Practice · Use submerged and exposed parts to find bar length · Sensim Math

Variant 1 easy

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 $\dfrac{4}{7}$ 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 $\dfrac{3}{7}$ m. Find the length of the pole.

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

Your answer

Show solution

1 · Understandwhat's really being asked

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

Givens

Dipping from one end wets a $\frac{4}{7}$ m length (equal to the water depth).
After flipping and dipping from the other end, the part wet twice is $\frac{3}{7}$ m.
The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, $\frac{4}{7}$ m.

Unknowns

The total length of the pole.

Constraints

Each dip wets a length equal to the water depth, $\frac{4}{7}$ m, measured from whichever end goes in.
The two wet regions overlap in the middle by $\frac{3}{7}$ 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 $\frac{4}{7}$ wet region from each end. The two regions overlap by $\frac{3}{7}$ . Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets $\frac{4}{7}$ 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 =

\frac{4}{7}

m. Flipping and dipping the other end wets another

\frac{4}{7}

m measured from that end.

\text{wet from each end}=\dfrac{4}{7}\text{ m}

Same pond, same depth, so each end gets wet over the same

\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

\frac{3}{7}

m.

\text{overlap}=\dfrac{3}{7}\text{ m}

The middle stretch counted in both dips is the

\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 =

\frac{4}{7}

+

\frac{4}{7}

-

\frac{3}{7}

=

\frac{4+4-3}{7}

=

\frac{5}{7}

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

\frac{3}{7}

overlap once.

Answer:

\frac{5}{7}

m

4 · Reviewdoes it hold up?

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

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

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 $\dfrac{5}{7}$ 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 $\dfrac{2}{7}$ m. Find the length of the pole.

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

Your answer

Show solution

1 · Understandwhat's really being asked

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

Givens

Dipping from one end wets a $\frac{5}{7}$ m length (equal to the water depth).
After flipping and dipping from the other end, the part wet twice is $\frac{2}{7}$ m.
The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, $\frac{5}{7}$ m.

Unknowns

The total length of the pole.

Constraints

Each dip wets a length equal to the water depth, $\frac{5}{7}$ m, measured from whichever end goes in.
The two wet regions overlap in the middle by $\frac{2}{7}$ 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 $\frac{5}{7}$ wet region from each end. The two regions overlap by $\frac{2}{7}$ . Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets $\frac{5}{7}$ 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 =

\frac{5}{7}

m. Flipping and dipping the other end wets another

\frac{5}{7}

m measured from that end.

\text{wet from each end}=\dfrac{5}{7}\text{ m}

Same pond, same depth, so each end gets wet over the same

\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

\frac{2}{7}

m.

\text{overlap}=\dfrac{2}{7}\text{ m}

The middle stretch counted in both dips is the

\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 =

\frac{5}{7}

+

\frac{5}{7}

-

\frac{2}{7}

=

\frac{5+5-2}{7}

=

\frac{8}{7}

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

\frac{2}{7}

overlap once.

Answer:

\frac{8}{7}

m

4 · Reviewdoes it hold up?

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

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

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 $\dfrac{5}{8}$ 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 $\dfrac{3}{8}$ m. Find the length of the pole.

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

Your answer

Show solution

1 · Understandwhat's really being asked

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

Givens

Dipping from one end wets a $\frac{5}{8}$ m length (equal to the water depth).
After flipping and dipping from the other end, the part wet twice is $\frac{3}{8}$ m.
The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, $\frac{5}{8}$ m.

Unknowns

The total length of the pole.

Constraints

Each dip wets a length equal to the water depth, $\frac{5}{8}$ m, measured from whichever end goes in.
The two wet regions overlap in the middle by $\frac{3}{8}$ 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 $\frac{5}{8}$ wet region from each end. The two regions overlap by $\frac{3}{8}$ . Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets $\frac{5}{8}$ 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 =

\frac{5}{8}

m. Flipping and dipping the other end wets another

\frac{5}{8}

m measured from that end.

\text{wet from each end}=\dfrac{5}{8}\text{ m}

Same pond, same depth, so each end gets wet over the same

\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

\frac{3}{8}

m.

\text{overlap}=\dfrac{3}{8}\text{ m}

The middle stretch counted in both dips is the

\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 =

\frac{5}{8}

+

\frac{5}{8}

-

\frac{3}{8}

=

\frac{5+5-3}{8}

=

\frac{7}{8}

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

\frac{3}{8}

overlap once.

Answer:

\frac{7}{8}

m

4 · Reviewdoes it hold up?

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

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

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 $\dfrac{6}{9}$ 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 $\dfrac{4}{9}$ m. Find the length of the pole.

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

Your answer

Show solution

1 · Understandwhat's really being asked

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

Givens

Dipping from one end wets a $\frac{6}{9}$ m length (equal to the water depth).
After flipping and dipping from the other end, the part wet twice is $\frac{4}{9}$ m.
The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, $\frac{6}{9}$ m.

Unknowns

The total length of the pole.

