(Quickly) finding the smallest fraction

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Please see these fractions:

(A) $\frac{33}{128}$ (B) $\frac{45}{138}$ (C) $\frac{53}{216}$ (D) $\frac{83}{324}$ (E) $\frac{15}{59}$.

I need to find out quickly (in about a minute) the smallest of these fractions. I am not allowed to use a calculator, though a little rough calculation is allowed.

As I see the problem, without a calculator, converting these to decimal values is not an option. Moreover, since the numbers involved are fairly large, finding the least common denominator and changing each fraction to make their denominators the same as the least common denominator is again next to impossible without a calculator. For the same reason, we can not try the method of making the numerators identical.

If we try the approximate method of changing the numerators and denominators to easy numbers, we can get something like this:

(A) $\frac{30}{120}$ (B) $\frac{45}{135}$ (C) $\frac{50}{200}$ (D) $\frac{80}{320}$ (E) $\frac{15}{60}$.

Unfortunately, this makes all of the fractions approximately equivalent to $\frac{1}{4}$ except B, which becomes $\frac{1}{3}$.

This is the point where I can not proceed any further.

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Reciprocate/flip them, divide (long division), and see which of these is the biggest. The biggest reciprocal should correspond to the smallest number.

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All fractions are greater than $1/4$ except for $\frac{53}{216}$

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If you have $\frac{a}{b}$ and $\frac cd$ then $\frac ab > \frac cd \iff ad> bc$. Multiplication isn't too slow.

Now pick your favourite fraction, compare it against another, discarding the larger. To ensure you are working with smaller numbers, you can discard any common factors between $ab$ and $cd$. You need 4 comparisons to check which is smallest. Each comparison requires 2 multiplications. I understand that 7.5 seconds is not enough to do a multiplication, but it does not seem too bad sometimes.

As both @lulu and @roddy point out, there are more tricks for these specific fractions, which speed up the process quite a bit. Practicing your multiplication as well as exploiting these tricks should make this problem easy.