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Use submerged and exposed parts to find bar length · 10 practice problems

4.NF.B.3

Generated variants — 10

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy answer: 57\frac{5}{7} m

A pole is pushed straight down until it touches the bottom of a pond and then pulled back out. The wet part of the pole measures 47\dfrac{4}{7} m. The pole is then turned upside down and again pushed straight down to the bottom of the pond and pulled out. This time the part that has now been wet twice measures 37\dfrac{3}{7} m. Find the length of the pole.

(The pole is always inserted vertically, and the bottom of the pond is flat.)

Show solution
1 · Understandwhat's really being asked

A pole is dipped to the pond bottom from one end, wetting 47\frac{4}{7} m. Flipped and dipped from the other end, the part now wet twice (the overlap of the two wet regions) measures 37\frac{3}{7} m. I must find the pole's full length.

Givens
  • Dipping from one end wets a 47\frac{4}{7} m length (equal to the water depth).
  • After flipping and dipping from the other end, the part wet twice is 37\frac{3}{7} m.
  • The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, 47\frac{4}{7} m.
Unknowns
  • The total length of the pole.
Constraints
  • Each dip wets a length equal to the water depth, 47\frac{4}{7} m, measured from whichever end goes in.
  • The two wet regions overlap in the middle by 37\frac{3}{7} m.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#11 Work Backwards

Draw the pole as a segment with a 47\frac{4}{7} wet region from each end. The two regions overlap by 37\frac{3}{7}. Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets 47\frac{4}{7} m

#1 Draw a Diagram 4.OA.A.3
Pushing the pole to the flat bottom wets exactly the part below the water surface, a length equal to the water depth = 47\frac{4}{7} m. Flipping and dipping the other end wets another 47\frac{4}{7} m measured from that end.
wet from each end=47 m\text{wet from each end}=\dfrac{4}{7}\text{ m}
Same pond, same depth, so each end gets wet over the same 47\frac{4}{7} m length.

2The overlap is the part wet twice

#1 Draw a Diagram 4.NF.B.3
Drawn on the pole, the wet region from the left and the wet region from the right meet in the middle. Where they meet is wet twice — that overlap is given as 37\frac{3}{7} m.
overlap=37 m\text{overlap}=\dfrac{3}{7}\text{ m}
The middle stretch counted in both dips is the 37\frac{3}{7} m wet-twice part.

3Combine with overlap subtraction

#16 Count the Complement 4.NF.B.3
Total length = left wet + right wet - overlap = 47\frac{4}{7} + 47\frac{4}{7} - 37\frac{3}{7} = 4+4−37\frac{4+4-3}{7} = 57\frac{5}{7} m.
47+47−37=57 m\dfrac{4}{7}+\dfrac{4}{7}-\dfrac{3}{7}=\dfrac{5}{7}\text{ m}
Adding both wet parts double-counts the middle, so subtract the 37\frac{3}{7} overlap once.
Answer: 57\frac{5}{7} m
4 · Reviewdoes it hold up?

The pole length 57\frac{5}{7} m must be longer than the wet depth 47\frac{4}{7} (true) but shorter than two full dips 87\frac{8}{7} (true), and the overlap 37\frac{3}{7} must be less than the depth 47\frac{4}{7} (true). Everything is consistent in 7ths of a meter.

Another way: Work from the overlap (tool 11): each wet region 47\frac{4}{7} minus the shared 37\frac{3}{7} leaves 17\frac{1}{7} of dry-once at each end; total = 17\frac{1}{7} (left only) + 37\frac{3}{7} (both) + 17\frac{1}{7} (right only) = 57\frac{5}{7} m.

Standardsmin grade 4
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Modeling the two dips and setting up the overlap-subtraction relationship.
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding and subtracting the like-denominator 7ths to get $\frac{5}{7}$ m.
💡Takeaway. This only needs Grade 4 fraction add/subtract — draw the wet parts from each end and subtract the wet-twice overlap once!
Variant 2 easy answer: 87\frac{8}{7} m

A pole is pushed straight down until it touches the bottom of a pond and then pulled back out. The wet part of the pole measures 57\dfrac{5}{7} m. The pole is then turned upside down and again pushed straight down to the bottom of the pond and pulled out. This time the part that has now been wet twice measures 27\dfrac{2}{7} m. Find the length of the pole.

(The pole is always inserted vertically, and the bottom of the pond is flat.)

Show solution
1 · Understandwhat's really being asked

A pole is dipped to the pond bottom from one end, wetting 57\frac{5}{7} m. Flipped and dipped from the other end, the part now wet twice (the overlap of the two wet regions) measures 27\frac{2}{7} m. I must find the pole's full length.

