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Bunuel niks18 amanvermagmat gmatbusters chetan2u @veritasprepakarishma


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If the numbers 19/36, 5/11, 12/25, 6/11, and 8/18 were arranged from least to greatest, which number would be in the middle?

19/36
12/25
6/11
5/11
8/18

Do we not make denominators same for comparing numerators in such a ratio problem?
What is the most efficient way to solve this?
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adkikani
Bunuel niks18 amanvermagmat gmatbusters chetan2u @veritasprepakarishma


Quote:
If the numbers 19/36, 5/11, 12/25, 6/11, and 8/18 were arranged from least to greatest, which number would be in the middle?

19/36
12/25
6/11
5/11
8/18

Do we not make denominators same for comparing numerators in such a ratio problem?
What is the most efficient way to solve this?

Hi, each question would require a approach suiting it....

Here you can easily see 2 of them, 19/36 and 6/11, to be >1/2 and other three are less than 1/2..
So our answer is greatest of remaining three..
Some similarity in numerator of 12/25 and 8/18..
12=8*3/2..
So 8/18=(8*3/2)/(18*3/2)=12/27<12/25

Finally check between 5/11 and 12/25..
Here you can cross Multiply..
5*25=125 and 11*12=132..
132 is greater, so 12/25 is greater and the middle value
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By scanning through the fractions, one can tell that two of them, \(\frac{19}{36}\) and \(\frac{6}{11}\), are slightly greater than \(\frac{1}{2}\), so we are left with \(\frac{5}{11}\), \(\frac{12}{25}\), and \(\frac{8}{18}\) (same as \(\frac{4}{9}\))
We can quickly manipulate the three fractions to have the same numerator as follows:
\(\frac{5}{11}\) x \(\frac{12}{12}\) = \(\frac{60}{132}\)
\(\frac{12}{25}\) x \(\frac{5}{5}\) = \(\frac{60}{125}\)
\(\frac{4}{9}\) x \(\frac{15}{15}\) = \(\frac{60}{135}\)

Therefore, \(\frac{4}{9}\) < \(\frac{5}{11}\) < \(\frac{12}{25}\) < ... < ... ====> Answer is \(\frac{12}{25}\) (B)
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Asked: If the numbers 19/36, 5/11, 12/25, 6/11, and 8/18 were arranged from least to greatest, which number would be in the middle?

5/11 = 0.45555
19/36 = 0.503
12/25 = 0.48
6/11 = 0.54444
8/18 = 4/9 = .4444

8/18 < 5/11 < 12/25 < 19/36 < 6/11

IMO B
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I finally have a good solution that wasn't mentioned before so:

Greater than 1/2: 6/11, 19/36
Less than 1/2: 5/11, 12/25, 8/18

The answer is one of the values less than 1/2.

- 5/11 is 0.5/11 away from 1/2
- 12/25 is 0.5/25 away from 1/2
- 8/18 or 4/9 is 0.5/9 away from 1/2

12/25 is the least distance away from 1/2, so it's the middle number.
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multiply the numerators by 2
38/36; 10/11; 24/25; 12/11; 16/18

Now group those fractions that are less than 1 and greater than 1

those < 1: 10/11; 24/25; 16/18
those > 1: 38/36; 12/11

now arrange them in order using the difference between the numerator and denominator

when the differences for two fractions are same, the fraction with the highest numerator is larger in value than the fraction with the lowest numerator

with that in mind, lets re-arrange only those < 1, coz those >1 will never be the middle one in the arrangement

those < 1: 16/18 < 10/11 < 24/25

Final ascending order: 16/18 < 10/11 < 24/25 < [38/36; 12/11]

Answer: 12/25
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Deconstructing the Question

We must arrange the fractions \(\frac{19}{36}\), \(\frac{5}{11}\), \(\frac{12}{25}\), \(\frac{6}{11}\), and \(\frac{8}{18}\) from least to greatest and find the middle value.

Since there are \(5\) numbers, the middle number will be the third after sorting.

A fast GMAT approach is to approximate the fractions as decimals.

Step-by-step

First simplify \(\frac{8}{18}\)

\(\frac{8}{18} = \frac{4}{9}\)

\(\frac{4}{9} \approx 0.444\)

Next approximate \(\frac{5}{11}\)

\(\frac{5}{11} \approx 0.455\)

Convert \(\frac{12}{25}\)

\(\frac{12}{25} = 0.48\)

Estimate \(\frac{19}{36}\)

\(\frac{18}{36} = 0.5\)

So

\(\frac{19}{36} \approx 0.528\)

Now estimate \(\frac{6}{11}\)

\(\frac{6}{11} \approx 0.545\)

Order from least to greatest

\(\frac{8}{18} \approx 0.444\)

\(\frac{5}{11} \approx 0.455\)

\(\frac{12}{25} = 0.48\)

\(\frac{19}{36} \approx 0.528\)

\(\frac{6}{11} \approx 0.545\)

The third number is \(\frac{12}{25}\)

Answer B: 12/25
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