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M21-01

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:09
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69% (00:57) correct 31% (00:46) wrong based on 149 sessions

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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. $$\frac{41}{81}$$

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16 Sep 2014, 01:09
1
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. $$\frac{9}{17} = \frac{1}{2} + \frac{0.5}{17} \gt \frac{31}{61} = \frac{1}{2} + \frac{0.5}{61} \gt \frac{41}{81} = \frac{1}{2} + \frac{0.5}{81}$$.

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03 Jan 2015, 08:58
1
1
Or you can just find a fraction that has the least difference between numerator and denominator.
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14 Jan 2015, 00:53
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. $$\frac{9}{17} = \frac{1}{2} + \frac{0.5}{17} \gt \frac{31}{61} = \frac{1}{2} + \frac{0.5}{61} \gt \frac{41}{81} = \frac{1}{2} + \frac{0.5}{81}$$.

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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Joined: 05 Dec 2013
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13 May 2015, 19:32
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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08 Jul 2015, 13:35
2
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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30 Jul 2017, 22:06
@Harley 1980
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Dr. Pratik

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05 Aug 2017, 07:51
3
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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05 Aug 2017, 08:11
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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03 Nov 2017, 06:44
+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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