Givens
  • Dipping from one end wets a 57\frac{5}{7} m length (equal to the water depth).
  • After flipping and dipping from the other end, the part wet twice is 27\frac{2}{7} m.
  • The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, 57\frac{5}{7} m.
Unknowns
  • The total length of the pole.
Constraints
  • Each dip wets a length equal to the water depth, 57\frac{5}{7} m, measured from whichever end goes in.
  • The two wet regions overlap in the middle by 27\frac{2}{7} m.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#11 Work Backwards

Draw the pole as a segment with a 57\frac{5}{7} wet region from each end. The two regions overlap by 27\frac{2}{7}. Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets 57\frac{5}{7} m

#1 Draw a Diagram 4.OA.A.3
Pushing the pole to the flat bottom wets exactly the part below the water surface, a length equal to the water depth = 57\frac{5}{7} m. Flipping and dipping the other end wets another 57\frac{5}{7} m measured from that end.
wet from each end=57 m\text{wet from each end}=\dfrac{5}{7}\text{ m}
Same pond, same depth, so each end gets wet over the same 57\frac{5}{7} m length.

2The overlap is the part wet twice

#1 Draw a Diagram 4.NF.B.3
Drawn on the pole, the wet region from the left and the wet region from the right meet in the middle. Where they meet is wet twice — that overlap is given as 27\frac{2}{7} m.
overlap=27 m\text{overlap}=\dfrac{2}{7}\text{ m}
The middle stretch counted in both dips is the 27\frac{2}{7} m wet-twice part.

3Combine with overlap subtraction

#16 Count the Complement 4.NF.B.3
Total length = left wet + right wet - overlap = 57\frac{5}{7} + 57\frac{5}{7} - 27\frac{2}{7} = 5+5−27\frac{5+5-2}{7} = 87\frac{8}{7} m.
57+57−27=87 m\dfrac{5}{7}+\dfrac{5}{7}-\dfrac{2}{7}=\dfrac{8}{7}\text{ m}
Adding both wet parts double-counts the middle, so subtract the 27\frac{2}{7} overlap once.
Answer: 87\frac{8}{7} m
4 · Reviewdoes it hold up?

The pole length 87\frac{8}{7} m must be longer than the wet depth 57\frac{5}{7} (true) but shorter than two full dips 107\frac{10}{7} (true), and the overlap 27\frac{2}{7} must be less than the depth 57\frac{5}{7} (true). Everything is consistent in 7ths of a meter.

Another way: Work from the overlap (tool 11): each wet region 57\frac{5}{7} minus the shared 27\frac{2}{7} leaves 37\frac{3}{7} of dry-once at each end; total = 37\frac{3}{7} (left only) + 27\frac{2}{7} (both) + 37\frac{3}{7} (right only) = 87\frac{8}{7} m.

Standardsmin grade 4
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Modeling the two dips and setting up the overlap-subtraction relationship.
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding and subtracting the like-denominator 7ths to get $\frac{8}{7}$ m.
💡Takeaway. This only needs Grade 4 fraction add/subtract — draw the wet parts from each end and subtract the wet-twice overlap once!
Variant 3 easy answer: 78\frac{7}{8} m

A pole is pushed straight down until it touches the bottom of a pond and then pulled back out. The wet part of the pole measures 58\dfrac{5}{8} m. The pole is then turned upside down and again pushed straight down to the bottom of the pond and pulled out. This time the part that has now been wet twice measures 38\dfrac{3}{8} m. Find the length of the pole.

(The pole is always inserted vertically, and the bottom of the pond is flat.)

Show solution
1 · Understandwhat's really being asked

A pole is dipped to the pond bottom from one end, wetting 58\frac{5}{8} m. Flipped and dipped from the other end, the part now wet twice (the overlap of the two wet regions) measures 38\frac{3}{8} m. I must find the pole's full length.

Givens
  • Dipping from one end wets a 58\frac{5}{8} m length (equal to the water depth).
  • After flipping and dipping from the other end, the part wet twice is 38\frac{3}{8} m.
  • The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, 58\frac{5}{8} m.
Unknowns
  • The total length of the pole.
Constraints
  • Each dip wets a length equal to the water depth, 58\frac{5}{8} m, measured from whichever end goes in.
  • The two wet regions overlap in the middle by 38\frac{3}{8} m.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#11 Work Backwards

Draw the pole as a segment with a 58\frac{5}{8} wet region from each end. The two regions overlap by 38\frac{3}{8}. Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets 58\frac{5}{8} m

#1 Draw a Diagram 4.OA.A.3
Pushing the pole to the flat bottom wets exactly the part below the water surface, a length equal to the water depth = 58\frac{5}{8} m. Flipping and dipping the other end wets another 58\frac{5}{8} m measured from that end.
wet from each end=58 m\text{wet from each end}=\dfrac{5}{8}\text{ m}
Same pond, same depth, so each end gets wet over the same 58\frac{5}{8} m length.

