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Build the largest proper fraction · 12 practice problems

4.NF.A.25.NF.A.1

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Generated variants — 12

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Variant 1 easy

From the number cards $2$ , $6$ , $8$ , $3$ , $7$ , choose $4$ of them and use each chosen card exactly once to build (proper fraction) $+$ (proper fraction). Find the largest possible value of this sum.

Your answer

Show solution

1 · Understandwhat's really being asked

Four of the five cards 2, 6, 8, 3, 7 fill two proper fractions, which are then added. We want the largest sum possible.

Givens

The cards are 2, 6, 8, 3, 7.
Four are chosen, each used once, and one card is left out.
Both fractions must be proper -- numerator below denominator.

Unknowns

The largest possible sum.

Constraints

A proper fraction is always less than 1, so the sum is under 2.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List#5 Look for a Pattern

A proper fraction is largest when its numerator sits just below its denominator. That narrows the sensible pairs to a few, which can then be added and compared.

3 · Execute4 carry out the plan

1Find what makes one proper fraction big

#5 Look for a Pattern 4.NF.A.2

The closer the numerator is to the denominator, the nearer the fraction is to 1. So the big cards want to be tops of fractions whose bottoms are only just bigger.

\frac{7}{8} \ \text{is close to}\ 1

Near-1 fractions come from near-equal pairs.

2List the strongest candidate fractions

#2 Make a Systematic List 4.NF.A.2

Pair the cards so each fraction is as near 1 as the cards allow, and leave out the card that helps least.

\frac{2}{3} + \frac{7}{8}

The card 6 is the one left out.

3Add the two fractions with a common denominator

#6 Guess and Check 5.NF.A.1

Put both over 24 and add the numerators.

\frac{2}{3} + \frac{7}{8} = \frac{16}{24} + \frac{21}{24} = \frac{37}{24}

The sum is

\frac{37}{24}

4Check against the next-best guess

#6 Guess and Check 4.NF.A.2

The nearest rival arrangement gives a smaller total, so this really is the largest.

\frac{37}{24} > \frac{32}{21}

No rearrangement beats it.

Answer:

\frac{37}{24}

4 · Reviewdoes it hold up?

Both fractions are proper, so the sum must be under 2 -- and $\frac{37}{24}$ is just under, which is what a maximum should look like.

Another way: Trying every choice of four cards and every arrangement confirms the same maximum; the near-1 argument just skips the ones that were never going to win.

Standardsmin grade 5

4.NF.A.2 Compare two fractions with different numerators and different denominators — Judging which pairings give fractions closest to 1.
5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two winning fractions over a common denominator.

💡Takeaway. A proper fraction is biggest when its top is just under its bottom -- so look for near-equal pairs before adding anything.

Variant 2 easy

From the number cards $4$ , $8$ , $5$ , $7$ , $2$ , choose $4$ of them and use each chosen card exactly once to build (proper fraction) $+$ (proper fraction). Find the largest possible value of this sum.

Your answer

Show solution

1 · Understandwhat's really being asked

Four of the five cards 4, 8, 5, 7, 2 fill two proper fractions, which are then added. We want the largest sum possible.

Givens

The cards are 4, 8, 5, 7, 2.
Four are chosen, each used once, and one card is left out.
Both fractions must be proper -- numerator below denominator.

Unknowns

The largest possible sum.

Constraints

A proper fraction is always less than 1, so the sum is under 2.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List#5 Look for a Pattern

A proper fraction is largest when its numerator sits just below its denominator. That narrows the sensible pairs to a few, which can then be added and compared.

3 · Execute4 carry out the plan

1Find what makes one proper fraction big

#5 Look for a Pattern 4.NF.A.2

The closer the numerator is to the denominator, the nearer the fraction is to 1. So the big cards want to be tops of fractions whose bottoms are only just bigger.

\frac{7}{8} \ \text{is close to}\ 1

Near-1 fractions come from near-equal pairs.

