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20 Dec 2019, 01:17
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
72% (02:18) correct
28%
(02:37)
wrong
based on 341
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Not Attempted Yet
If a and b are positive and \(5ab=24-a^2b^2\), then \(b^2=\)
(A) \(\frac{81}{a^2}\)
(B) \(\frac{64}{a^2}\)
(C) \(\frac{35}{a^2}\)
(D) \(\frac{15}{a^2}\)
(E) \(\frac{9}{a^2}\)
(A) \(\frac{81}{a^2}\)
(B) \(\frac{64}{a^2}\)
(C) \(\frac{35}{a^2}\)
(D) \(\frac{15}{a^2}\)
(E) \(\frac{9}{a^2}\)
20 Dec 2019, 01:42
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a,b — positive numbers
—>\(( ab)^{2} + 5(ab) —24= 0\)
(ab—3)(ab+ 8) = 0
ab= 3, ab = —8
—> since both a and b are positive numbers, ab = 3
—>\(b= \frac{3}{a}\)—> \(b^{2} = \frac{9}{a^{2}}\)
The answer is E .
Posted from my mobile device
—>\(( ab)^{2} + 5(ab) —24= 0\)
(ab—3)(ab+ 8) = 0
ab= 3, ab = —8
—> since both a and b are positive numbers, ab = 3
—>\(b= \frac{3}{a}\)—> \(b^{2} = \frac{9}{a^{2}}\)
The answer is E .
Posted from my mobile device
28 Jan 2020, 18:43
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Anyone willing to post an alternate approach?
I got it right but not sure if that is the best approach.
I basically could not visualize the path to the solution.
GMATPrepNow may be you can help?
I got it right but not sure if that is the best approach.
I basically could not visualize the path to the solution.
GMATPrepNow may be you can help?
