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# Rank those three in order from smallest to biggest.

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Magoosh GMAT Instructor
Joined: 28 Dec 2011
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Kudos [?]: 4924 [1] , given: 54

Rank those three in order from smallest to biggest. [#permalink]

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05 Dec 2012, 12:32
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Difficulty:

55% (hard)

Question Stats:

57% (02:21) correct 43% (01:21) wrong based on 147 sessions

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Remember, like all GMAT Quantitative problems, this is a no calculator problem.

I. $$\frac{47}{150}$$

II. $$\frac{111}{330}$$

III. $$\frac{299}{900}$$

Rank those three in order from smallest to biggest.
(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, I, II

For a full discussion of this question and the topic of number sense, see this blog:
http://magoosh.com/gmat/2012/number-sense-for-the-gmat/
Experts, feel free to share you thoughts on what number sense is and how to develop it.
[Reveal] Spoiler: OA

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Mike McGarry
Magoosh Test Prep

Last edited by mikemcgarry on 05 Dec 2012, 12:52, edited 1 time in total.
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Re: number sense problem, estimation [#permalink]

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05 Dec 2012, 12:44
all fractions are close to 1/3
So if we take each
$$I. 47/150<50/150=1/3$$
$$II. 111/330>110/330=1/3$$
$$III. 299/900<300/900=1/3$$
Now between II and III we can easily figure out that 299 is closer to 300 than $$47*3$$. So III>I
So the correct order from smallest to largest is : $$I>III>II$$
So shouldn't it be B?
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Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 3305
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Kudos [?]: 4924 [1] , given: 54

Re: number sense problem, estimation [#permalink]

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05 Dec 2012, 12:53
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souvik101990 wrote:
So the correct order from smallest to largest is : $$I>III>II$$
So shouldn't it be B?

Yes, you are 100% correct. I changed this in the problem above. I said "smallest to largest", but then when I did it in my head, I put them in order from largest to smallest! Sorry about any confusion.
Mike
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Mike McGarry
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Re: Rank those three in order from smallest to biggest. [#permalink]

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19 Dec 2012, 05:28
I. 47/150 is almost 50/150 = 1/3. It is (50-47)/150 = 3/150 = 1/50 away from 1/3
II. 111/330 is beyond 110/330 = 1/3
III. 299/900 is almost 300/900 = 1/3. 299/900 is (300-299)/900 = 1/900 away from 1/3

I is the farther to the left of 1/3 than III while II is the largest.

Rank: I,III,II
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Re: Rank those three in order from smallest to biggest. [#permalink]

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19 Dec 2012, 06:54
Try and get all the denominators the same, i.e. in terms of 900:

I: 47x6/150x6 = 282/900

II: 111x3/330x3 =:(approx) 333/990

III: 299/900

From the above, we know that I is the smallest (as it has the smallest numerator) and hence comes first, eliminate options C,D and E. Only left with A & B.

299/900 will definitely be greater than 333/990 as if we find the accurate multiple by which to get 900 in II, it will give us a numerator being less than 299. Hence III<I

Final order: I,III,II

Option B
Re: Rank those three in order from smallest to biggest.   [#permalink] 19 Dec 2012, 06:54
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