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# If a=1+1/4+1/16+1/64 and b=1+(1/4)a

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Intern
Joined: 23 Jul 2017
Posts: 27
Location: United Arab Emirates
Concentration: Finance, Entrepreneurship
WE: Analyst (Commercial Banking)

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Updated on: 04 Aug 2018, 23:47
00:00

Difficulty:

45% (medium)

Question Stats:

65% (01:17) correct 35% (02:58) wrong based on 20 sessions

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If a=1+$$\frac{1}{4}$$+$$\frac{1}{16}$$+$$\frac{1}{64}$$and b= 1+$$\frac{1}{4}$$a, then what is the value of a-b?

A)$$\frac{1}{256}$$
B)$$\frac{1}{128}$$
C)$$0$$
D)$$\frac{-1}{128}$$
E)$$\frac{-1}{256}$$

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Originally posted by delta23 on 04 Aug 2018, 23:15.
Last edited by delta23 on 04 Aug 2018, 23:47, edited 1 time in total.
Director
Status: Learning stage
Joined: 01 Oct 2017
Posts: 952
WE: Supply Chain Management (Energy and Utilities)
Re: If a=1+1/4+1/16+1/64 and b=1+(1/4)a  [#permalink]

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04 Aug 2018, 23:38
1
1
delta23 wrote:
If a=1+$$\frac{1}{4}$$+$$\frac{1}{16}$$+$$\frac{1}{64}$$and b= 1+$$\frac{1}{4}$$a, then what is the value of a-b?

A)$$\frac{1}{256}$$
B)$$\frac{1}{128}$$
C)$$0$$
D)$$\frac{-1}{128}$$
E)$$\frac{-1}{256}$$

Given, a=1+$$\frac{1}{4}$$+$$\frac{1}{16}$$+$$\frac{1}{64}$$
So, $$\frac{1}{4}$$a=$$\frac{1}{4}$$+$$\frac{1}{16}$$+$$\frac{1}{64}$$+$$\frac{1}{256}$$
So, b=1+$$\frac{1}{4}$$a=1+$$\frac{1}{4}$$+$$\frac{1}{16}$$+$$\frac{1}{64}$$+$$\frac{1}{256}$$

Hence, a-b=(1+$$\frac{1}{4}$$+$$\frac{1}{16}$$+$$\frac{1}{64}$$)-(1+$$\frac{1}{4}$$+$$\frac{1}{16}$$+$$\frac{1}{64}$$+$$\frac{1}{256}$$)
Or, $$a-b=-\frac{1}{256}$$

Ans. (E)

Hi delta23,
If a=1+$$\frac{1}{4}$$+$$\frac{1}{16}$$+$$\frac{1}{64}$$
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Regards,

PKN

Rise above the storm, you will find the sunshine

Intern
Joined: 23 Jul 2017
Posts: 27
Location: United Arab Emirates
Concentration: Finance, Entrepreneurship
WE: Analyst (Commercial Banking)
Re: If a=1+1/4+1/16+1/64 and b=1+(1/4)a  [#permalink]

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04 Aug 2018, 23:49
PKN,

Thanks for pointing this out. It's rectified.
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Math Expert
Joined: 02 Sep 2009
Posts: 52386
Re: If a=1+1/4+1/16+1/64 and b=1+(1/4)a  [#permalink]

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05 Aug 2018, 03:25
delta23 wrote:
If a=1+$$\frac{1}{4}$$+$$\frac{1}{16}$$+$$\frac{1}{64}$$and b= 1+$$\frac{1}{4}$$a, then what is the value of a-b?

A)$$\frac{1}{256}$$
B)$$\frac{1}{128}$$
C)$$0$$
D)$$\frac{-1}{128}$$
E)$$\frac{-1}{256}$$

Discussed here: https://gmatclub.com/forum/if-a-1-1-4-1 ... 72041.html
_________________
Re: If a=1+1/4+1/16+1/64 and b=1+(1/4)a &nbs [#permalink] 05 Aug 2018, 03:25
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