dfrac{1}{4}

\dfrac{1}{4}

of the class likes both sports.

Answer:

\frac{1}{4}

4 · Reviewdoes it hold up?

Add the four parts of the picture back up: soccer only $\dfrac{1}{3}$ , baseball only $\dfrac{1}{4}$ , both $\dfrac{1}{4}$ , neither $\dfrac{1}{6}$ . Together they make 1, the whole class.

Another way: Or start from the overlap: if $\dfrac{1}{4}$ like both, then soccer-only is $\dfrac{1}{3}$ and baseball-only is $\dfrac{1}{4}$ , and adding those two to the overlap gives $\dfrac{5}{6}$ -- matching the part that is not 'neither'.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two sport fractions over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Taking 'neither' off the whole and reading the leftover as the overlap.

💡Takeaway. Adding two groups counts whoever is in both of them twice, so the extra you get is exactly the overlap.

Variant 4 easy

In a class, $\dfrac{5}{8}$ of all the students like soccer, $\dfrac{5}{12}$ of all the students like baseball, and $\dfrac{1}{8}$ of all the students like neither soccer nor baseball. What fraction of all the students like both soccer and baseball?

Your answer

Show solution

1 · Understandwhat's really being asked

A class is split by two sports. $\dfrac{5}{8}$ like soccer, $\dfrac{5}{12}$ like baseball, and $\dfrac{1}{8}$ like neither. We need the fraction who like both.

Givens

Soccer: $\dfrac{5}{8}$ of the class.
Baseball: $\dfrac{5}{12}$ of the class.
Neither sport: $\dfrac{1}{8}$ of the class.

Unknowns

The fraction of the class who like both sports.

Constraints

Every fraction is of the same whole -- the class -- and the whole class is 1.

2 · Planchoose the strategy

#12 Draw a Venn Diagram · also uses: #7 Identify Subproblems

Two overlapping circles inside a box. The part outside both circles is given, so the part inside at least one circle is what is left. Adding the two circles counts the overlap twice, and the difference is the overlap.

3 · Execute3 carry out the plan

1Find the 'soccer or baseball' part

#12 Draw a Venn Diagram 5.NF.A.2

The whole class is 1. The students who like neither sport are

\dfrac{1}{8}

, so everyone else likes soccer, baseball, or both. Subtract the 'neither' fraction from the whole.

1 - \dfrac{1}{8} = \dfrac{24}{24} - \dfrac{3}{24} = \dfrac{7}{8}

\dfrac{7}{8}

of the class is inside at least one circle.

2Add the two circles with a common denominator

#7 Identify Subproblems 5.NF.A.1

If I add the soccer fraction and the baseball fraction, the overlap (students who like both) gets counted twice. First write both over 24, then add.

\dfrac{5}{8} + \dfrac{5}{12} = \dfrac{15}{24} + \dfrac{10}{24} = \dfrac{25}{24}

\dfrac{25}{24}

-- bigger than the class allows, because of the double count.

3The double-counted part is 'both'

#7 Identify Subproblems 5.NF.A.2

The true 'soccer or baseball' part is

\dfrac{7}{8}

, but adding the two circles gave

\dfrac{25}{24}

. The extra is exactly the overlap that was counted twice, so it is the fraction who like both.

\dfrac{25}{24} - \dfrac{7}{8} = \dfrac{25}{24} - \dfrac{21}{24} = \dfrac{1}{6}

\dfrac{1}{6}

of the class likes both sports.

Answer:

\frac{1}{6}

4 · Reviewdoes it hold up?

Add the four parts of the picture back up: soccer only $\dfrac{11}{24}$ , baseball only $\dfrac{1}{4}$ , both $\dfrac{1}{6}$ , neither $\dfrac{1}{8}$ . Together they make 1, the whole class.

