If n is one of the numbers in \(\frac{1}{3}\), \(\frac{3}{16}\), \(\frac{4}{7}\), \(\frac{3}{5}\) then what is the value of n?

The method of cross multiplication:
Suppose we want to know which positive fraction is greater \(\frac{4}{7}\) or \(\frac{7}{12}\). Cross-multiply --> \(4*12=48\) and \(7*7=49\) --> \(48<49\). Now, ask yourself, which fraction contributed nominator for the larger value? \(\frac{7}{12}\)! Thus \(\frac{4}{7}<\frac{7}{12}\).


(1) \(\frac{5}{16}<n<\frac{7}{12}\) --> \(\frac{5}{16}<(\frac{1}{3}=\frac{5}{15})<\frac{7}{12}\), hence \(\frac{1}{3}\) is obviously in the given range (notice also that \(\frac{1}{2}<\frac{7}{12}\)). Next, from our example above we know that \(\frac{4}{7}<\frac{7}{12}\) so \(\frac{4}{7}\) is also in the given range. Not sufficient.


(2) \(\frac{7}{13}<n<\frac{19}{33}\) --> only two values might be in this range: \(\frac{4}{7}\approx{5.7}\) and \(\frac{3}{5}=0.6\) (other possible values of n are less than 1/2 and are clearly out of the range). As the second one is larger, then let's compare it with \(\frac{19}{33}\) (the upper limit of the range). So we are comparing \(\frac{19}{33}\) and \(\frac{3}{5}\): cross-multiply --> \(3*33=99\) and \(19*5=95\) --> \(99>95\). Which fraction contributed nominator for the larger value? \(\frac{3}{5}\)! Thus \(\frac{3}{5}>\frac{19}{33}\), which means that \(\frac{3}{5}\) is out of the range. \(n\) can only be \(\frac{4}{7}\). Sufficient.

Answer: B.

Hope it's clear.
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