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Bunuel
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Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
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We can take out (2^2)(5^2) common from all denominators (this is the HCF of all denominators).
Then we are left with:

A) 8/2 = 4
B) 10/5 = 2
C) 28/(2*5) = 2.8
D) 16/1 = 16
E) 140/(2^2 * 5) = 7

Clearly B is the smallest. Hence B
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Bunuel
Which has the least value?

A. \(\frac{8}{(2^3)(5^2)}\)
B. \(\frac{10}{(2^2)(5^3)}\)
C. \(\frac{28}{(2^3)(5^3)}\)
D. \(\frac{16}{(2^2)(5^2)}\)
E. \(\frac{140}{(2^4)(5^3)}\)

A. \(\frac{8}{(2^3)(5^2)}=\frac{8}{(10^2)(2)}…=\frac{40}{(10^3)}\)
B. \(\frac{10}{(2^2)(5^3)}=\frac{10}{(10^2)(5)}…=\frac{20}{(10^3)}\)
C. \(\frac{28}{(2^3)(5^3)}=\frac{28}{(10^3)}\)
D. \(\frac{16}{(2^2)(5^2)}=\frac{16}{(10^2)}=\frac{160}{(10^3)}\)
E. \(\frac{140}{(2^4)(5^3)}=\frac{140}{(10^3)(2)}=\frac{70}{(10^3)}\)

Answer (B).
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easiest way to do this is to convert all the denominators to (2^3)*(5^3) and compare numerators
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