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Algebra Level 2

$\frac{79}{32} \quad, \quad\frac{59}{24} \quad, \quad \frac{99}{40} \quad, \quad \frac{149}{60}$

Which of these fractions is the smallest?

\frac{79}{32}

\frac{149}{60}

\frac{59}{24}

\frac{99}{40}

8 solutions

Brian Charlesworth
Nov 11, 2015

We can write the given options as follows:

(i) $\dfrac{149}{60} = 2 + \dfrac{29}{60} = 2 + \dfrac{1}{\dfrac{60}{29}} = 2 + \dfrac{1}{2 + \dfrac{2}{29}}$

(ii) $\dfrac{59}{24} = 2 + \dfrac{11}{24} = 2 + \dfrac{1}{\dfrac{24}{11}} = 2 + \dfrac{1}{2 + \dfrac{2}{11}}$

(iii) $\dfrac{99}{40} = 2 + \dfrac{19}{40} = 2 + \dfrac{1}{\dfrac{40}{19}} = 2 + \dfrac{1}{2 + \dfrac{2}{19}}$

(iv) $\dfrac{79}{32} = 2 + \dfrac{15}{32} = 2 + \dfrac{1}{\dfrac{32}{15}} = 2 + \dfrac{1}{2 + \dfrac{2}{15}}.$

As $2 + \dfrac{2}{11}$ is greater than each of $2 + \dfrac{2}{29}, 2 + \dfrac{2}{19}$ and $2 + \dfrac{2}{15}$ we can conclude that fraction (ii), namely $\boxed{\dfrac{59}{24}},$ is the smallest.

149/60 + 1/60 =5/2

99/40 + 1/40 =5/2

79/32 + 1/32 =5/2

59/24 + 1/24 = 5/2

And 1/24 > 1/32 > 1/40 > 1/60

Hence 59/24 < 79/32 < 99/40 < 149/60

Arjun SivaÞrasadam - 5 years, 7 months ago

This is most appropriate solution I too done this way

ankit raj - 5 years, 6 months ago

Nice solution! However, you might have noticed that when the fractions were written in the form $2+x,$ the $x$ were very close to $\frac{1}{2}.$ How could this idea be used to simplify the solution?

Eli Ross Staff - 5 years, 7 months ago

Oh, right. We could also have written the options as

(i) $\dfrac{149}{60} = 2 + \dfrac{29}{60} = 2 + \dfrac{1}{2} - \dfrac{1}{60}$

(ii) $\dfrac{59}{24} = 2 + \dfrac{11}{24} = 2 + \dfrac{1}{2} - \dfrac{1}{24}$

(iii) $\dfrac{99}{40} = 2 + \dfrac{19}{40} = 2 + \dfrac{1}{2} - \dfrac{1}{40}$

(iv) $\dfrac{79}{32} = 2 + \dfrac{15}{32} = 2 + \dfrac{1}{2} - \dfrac{1}{32}$

Then as $\dfrac{1}{24}$ is greater than each of $\dfrac{1}{60}, \dfrac{1}{40}$ and $\dfrac{1}{32}$ we can conclude that fraction (ii) is indeed the smallest. I guess I was getting too "cute" with my posted solution. :)

Brian Charlesworth - 5 years, 7 months ago

The rest of all fractions when substracted by 2, respectively: 15/32, 11/24, 29/60, and 19/40.

All of them are in format of (n-1)/2n.

The smaller n is, then the smaller (n-1)/2n as well. This is concluded by logic (example: 2/6 or 1/3, is smaller than 3/8. Compare them: 8/24 and 9/24).

Then just look for the smallest divisor among all fractions, and then we find 59/24 as the smallest fraction.

This was what I did to solve this problem, actually.

Henny Lim - 5 years, 7 months ago

That's a really cool solution! I don't think I would have ever thought about it via continuous fractions.

Nick Benallo - 5 years, 7 months ago

Thanks! It was a more involved method than was necessary to solve the problem but I thought that it looked kind of cool. :)

Brian Charlesworth - 5 years, 7 months ago

Xavier Mendoza
Nov 12, 2015

Since the point is to find the smallest, you could calculate the difference between numerator and denominator and find the closest one to 0, based on proximity to 1. The bigger the difference, the further will be from 1.

That's the simplest and Most efficient approach in this case.

Hero Miles - 5 years, 7 months ago

Randy Petryk
Nov 14, 2015

I really just did it all mentally. Solved each one to three decimal points, wasn't needed really. Too lazy to use another method.

Ayobamzz Ola
Nov 16, 2015

Since each division produces a result between 2 and 3, You can use the remainder of each division.
% is modulo.
149%60 = 29.
54%24 = 6.
99%40 = 19.
79%32 = 15.
Pick the smallest. Worked 👌

Nick Benallo
Nov 15, 2015

The Least Common Multiple of 24, 32, 40, and 60 is 480. (LCM(24, 32, 40, 60) = 480.)

So we multiply the numerator and the denominator of each fraction by the appropriate factor to get the common denominator of 480.

$\dfrac{79}{32} = \left(\dfrac{15}{15}\right) \cdot \dfrac{79}{32} = \dfrac{15 \cdot 79}{15 \cdot 32} = \dfrac{1185}{480}$ ,

$\dfrac{59}{24} = \left(\dfrac{20}{20}\right) \cdot \dfrac{59}{24} = \dfrac{20 \cdot 59}{20 \cdot 24} = \dfrac{1180}{480}$ ,

$\dfrac{99}{40} = \left(\dfrac{12}{12}\right) \cdot \dfrac{99}{40} = \dfrac{12 \cdot 99}{12 \cdot 40} = \dfrac{1188}{480}$ , and

$\dfrac{149}{60} = \left(\dfrac{8}{8}\right) \cdot \dfrac{149}{60} = \dfrac{8 \cdot 149}{8 \cdot 60} = \dfrac{1192}{480}$ .

By comparing the numerators of each fraction, {1185, 1180, 1188, 1192}, we see that 1180 is the smallest one.

Therefore, $\dfrac{1180}{480} = \boxed{\dfrac{59}{24} \text{is the smallest fraction}}$ .

Raymond Lei
Nov 15, 2015

They are all equal to 5/2-1/x, so the smallest one must be the with x equal to 24 since 1/24 is greater than all other 3 fractions.

Tasnim Ferdous
Nov 15, 2015

{59-(24 . 2)=11}<{79-(32 . 2)=15}< {99-(40 . 2)=19}< {149-(60 . 2)=29}

so, 59/24 < 79/32 < 99/40 < 149/60

here, (.) indicates the multiplication

Pamela Barnes
Nov 14, 2015

Look at the fraction more than 2 for each value.

19/40, 29/60, 15/32, and 11/24.

I reasoned that, since each was just less than half, which was the "most" less than half? 1/24 is a larger fraction than 1/40, 1/60, and 1/32, so subtracting 1/24 from one half gives a smaller value than the others.

To check myself before answering, I then thought visually and imagined the number line: which fraction gave the largest gap between itself and one half? It seemed obvious to me that 11/24 had the largest gap, making it the smallest value. (Twenty-fourths are bigger pieces than thirty-seconds, fortieths, and sixtieths.)

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