Another way: Or start from the overlap: if $\dfrac{1}{6}$ like both, then soccer-only is $\dfrac{11}{24}$ and baseball-only is $\dfrac{1}{4}$ , and adding those two to the overlap gives $\dfrac{7}{8}$ -- matching the part that is not 'neither'.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two sport fractions over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Taking 'neither' off the whole and reading the leftover as the overlap.

💡Takeaway. Adding two groups counts whoever is in both of them twice, so the extra you get is exactly the overlap.

Variant 5 medium

In a class, $\dfrac{9}{14}$ of all the students like soccer, $\dfrac{4}{7}$ of all the students like baseball, and $\dfrac{3}{14}$ of all the students like neither soccer nor baseball. What fraction of all the students like both soccer and baseball?

Your answer

Show solution

1 · Understandwhat's really being asked

A class is split by two sports. $\dfrac{9}{14}$ like soccer, $\dfrac{4}{7}$ like baseball, and $\dfrac{3}{14}$ like neither. We need the fraction who like both.

Givens

Soccer: $\dfrac{9}{14}$ of the class.
Baseball: $\dfrac{4}{7}$ of the class.
Neither sport: $\dfrac{3}{14}$ of the class.

Unknowns

The fraction of the class who like both sports.

Constraints

Every fraction is of the same whole -- the class -- and the whole class is 1.

2 · Planchoose the strategy

#12 Draw a Venn Diagram · also uses: #7 Identify Subproblems

Two overlapping circles inside a box. The part outside both circles is given, so the part inside at least one circle is what is left. Adding the two circles counts the overlap twice, and the difference is the overlap.

3 · Execute3 carry out the plan

1Find the 'soccer or baseball' part

#12 Draw a Venn Diagram 5.NF.A.2

The whole class is 1. The students who like neither sport are

\dfrac{3}{14}

, so everyone else likes soccer, baseball, or both. Subtract the 'neither' fraction from the whole.

1 - \dfrac{3}{14} = \dfrac{14}{14} - \dfrac{3}{14} = \dfrac{11}{14}

\dfrac{11}{14}

of the class is inside at least one circle.

2Add the two circles with a common denominator

#7 Identify Subproblems 5.NF.A.1

If I add the soccer fraction and the baseball fraction, the overlap (students who like both) gets counted twice. First write both over 14, then add.

\dfrac{9}{14} + \dfrac{4}{7} = \dfrac{9}{14} + \dfrac{8}{14} = \dfrac{17}{14}

\dfrac{17}{14}

-- bigger than the class allows, because of the double count.

3The double-counted part is 'both'

#7 Identify Subproblems 5.NF.A.2

The true 'soccer or baseball' part is

\dfrac{11}{14}

, but adding the two circles gave

\dfrac{17}{14}

. The extra is exactly the overlap that was counted twice, so it is the fraction who like both.

\dfrac{17}{14} - \dfrac{11}{14} = \dfrac{17}{14} - \dfrac{11}{14} = \dfrac{3}{7}

\dfrac{3}{7}

of the class likes both sports.

Answer:

\frac{3}{7}

4 · Reviewdoes it hold up?

Add the four parts of the picture back up: soccer only $\dfrac{3}{14}$ , baseball only $\dfrac{1}{7}$ , both $\dfrac{3}{7}$ , neither $\dfrac{3}{14}$ . Together they make 1, the whole class.

Another way: Or start from the overlap: if $\dfrac{3}{7}$ like both, then soccer-only is $\dfrac{3}{14}$ and baseball-only is $\dfrac{1}{7}$ , and adding those two to the overlap gives $\dfrac{11}{14}$ -- matching the part that is not 'neither'.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two sport fractions over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Taking 'neither' off the whole and reading the leftover as the overlap.

💡Takeaway. Adding two groups counts whoever is in both of them twice, so the extra you get is exactly the overlap.

Variant 6 medium

In a class, $\dfrac{2}{5}$ of all the students like soccer, $\dfrac{8}{15}$ of all the students like baseball, and $\dfrac{4}{15}$ of all the students like neither soccer nor baseball. What fraction of all the students like both soccer and baseball?

