Carcass wrote:
Quantity A |
Quantity B |
\(\frac{75}{4^2} \times \frac{3^2}{45} \times \frac{2^4}{45}\) |
\(\frac{3^2}{4^2} \times \frac{2^2}{5^2} \times \frac{10}{3}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Break every number into the product of prime numbers...
Quantity A: \(\frac{75}{4^2} \times \frac{3^2}{45} \times \frac{2^4}{45}=\frac{(3)(5^2)}{(2^2)^2} \times \frac{3^2}{(3^2)(5)} \times \frac{2^4}{(3^2)(5)}=\frac{(3)(5^2)}{2^4} \times \frac{1}{5} \times \frac{2^4}{(3^2)(5)}=\frac{(2^4)(3)(5^2)}{(2^4)(3^2)(5^2)}=\frac{1}{3}\)
Quantity B: \(\frac{3^2}{4^2} \times \frac{2^2}{5^2} \times \frac{10}{3}=\frac{3^2}{(2^2)^2} \times \frac{2^2}{5^2} \times \frac{(2)(5)}{3}=\frac{3^2}{2^4} \times \frac{2^2}{5^2} \times \frac{(2)(5)}{3}=\frac{(2^3)(3^2)(5)}{(2^4)(3)(5^2)}=\frac{3}{(2)(5)}=\frac{3}{10}\)
Since \(\frac{1}{3}>\frac{3}{10}\), the correct answer is A.
Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com
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