2The overlap is the part wet twice

#1 Draw a Diagram 4.NF.B.3
Drawn on the pole, the wet region from the left and the wet region from the right meet in the middle. Where they meet is wet twice — that overlap is given as 38\frac{3}{8} m.
overlap=38 m\text{overlap}=\dfrac{3}{8}\text{ m}
The middle stretch counted in both dips is the 38\frac{3}{8} m wet-twice part.

3Combine with overlap subtraction

#16 Count the Complement 4.NF.B.3
Total length = left wet + right wet - overlap = 58\frac{5}{8} + 58\frac{5}{8} - 38\frac{3}{8} = 5+5−38\frac{5+5-3}{8} = 78\frac{7}{8} m.
58+58−38=78 m\dfrac{5}{8}+\dfrac{5}{8}-\dfrac{3}{8}=\dfrac{7}{8}\text{ m}
Adding both wet parts double-counts the middle, so subtract the 38\frac{3}{8} overlap once.
Answer: 78\frac{7}{8} m
4 · Reviewdoes it hold up?

The pole length 78\frac{7}{8} m must be longer than the wet depth 58\frac{5}{8} (true) but shorter than two full dips 108\frac{10}{8} (true), and the overlap 38\frac{3}{8} must be less than the depth 58\frac{5}{8} (true). Everything is consistent in 8ths of a meter.

Another way: Work from the overlap (tool 11): each wet region 58\frac{5}{8} minus the shared 38\frac{3}{8} leaves 28\frac{2}{8} of dry-once at each end; total = 28\frac{2}{8} (left only) + 38\frac{3}{8} (both) + 28\frac{2}{8} (right only) = 78\frac{7}{8} m.

Standardsmin grade 4
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Modeling the two dips and setting up the overlap-subtraction relationship.
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding and subtracting the like-denominator 8ths to get $\frac{7}{8}$ m.
💡Takeaway. This only needs Grade 4 fraction add/subtract — draw the wet parts from each end and subtract the wet-twice overlap once!
Variant 4 medium answer: 89\frac{8}{9} m

A pole is pushed straight down until it touches the bottom of a pond and then pulled back out. The wet part of the pole measures 69\dfrac{6}{9} m. The pole is then turned upside down and again pushed straight down to the bottom of the pond and pulled out. This time the part that has now been wet twice measures 49\dfrac{4}{9} m. Find the length of the pole.

(The pole is always inserted vertically, and the bottom of the pond is flat.)

Show solution
1 · Understandwhat's really being asked

A pole is dipped to the pond bottom from one end, wetting 69\frac{6}{9} m. Flipped and dipped from the other end, the part now wet twice (the overlap of the two wet regions) measures 49\frac{4}{9} m. I must find the pole's full length.

Givens
  • Dipping from one end wets a 69\frac{6}{9} m length (equal to the water depth).
  • After flipping and dipping from the other end, the part wet twice is 49\frac{4}{9} m.
  • The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, 69\frac{6}{9} m.
Unknowns
  • The total length of the pole.
Constraints
  • Each dip wets a length equal to the water depth, 69\frac{6}{9} m, measured from whichever end goes in.
  • The two wet regions overlap in the middle by 49\frac{4}{9} m.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#11 Work Backwards

Draw the pole as a segment with a 69\frac{6}{9} wet region from each end. The two regions overlap by 49\frac{4}{9}. Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets 69\frac{6}{9} m

#1 Draw a Diagram 4.OA.A.3
Pushing the pole to the flat bottom wets exactly the part below the water surface, a length equal to the water depth = 69\frac{6}{9} m. Flipping and dipping the other end wets another 69\frac{6}{9} m measured from that end.
wet from each end=69 m\text{wet from each end}=\dfrac{6}{9}\text{ m}
Same pond, same depth, so each end gets wet over the same 69\frac{6}{9} m length.

2The overlap is the part wet twice

#1 Draw a Diagram 4.NF.B.3
Drawn on the pole, the wet region from the left and the wet region from the right meet in the middle. Where they meet is wet twice — that overlap is given as 49\frac{4}{9} m.
overlap=49 m\text{overlap}=\dfrac{4}{9}\text{ m}
The middle stretch counted in both dips is the 49\frac{4}{9} m wet-twice part.

3Combine with overlap subtraction

#16 Count the Complement 4.NF.B.3
Total length = left wet + right wet - overlap = 69\frac{6}{9} + 69\frac{6}{9} - 49\frac{4}{9} = 6+6−49\frac{6+6-4}{9} = 89\frac{8}{9} m.
69+69−49=89 m\dfrac{6}{9}+\dfrac{6}{9}-\dfrac{4}{9}=\dfrac{8}{9}\text{ m}
Adding both wet parts double-counts the middle, so subtract the 49\frac{4}{9} overlap once.
Answer: 89\frac{8}{9} m
4 · Reviewdoes it hold up?