Your answer

Show solution

1 · Understandwhat's really being asked

A class is split by two sports. $\dfrac{2}{5}$ like soccer, $\dfrac{8}{15}$ like baseball, and $\dfrac{4}{15}$ like neither. We need the fraction who like both.

Givens

Soccer: $\dfrac{2}{5}$ of the class.
Baseball: $\dfrac{8}{15}$ of the class.
Neither sport: $\dfrac{4}{15}$ of the class.

Unknowns

The fraction of the class who like both sports.

Constraints

Every fraction is of the same whole -- the class -- and the whole class is 1.

2 · Planchoose the strategy

#12 Draw a Venn Diagram · also uses: #7 Identify Subproblems

Two overlapping circles inside a box. The part outside both circles is given, so the part inside at least one circle is what is left. Adding the two circles counts the overlap twice, and the difference is the overlap.

3 · Execute3 carry out the plan

1Find the 'soccer or baseball' part

#12 Draw a Venn Diagram 5.NF.A.2

The whole class is 1. The students who like neither sport are

\dfrac{4}{15}

, so everyone else likes soccer, baseball, or both. Subtract the 'neither' fraction from the whole.

1 - \dfrac{4}{15} = \dfrac{15}{15} - \dfrac{4}{15} = \dfrac{11}{15}

\dfrac{11}{15}

of the class is inside at least one circle.

2Add the two circles with a common denominator

#7 Identify Subproblems 5.NF.A.1

If I add the soccer fraction and the baseball fraction, the overlap (students who like both) gets counted twice. First write both over 15, then add.

\dfrac{2}{5} + \dfrac{8}{15} = \dfrac{6}{15} + \dfrac{8}{15} = \dfrac{14}{15}

\dfrac{14}{15}

-- bigger than the class allows, because of the double count.

3The double-counted part is 'both'

#7 Identify Subproblems 5.NF.A.2

The true 'soccer or baseball' part is

\dfrac{11}{15}

, but adding the two circles gave

\dfrac{14}{15}

. The extra is exactly the overlap that was counted twice, so it is the fraction who like both.

\dfrac{14}{15} - \dfrac{11}{15} = \dfrac{14}{15} - \dfrac{11}{15} = \dfrac{1}{5}

\dfrac{1}{5}

of the class likes both sports.

Answer:

\frac{1}{5}

4 · Reviewdoes it hold up?

Add the four parts of the picture back up: soccer only $\dfrac{1}{5}$ , baseball only $\dfrac{1}{3}$ , both $\dfrac{1}{5}$ , neither $\dfrac{4}{15}$ . Together they make 1, the whole class.

Another way: Or start from the overlap: if $\dfrac{1}{5}$ like both, then soccer-only is $\dfrac{1}{5}$ and baseball-only is $\dfrac{1}{3}$ , and adding those two to the overlap gives $\dfrac{11}{15}$ -- matching the part that is not 'neither'.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two sport fractions over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Taking 'neither' off the whole and reading the leftover as the overlap.

💡Takeaway. Adding two groups counts whoever is in both of them twice, so the extra you get is exactly the overlap.

Variant 7 medium

In a class, $\dfrac{9}{16}$ of all the students like soccer, $\dfrac{11}{16}$ of all the students like baseball, and $\dfrac{1}{4}$ of all the students like neither soccer nor baseball. What fraction of all the students like both soccer and baseball?

Your answer

Show solution

1 · Understandwhat's really being asked

A class is split by two sports. $\dfrac{9}{16}$ like soccer, $\dfrac{11}{16}$ like baseball, and $\dfrac{1}{4}$ like neither. We need the fraction who like both.

Givens

Soccer: $\dfrac{9}{16}$ of the class.
Baseball: $\dfrac{11}{16}$ of the class.
Neither sport: $\dfrac{1}{4}$ of the class.

