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tracyyahoo
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Pls help me to solve this problem.(3) 31 Jul 2011, 03:35
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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Question Stats:

100% (01:57) correct 0% (00:00) wrong based on 1 sessions

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1/2^10+1/2^11+1/2^12+1/2^12 =


a)1/2^7
b)1/2^8
c)1/2^9
d)1/2^13
e)1/2^45

is there any simple way to solve this kind of problem?



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C
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jamifahad
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Pls help me to solve this problem.(3) 31 Jul 2011, 05:39
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tracyyahoo
is there any simple way to solve this kind of problem?
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Yes.

\(1/2^1^0+1/2^1^1+1/2^1^2+1/2^1^2\)
=\(1/2^1^0 + 1/2^1^1 + 2/2^1^2\)
= \(1/2^1^0 + 1/2^1^1 + 1/2^1^1\)
= \(1/2^1^0 + 2/2^1^1\)
= \(1/2^1^0 + 1/2^1^0\)
= \(2/2^1^0\)
= \(1/2^9\)

Mere addition.
OA C
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tracyyahoo
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Pls help me to solve this problem.(3) 31 Jul 2011, 18:25
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Yeah, Thank you. It is a simple way.


jamifahad
tracyyahoo
is there any simple way to solve this kind of problem?

Yes.

\(1/2^1^0+1/2^1^1+1/2^1^2+1/2^1^2\)
=\(1/2^1^0 + 1/2^1^1 + 2/2^1^2\)
= \(1/2^1^0 + 1/2^1^1 + 1/2^1^1\)
= \(1/2^1^0 + 2/2^1^1\)
= \(1/2^1^0 + 1/2^1^0\)
= \(2/2^1^0\)
= \(1/2^9\)

Mere addition.
OA C
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Sudhanshuacharya
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Pls help me to solve this problem.(3) 09 Aug 2011, 07:20
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tracyyahoo
1/2^10+1/2^11+1/2^12+1/2^12 =


a)1/2^7
b)1/2^8
c)1/2^9
d)1/2^13
e)1/2^45

is there any simple way to solve this kind of problem?
Show more

HI Tracee,
You can also solve the problem by taking factors like shown below:

1/2^10+1/2^11+1/2^12+1/2^12
= 1/2^12[2^2+2+1+1] (taking 1/2^12 common)
= 1/2^12(8)
=1/2^9

Hence C



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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