2List the strongest candidate fractions

#2 Make a Systematic List 4.NF.A.2

Pair the cards so each fraction is as near 1 as the cards allow, and leave out the card that helps least.

\frac{4}{5} + \frac{7}{8}

The card 2 is the one left out.

3Add the two fractions with a common denominator

#6 Guess and Check 5.NF.A.1

Put both over 40 and add the numerators.

\frac{4}{5} + \frac{7}{8} = \frac{32}{40} + \frac{35}{40} = \frac{67}{40}

The sum is

\frac{67}{40}

4Check against the next-best guess

#6 Guess and Check 4.NF.A.2

The nearest rival arrangement gives a smaller total, so this really is the largest.

\frac{67}{40} > \frac{11}{8}

No rearrangement beats it.

Answer:

\frac{67}{40}

4 · Reviewdoes it hold up?

Both fractions are proper, so the sum must be under 2 -- and $\frac{67}{40}$ is just under, which is what a maximum should look like.

Another way: Trying every choice of four cards and every arrangement confirms the same maximum; the near-1 argument just skips the ones that were never going to win.

Standardsmin grade 5

4.NF.A.2 Compare two fractions with different numerators and different denominators — Judging which pairings give fractions closest to 1.
5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two winning fractions over a common denominator.

💡Takeaway. A proper fraction is biggest when its top is just under its bottom -- so look for near-equal pairs before adding anything.

Variant 3 easy

From the number cards $8$ , $1$ , $6$ , $3$ , $7$ , choose $4$ of them and use each chosen card exactly once to build (proper fraction) $+$ (proper fraction). Find the largest possible value of this sum.

Your answer

Show solution

1 · Understandwhat's really being asked

Four of the five cards 8, 1, 6, 3, 7 fill two proper fractions, which are then added. We want the largest sum possible.

Givens

The cards are 8, 1, 6, 3, 7.
Four are chosen, each used once, and one card is left out.
Both fractions must be proper -- numerator below denominator.

Unknowns

The largest possible sum.

Constraints

A proper fraction is always less than 1, so the sum is under 2.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List#5 Look for a Pattern

A proper fraction is largest when its numerator sits just below its denominator. That narrows the sensible pairs to a few, which can then be added and compared.

3 · Execute4 carry out the plan

1Find what makes one proper fraction big

#5 Look for a Pattern 4.NF.A.2

The closer the numerator is to the denominator, the nearer the fraction is to 1. So the big cards want to be tops of fractions whose bottoms are only just bigger.

\frac{7}{8} \ \text{is close to}\ 1

Near-1 fractions come from near-equal pairs.

2List the strongest candidate fractions

#2 Make a Systematic List 4.NF.A.2

Pair the cards so each fraction is as near 1 as the cards allow, and leave out the card that helps least.

\frac{1}{2} + \frac{7}{8}

The card 1 is the one left out.

3Add the two fractions with a common denominator

#6 Guess and Check 5.NF.A.1

Put both over 48 and add the numerators.

\frac{1}{2} + \frac{7}{8} = \frac{24}{48} + \frac{42}{48} = \frac{11}{8}

The sum is

\frac{11}{8}

4Check against the next-best guess

#6 Guess and Check 4.NF.A.2

The nearest rival arrangement gives a smaller total, so this really is the largest.

\frac{11}{8} > \frac{69}{56}

No rearrangement beats it.

Answer:

\frac{11}{8}

4 · Reviewdoes it hold up?

Both fractions are proper, so the sum must be under 2 -- and $\frac{11}{8}$ is just under, which is what a maximum should look like.

Another way: Trying every choice of four cards and every arrangement confirms the same maximum; the near-1 argument just skips the ones that were never going to win.

Standardsmin grade 5

4.NF.A.2 Compare two fractions with different numerators and different denominators — Judging which pairings give fractions closest to 1.
5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two winning fractions over a common denominator.