Constraints

Each dip wets a length equal to the water depth, $\frac{6}{9}$ m, measured from whichever end goes in.
The two wet regions overlap in the middle by $\frac{4}{9}$ 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 $\frac{6}{9}$ wet region from each end. The two regions overlap by $\frac{4}{9}$ . Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets $\frac{6}{9}$ 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 =

\frac{6}{9}

m. Flipping and dipping the other end wets another

\frac{6}{9}

m measured from that end.

\text{wet from each end}=\dfrac{6}{9}\text{ m}

Same pond, same depth, so each end gets wet over the same

\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

\frac{4}{9}

m.

\text{overlap}=\dfrac{4}{9}\text{ m}

The middle stretch counted in both dips is the

\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 =

\frac{6}{9}

+

\frac{6}{9}

-

\frac{4}{9}

=

\frac{6+6-4}{9}

=

\frac{8}{9}

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

\frac{4}{9}

overlap once.

Answer:

\frac{8}{9}

m

4 · Reviewdoes it hold up?

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

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

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 $\dfrac{5}{9}$ 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 $\dfrac{4}{9}$ m. Find the length of the pole.

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

Your answer

Show solution

1 · Understandwhat's really being asked

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

Givens

Dipping from one end wets a $\frac{5}{9}$ m length (equal to the water depth).
After flipping and dipping from the other end, the part wet twice is $\frac{4}{9}$ m.
The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, $\frac{5}{9}$ m.

Unknowns

The total length of the pole.

Constraints

Each dip wets a length equal to the water depth, $\frac{5}{9}$ m, measured from whichever end goes in.
The two wet regions overlap in the middle by $\frac{4}{9}$ 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 $\frac{5}{9}$ wet region from each end. The two regions overlap by $\frac{4}{9}$ . Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets $\frac{5}{9}$ 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 =

\frac{5}{9}

m. Flipping and dipping the other end wets another

\frac{5}{9}

m measured from that end.

\text{wet from each end}=\dfrac{5}{9}\text{ m}

Same pond, same depth, so each end gets wet over the same

\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

\frac{4}{9}

m.

\text{overlap}=\dfrac{4}{9}\text{ m}

The middle stretch counted in both dips is the

\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 =

\frac{5}{9}

+

\frac{5}{9}

-

\frac{4}{9}

=

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

=

\frac{6}{9}

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

\frac{4}{9}

overlap once.

Answer:

\frac{6}{9}

m

4 · Reviewdoes it hold up?

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

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

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 $\dfrac{6}{10}$ 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 $\dfrac{4}{10}$ m. Find the length of the pole.

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

Your answer

Show solution

1 · Understandwhat's really being asked

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

Givens

Dipping from one end wets a $\frac{6}{10}$ m length (equal to the water depth).
After flipping and dipping from the other end, the part wet twice is $\frac{4}{10}$ m.
The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, $\frac{6}{10}$ m.

Unknowns

The total length of the pole.

Constraints

Each dip wets a length equal to the water depth, $\frac{6}{10}$ m, measured from whichever end goes in.
The two wet regions overlap in the middle by $\frac{4}{10}$ 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 $\frac{6}{10}$ wet region from each end. The two regions overlap by $\frac{4}{10}$ . Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets $\frac{6}{10}$ 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 =

\frac{6}{10}

m. Flipping and dipping the other end wets another

\frac{6}{10}

m measured from that end.

\text{wet from each end}=\dfrac{6}{10}\text{ m}

Same pond, same depth, so each end gets wet over the same

\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

\frac{4}{10}

m.

\text{overlap}=\dfrac{4}{10}\text{ m}

The middle stretch counted in both dips is the

\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 =

\frac{6}{10}

+

\frac{6}{10}

-

\frac{4}{10}

=

\frac{6+6-4}{10}

=

\frac{8}{10}

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

\frac{4}{10}

overlap once.

Answer:

\frac{8}{10}

m

4 · Reviewdoes it hold up?

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

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

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 $\dfrac{6}{11}$ 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 $\dfrac{5}{11}$ m. Find the length of the pole.

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

Your answer

Show solution

1 · Understandwhat's really being asked

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

Givens

Dipping from one end wets a $\frac{6}{11}$ m length (equal to the water depth).
After flipping and dipping from the other end, the part wet twice is $\frac{5}{11}$ m.
The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, $\frac{6}{11}$ m.

Unknowns

The total length of the pole.

Constraints

Each dip wets a length equal to the water depth, $\frac{6}{11}$ m, measured from whichever end goes in.
The two wet regions overlap in the middle by $\frac{5}{11}$ 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 $\frac{6}{11}$ wet region from each end. The two regions overlap by $\frac{5}{11}$ . Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets $\frac{6}{11}$ 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 =

\frac{6}{11}

m. Flipping and dipping the other end wets another

\frac{6}{11}

m measured from that end.

\text{wet from each end}=\dfrac{6}{11}\text{ m}

Same pond, same depth, so each end gets wet over the same

\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

\frac{5}{11}

m.

