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M21-01

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M21-01 [#permalink]

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Which of the following fractions is the largest?

A. \(\frac{9}{17}\)
B. \(\frac{17}{35}\)
C. \(\frac{31}{61}\)
D. \(\frac{25}{52}\)
E. \(\frac{41}{81}\)
[Reveal] Spoiler: OA

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Official Solution:

Which of the following fractions is the largest?

A. \(\frac{9}{17}\)
B. \(\frac{17}{35}\)
C. \(\frac{31}{61}\)
D. \(\frac{25}{52}\)
E. \(\frac{41}{81}\)

\(\frac{17}{35}\) and \(\frac{25}{52}\) are smaller than \(\frac{1}{2}\) and so are out. \(\frac{9}{17} = \frac{1}{2} + \frac{0.5}{17} \gt \frac{31}{61} = \frac{1}{2} + \frac{0.5}{61} \gt \frac{41}{81} = \frac{1}{2} + \frac{0.5}{81}\).

Answer: A
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Re: M21-01 [#permalink]

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Or you can just find a fraction that has the least difference between numerator and denominator.

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Re: M21-01 [#permalink]

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New post 13 Jan 2015, 23:53
Bunuel wrote:
Official Solution:

Which of the following fractions is the largest?

A. \(\frac{9}{17}\)
B. \(\frac{17}{35}\)
C. \(\frac{31}{61}\)
D. \(\frac{25}{52}\)
E. \(\frac{41}{81}\)

\(\frac{17}{35}\) and \(\frac{25}{52}\) are smaller than \(\frac{1}{2}\) and so are out. \(\frac{9}{17} = \frac{1}{2} + \frac{0.5}{17} \gt \frac{31}{61} = \frac{1}{2} + \frac{0.5}{61} \gt \frac{41}{81} = \frac{1}{2} + \frac{0.5}{81}\).

Answer: A


I am unable to understand the above approach. As in, i don't think i would be able to apply it while in exam mode.

Is there an alternative approach to the above problem?

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Re: M21-01 [#permalink]

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New post 13 May 2015, 18:32
filipTGIM wrote:
Or you can just find a fraction that has the least difference between numerator and denominator.


Can you please explain the concept behind this approach.

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M21-01 [#permalink]

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Randude wrote:
filipTGIM wrote:
Or you can just find a fraction that has the least difference between numerator and denominator.


Can you please explain the concept behind this approach.


Hello Randude

\(\frac{9}{17}\), \(\frac{31}{61}\), \(\frac{41}{81}\)

We can see common thing in these fractions. Their nominator mutipled on 2 equal to denominator - 1
9*2 = 18 = 17-1
31*2 = 62 = 62-1
41*2 = 82 = 82-1

so 9/17 --> 17/2 = 8.5 this number on 0.5 less than 9 and we can write this fraction as half of 17 + 0.5 divide by 17 --> 1/2 + 0.5/17
so 31/61 --> 61/2 = 30.5 this number on 0.5 less than 31 and we can write this fraction as half of 61 + 0.5 divide by 61 --> 1/2 + 0.5/61
so 41/81 --> 81/2 = 40.5 this number on 0.5 less than 41 and we can write this fraction as half of 81 + 0.5 divide by 81 --> 1/2 + 0.5/81
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Re: M21-01 [#permalink]

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New post 30 Jul 2017, 21:06
@Harley 1980
I did not follow your explanation, can you elaborate?
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Multiplying \(\frac{9}{17}\) with 2 gives\(\frac{18}{34}\)
which can be written as\(\frac{17}{34}\)+ \(\frac{1}{34}\)
that gives us\(\frac{1}{2}\) + \(\frac{0.5}{17}\)

same way it is for \(\frac{31}{61}\)--> \(\frac{62}{61*2}\)= \(\frac{61}{61*2} + \frac{1}{61*2}\) = \(\frac{1}{2} + \frac{0.5}{61}\)

Hope it is clear now :)

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Re: M21-01 [#permalink]

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New post 05 Aug 2017, 07:11
onamarif wrote:
Multiplying \(\frac{9}{17}\) with 2 gives\(\frac{18}{34}\)
which can be written as\(\frac{17}{34}\)+ \(\frac{1}{34}\)
that gives us\(\frac{1}{2}\) + \(\frac{0.5}{17}\)

same way it is for \(\frac{31}{61}\)--> \(\frac{62}{61*2}\)= \(\frac{61}{61*2} + \frac{1}{61*2}\) = \(\frac{1}{2} + \frac{0.5}{61}\)

Hope it is clear now :)


Now I get it.
Thanks onamarif
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Re: M21-01 [#permalink]

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New post 03 Nov 2017, 05:44
+1 for A. I did it using a more circuitous way instead. The denominators are prime with respect to one another. Use LCM concept and compare. Option A comes out to be the largest.
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Re: M21-01   [#permalink] 03 Nov 2017, 05:44
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