💡Takeaway. A proper fraction is biggest when its top is just under its bottom -- so look for near-equal pairs before adding anything.

Variant 4 easy

From the number cards $3$ , $8$ , $2$ , $7$ , $5$ , choose $4$ of them and use each chosen card exactly once to build (proper fraction) $+$ (proper fraction). Find the largest possible value of this sum.

Your answer

Show solution

1 · Understandwhat's really being asked

Four of the five cards 3, 8, 2, 7, 5 fill two proper fractions, which are then added. We want the largest sum possible.

Givens

The cards are 3, 8, 2, 7, 5.
Four are chosen, each used once, and one card is left out.
Both fractions must be proper -- numerator below denominator.

Unknowns

The largest possible sum.

Constraints

A proper fraction is always less than 1, so the sum is under 2.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List#5 Look for a Pattern

A proper fraction is largest when its numerator sits just below its denominator. That narrows the sensible pairs to a few, which can then be added and compared.

3 · Execute4 carry out the plan

1Find what makes one proper fraction big

#5 Look for a Pattern 4.NF.A.2

The closer the numerator is to the denominator, the nearer the fraction is to 1. So the big cards want to be tops of fractions whose bottoms are only just bigger.

\frac{7}{8} \ \text{is close to}\ 1

Near-1 fractions come from near-equal pairs.

2List the strongest candidate fractions

#2 Make a Systematic List 4.NF.A.2

Pair the cards so each fraction is as near 1 as the cards allow, and leave out the card that helps least.

\frac{2}{3} + \frac{7}{8}

The card 5 is the one left out.

3Add the two fractions with a common denominator

#6 Guess and Check 5.NF.A.1

Put both over 24 and add the numerators.

\frac{2}{3} + \frac{7}{8} = \frac{16}{24} + \frac{21}{24} = \frac{37}{24}

The sum is

\frac{37}{24}

4Check against the next-best guess

#6 Guess and Check 4.NF.A.2

The nearest rival arrangement gives a smaller total, so this really is the largest.

\frac{37}{24} > \frac{59}{40}

No rearrangement beats it.

Answer:

\frac{37}{24}

4 · Reviewdoes it hold up?

Both fractions are proper, so the sum must be under 2 -- and $\frac{37}{24}$ is just under, which is what a maximum should look like.

Another way: Trying every choice of four cards and every arrangement confirms the same maximum; the near-1 argument just skips the ones that were never going to win.

Standardsmin grade 5

4.NF.A.2 Compare two fractions with different numerators and different denominators — Judging which pairings give fractions closest to 1.
5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two winning fractions over a common denominator.

💡Takeaway. A proper fraction is biggest when its top is just under its bottom -- so look for near-equal pairs before adding anything.

Variant 5 hard

From the number cards $3$ , $9$ , $6$ , $1$ , $8$ , choose $4$ of them and use each chosen card exactly once to build (proper fraction) $+$ (proper fraction). Find the largest possible value of this sum.

Your answer

Show solution

1 · Understandwhat's really being asked

Four of the five cards 3, 9, 6, 1, 8 fill two proper fractions, which are then added. We want the largest sum possible.

Givens

The cards are 3, 9, 6, 1, 8.
Four are chosen, each used once, and one card is left out.
Both fractions must be proper -- numerator below denominator.

Unknowns

The largest possible sum.

Constraints

A proper fraction is always less than 1, so the sum is under 2.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List#5 Look for a Pattern

A proper fraction is largest when its numerator sits just below its denominator. That narrows the sensible pairs to a few, which can then be added and compared.

3 · Execute4 carry out the plan

1Find what makes one proper fraction big

#5 Look for a Pattern 4.NF.A.2

The closer the numerator is to the denominator, the nearer the fraction is to 1. So the big cards want to be tops of fractions whose bottoms are only just bigger.

\frac{8}{9} \ \text{is close to}\ 1

Near-1 fractions come from near-equal pairs.