The pole length 89\frac{8}{9} m must be longer than the wet depth 69\frac{6}{9} (true) but shorter than two full dips 129\frac{12}{9} (true), and the overlap 49\frac{4}{9} must be less than the depth 69\frac{6}{9} (true). Everything is consistent in 9ths of a meter.

Another way: Work from the overlap (tool 11): each wet region 69\frac{6}{9} minus the shared 49\frac{4}{9} leaves 29\frac{2}{9} of dry-once at each end; total = 29\frac{2}{9} (left only) + 49\frac{4}{9} (both) + 29\frac{2}{9} (right only) = 89\frac{8}{9} m.

Standardsmin grade 4
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Modeling the two dips and setting up the overlap-subtraction relationship.
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding and subtracting the like-denominator 9ths to get $\frac{8}{9}$ m.
💡Takeaway. This only needs Grade 4 fraction add/subtract — draw the wet parts from each end and subtract the wet-twice overlap once!
Variant 5 medium answer: 69\frac{6}{9} m

A pole is pushed straight down until it touches the bottom of a pond and then pulled back out. The wet part of the pole measures 59\dfrac{5}{9} m. The pole is then turned upside down and again pushed straight down to the bottom of the pond and pulled out. This time the part that has now been wet twice measures 49\dfrac{4}{9} m. Find the length of the pole.

(The pole is always inserted vertically, and the bottom of the pond is flat.)

Show solution
1 · Understandwhat's really being asked

A pole is dipped to the pond bottom from one end, wetting 59\frac{5}{9} m. Flipped and dipped from the other end, the part now wet twice (the overlap of the two wet regions) measures 49\frac{4}{9} m. I must find the pole's full length.

Givens
  • Dipping from one end wets a 59\frac{5}{9} m length (equal to the water depth).
  • After flipping and dipping from the other end, the part wet twice is 49\frac{4}{9} m.
  • The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, 59\frac{5}{9} m.
Unknowns
  • The total length of the pole.
Constraints
  • Each dip wets a length equal to the water depth, 59\frac{5}{9} m, measured from whichever end goes in.
  • The two wet regions overlap in the middle by 49\frac{4}{9} m.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#11 Work Backwards

Draw the pole as a segment with a 59\frac{5}{9} wet region from each end. The two regions overlap by 49\frac{4}{9}. Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets 59\frac{5}{9} m

#1 Draw a Diagram 4.OA.A.3
Pushing the pole to the flat bottom wets exactly the part below the water surface, a length equal to the water depth = 59\frac{5}{9} m. Flipping and dipping the other end wets another 59\frac{5}{9} m measured from that end.
wet from each end=59 m\text{wet from each end}=\dfrac{5}{9}\text{ m}
Same pond, same depth, so each end gets wet over the same 59\frac{5}{9} m length.

2The overlap is the part wet twice

#1 Draw a Diagram 4.NF.B.3
Drawn on the pole, the wet region from the left and the wet region from the right meet in the middle. Where they meet is wet twice — that overlap is given as 49\frac{4}{9} m.
overlap=49 m\text{overlap}=\dfrac{4}{9}\text{ m}
The middle stretch counted in both dips is the 49\frac{4}{9} m wet-twice part.

3Combine with overlap subtraction

#16 Count the Complement 4.NF.B.3
Total length = left wet + right wet - overlap = 59\frac{5}{9} + 59\frac{5}{9} - 49\frac{4}{9} = 5+5−49\frac{5+5-4}{9} = 69\frac{6}{9} m.
59+59−49=69 m\dfrac{5}{9}+\dfrac{5}{9}-\dfrac{4}{9}=\dfrac{6}{9}\text{ m}
Adding both wet parts double-counts the middle, so subtract the 49\frac{4}{9} overlap once.
Answer: 69\frac{6}{9} m
4 · Reviewdoes it hold up?

The pole length 69\frac{6}{9} m must be longer than the wet depth 59\frac{5}{9} (true) but shorter than two full dips 109\frac{10}{9} (true), and the overlap 49\frac{4}{9} must be less than the depth 59\frac{5}{9} (true). Everything is consistent in 9ths of a meter.

Another way: Work from the overlap (tool 11): each wet region 59\frac{5}{9} minus the shared 49\frac{4}{9} leaves 19\frac{1}{9} of dry-once at each end; total = 19\frac{1}{9} (left only) + 49\frac{4}{9} (both) + 19\frac{1}{9} (right only) = 69\frac{6}{9} m.