Unknowns

The fraction of the class who like both sports.

Constraints

Every fraction is of the same whole -- the class -- and the whole class is 1.

2 · Planchoose the strategy

#12 Draw a Venn Diagram · also uses: #7 Identify Subproblems

Two overlapping circles inside a box. The part outside both circles is given, so the part inside at least one circle is what is left. Adding the two circles counts the overlap twice, and the difference is the overlap.

3 · Execute3 carry out the plan

1Find the 'soccer or baseball' part

#12 Draw a Venn Diagram 5.NF.A.2

The whole class is 1. The students who like neither sport are

\dfrac{1}{4}

, so everyone else likes soccer, baseball, or both. Subtract the 'neither' fraction from the whole.

1 - \dfrac{1}{4} = \dfrac{16}{16} - \dfrac{4}{16} = \dfrac{3}{4}

\dfrac{3}{4}

of the class is inside at least one circle.

2Add the two circles with a common denominator

#7 Identify Subproblems 5.NF.A.1

If I add the soccer fraction and the baseball fraction, the overlap (students who like both) gets counted twice. First write both over 16, then add.

\dfrac{9}{16} + \dfrac{11}{16} = \dfrac{9}{16} + \dfrac{11}{16} = \dfrac{5}{4}

\dfrac{5}{4}

-- bigger than the class allows, because of the double count.

3The double-counted part is 'both'

#7 Identify Subproblems 5.NF.A.2

The true 'soccer or baseball' part is

\dfrac{3}{4}

, but adding the two circles gave

\dfrac{5}{4}

. The extra is exactly the overlap that was counted twice, so it is the fraction who like both.

\dfrac{5}{4} - \dfrac{3}{4} = \dfrac{20}{16} - \dfrac{12}{16} = \dfrac{1}{2}

\dfrac{1}{2}

of the class likes both sports.

Answer:

\frac{1}{2}

4 · Reviewdoes it hold up?

Add the four parts of the picture back up: soccer only $\dfrac{1}{16}$ , baseball only $\dfrac{3}{16}$ , both $\dfrac{1}{2}$ , neither $\dfrac{1}{4}$ . Together they make 1, the whole class.

Another way: Or start from the overlap: if $\dfrac{1}{2}$ like both, then soccer-only is $\dfrac{1}{16}$ and baseball-only is $\dfrac{3}{16}$ , and adding those two to the overlap gives $\dfrac{3}{4}$ -- matching the part that is not 'neither'.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two sport fractions over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Taking 'neither' off the whole and reading the leftover as the overlap.

💡Takeaway. Adding two groups counts whoever is in both of them twice, so the extra you get is exactly the overlap.

Variant 8 medium

In a class, $\dfrac{11}{18}$ of all the students like soccer, $\dfrac{5}{9}$ of all the students like baseball, and $\dfrac{1}{6}$ of all the students like neither soccer nor baseball. What fraction of all the students like both soccer and baseball?

Your answer

Show solution

1 · Understandwhat's really being asked

A class is split by two sports. $\dfrac{11}{18}$ like soccer, $\dfrac{5}{9}$ like baseball, and $\dfrac{1}{6}$ like neither. We need the fraction who like both.

Givens

Soccer: $\dfrac{11}{18}$ of the class.
Baseball: $\dfrac{5}{9}$ of the class.
Neither sport: $\dfrac{1}{6}$ of the class.

Unknowns

The fraction of the class who like both sports.

Constraints

Every fraction is of the same whole -- the class -- and the whole class is 1.

2 · Planchoose the strategy

#12 Draw a Venn Diagram · also uses: #7 Identify Subproblems

Two overlapping circles inside a box. The part outside both circles is given, so the part inside at least one circle is what is left. Adding the two circles counts the overlap twice, and the difference is the overlap.

