Baker's Parallax

Let the volume of the loaf be $1$ , $x=40\%=0.4$ and $y=60\%=0.6$ . Then after the first cut the two pieces are $x : y$ . After the second cut $x^2:xy:yx: y^2$ . Note that the sizes of the pieces follow binomial expansion $(x+y)^1 = x+y$ , $(x+y)^2 = x^2 + 2xy + y^2$ , $(x+y)^3$ .... Therefore after the fifth cut, we have $(x+y)^5 = x^5 + 5x^4y+10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5$ . The smallest piece is $x^5$ and the largest $y^5$ . And their ratio is:

$\frac {x^5}{y^5} = \left(\frac {0.4}{0.6}\right)^5 = \left(\frac 46 \right)^5 = \frac {1024}{7776} \text{ or } \boxed{1024:7776}$

The ratio is of the form

$\boxed{0.4^n : 0.6^n}$

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$0.4^5:0.6^5$ $= 0.01024: 0.07776$ $=1024:7776$

After each cut, the smallest piece of the loaf will be $(\dfrac{40}{100})^{th}$ or $(\dfrac{2}{5})^{th}$ of the previous one, and the largest will be $(\dfrac{3}{5})^{th}$ of the previous one. So, after $n$ cuts, the smallest slice will be $(\dfrac{2^n}{5^n})^{th}$ and the largest one will be $(\dfrac{3^n}{5^n})^{th}$ of the original size of the loaf. Hence the required ratio is $\dfrac{2^n}{3^n}$ . For $n=5$ , the ratio is $\dfrac{2^5}{3^5}=\boxed {\dfrac{32}{243}=\dfrac{1024}{7776}}$