2List the strongest candidate fractions

#2 Make a Systematic List 4.NF.A.2

Pair the cards so each fraction is as near 1 as the cards allow, and leave out the card that helps least.

\frac{1}{2} + \frac{8}{9}

The card 1 is the one left out.

3Add the two fractions with a common denominator

#6 Guess and Check 5.NF.A.1

Put both over 54 and add the numerators.

\frac{1}{2} + \frac{8}{9} = \frac{27}{54} + \frac{48}{54} = \frac{25}{18}

The sum is

\frac{25}{18}

4Check against the next-best guess

#6 Guess and Check 4.NF.A.2

The nearest rival arrangement gives a smaller total, so this really is the largest.

\frac{25}{18} > \frac{11}{9}

No rearrangement beats it.

Answer:

\frac{25}{18}

4 · Reviewdoes it hold up?

Both fractions are proper, so the sum must be under 2 -- and $\frac{25}{18}$ is just under, which is what a maximum should look like.

Another way: Trying every choice of four cards and every arrangement confirms the same maximum; the near-1 argument just skips the ones that were never going to win.

Standardsmin grade 5

4.NF.A.2 Compare two fractions with different numerators and different denominators — Judging which pairings give fractions closest to 1.
5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two winning fractions over a common denominator.

💡Takeaway. A proper fraction is biggest when its top is just under its bottom -- so look for near-equal pairs before adding anything.

Variant 6 hard

From the number cards $2$ , $4$ , $9$ , $6$ , $5$ , choose $4$ of them and use each chosen card exactly once to build (proper fraction) $+$ (proper fraction). Find the largest possible value of this sum.

Your answer

Show solution

1 · Understandwhat's really being asked

Four of the five cards 2, 4, 9, 6, 5 fill two proper fractions, which are then added. We want the largest sum possible.

Givens

The cards are 2, 4, 9, 6, 5.
Four are chosen, each used once, and one card is left out.
Both fractions must be proper -- numerator below denominator.

Unknowns

The largest possible sum.

Constraints

A proper fraction is always less than 1, so the sum is under 2.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List#5 Look for a Pattern

A proper fraction is largest when its numerator sits just below its denominator. That narrows the sensible pairs to a few, which can then be added and compared.

3 · Execute4 carry out the plan

1Find what makes one proper fraction big

#5 Look for a Pattern 4.NF.A.2

The closer the numerator is to the denominator, the nearer the fraction is to 1. So the big cards want to be tops of fractions whose bottoms are only just bigger.

\frac{6}{9} \ \text{is close to}\ 1

Near-1 fractions come from near-equal pairs.

2List the strongest candidate fractions

#2 Make a Systematic List 4.NF.A.2

Pair the cards so each fraction is as near 1 as the cards allow, and leave out the card that helps least.

\frac{4}{5} + \frac{2}{3}

The card 2 is the one left out.

3Add the two fractions with a common denominator

#6 Guess and Check 5.NF.A.1

Put both over 45 and add the numerators.

\frac{4}{5} + \frac{2}{3} = \frac{36}{45} + \frac{30}{45} = \frac{22}{15}

The sum is

\frac{22}{15}

4Check against the next-best guess

#6 Guess and Check 4.NF.A.2

The nearest rival arrangement gives a smaller total, so this really is the largest.

\frac{22}{15} > \frac{4}{3}

No rearrangement beats it.

Answer:

\frac{22}{15}

4 · Reviewdoes it hold up?

Both fractions are proper, so the sum must be under 2 -- and $\frac{22}{15}$ is just under, which is what a maximum should look like.

Another way: Trying every choice of four cards and every arrangement confirms the same maximum; the near-1 argument just skips the ones that were never going to win.

Standardsmin grade 5

4.NF.A.2 Compare two fractions with different numerators and different denominators — Judging which pairings give fractions closest to 1.
5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two winning fractions over a common denominator.

💡Takeaway. A proper fraction is biggest when its top is just under its bottom -- so look for near-equal pairs before adding anything.