\text{overlap}=\dfrac{5}{11}\text{ m}

The middle stretch counted in both dips is the

\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 =

\frac{6}{11}

+

\frac{6}{11}

-

\frac{5}{11}

=

\frac{6+6-5}{11}

=

\frac{7}{11}

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

\frac{5}{11}

overlap once.

Answer:

\frac{7}{11}

m

4 · Reviewdoes it hold up?

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

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

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 $\dfrac{7}{12}$ 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 $\dfrac{5}{12}$ m. Find the length of the pole.

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

Your answer

Show solution

1 · Understandwhat's really being asked

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

Givens

Dipping from one end wets a $\frac{7}{12}$ m length (equal to the water depth).
After flipping and dipping from the other end, the part wet twice is $\frac{5}{12}$ m.
The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, $\frac{7}{12}$ m.

Unknowns

The total length of the pole.

Constraints

Each dip wets a length equal to the water depth, $\frac{7}{12}$ m, measured from whichever end goes in.
The two wet regions overlap in the middle by $\frac{5}{12}$ 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 $\frac{7}{12}$ wet region from each end. The two regions overlap by $\frac{5}{12}$ . Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets $\frac{7}{12}$ 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 =

\frac{7}{12}

m. Flipping and dipping the other end wets another

\frac{7}{12}

m measured from that end.

\text{wet from each end}=\dfrac{7}{12}\text{ m}

Same pond, same depth, so each end gets wet over the same

\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

\frac{5}{12}

m.

\text{overlap}=\dfrac{5}{12}\text{ m}

The middle stretch counted in both dips is the

\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 =

\frac{7}{12}

+

\frac{7}{12}

-

\frac{5}{12}

=

\frac{7+7-5}{12}

=

\frac{9}{12}

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

\frac{5}{12}

overlap once.

Answer:

\frac{9}{12}

m

4 · Reviewdoes it hold up?

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

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

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 $\dfrac{7}{13}$ 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 $\dfrac{6}{13}$ m. Find the length of the pole.

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

Your answer

Show solution

1 · Understandwhat's really being asked

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

Givens

Dipping from one end wets a $\frac{7}{13}$ m length (equal to the water depth).
After flipping and dipping from the other end, the part wet twice is $\frac{6}{13}$ m.
The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, $\frac{7}{13}$ m.

Unknowns

The total length of the pole.

Constraints

Each dip wets a length equal to the water depth, $\frac{7}{13}$ m, measured from whichever end goes in.
The two wet regions overlap in the middle by $\frac{6}{13}$ 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 $\frac{7}{13}$ wet region from each end. The two regions overlap by $\frac{6}{13}$ . Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets $\frac{7}{13}$ 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 =

\frac{7}{13}

m. Flipping and dipping the other end wets another

\frac{7}{13}

m measured from that end.

\text{wet from each end}=\dfrac{7}{13}\text{ m}

Same pond, same depth, so each end gets wet over the same

\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

\frac{6}{13}

m.

\text{overlap}=\dfrac{6}{13}\text{ m}

The middle stretch counted in both dips is the

\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 =

\frac{7}{13}

+

\frac{7}{13}

-

\frac{6}{13}

=

\frac{7+7-6}{13}

=

\frac{8}{13}

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

\frac{6}{13}

overlap once.

Answer:

\frac{8}{13}

m

4 · Reviewdoes it hold up?

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

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

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 $\dfrac{8}{15}$ 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 $\dfrac{7}{15}$ m. Find the length of the pole.

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

Your answer

Show solution

1 · Understandwhat's really being asked

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

Givens

Dipping from one end wets a $\frac{8}{15}$ m length (equal to the water depth).
After flipping and dipping from the other end, the part wet twice is $\frac{7}{15}$ m.
The pole is inserted vertically and the bottom is flat, so each dip wets the same depth, $\frac{8}{15}$ m.

Unknowns

The total length of the pole.

Constraints

Each dip wets a length equal to the water depth, $\frac{8}{15}$ m, measured from whichever end goes in.
The two wet regions overlap in the middle by $\frac{7}{15}$ 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 $\frac{8}{15}$ wet region from each end. The two regions overlap by $\frac{7}{15}$ . Length = (left wet) + (right wet) - (overlap), the classic overlap subtraction.

3 · Execute3 carry out the plan

1Each dip wets $\frac{8}{15}$ 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 =

\frac{8}{15}

m. Flipping and dipping the other end wets another

\frac{8}{15}

m measured from that end.

\text{wet from each end}=\dfrac{8}{15}\text{ m}

Same pond, same depth, so each end gets wet over the same

\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

\frac{7}{15}

m.

\text{overlap}=\dfrac{7}{15}\text{ m}

The middle stretch counted in both dips is the

\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 =

\frac{8}{15}

+

\frac{8}{15}

-

\frac{7}{15}

=

\frac{8+8-7}{15}

=

\frac{9}{15}

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

\frac{7}{15}

overlap once.

Answer:

\frac{9}{15}

m

4 · Reviewdoes it hold up?

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

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

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

Generated variants — 10

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