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# Pls help me to solve this problem.(3)

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Manager
Status: Single
Joined: 05 Jun 2011
Posts: 125
Location: Shanghai China
Followers: 2

Kudos [?]: 25 [0], given: 0

Pls help me to solve this problem.(3) [#permalink]

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31 Jul 2011, 03:35
00:00

Difficulty:

(N/A)

Question Stats:

80% (01:29) correct 20% (00:00) wrong based on 5 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?
[Reveal] Spoiler: OA
Senior Manager
Joined: 03 Mar 2010
Posts: 440
Schools: Simon '16 (M)
Followers: 5

Kudos [?]: 220 [0], given: 22

Re: Pls help me to solve this problem.(3) [#permalink]

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31 Jul 2011, 05:39
tracyyahoo wrote:
is there any simple way to solve this kind of problem?

Yes.

$$1/2^1^0+1/2^1^1+1/2^1^2+1/2^1^2$$
=$$1/2^1^0 + 1/2^1^1 + 2/2^1^2$$
= $$1/2^1^0 + 1/2^1^1 + 1/2^1^1$$
= $$1/2^1^0 + 2/2^1^1$$
= $$1/2^1^0 + 1/2^1^0$$
= $$2/2^1^0$$
= $$1/2^9$$

OA C
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Manager
Status: Single
Joined: 05 Jun 2011
Posts: 125
Location: Shanghai China
Followers: 2

Kudos [?]: 25 [0], given: 0

Re: Pls help me to solve this problem.(3) [#permalink]

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31 Jul 2011, 18:25
Yeah, Thank you. It is a simple way.

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

Yes.

$$1/2^1^0+1/2^1^1+1/2^1^2+1/2^1^2$$
=$$1/2^1^0 + 1/2^1^1 + 2/2^1^2$$
= $$1/2^1^0 + 1/2^1^1 + 1/2^1^1$$
= $$1/2^1^0 + 2/2^1^1$$
= $$1/2^1^0 + 1/2^1^0$$
= $$2/2^1^0$$
= $$1/2^9$$

OA C
Manager
Joined: 31 May 2011
Posts: 88
Location: India
GMAT Date: 12-07-2011
GPA: 3.22
WE: Information Technology (Computer Software)
Followers: 1

Kudos [?]: 45 [0], given: 4

Re: Pls help me to solve this problem.(3) [#permalink]

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09 Aug 2011, 07:20
tracyyahoo wrote:
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?

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
Re: Pls help me to solve this problem.(3)   [#permalink] 09 Aug 2011, 07:20
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