Variant 7 medium

From the number cards $5$ , $3$ , $7$ , $1$ , $9$ , choose $4$ of them and use each chosen card exactly once to build (proper fraction) $+$ (proper fraction). Find the largest possible value of this sum.

Your answer

Show solution

1 · Understandwhat's really being asked

Four of the five cards 5, 3, 7, 1, 9 fill two proper fractions, which are then added. We want the largest sum possible.

Givens

The cards are 5, 3, 7, 1, 9.
Four are chosen, each used once, and one card is left out.
Both fractions must be proper -- numerator below denominator.

Unknowns

The largest possible sum.

Constraints

A proper fraction is always less than 1, so the sum is under 2.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List#5 Look for a Pattern

A proper fraction is largest when its numerator sits just below its denominator. That narrows the sensible pairs to a few, which can then be added and compared.

3 · Execute4 carry out the plan

1Find what makes one proper fraction big

#5 Look for a Pattern 4.NF.A.2

The closer the numerator is to the denominator, the nearer the fraction is to 1. So the big cards want to be tops of fractions whose bottoms are only just bigger.

\frac{7}{9} \ \text{is close to}\ 1

Near-1 fractions come from near-equal pairs.

2List the strongest candidate fractions

#2 Make a Systematic List 4.NF.A.2

Pair the cards so each fraction is as near 1 as the cards allow, and leave out the card that helps least.

\frac{3}{5} + \frac{7}{9}

The card 1 is the one left out.

3Add the two fractions with a common denominator

#6 Guess and Check 5.NF.A.1

Put both over 45 and add the numerators.

\frac{3}{5} + \frac{7}{9} = \frac{27}{45} + \frac{35}{45} = \frac{62}{45}

The sum is

\frac{62}{45}

4Check against the next-best guess

#6 Guess and Check 4.NF.A.2

The nearest rival arrangement gives a smaller total, so this really is the largest.

\frac{62}{45} > \frac{10}{9}

No rearrangement beats it.

Answer:

\frac{62}{45}

4 · Reviewdoes it hold up?

Both fractions are proper, so the sum must be under 2 -- and $\frac{62}{45}$ is just under, which is what a maximum should look like.

Another way: Trying every choice of four cards and every arrangement confirms the same maximum; the near-1 argument just skips the ones that were never going to win.

Standardsmin grade 5

4.NF.A.2 Compare two fractions with different numerators and different denominators — Judging which pairings give fractions closest to 1.
5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two winning fractions over a common denominator.

💡Takeaway. A proper fraction is biggest when its top is just under its bottom -- so look for near-equal pairs before adding anything.

Variant 8 hard

From the number cards $7$ , $5$ , $2$ , $9$ , $4$ , choose $4$ of them and use each chosen card exactly once to build (proper fraction) $+$ (proper fraction). Find the largest possible value of this sum.

Your answer

Show solution

1 · Understandwhat's really being asked

Four of the five cards 7, 5, 2, 9, 4 fill two proper fractions, which are then added. We want the largest sum possible.

Givens

The cards are 7, 5, 2, 9, 4.
Four are chosen, each used once, and one card is left out.
Both fractions must be proper -- numerator below denominator.

Unknowns

The largest possible sum.

Constraints

A proper fraction is always less than 1, so the sum is under 2.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List#5 Look for a Pattern

A proper fraction is largest when its numerator sits just below its denominator. That narrows the sensible pairs to a few, which can then be added and compared.

3 · Execute4 carry out the plan

1Find what makes one proper fraction big

#5 Look for a Pattern 4.NF.A.2

The closer the numerator is to the denominator, the nearer the fraction is to 1. So the big cards want to be tops of fractions whose bottoms are only just bigger.

\frac{7}{9} \ \text{is close to}\ 1

Near-1 fractions come from near-equal pairs.

