How do we compare fraction without changing to a similar denominator?

Question

This is Singapore Mathematical Olympiad 2015 Grade 8/Secondary 2 Junior Round 1 Question 1.

1.Among the five numbers, $\frac{5}{9},\frac{4}{7},\frac{3}{5},\frac{6}{11}$ and $\frac{13}{21}$, which one has the smallest value?

So the obvious method is to either make the same denominator (Which is 3465) OR to compare the 2 fractions at a time and determine the smallest, which definitely you all know how to do.(Pretty sure I don't need to show how to do that right?)

My question is, because this is a Maths Olympiad question, you are deemed to do the question quickly and obvious by doing these 2 methods will ruin time, so how do we compare fraction very quickly? Or comparing fraction simply just by looking at the numbers?

Otherwise, is that the only way?

Answer 1

$35\lt 36$ so $\frac 59\lt \frac 47$; $25\lt 27$ so $\frac 59\lt \frac 35$; $55\gt 54$ so $\frac 59\gt \frac 6{11}$ and $126\lt 143$ so $\frac 6{11}\lt \frac {13}{21}$

Answer 2

Order the numbers by increasing denominator,

$${3\over5},{4\over7},{5\over9},{6\over11},{13\over21}$$

and then think of them "free throw" statistics, i.e., at some point your (basketball) team has made $3$ of $5$ free throws, a little later it's made $4$ of $7$, and so forth. Note that each stat, by itself, is better than $50$%, but for the first four numbers, you're adding $1$ out of $2$, which means you're bringing your average down. Thus

$${3\over5}\gt{4\over7}\gt{5\over9}\gt{6\over11}$$

For the final comparison, note that if you maintained the $6/11$ for another $11$ attempts, you'd be $12$ for $22$, which is clearly not as good as going $13$ for $21$. Thus we have

$${3\over5}\gt{4\over7}\gt{5\over9}\gt{6\over11}\lt{13\over21}$$

so ${6\over11}$ is the smallest fraction.