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

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

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New post Updated on: 05 Aug 2018, 00:47
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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 05 Aug 2018, 00:15.
Last edited by delta23 on 05 Aug 2018, 00:47, edited 1 time in total.
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Re: If a=1+1/4+1/16+1/64 and b=1+(1/4)a  [#permalink]

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New post 05 Aug 2018, 00:38
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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,
Please correct the question.
If a=1+\(\frac{1}{4}\)+\(\frac{1}{16}\)+\(\frac{1}{64}\)
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Re: If a=1+1/4+1/16+1/64 and b=1+(1/4)a  [#permalink]

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New post 05 Aug 2018, 00:49
PKN,

Thanks for pointing this out. It's rectified.
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Re: If a=1+1/4+1/16+1/64 and b=1+(1/4)a  [#permalink]

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New post 05 Aug 2018, 04:25
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
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Re: If a=1+1/4+1/16+1/64 and b=1+(1/4)a &nbs [#permalink] 05 Aug 2018, 04:25
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