Standardsmin grade 4
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Modeling the two dips and setting up the overlap-subtraction relationship.
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding and subtracting the like-denominator 9ths to get $\frac{6}{9}$ m.
💡Takeaway. This only needs Grade 4 fraction add/subtract — draw the wet parts from each end and subtract the wet-twice overlap once!
Variant 6 medium answer: 810\frac{8}{10} m

A pole is pushed straight down until it touches the bottom of a pond and then pulled back out. The wet part of the pole measures 610\dfrac{6}{10} m. The pole is then turned upside down and again pushed straight down to the bottom of the pond and pulled out. This time the part that has now been wet twice measures 410\dfrac{4}{10} m. Find the length of the pole.

(The pole is always inserted vertically, and the bottom of the pond is flat.)

Show solution
1 · Understandwhat's really being asked

A pole is dipped to the pond bottom from one end, wetting 610\frac{6}{10} m. Flipped and dipped from the other end, the part now wet twice (the overlap of the two wet regions) measures 410\frac{4}{10} m. I must find the pole's full length.

Givens
  • Dipping from one end wets a 610\frac{6}{10} m length (equal to the water depth).
  • After flipping and dipping from the other end, the part wet twice is 410\frac{4}{10} m.
  • The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, 610\frac{6}{10} m.
Unknowns
  • The total length of the pole.
Constraints
  • Each dip wets a length equal to the water depth, 610\frac{6}{10} m, measured from whichever end goes in.
  • The two wet regions overlap in the middle by 410\frac{4}{10} m.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#11 Work Backwards

Draw the pole as a segment with a 610\frac{6}{10} wet region from each end. The two regions overlap by 410\frac{4}{10}. Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets 610\frac{6}{10} m

#1 Draw a Diagram 4.OA.A.3
Pushing the pole to the flat bottom wets exactly the part below the water surface, a length equal to the water depth = 610\frac{6}{10} m. Flipping and dipping the other end wets another 610\frac{6}{10} m measured from that end.
wet from each end=610 m\text{wet from each end}=\dfrac{6}{10}\text{ m}
Same pond, same depth, so each end gets wet over the same 610\frac{6}{10} m length.

2The overlap is the part wet twice

#1 Draw a Diagram 4.NF.B.3
Drawn on the pole, the wet region from the left and the wet region from the right meet in the middle. Where they meet is wet twice — that overlap is given as 410\frac{4}{10} m.
overlap=410 m\text{overlap}=\dfrac{4}{10}\text{ m}
The middle stretch counted in both dips is the 410\frac{4}{10} m wet-twice part.

3Combine with overlap subtraction

#16 Count the Complement 4.NF.B.3
Total length = left wet + right wet - overlap = 610\frac{6}{10} + 610\frac{6}{10} - 410\frac{4}{10} = 6+6−410\frac{6+6-4}{10} = 810\frac{8}{10} m.
610+610−410=810 m\dfrac{6}{10}+\dfrac{6}{10}-\dfrac{4}{10}=\dfrac{8}{10}\text{ m}
Adding both wet parts double-counts the middle, so subtract the 410\frac{4}{10} overlap once.
Answer: 810\frac{8}{10} m
4 · Reviewdoes it hold up?

The pole length 810\frac{8}{10} m must be longer than the wet depth 610\frac{6}{10} (true) but shorter than two full dips 1210\frac{12}{10} (true), and the overlap 410\frac{4}{10} must be less than the depth 610\frac{6}{10} (true). Everything is consistent in 10ths of a meter.

Another way: Work from the overlap (tool 11): each wet region 610\frac{6}{10} minus the shared 410\frac{4}{10} leaves 210\frac{2}{10} of dry-once at each end; total = 210\frac{2}{10} (left only) + 410\frac{4}{10} (both) + 210\frac{2}{10} (right only) = 810\frac{8}{10} m.

Standardsmin grade 4
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Modeling the two dips and setting up the overlap-subtraction relationship.
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding and subtracting the like-denominator 10ths to get $\frac{8}{10}$ m.
💡Takeaway. This only needs Grade 4 fraction add/subtract — draw the wet parts from each end and subtract the wet-twice overlap once!
Variant 7 medium answer: 711\frac{7}{11} m

A pole is pushed straight down until it touches the bottom of a pond and then pulled back out. The wet part of the pole measures 611\dfrac{6}{11} m. The pole is then turned upside down and again pushed straight down to the bottom of the pond and pulled out. This time the part that has now been wet twice measures 511\dfrac{5}{11} m. Find the length of the pole.

(The pole is always inserted vertically, and the bottom of the pond is flat.)

Show solution
1 · Understandwhat's really being asked

A pole is dipped to the pond bottom from one end, wetting 611\frac{6}{11} m. Flipped and dipped from the other end, the part now wet twice (the overlap of the two wet regions) measures 511\frac{5}{11} m. I must find the pole's full length.