3 · Execute3 carry out the plan

1Find the 'soccer or baseball' part

#12 Draw a Venn Diagram 5.NF.A.2

The whole class is 1. The students who like neither sport are

\dfrac{1}{6}

, so everyone else likes soccer, baseball, or both. Subtract the 'neither' fraction from the whole.

1 - \dfrac{1}{6} = \dfrac{18}{18} - \dfrac{3}{18} = \dfrac{5}{6}

\dfrac{5}{6}

of the class is inside at least one circle.

2Add the two circles with a common denominator

#7 Identify Subproblems 5.NF.A.1

If I add the soccer fraction and the baseball fraction, the overlap (students who like both) gets counted twice. First write both over 18, then add.

\dfrac{11}{18} + \dfrac{5}{9} = \dfrac{11}{18} + \dfrac{10}{18} = \dfrac{7}{6}

\dfrac{7}{6}

-- bigger than the class allows, because of the double count.

3The double-counted part is 'both'

#7 Identify Subproblems 5.NF.A.2

The true 'soccer or baseball' part is

\dfrac{5}{6}

, but adding the two circles gave

\dfrac{7}{6}

. The extra is exactly the overlap that was counted twice, so it is the fraction who like both.

\dfrac{7}{6} - \dfrac{5}{6} = \dfrac{21}{18} - \dfrac{15}{18} = \dfrac{1}{3}

\dfrac{1}{3}

of the class likes both sports.

Answer:

\frac{1}{3}

4 · Reviewdoes it hold up?

Add the four parts of the picture back up: soccer only $\dfrac{5}{18}$ , baseball only $\dfrac{2}{9}$ , both $\dfrac{1}{3}$ , neither $\dfrac{1}{6}$ . Together they make 1, the whole class.

Another way: Or start from the overlap: if $\dfrac{1}{3}$ like both, then soccer-only is $\dfrac{5}{18}$ and baseball-only is $\dfrac{2}{9}$ , and adding those two to the overlap gives $\dfrac{5}{6}$ -- matching the part that is not 'neither'.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two sport fractions over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Taking 'neither' off the whole and reading the leftover as the overlap.

💡Takeaway. Adding two groups counts whoever is in both of them twice, so the extra you get is exactly the overlap.

Variant 9 hard

In a class, $\dfrac{9}{20}$ of all the students like soccer, $\dfrac{7}{10}$ of all the students like baseball, and $\dfrac{1}{4}$ of all the students like neither soccer nor baseball. What fraction of all the students like both soccer and baseball?

Your answer

Show solution

1 · Understandwhat's really being asked

A class is split by two sports. $\dfrac{9}{20}$ like soccer, $\dfrac{7}{10}$ like baseball, and $\dfrac{1}{4}$ like neither. We need the fraction who like both.

Givens

Soccer: $\dfrac{9}{20}$ of the class.
Baseball: $\dfrac{7}{10}$ of the class.
Neither sport: $\dfrac{1}{4}$ of the class.

Unknowns

The fraction of the class who like both sports.

Constraints

Every fraction is of the same whole -- the class -- and the whole class is 1.

2 · Planchoose the strategy

#12 Draw a Venn Diagram · also uses: #7 Identify Subproblems

Two overlapping circles inside a box. The part outside both circles is given, so the part inside at least one circle is what is left. Adding the two circles counts the overlap twice, and the difference is the overlap.

3 · Execute3 carry out the plan

1Find the 'soccer or baseball' part

#12 Draw a Venn Diagram 5.NF.A.2

The whole class is 1. The students who like neither sport are

\dfrac{1}{4}

, so everyone else likes soccer, baseball, or both. Subtract the 'neither' fraction from the whole.

1 - \dfrac{1}{4} = \dfrac{20}{20} - \dfrac{5}{20} = \dfrac{3}{4}

\dfrac{3}{4}

of the class is inside at least one circle.

2Add the two circles with a common denominator

#7 Identify Subproblems 5.NF.A.1

If I add the soccer fraction and the baseball fraction, the overlap (students who like both) gets counted twice. First write both over 20, then add.

