Operations & Word Problems

Problem

Use submerged and exposed parts to find bar length

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.
Fractions
Your answer
m
How to solve
Strategy Draw a Diagram — 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.
1STEP 1

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

Each dip wets a length equal to the water depth: 47\frac{4}{7} m from either end.

wet from each end=47 m\text{wet from each end} = \frac{4}{7} \text{ m}
2STEP 2

The overlap is the part wet twice

The wet parts from each end overlap in the middle — the overlap is 37\frac{3}{7} m.

overlap=37 m\text{overlap} = \frac{3}{7} \text{ m}
3STEP 3

Combine with overlap subtraction

Total length = left wet + right wet - overlap = 47\frac{4}{7} + 47\frac{4}{7} - 37\frac{3}{7} = 4+437\frac{4+4-3}{7} = 57\frac{5}{7} m.

47+4737=57 m\frac{4}{7}+\frac{4}{7}-\frac{3}{7}=\frac{5}{7} \text{ m}
Answer
57\frac{5}{7} m
47+4737=57 m\frac{4}{7}+\frac{4}{7}-\frac{3}{7}=\frac{5}{7} \text{ m}
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 sevenths of a meter.
Takeaway

This only needs Grade 4 fraction add/subtract — draw the wet parts from each end and subtract the wet-twice overlap once!

  • Each dip wets 47\frac{4}{7} m
  • The overlap is the part wet twice
  • Combine with overlap subtraction
Where next?
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