## Fraction comparison techniques for solving difficult fraction problems

*Fraction comparison is needed to solve difficult fraction problems.*

The fraction comparison techniques are explained by solving carefully selected fraction problems.

For fractions concepts, go through our article,

**Fractions and decimals, basic concepts part 1.**

### Fraction comparison techniques for solving difficult fraction problems of different types

First method is the conventional one—subtracting one from the other.

#### Method 1: Fraction comparison by subtracting one fraction from the other

**Example problem 1.**

Compare $\displaystyle\frac{6}{7}$ with $\displaystyle\frac{101}{120}$ to find which one is larger.

**Solution to example problem 1.**

Fact is, you don't know which one is the larger and which one smaller.

So take any of the two fractions and subtract it from the other.

Subtract the second frction from the first for example,

$\displaystyle\frac{6}{7}-\displaystyle\frac{101}{120}$

$=\displaystyle\frac{6\times{120}-7\times{101}}{7\times{120}}$

$=\displaystyle\frac{720-707}{840}$

$=\displaystyle\frac{13}{840}$.

Result is positive.

So, the fraction subtracted from must be the larger. Rule is simple,

In a fraction subtraction (just like in a subtraction of two integers) $a-b$, where $b$ is the fraction subtracted from fraction $a$,

$\qquad$ If $a-b \gt 0$, $a \gt b$, the fraction subtracted $b$ is smaller of the two.

$\qquad$ If $a-b \lt 0$, $a \lt b$, the fraction subtracted $b$ is larger of the two.

The rule is based on the most basic number system concepts.

And in our problem, as $\displaystyle\frac{6}{7}-\displaystyle\frac{101}{120}$ is positive,

$\displaystyle\frac{6}{7} \gt \displaystyle\frac{101}{120}$.

**Question is:** how quickly can we say which fraction is larger?

**This is the main problem with fraction comparison.** It is not as simple as comparing two integers.

So you have to go over the concept of fraction subtraction itself. How do you subtract one fraction from another?

By making the denominators equal, so that the numerators can be combined into a subtraction operation.

In this simple method of fraction comparison by subtraction, you must equalize the denominators to their LCM and suitably modify the numerators.

In our problem,

$\displaystyle\frac{6}{7}-\displaystyle\frac{101}{120}=\displaystyle\frac{6\times{120}-7\times{101}}{7\times{120}}$.

The denominators having no common factor LCM of the two is $7\times{120}$. We do not need to calculate it. We need the numerators only. Right?

The suitably modified numerators are, $6\times{120}=720$ and $7\times{101}=707$,

With equal denominators of LCM of the two, fraction with larger modified numerator, the first fraction, is the larger one.

Simple. Is it?

Yes, it is simple no doubt. But like all simple methods, it is not good enough for many of the fraction comparisons.

**Overheads are,**

Calculation time for finding the LCM.

Calculation time for finding the two modified numerators.

If there is no other way to determine which of the two fractions is larger, this is the basic method that has to be used.

Its obvious **disadvantage is its calculation load involved.**

In each of the methods that follow, you will find that the calculation overhead or amount of calculation is reduced. But it will always be true that,

Which method will compare two fraction quickest will depend on the fractions themselves.

If finding LCM of two denominators is easy and quick, go for the simplest subtraction method without any hesitation.

Let's see the other methods of fraction comparison.

#### Method 2: Fraction comparison by Numerator equalization

This is just the reverse of the subtraction method of equalizing two denominators. Let us see where and how to use the method by solving a problem

**Problem example 2.**

Find the larger of the two fractions, $\displaystyle\frac{12}{17}$ and $\displaystyle\frac{30}{37}$.

**Solution to problem example 2.**

The two denominators 17 and 37 are both prime numbers. So LCM of the two is their product. But more important is the fact that, to get the numerators you have to calculate,

$12\times{37}$ and $30\times{17}$.

Instead, identifying the LCM 60 of the two numerators, 12 and 30 takes a few seconds at most.

The denominators are modified same way as numerators modified in case of subtration.

The denominators in this case aregiven by,

$\displaystyle\frac{5\times{12}}{5\times{17}}$ and $\displaystyle\frac{2\times{30}}{2\times{37}}$,

Or, $\displaystyle\frac{60}{85}$ and $\displaystyle\frac{60}{74}$.

Rule for fraction comparison by numerator equalization is,

Same number divided by a smaller number will give larger result than if divided by a larger number.

With equal numerators, fraction with smaller denominator will be the larger one.

**Answer:** $\displaystyle\frac{30}{37}$ is the larger of the two fractions.

For this problem, numerator equalization takes lesser time than conventional subtraction method.

So this second method is the right method of fraction comparison for this problem.

#### Method 3: Fraction comparison by Equal denominator numerator difference pattern

Occasionally in a fraction comparison,

The differences between the denominator and numerator for both fractions are same.

This is a specific pattern to be identified and exploited.

If this equal difference pattern is satisfied,

The fraction with larger numerator will be the larger one.

Let us solve the third problem example to show how the method works.

**Problem example 3.**

Find the larger of the two fractions, $\displaystyle\frac{31}{37}$ and $\displaystyle\frac{17}{23}$.

For both the fractions, denominator numerator difference is 6.

The first fraction with larger numerator 31 is the larger one. Answer is nearly instantaneous with no calculation involved.