\dfrac{9}{20} + \dfrac{7}{10} = \dfrac{9}{20} + \dfrac{14}{20} = \dfrac{23}{20}

\dfrac{23}{20}

-- bigger than the class allows, because of the double count.

3The double-counted part is 'both'

#7 Identify Subproblems 5.NF.A.2

The true 'soccer or baseball' part is

\dfrac{3}{4}

, but adding the two circles gave

\dfrac{23}{20}

. The extra is exactly the overlap that was counted twice, so it is the fraction who like both.

\dfrac{23}{20} - \dfrac{3}{4} = \dfrac{23}{20} - \dfrac{15}{20} = \dfrac{2}{5}

\dfrac{2}{5}

of the class likes both sports.

Answer:

\frac{2}{5}

4 · Reviewdoes it hold up?

Add the four parts of the picture back up: soccer only $\dfrac{1}{20}$ , baseball only $\dfrac{3}{10}$ , both $\dfrac{2}{5}$ , neither $\dfrac{1}{4}$ . Together they make 1, the whole class.

Another way: Or start from the overlap: if $\dfrac{2}{5}$ like both, then soccer-only is $\dfrac{1}{20}$ and baseball-only is $\dfrac{3}{10}$ , and adding those two to the overlap gives $\dfrac{3}{4}$ -- matching the part that is not 'neither'.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two sport fractions over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Taking 'neither' off the whole and reading the leftover as the overlap.

💡Takeaway. Adding two groups counts whoever is in both of them twice, so the extra you get is exactly the overlap.

Variant 10 hard

In a class, $\dfrac{13}{21}$ of all the students like soccer, $\dfrac{4}{7}$ of all the students like baseball, and $\dfrac{4}{21}$ of all the students like neither soccer nor baseball. What fraction of all the students like both soccer and baseball?

Your answer

Show solution

1 · Understandwhat's really being asked

A class is split by two sports. $\dfrac{13}{21}$ like soccer, $\dfrac{4}{7}$ like baseball, and $\dfrac{4}{21}$ like neither. We need the fraction who like both.

Givens

Soccer: $\dfrac{13}{21}$ of the class.
Baseball: $\dfrac{4}{7}$ of the class.
Neither sport: $\dfrac{4}{21}$ of the class.

Unknowns

The fraction of the class who like both sports.

Constraints

Every fraction is of the same whole -- the class -- and the whole class is 1.

2 · Planchoose the strategy

#12 Draw a Venn Diagram · also uses: #7 Identify Subproblems

Two overlapping circles inside a box. The part outside both circles is given, so the part inside at least one circle is what is left. Adding the two circles counts the overlap twice, and the difference is the overlap.

3 · Execute3 carry out the plan

1Find the 'soccer or baseball' part

#12 Draw a Venn Diagram 5.NF.A.2

The whole class is 1. The students who like neither sport are

\dfrac{4}{21}

, so everyone else likes soccer, baseball, or both. Subtract the 'neither' fraction from the whole.

1 - \dfrac{4}{21} = \dfrac{21}{21} - \dfrac{4}{21} = \dfrac{17}{21}

\dfrac{17}{21}

of the class is inside at least one circle.

2Add the two circles with a common denominator

#7 Identify Subproblems 5.NF.A.1

If I add the soccer fraction and the baseball fraction, the overlap (students who like both) gets counted twice. First write both over 21, then add.

\dfrac{13}{21} + \dfrac{4}{7} = \dfrac{13}{21} + \dfrac{12}{21} = \dfrac{25}{21}

\dfrac{25}{21}

-- bigger than the class allows, because of the double count.

3The double-counted part is 'both'

#7 Identify Subproblems 5.NF.A.2

The true 'soccer or baseball' part is

\dfrac{17}{21}

, but adding the two circles gave

\dfrac{25}{21}

. The extra is exactly the overlap that was counted twice, so it is the fraction who like both.