2List the strongest candidate fractions

#2 Make a Systematic List 4.NF.A.2

Pair the cards so each fraction is as near 1 as the cards allow, and leave out the card that helps least.

\frac{7}{9} + \frac{4}{5}

The card 2 is the one left out.

3Add the two fractions with a common denominator

#6 Guess and Check 5.NF.A.1

Put both over 45 and add the numerators.

\frac{7}{9} + \frac{4}{5} = \frac{35}{45} + \frac{36}{45} = \frac{71}{45}

The sum is

\frac{71}{45}

4Check against the next-best guess

#6 Guess and Check 4.NF.A.2

The nearest rival arrangement gives a smaller total, so this really is the largest.

\frac{71}{45} > \frac{23}{18}

No rearrangement beats it.

Answer:

\frac{71}{45}

4 · Reviewdoes it hold up?

Both fractions are proper, so the sum must be under 2 -- and $\frac{71}{45}$ is just under, which is what a maximum should look like.

Another way: Trying every choice of four cards and every arrangement confirms the same maximum; the near-1 argument just skips the ones that were never going to win.

Standardsmin grade 5

4.NF.A.2 Compare two fractions with different numerators and different denominators — Judging which pairings give fractions closest to 1.
5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two winning fractions over a common denominator.

💡Takeaway. A proper fraction is biggest when its top is just under its bottom -- so look for near-equal pairs before adding anything.

Variant 9 medium

From the number cards $1$ , $5$ , $9$ , $4$ , $6$ , choose $4$ of them and use each chosen card exactly once to build (proper fraction) $+$ (proper fraction). Find the largest possible value of this sum.

Your answer

Show solution

1 · Understandwhat's really being asked

Four of the five cards 1, 5, 9, 4, 6 fill two proper fractions, which are then added. We want the largest sum possible.

Givens

The cards are 1, 5, 9, 4, 6.
Four are chosen, each used once, and one card is left out.
Both fractions must be proper -- numerator below denominator.

Unknowns

The largest possible sum.

Constraints

A proper fraction is always less than 1, so the sum is under 2.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List#5 Look for a Pattern

A proper fraction is largest when its numerator sits just below its denominator. That narrows the sensible pairs to a few, which can then be added and compared.

3 · Execute4 carry out the plan

1Find what makes one proper fraction big

#5 Look for a Pattern 4.NF.A.2

The closer the numerator is to the denominator, the nearer the fraction is to 1. So the big cards want to be tops of fractions whose bottoms are only just bigger.

\frac{6}{9} \ \text{is close to}\ 1

Near-1 fractions come from near-equal pairs.

2List the strongest candidate fractions

#2 Make a Systematic List 4.NF.A.2

Pair the cards so each fraction is as near 1 as the cards allow, and leave out the card that helps least.

\frac{4}{5} + \frac{2}{3}

The card 1 is the one left out.

3Add the two fractions with a common denominator

#6 Guess and Check 5.NF.A.1

Put both over 45 and add the numerators.

\frac{4}{5} + \frac{2}{3} = \frac{36}{45} + \frac{30}{45} = \frac{22}{15}

The sum is

\frac{22}{15}

4Check against the next-best guess

#6 Guess and Check 4.NF.A.2

The nearest rival arrangement gives a smaller total, so this really is the largest.

\frac{22}{15} > \frac{23}{18}

No rearrangement beats it.

Answer:

\frac{22}{15}

4 · Reviewdoes it hold up?

Both fractions are proper, so the sum must be under 2 -- and $\frac{22}{15}$ is just under, which is what a maximum should look like.

Another way: Trying every choice of four cards and every arrangement confirms the same maximum; the near-1 argument just skips the ones that were never going to win.

Standardsmin grade 5

4.NF.A.2 Compare two fractions with different numerators and different denominators — Judging which pairings give fractions closest to 1.
5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two winning fractions over a common denominator.

💡Takeaway. A proper fraction is biggest when its top is just under its bottom -- so look for near-equal pairs before adding anything.

