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Updated on: 14 Nov 2014, 02:08
Originally posted by devmillenium2k on 13 Nov 2014, 19:02.
Last edited by Bunuel on 14 Nov 2014, 02:08, edited 2 times in total.
Last edited by Bunuel on 14 Nov 2014, 02:08, edited 2 times in total.
Renamed the topic and edited the question.
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E
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Difficulty:
95%
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Question Stats:
41% (02:09) correct
59%
(02:05)
wrong
based on 161
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If \(R =\frac{30^{65} - 29^{65}}{30^{64} + 29^{64}}\) then
(A) \(0<R\leq{0.1}\)
(B) \(0.1<R\leq{0.5}\)
(C) \(0.5<R\leq{1.0}\)
(D) R<1.0
(E) R>1.0
(A) \(0<R\leq{0.1}\)
(B) \(0.1<R\leq{0.5}\)
(C) \(0.5<R\leq{1.0}\)
(D) R<1.0
(E) R>1.0
13 Nov 2014, 21:05
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devmillenium2k
If R ={\(30^{65}\)-\(29^{65}\)}/{\(30^{64}\)+\(29^{64}\)} then
(A) 0<\(R\leq{0.1}\)
(B) 0.1<\(R\leq{0.5}\)
(C) 0.5<\(R\leq{1.0}\)
(D) R<1.0
(E) R>1.0
(A) 0<\(R\leq{0.1}\)
(B) 0.1<\(R\leq{0.5}\)
(C) 0.5<\(R\leq{1.0}\)
(D) R<1.0
(E) R>1.0
This question is based on your understanding of number properties. You don't have to do any calculations actually.
At higher powers, a difference of even 1 in the base creates creates a huge difference in the number. For example, \(2^{10} = 1024\) but \(3^{10} = 59049\).
So \(30^{65}\) will not be very different from \(30^{65}\)-\(29^{65}\).
Similarly, \(30^{64}\) will not be very different from \(30^{64}\)+\(29^{64}\)
So the question boils down to approx \(\frac{30^{65}}{30^{64}}\). This will be something less than 30 but certainly greater than 1.
Answer (E)
General Discussion
14 Nov 2014, 01:18
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\(R = \frac{30^{65} - 29^{65}}{30^{64} + 29^{64}} \approx{\frac{30^{65}}{30^{64}}} \approx{30}\)
Answer = E
\(30^{65} or 30^{64}\) is a very big number; any addition/subtraction of "equivalent value" would hardly make a big difference
Answer = E
\(30^{65} or 30^{64}\) is a very big number; any addition/subtraction of "equivalent value" would hardly make a big difference