Givens
  • Dipping from one end wets a 611\frac{6}{11} m length (equal to the water depth).
  • After flipping and dipping from the other end, the part wet twice is 511\frac{5}{11} m.
  • The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, 611\frac{6}{11} m.
Unknowns
  • The total length of the pole.
Constraints
  • Each dip wets a length equal to the water depth, 611\frac{6}{11} m, measured from whichever end goes in.
  • The two wet regions overlap in the middle by 511\frac{5}{11} m.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#11 Work Backwards

Draw the pole as a segment with a 611\frac{6}{11} wet region from each end. The two regions overlap by 511\frac{5}{11}. Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets 611\frac{6}{11} m

#1 Draw a Diagram 4.OA.A.3
Pushing the pole to the flat bottom wets exactly the part below the water surface, a length equal to the water depth = 611\frac{6}{11} m. Flipping and dipping the other end wets another 611\frac{6}{11} m measured from that end.
wet from each end=611 m\text{wet from each end}=\dfrac{6}{11}\text{ m}
Same pond, same depth, so each end gets wet over the same 611\frac{6}{11} m length.

2The overlap is the part wet twice

#1 Draw a Diagram 4.NF.B.3
Drawn on the pole, the wet region from the left and the wet region from the right meet in the middle. Where they meet is wet twice — that overlap is given as 511\frac{5}{11} m.
overlap=511 m\text{overlap}=\dfrac{5}{11}\text{ m}
The middle stretch counted in both dips is the 511\frac{5}{11} m wet-twice part.

3Combine with overlap subtraction

#16 Count the Complement 4.NF.B.3
Total length = left wet + right wet - overlap = 611\frac{6}{11} + 611\frac{6}{11} - 511\frac{5}{11} = 6+6−511\frac{6+6-5}{11} = 711\frac{7}{11} m.
611+611−511=711 m\dfrac{6}{11}+\dfrac{6}{11}-\dfrac{5}{11}=\dfrac{7}{11}\text{ m}
Adding both wet parts double-counts the middle, so subtract the 511\frac{5}{11} overlap once.
Answer: 711\frac{7}{11} m
4 · Reviewdoes it hold up?

The pole length 711\frac{7}{11} m must be longer than the wet depth 611\frac{6}{11} (true) but shorter than two full dips 1211\frac{12}{11} (true), and the overlap 511\frac{5}{11} must be less than the depth 611\frac{6}{11} (true). Everything is consistent in 11ths of a meter.

Another way: Work from the overlap (tool 11): each wet region 611\frac{6}{11} minus the shared 511\frac{5}{11} leaves 111\frac{1}{11} of dry-once at each end; total = 111\frac{1}{11} (left only) + 511\frac{5}{11} (both) + 111\frac{1}{11} (right only) = 711\frac{7}{11} m.

Standardsmin grade 4
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Modeling the two dips and setting up the overlap-subtraction relationship.
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding and subtracting the like-denominator 11ths to get $\frac{7}{11}$ m.
💡Takeaway. This only needs Grade 4 fraction add/subtract — draw the wet parts from each end and subtract the wet-twice overlap once!
Variant 8 hard answer: 912\frac{9}{12} m

A pole is pushed straight down until it touches the bottom of a pond and then pulled back out. The wet part of the pole measures 712\dfrac{7}{12} m. The pole is then turned upside down and again pushed straight down to the bottom of the pond and pulled out. This time the part that has now been wet twice measures 512\dfrac{5}{12} m. Find the length of the pole.

(The pole is always inserted vertically, and the bottom of the pond is flat.)

Show solution
1 · Understandwhat's really being asked

A pole is dipped to the pond bottom from one end, wetting 712\frac{7}{12} m. Flipped and dipped from the other end, the part now wet twice (the overlap of the two wet regions) measures 512\frac{5}{12} m. I must find the pole's full length.

Givens
  • Dipping from one end wets a 712\frac{7}{12} m length (equal to the water depth).
  • After flipping and dipping from the other end, the part wet twice is 512\frac{5}{12} m.
  • The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, 712\frac{7}{12} m.
Unknowns
  • The total length of the pole.
Constraints
  • Each dip wets a length equal to the water depth, 712\frac{7}{12} m, measured from whichever end goes in.
  • The two wet regions overlap in the middle by 512\frac{5}{12} m.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#11 Work Backwards

Draw the pole as a segment with a 712\frac{7}{12} wet region from each end. The two regions overlap by 512\frac{5}{12}. Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets 712\frac{7}{12} m

#1 Draw a Diagram 4.OA.A.3
Pushing the pole to the flat bottom wets exactly the part below the water surface, a length equal to the water depth = 712\frac{7}{12} m. Flipping and dipping the other end wets another 712\frac{7}{12} m measured from that end.
wet from each end=712 m\text{wet from each end}=\dfrac{7}{12}\text{ m}
Same pond, same depth, so each end gets wet over the same 712\frac{7}{12} m length.