Variant 10 medium

From the number cards $4$ , $7$ , $3$ , $5$ , $9$ , choose $4$ of them and use each chosen card exactly once to build (proper fraction) $+$ (proper fraction). Find the largest possible value of this sum.

Your answer

Show solution

1 · Understandwhat's really being asked

Four of the five cards 4, 7, 3, 5, 9 fill two proper fractions, which are then added. We want the largest sum possible.

Givens

The cards are 4, 7, 3, 5, 9.
Four are chosen, each used once, and one card is left out.
Both fractions must be proper -- numerator below denominator.

Unknowns

The largest possible sum.

Constraints

A proper fraction is always less than 1, so the sum is under 2.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List#5 Look for a Pattern

A proper fraction is largest when its numerator sits just below its denominator. That narrows the sensible pairs to a few, which can then be added and compared.

3 · Execute4 carry out the plan

1Find what makes one proper fraction big

#5 Look for a Pattern 4.NF.A.2

The closer the numerator is to the denominator, the nearer the fraction is to 1. So the big cards want to be tops of fractions whose bottoms are only just bigger.

\frac{7}{9} \ \text{is close to}\ 1

Near-1 fractions come from near-equal pairs.

2List the strongest candidate fractions

#2 Make a Systematic List 4.NF.A.2

Pair the cards so each fraction is as near 1 as the cards allow, and leave out the card that helps least.

\frac{4}{5} + \frac{7}{9}

The card 3 is the one left out.

3Add the two fractions with a common denominator

#6 Guess and Check 5.NF.A.1

Put both over 45 and add the numerators.

\frac{4}{5} + \frac{7}{9} = \frac{36}{45} + \frac{35}{45} = \frac{71}{45}

The sum is

\frac{71}{45}

4Check against the next-best guess

#6 Guess and Check 4.NF.A.2

The nearest rival arrangement gives a smaller total, so this really is the largest.

\frac{71}{45} > \frac{55}{36}

No rearrangement beats it.

Answer:

\frac{71}{45}

4 · Reviewdoes it hold up?

Both fractions are proper, so the sum must be under 2 -- and $\frac{71}{45}$ is just under, which is what a maximum should look like.

Another way: Trying every choice of four cards and every arrangement confirms the same maximum; the near-1 argument just skips the ones that were never going to win.

Standardsmin grade 5

4.NF.A.2 Compare two fractions with different numerators and different denominators — Judging which pairings give fractions closest to 1.
5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two winning fractions over a common denominator.

💡Takeaway. A proper fraction is biggest when its top is just under its bottom -- so look for near-equal pairs before adding anything.

Variant 11 medium

From the number cards $6$ , $9$ , $4$ , $2$ , $8$ , choose $4$ of them and use each chosen card exactly once to build (proper fraction) $+$ (proper fraction). Find the largest possible value of this sum.

Your answer

Show solution

1 · Understandwhat's really being asked

Four of the five cards 6, 9, 4, 2, 8 fill two proper fractions, which are then added. We want the largest sum possible.

Givens

The cards are 6, 9, 4, 2, 8.
Four are chosen, each used once, and one card is left out.
Both fractions must be proper -- numerator below denominator.

Unknowns

The largest possible sum.

Constraints

A proper fraction is always less than 1, so the sum is under 2.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List#5 Look for a Pattern

A proper fraction is largest when its numerator sits just below its denominator. That narrows the sensible pairs to a few, which can then be added and compared.

3 · Execute4 carry out the plan

1Find what makes one proper fraction big

#5 Look for a Pattern 4.NF.A.2

The closer the numerator is to the denominator, the nearer the fraction is to 1. So the big cards want to be tops of fractions whose bottoms are only just bigger.

\frac{8}{9} \ \text{is close to}\ 1

Near-1 fractions come from near-equal pairs.

2List the strongest candidate fractions

#2 Make a Systematic List 4.NF.A.2

Pair the cards so each fraction is as near 1 as the cards allow, and leave out the card that helps least.