\dfrac{25}{21} - \dfrac{17}{21} = \dfrac{25}{21} - \dfrac{17}{21} = \dfrac{8}{21}

\dfrac{8}{21}

of the class likes both sports.

Answer:

\frac{8}{21}

4 · Reviewdoes it hold up?

Add the four parts of the picture back up: soccer only $\dfrac{5}{21}$ , baseball only $\dfrac{4}{21}$ , both $\dfrac{8}{21}$ , neither $\dfrac{4}{21}$ . Together they make 1, the whole class.

Another way: Or start from the overlap: if $\dfrac{8}{21}$ like both, then soccer-only is $\dfrac{5}{21}$ and baseball-only is $\dfrac{4}{21}$ , and adding those two to the overlap gives $\dfrac{17}{21}$ -- matching the part that is not 'neither'.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two sport fractions over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Taking 'neither' off the whole and reading the leftover as the overlap.

💡Takeaway. Adding two groups counts whoever is in both of them twice, so the extra you get is exactly the overlap.

Variant 11 hard

In a class, $\dfrac{16}{25}$ of all the students like soccer, $\dfrac{14}{25}$ of all the students like baseball, and $\dfrac{1}{5}$ of all the students like neither soccer nor baseball. What fraction of all the students like both soccer and baseball?

Your answer

Show solution

1 · Understandwhat's really being asked

A class is split by two sports. $\dfrac{16}{25}$ like soccer, $\dfrac{14}{25}$ like baseball, and $\dfrac{1}{5}$ like neither. We need the fraction who like both.

Givens

Soccer: $\dfrac{16}{25}$ of the class.
Baseball: $\dfrac{14}{25}$ of the class.
Neither sport: $\dfrac{1}{5}$ of the class.

Unknowns

The fraction of the class who like both sports.

Constraints

Every fraction is of the same whole -- the class -- and the whole class is 1.

2 · Planchoose the strategy

#12 Draw a Venn Diagram · also uses: #7 Identify Subproblems

Two overlapping circles inside a box. The part outside both circles is given, so the part inside at least one circle is what is left. Adding the two circles counts the overlap twice, and the difference is the overlap.

3 · Execute3 carry out the plan

1Find the 'soccer or baseball' part

#12 Draw a Venn Diagram 5.NF.A.2

The whole class is 1. The students who like neither sport are

\dfrac{1}{5}

, so everyone else likes soccer, baseball, or both. Subtract the 'neither' fraction from the whole.

1 - \dfrac{1}{5} = \dfrac{25}{25} - \dfrac{5}{25} = \dfrac{4}{5}

\dfrac{4}{5}

of the class is inside at least one circle.

2Add the two circles with a common denominator

#7 Identify Subproblems 5.NF.A.1

If I add the soccer fraction and the baseball fraction, the overlap (students who like both) gets counted twice. First write both over 25, then add.

\dfrac{16}{25} + \dfrac{14}{25} = \dfrac{16}{25} + \dfrac{14}{25} = \dfrac{6}{5}

\dfrac{6}{5}

-- bigger than the class allows, because of the double count.

3The double-counted part is 'both'

#7 Identify Subproblems 5.NF.A.2

The true 'soccer or baseball' part is

\dfrac{4}{5}

, but adding the two circles gave

\dfrac{6}{5}

. The extra is exactly the overlap that was counted twice, so it is the fraction who like both.

\dfrac{6}{5} - \dfrac{4}{5} = \dfrac{30}{25} - \dfrac{20}{25} = \dfrac{2}{5}

\dfrac{2}{5}

of the class likes both sports.

Answer:

\frac{2}{5}

4 · Reviewdoes it hold up?

Add the four parts of the picture back up: soccer only $\dfrac{6}{25}$ , baseball only $\dfrac{4}{25}$ , both $\dfrac{2}{5}$ , neither $\dfrac{1}{5}$ . Together they make 1, the whole class.