2The overlap is the part wet twice

#1 Draw a Diagram 4.NF.B.3
Drawn on the pole, the wet region from the left and the wet region from the right meet in the middle. Where they meet is wet twice — that overlap is given as 512\frac{5}{12} m.
overlap=512 m\text{overlap}=\dfrac{5}{12}\text{ m}
The middle stretch counted in both dips is the 512\frac{5}{12} m wet-twice part.

3Combine with overlap subtraction

#16 Count the Complement 4.NF.B.3
Total length = left wet + right wet - overlap = 712\frac{7}{12} + 712\frac{7}{12} - 512\frac{5}{12} = 7+7−512\frac{7+7-5}{12} = 912\frac{9}{12} m.
712+712−512=912 m\dfrac{7}{12}+\dfrac{7}{12}-\dfrac{5}{12}=\dfrac{9}{12}\text{ m}
Adding both wet parts double-counts the middle, so subtract the 512\frac{5}{12} overlap once.
Answer: 912\frac{9}{12} m
4 · Reviewdoes it hold up?

The pole length 912\frac{9}{12} m must be longer than the wet depth 712\frac{7}{12} (true) but shorter than two full dips 1412\frac{14}{12} (true), and the overlap 512\frac{5}{12} must be less than the depth 712\frac{7}{12} (true). Everything is consistent in 12ths of a meter.

Another way: Work from the overlap (tool 11): each wet region 712\frac{7}{12} minus the shared 512\frac{5}{12} leaves 212\frac{2}{12} of dry-once at each end; total = 212\frac{2}{12} (left only) + 512\frac{5}{12} (both) + 212\frac{2}{12} (right only) = 912\frac{9}{12} m.

Standardsmin grade 4
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Modeling the two dips and setting up the overlap-subtraction relationship.
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding and subtracting the like-denominator 12ths to get $\frac{9}{12}$ m.
💡Takeaway. This only needs Grade 4 fraction add/subtract — draw the wet parts from each end and subtract the wet-twice overlap once!
Variant 9 hard answer: 813\frac{8}{13} m

A pole is pushed straight down until it touches the bottom of a pond and then pulled back out. The wet part of the pole measures 713\dfrac{7}{13} m. The pole is then turned upside down and again pushed straight down to the bottom of the pond and pulled out. This time the part that has now been wet twice measures 613\dfrac{6}{13} m. Find the length of the pole.

(The pole is always inserted vertically, and the bottom of the pond is flat.)

Show solution
1 · Understandwhat's really being asked

A pole is dipped to the pond bottom from one end, wetting 713\frac{7}{13} m. Flipped and dipped from the other end, the part now wet twice (the overlap of the two wet regions) measures 613\frac{6}{13} m. I must find the pole's full length.

Givens
  • Dipping from one end wets a 713\frac{7}{13} m length (equal to the water depth).
  • After flipping and dipping from the other end, the part wet twice is 613\frac{6}{13} m.
  • The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, 713\frac{7}{13} m.
Unknowns
  • The total length of the pole.
Constraints
  • Each dip wets a length equal to the water depth, 713\frac{7}{13} m, measured from whichever end goes in.
  • The two wet regions overlap in the middle by 613\frac{6}{13} m.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#11 Work Backwards

Draw the pole as a segment with a 713\frac{7}{13} wet region from each end. The two regions overlap by 613\frac{6}{13}. Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets 713\frac{7}{13} m

#1 Draw a Diagram 4.OA.A.3
Pushing the pole to the flat bottom wets exactly the part below the water surface, a length equal to the water depth = 713\frac{7}{13} m. Flipping and dipping the other end wets another 713\frac{7}{13} m measured from that end.
wet from each end=713 m\text{wet from each end}=\dfrac{7}{13}\text{ m}
Same pond, same depth, so each end gets wet over the same 713\frac{7}{13} m length.

2The overlap is the part wet twice

#1 Draw a Diagram 4.NF.B.3
Drawn on the pole, the wet region from the left and the wet region from the right meet in the middle. Where they meet is wet twice — that overlap is given as 613\frac{6}{13} m.
overlap=613 m\text{overlap}=\dfrac{6}{13}\text{ m}
The middle stretch counted in both dips is the 613\frac{6}{13} m wet-twice part.

3Combine with overlap subtraction

#16 Count the Complement 4.NF.B.3
Total length = left wet + right wet - overlap = 713\frac{7}{13} + 713\frac{7}{13} - 613\frac{6}{13} = 7+7−613\frac{7+7-6}{13} = 813\frac{8}{13} m.
713+713−613=813 m\dfrac{7}{13}+\dfrac{7}{13}-\dfrac{6}{13}=\dfrac{8}{13}\text{ m}
Adding both wet parts double-counts the middle, so subtract the 613\frac{6}{13} overlap once.
Answer: 813\frac{8}{13} m
4 · Reviewdoes it hold up?