\frac{2}{3} + \frac{8}{9}

The card 2 is the one left out.

3Add the two fractions with a common denominator

#6 Guess and Check 5.NF.A.1

Put both over 54 and add the numerators.

\frac{2}{3} + \frac{8}{9} = \frac{36}{54} + \frac{48}{54} = \frac{14}{9}

The sum is

\frac{14}{9}

4Check against the next-best guess

#6 Guess and Check 4.NF.A.2

The nearest rival arrangement gives a smaller total, so this really is the largest.

\frac{14}{9} > \frac{25}{18}

No rearrangement beats it.

Answer:

\frac{14}{9}

4 · Reviewdoes it hold up?

Both fractions are proper, so the sum must be under 2 -- and $\frac{14}{9}$ is just under, which is what a maximum should look like.

Another way: Trying every choice of four cards and every arrangement confirms the same maximum; the near-1 argument just skips the ones that were never going to win.

Standardsmin grade 5

4.NF.A.2 Compare two fractions with different numerators and different denominators — Judging which pairings give fractions closest to 1.
5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two winning fractions over a common denominator.

💡Takeaway. A proper fraction is biggest when its top is just under its bottom -- so look for near-equal pairs before adding anything.

Variant 12 hard

From the number cards $6$ , $2$ , $3$ , $9$ , $5$ , choose $4$ of them and use each chosen card exactly once to build (proper fraction) $+$ (proper fraction). Find the largest possible value of this sum.

Your answer

Show solution

1 · Understandwhat's really being asked

Four of the five cards 6, 2, 3, 9, 5 fill two proper fractions, which are then added. We want the largest sum possible.

Givens

The cards are 6, 2, 3, 9, 5.
Four are chosen, each used once, and one card is left out.
Both fractions must be proper -- numerator below denominator.

Unknowns

The largest possible sum.

Constraints

A proper fraction is always less than 1, so the sum is under 2.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List#5 Look for a Pattern

A proper fraction is largest when its numerator sits just below its denominator. That narrows the sensible pairs to a few, which can then be added and compared.

3 · Execute4 carry out the plan

1Find what makes one proper fraction big

#5 Look for a Pattern 4.NF.A.2

The closer the numerator is to the denominator, the nearer the fraction is to 1. So the big cards want to be tops of fractions whose bottoms are only just bigger.

\frac{6}{9} \ \text{is close to}\ 1

Near-1 fractions come from near-equal pairs.

2List the strongest candidate fractions

#2 Make a Systematic List 4.NF.A.2

Pair the cards so each fraction is as near 1 as the cards allow, and leave out the card that helps least.

\frac{2}{3} + \frac{5}{6}

The card 9 is the one left out.

3Add the two fractions with a common denominator

#6 Guess and Check 5.NF.A.1

Put both over 18 and add the numerators.

\frac{2}{3} + \frac{5}{6} = \frac{12}{18} + \frac{15}{18} = \frac{3}{2}

The sum is

\frac{3}{2}

4Check against the next-best guess

#6 Guess and Check 4.NF.A.2

The nearest rival arrangement gives a smaller total, so this really is the largest.

\frac{3}{2} > \frac{4}{3}

No rearrangement beats it.

Answer:

\frac{3}{2}

4 · Reviewdoes it hold up?

Both fractions are proper, so the sum must be under 2 -- and $\frac{3}{2}$ is just under, which is what a maximum should look like.

Another way: Trying every choice of four cards and every arrangement confirms the same maximum; the near-1 argument just skips the ones that were never going to win.

Standardsmin grade 5

4.NF.A.2 Compare two fractions with different numerators and different denominators — Judging which pairings give fractions closest to 1.
5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two winning fractions over a common denominator.

💡Takeaway. A proper fraction is biggest when its top is just under its bottom -- so look for near-equal pairs before adding anything.

Where next?

← Back to the problem: Build the largest proper fraction