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If R =(30^65 - 29^65)/(30^64 + 29^64) then

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If R =(30^65 - 29^65)/(30^64 + 29^64) then  [#permalink]

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New post Updated on: 14 Nov 2014, 01:08
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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

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Originally posted by devmillenium2k on 13 Nov 2014, 18:02.
Last edited by Bunuel on 14 Nov 2014, 01:08, edited 2 times in total.
Renamed the topic and edited the question.
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Re: If R =(30^65 - 29^65)/(30^64 + 29^64) then  [#permalink]

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New post 13 Nov 2014, 20:05
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devmillenium2k wrote:
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



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)
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Re: If R =(30^65 - 29^65)/(30^64 + 29^64) then  [#permalink]

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New post 14 Nov 2014, 00:18
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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
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If R =(30^65 - 29^65)/(30^64 + 29^64) then  [#permalink]

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New post Updated on: 17 Nov 2016, 21:44
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devmillenium2k wrote:
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


\(a^n – b^n = (a – b)(a^{n-1} - a^{n-2}*b + a^{n-3}*b^2 …… + b^{n-1})\)

Plugging in our values:

\(1*(30^{64} - 30^{63}*29 + 30^{62}*29^2 - ….. + 29^{64}) = 30^{64} + 29^{64} + k, k > 1.\)

Hence \(30^{65} – 29^{65} = 30^{64} + 29^{64} + k > 30^{64} + 29^{64}\)

In general: \(a^n - b^n = a^{n-1} + b^{n-1} +k > a^{n-1} + b^{n-1}\), because \(k\) is always \(> 1\)

And our fraction is \(> 1\).

If power of polynomial in numerator is greater than power of polynomial in denominator the value of fraction will always be >1. We can prove this with application of simple calculus. Just by taking consecutive derivatives of numerator and denominator.

\(\frac{65*64*63*62* ….. *x}{64*63*62* …*1}\)

Which is always \(> 1.\)

Originally posted by vitaliyGMAT on 17 Nov 2016, 01:01.
Last edited by vitaliyGMAT on 17 Nov 2016, 21:44, edited 1 time in total.
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Re: If R =(30^65 - 29^65)/(30^64 + 29^64) then  [#permalink]

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New post 17 Nov 2016, 13:35
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You don't need to solve using the exact numbers in the problem-

You can use for example: (3^3 - 2^3)/(3^2 + 2^2) = 19/13

Hence, R>1 --> E is correct
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Re: If R =(30^65 - 29^65)/(30^64 + 29^64) then  [#permalink]

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New post 26 Nov 2018, 00:26
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devmillenium2k wrote:
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


R= 30^{65} - 29^{65}/30^{64} + 29^{64}
=>[ 30^{64} + 29^{64} ]R= 30^{65} - 29^{65}
=> 30^{64} [30-R] = 29^{64} [29+R]
=> (30/29)^{64) = (29+R)/(30-R)
Now, LHS > 1 => 29+R> 30-R => R > 0.5
Therefore, R= ( 0.5, ∞ )

Ans. only E satisfy this condt.
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Re: If R =(30^65 - 29^65)/(30^64 + 29^64) then &nbs [#permalink] 26 Nov 2018, 00:26
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