The pole length 813\frac{8}{13} m must be longer than the wet depth 713\frac{7}{13} (true) but shorter than two full dips 1413\frac{14}{13} (true), and the overlap 613\frac{6}{13} must be less than the depth 713\frac{7}{13} (true). Everything is consistent in 13ths of a meter.

Another way: Work from the overlap (tool 11): each wet region 713\frac{7}{13} minus the shared 613\frac{6}{13} leaves 113\frac{1}{13} of dry-once at each end; total = 113\frac{1}{13} (left only) + 613\frac{6}{13} (both) + 113\frac{1}{13} (right only) = 813\frac{8}{13} m.

Standardsmin grade 4
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Modeling the two dips and setting up the overlap-subtraction relationship.
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding and subtracting the like-denominator 13ths to get $\frac{8}{13}$ m.
💡Takeaway. This only needs Grade 4 fraction add/subtract — draw the wet parts from each end and subtract the wet-twice overlap once!
Variant 10 hard answer: 915\frac{9}{15} m

A pole is pushed straight down until it touches the bottom of a pond and then pulled back out. The wet part of the pole measures 815\dfrac{8}{15} m. The pole is then turned upside down and again pushed straight down to the bottom of the pond and pulled out. This time the part that has now been wet twice measures 715\dfrac{7}{15} m. Find the length of the pole.

(The pole is always inserted vertically, and the bottom of the pond is flat.)

Show solution
1 · Understandwhat's really being asked

A pole is dipped to the pond bottom from one end, wetting 815\frac{8}{15} m. Flipped and dipped from the other end, the part now wet twice (the overlap of the two wet regions) measures 715\frac{7}{15} m. I must find the pole's full length.

Givens
  • Dipping from one end wets a 815\frac{8}{15} m length (equal to the water depth).
  • After flipping and dipping from the other end, the part wet twice is 715\frac{7}{15} m.
  • The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, 815\frac{8}{15} m.
Unknowns
  • The total length of the pole.
Constraints
  • Each dip wets a length equal to the water depth, 815\frac{8}{15} m, measured from whichever end goes in.
  • The two wet regions overlap in the middle by 715\frac{7}{15} m.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#11 Work Backwards

Draw the pole as a segment with a 815\frac{8}{15} wet region from each end. The two regions overlap by 715\frac{7}{15}. Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets 815\frac{8}{15} m

#1 Draw a Diagram 4.OA.A.3
Pushing the pole to the flat bottom wets exactly the part below the water surface, a length equal to the water depth = 815\frac{8}{15} m. Flipping and dipping the other end wets another 815\frac{8}{15} m measured from that end.
wet from each end=815 m\text{wet from each end}=\dfrac{8}{15}\text{ m}
Same pond, same depth, so each end gets wet over the same 815\frac{8}{15} m length.

2The overlap is the part wet twice

#1 Draw a Diagram 4.NF.B.3
Drawn on the pole, the wet region from the left and the wet region from the right meet in the middle. Where they meet is wet twice — that overlap is given as 715\frac{7}{15} m.
overlap=715 m\text{overlap}=\dfrac{7}{15}\text{ m}
The middle stretch counted in both dips is the 715\frac{7}{15} m wet-twice part.

3Combine with overlap subtraction

#16 Count the Complement 4.NF.B.3
Total length = left wet + right wet - overlap = 815\frac{8}{15} + 815\frac{8}{15} - 715\frac{7}{15} = 8+8−715\frac{8+8-7}{15} = 915\frac{9}{15} m.
815+815−715=915 m\dfrac{8}{15}+\dfrac{8}{15}-\dfrac{7}{15}=\dfrac{9}{15}\text{ m}
Adding both wet parts double-counts the middle, so subtract the 715\frac{7}{15} overlap once.
Answer: 915\frac{9}{15} m
4 · Reviewdoes it hold up?

The pole length 915\frac{9}{15} m must be longer than the wet depth 815\frac{8}{15} (true) but shorter than two full dips 1615\frac{16}{15} (true), and the overlap 715\frac{7}{15} must be less than the depth 815\frac{8}{15} (true). Everything is consistent in 15ths of a meter.

Another way: Work from the overlap (tool 11): each wet region 815\frac{8}{15} minus the shared 715\frac{7}{15} leaves 115\frac{1}{15} of dry-once at each end; total = 115\frac{1}{15} (left only) + 715\frac{7}{15} (both) + 115\frac{1}{15} (right only) = 915\frac{9}{15} m.

Standardsmin grade 4
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Modeling the two dips and setting up the overlap-subtraction relationship.
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding and subtracting the like-denominator 15ths to get $\frac{9}{15}$ m.
💡Takeaway. This only needs Grade 4 fraction add/subtract — draw the wet parts from each end and subtract the wet-twice overlap once!