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Bunuel
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filipTGIM
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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Randude
filipTGIM
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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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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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Is there any shortcut? I used the following technique but failed:

(a+1)/(b+1) > a/b
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KarishmaB
Is there any shortcut? I used the following technique but failed:

(a+1)/(b+1) > a/b

Exactly what Bunuel did above. With proper fractions, the first point of comparison is 'less than 1/2' or 'more than 1/2'
17/35 and 25/52 are less than 1/2.

9/17, 31/61 and 41/81 are slightly more than 1/2.

In fact, each numerator is 0.5 more.

\(\frac{9}{17} = \frac{8.5}{17} + \frac{0.5}{17}= \frac{1}{2} + \frac{0.5}{17}\)

\(\frac{31}{61} = \frac{30.5}{61} + \frac{0.5}{61} = \frac{1}{2} + \frac{0.5}{61}\)

\(\frac{41}{81} = \frac{40.5}{81} + \frac{0.5}{81}=\frac{1}{2} + \frac{0.5}{81}\)

We can now compare the additional amounts since each has the same numerator of 0.5. Since 9/17 has the smallest denominator, it must be the greatest.

Answer (A)

Here is a discussion on another such question: https://youtu.be/6AYRmxkAlrE
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can we solve it like this

9/17 = x-8/x = 1 - 8/x
17/35 = x-18/x = 1 - 18/x
31/61 = x-30/x = 1-30/x
25/52 = x-27/x = 1-27/x
41/81 = x-40/x = z-40/x

as we need the largest the number after "-" has to be smallest.
8 in numerator is smallest.

so A
Bunuel
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. \(\frac{41}{81}\)


The fractions \(\frac{17}{35}\) and \(\frac{25}{52}\) are both less than \(\frac{1}{2}\), so they can be eliminated.

Comparing the other fractions:

\(\frac{9}{17}\) can be broken down as \(\frac{8.5}{17} + \frac{0.5}{17}\), which equals \(\frac{1}{2} + \frac{0.5}{17}\).

\(\frac{31}{61}\) can be broken down as \(\frac{30.5}{61} + \frac{0.5}{61}\), or \(\frac{1}{2} + \frac{0.5}{61}\).

\(\frac{41}{81}\) can be broken down as \(\frac{40.5}{81} + \frac{0.5}{81}\), equating to \(\frac{1}{2} + \frac{0.5}{81}\).

Since \((\frac{1}{2} + \frac{0.5}{17}) > (\frac{1}{2} + \frac{0.5}{61}) > (\frac{1}{2} + \frac{0.5}{81})\), then option A has the largest value among the given fractions.


Answer: A
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Deconstructing the Question
We compare:
\(\frac{9}{17}, \frac{17}{35}, \frac{31}{61}, \frac{25}{52}, \frac{41}{81}\)

Key idea: compare each fraction to \(\frac12\).

Step-by-step

\(\frac{9}{17}\) → 9 is greater than half of 17 → greater than 1/2
\(\frac{17}{35}\) → less than 1/2
\(\frac{31}{61}\) → greater than 1/2
\(\frac{25}{52}\) → less than 1/2
\(\frac{41}{81}\) → greater than 1/2

Now compare the ones above 1/2:

\(\frac{9}{17}≈0.529\)
\(\frac{31}{61}≈0.508\)
\(\frac{41}{81}≈0.506\)

The largest is:

\(\frac{9}{17}\)

Answer: 9/17
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