[GMAT math practice question] 6.7
If 0.001 < a < 0.002 and 1,000 < b < 2,000, which of the following could be the value of a^3b^2+a^2b^3/a^3b^3?
A. 15
B. 125
C. 625
D. 1250
E. 6250
If 0.001 < a < 0.002 and 1,000 < b < 2,000, whic
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- Max@Math Revolution
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=>
$$\frac{a^3b^2+a^2b^3}{a^3b^3}=\frac{1}{a}+\frac{1}{b}$$
Since 0.001 < a < 0.002 and 1,000 < b < 2,000, we have 500 < 1/a < 1000 and 1/2000 < 1/b < 1/1000.
Thus
$$500+\frac{1}{2000}\ <\ \frac{a^3b^2+a^2b^3}{a^3b^3}\ <\ 1000+\frac{1\ }{1000}$$
Therefore, C is the only possible answer.
Answer: C
$$\frac{a^3b^2+a^2b^3}{a^3b^3}=\frac{1}{a}+\frac{1}{b}$$
Since 0.001 < a < 0.002 and 1,000 < b < 2,000, we have 500 < 1/a < 1000 and 1/2000 < 1/b < 1/1000.
Thus
$$500+\frac{1}{2000}\ <\ \frac{a^3b^2+a^2b^3}{a^3b^3}\ <\ 1000+\frac{1\ }{1000}$$
Therefore, C is the only possible answer.
Answer: C
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