Another way: Or start from the overlap: if $\dfrac{2}{5}$ like both, then soccer-only is $\dfrac{6}{25}$ and baseball-only is $\dfrac{4}{25}$ , and adding those two to the overlap gives $\dfrac{4}{5}$ -- matching the part that is not 'neither'.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two sport fractions over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Taking 'neither' off the whole and reading the leftover as the overlap.

💡Takeaway. Adding two groups counts whoever is in both of them twice, so the extra you get is exactly the overlap.

Variant 12 hard

In a class, $\dfrac{17}{30}$ of all the students like soccer, $\dfrac{2}{3}$ of all the students like baseball, and $\dfrac{7}{30}$ of all the students like neither soccer nor baseball. What fraction of all the students like both soccer and baseball?

Your answer

Show solution

1 · Understandwhat's really being asked

A class is split by two sports. $\dfrac{17}{30}$ like soccer, $\dfrac{2}{3}$ like baseball, and $\dfrac{7}{30}$ like neither. We need the fraction who like both.

Givens

Soccer: $\dfrac{17}{30}$ of the class.
Baseball: $\dfrac{2}{3}$ of the class.
Neither sport: $\dfrac{7}{30}$ of the class.

Unknowns

The fraction of the class who like both sports.

Constraints

Every fraction is of the same whole -- the class -- and the whole class is 1.

2 · Planchoose the strategy

#12 Draw a Venn Diagram · also uses: #7 Identify Subproblems

Two overlapping circles inside a box. The part outside both circles is given, so the part inside at least one circle is what is left. Adding the two circles counts the overlap twice, and the difference is the overlap.

3 · Execute3 carry out the plan

1Find the 'soccer or baseball' part

#12 Draw a Venn Diagram 5.NF.A.2

The whole class is 1. The students who like neither sport are

\dfrac{7}{30}

, so everyone else likes soccer, baseball, or both. Subtract the 'neither' fraction from the whole.

1 - \dfrac{7}{30} = \dfrac{30}{30} - \dfrac{7}{30} = \dfrac{23}{30}

\dfrac{23}{30}

of the class is inside at least one circle.

2Add the two circles with a common denominator

#7 Identify Subproblems 5.NF.A.1

If I add the soccer fraction and the baseball fraction, the overlap (students who like both) gets counted twice. First write both over 30, then add.

\dfrac{17}{30} + \dfrac{2}{3} = \dfrac{17}{30} + \dfrac{20}{30} = \dfrac{37}{30}

\dfrac{37}{30}

-- bigger than the class allows, because of the double count.

3The double-counted part is 'both'

#7 Identify Subproblems 5.NF.A.2

The true 'soccer or baseball' part is

\dfrac{23}{30}

, but adding the two circles gave

\dfrac{37}{30}

. The extra is exactly the overlap that was counted twice, so it is the fraction who like both.

\dfrac{37}{30} - \dfrac{23}{30} = \dfrac{37}{30} - \dfrac{23}{30} = \dfrac{7}{15}

\dfrac{7}{15}

of the class likes both sports.

Answer:

\frac{7}{15}

4 · Reviewdoes it hold up?

Add the four parts of the picture back up: soccer only $\dfrac{1}{10}$ , baseball only $\dfrac{1}{5}$ , both $\dfrac{7}{15}$ , neither $\dfrac{7}{30}$ . Together they make 1, the whole class.

Another way: Or start from the overlap: if $\dfrac{7}{15}$ like both, then soccer-only is $\dfrac{1}{10}$ and baseball-only is $\dfrac{1}{5}$ , and adding those two to the overlap gives $\dfrac{23}{30}$ -- matching the part that is not 'neither'.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two sport fractions over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Taking 'neither' off the whole and reading the leftover as the overlap.

💡Takeaway. Adding two groups counts whoever is in both of them twice, so the extra you get is exactly the overlap.

The overlap is what double-counting reveals · 12 practice problems

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