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# In the sequence above, each term after the first 1/2 of the preceding

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Manager
Joined: 09 Jun 2010
Posts: 55
In the sequence above, each term after the first 1/2 of the preceding  [#permalink]

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29 Jul 2010, 23:37
2
1
00:00

Difficulty:

15% (low)

Question Stats:

79% (01:12) correct 21% (01:43) wrong based on 119 sessions

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240,120,60,30

In the sequence above, each term after the first 1/2 of the preceding term. What is the least term of the sequence is greater than 1 ?

A. 32/15
B. 16/15
C. 15/8
D. 15/4
E. 15/2
Math Expert
Joined: 02 Sep 2009
Posts: 56307
Re: In the sequence above, each term after the first 1/2 of the preceding  [#permalink]

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29 Jul 2010, 23:57
xmagedo wrote:
240,120,60,30

In the sequnce above, each term after the first 1/2 of the preceding term. What is the least term of the sequnce is greater than 1 ?

a 32/15
b 16/15
c 15/8
d 15/4
e 15/2

pleas help with a formula

There is no need for formula:
240, 120, 60, 30, 30/2, 30/4, 30/8, 30/16, 30/32<1. So the leas term more than 1 is 30/16=15/8.

But still if you need: $$\frac{30}{2^n}>1$$, where n is integer --> $$2^n<30$$ --> $$n<5$$ --> $$n=4$$ --> $$\frac{30}{2^n}=\frac{30}{2^4}=\frac{15}{8}$$.

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Re: In the sequence above, each term after the first 1/2 of the preceding  [#permalink]

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29 Nov 2016, 09:19
xmagedo wrote:
240,120,60,30

In the sequence above, each term after the first 1/2 of the preceding term. What is the least term of the sequence is greater than 1 ?

A. 32/15
B. 16/15
C. 15/8
D. 15/4
E. 15/2

Greater than 1 = Numerator > Denominator

Now, go fearlessly- 30, 15 , 15/2 , 15/4 , 15/8, 15/16

Hence, the answer can be found in 10 seconds and it is (C)

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Re: In the sequence above, each term after the first 1/2 of the preceding  [#permalink]

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10 Apr 2018, 19:02
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In the sequence above, each term after the first 1/2 of the preceding   [#permalink] 10 Apr 2018, 19:02
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