**Answer:** $\displaystyle\frac{31}{37}$ is the larger fraction.

Similarly,

$\displaystyle\frac{6}{7} \gt \displaystyle\frac{5}{6}$, or,

$\displaystyle\frac{143}{217} \gt \displaystyle\frac{131}{205}$.

To state this powerful pattern based fraction comparison method formally,

If difference between the denominator and the numerator of each fraction in a set of fractions is same, the fractions with largest numerator will be the largest one.

Let us see why this happens, or whether this is a correct method at all!

#### Mechanism of fraction comparison by equal denominator numerator difference

For comparison let us take two generalized fractions,

$ \displaystyle\frac{a}{b}$, and $\displaystyle\frac{c}{d}$,

where $b-a=d-c$ and $a \gt c$.

To prove,

$ \displaystyle\frac{a}{b} \gt \displaystyle\frac{c}{d}$.

**Proof of fraction comparison by equal denominator numerator method:**

Subtract the second fraction from the first with no bias,

$ \displaystyle\frac{a}{b} - \displaystyle\frac{c}{d}$.

Subtract 1 from the first fraction and to compensate add 1,

$\displaystyle\frac{a}{b}-\displaystyle\frac{c}{d}$

$=\left(\displaystyle\frac{a}{b}-1\right)+\left(1-\displaystyle\frac{c}{d}\right)$

$=\displaystyle\frac{d-c}{d}-\displaystyle\frac{b-a}{b}$

$=(b-a)\left[\displaystyle\frac{1}{d}-\displaystyle\frac{1}{b}\right]$, as $d-c=b-a$,

$=(b-a)\displaystyle\frac{b-d}{bd}$.

This technique of adding 1 and subtracting 1 eliminates the numerator of both the fractions from further consideration.

It happens because denominator numerator difference of the two fractions are equal.

Now, **only the sign of the expression, $(b-d)$ will determine** which fraction is the larger one.

Again taking up the equality of differences,

$b-a=d-c$,

Or, $b-d=a-c$,

Or, $b-d \gt 0$, as $a \gt c$.

So,

$\displaystyle\frac{a}{b} - \displaystyle\frac{c}{d} \gt 0$,

Or, $\displaystyle\frac{a}{b} \gt \displaystyle\frac{c}{d}$.

By the nature of the pattern this will always be true.

Let us take another example to highlight the power of this technique.

**Problem Example 4.**

Order the following three fractions in ascending order,

$\displaystyle\frac{5}{6}$, $\displaystyle\frac{6}{7}$, $\displaystyle\frac{3}{4}$.

**Solution example 4.**

This is an example of **comparing more than two fractions**.

Usually in such cases, we compare the fractions pair by pair. This is the natural way of **ordering or sorting a set of numbers that are **integers or fractions.

By this approach any suitable two fractions will be compared first and then the smaller one will compared with the third to identify the smallest one.

Finally, the larger fraction of the second comparison has to be compared with the larger of the first comparison to finalize the ordering.

**It would involve three comparisons.**

By this approach, we apply **numerator equalization** between $\displaystyle\frac{6}{7}$ and $\displaystyle\frac{3}{4}$ and decide quickly that,

$\displaystyle\frac{6}{7} \gt \displaystyle\frac{3}{4}$.

This step is quick.

Next we have to compare $\displaystyle\frac{5}{6}$ with $\displaystyle\frac{6}{7}$.

The second fraction comes out larger again, but at the last stage, $\displaystyle\frac{5}{6}$ needs to be compared with $\displaystyle\frac{3}{4}$.

It is always a tedious process for ordering a set of numbers by conventional basic pair comparison approach.

Instead, identify **key pattern,**

A

ll three fractionshave same denominator numerator difference 1.

The three fractions are related in a few seconds by **denominator numerator equal difference technique** *in the order of their numerators*—a simple task,

$\displaystyle\frac{3}{4} \lt \displaystyle\frac{5}{6} \lt \displaystyle\frac{6}{7}$.

We will present now a more general and so more powerful method which combines the concept of denominator numerator difference and numerator superiority.

#### Method 4. Fraction with lesser denominator numerator difference as well as larger numerator will be larger

To show how this method works let us compare the fractions,

$\displaystyle\frac{29}{41}$ and $\displaystyle\frac{31}{42}$.

Denominator numerator difference for the first is 12 with numerator 29, and

Denominator numerator difference for the second is 11, smaller than the first and with numerator 31, larger than the first.

So the second fraction $\displaystyle\frac{31}{42}$ will be the larger of the two.

Formally expressed, the rule is,

Between two fractions, if the fraction with smaller denominator numerator difference has its numerator as larger of the two, the fraction itself will be the larger.

If you think for a little while, you will find that the reason why this technique works is based on the knowledge we have gained till now.

**Try to understand the reason yourself.**

We will explain though why this technique works. If you are not interested, you may skip the following section.

#### Reason why the equal denominator numerator difference and larger numerator pattern works

The larger fraction has smaller denominator numerator difference. Increase its denominator keeping the numerator same to make its denominator numerator difference equal to the other fraction being compared. Now the new fraction will be larger than the first fraction with same denominator numerator difference but with larger numerator of the two (as we have kept the numerator unchanged). This is **application of method 3**.

Now compare the new fraction with its mother fraction from which it was created. Numerators of the two are equal, but denominator of the new one is larger (as we have increased it to make the difference equal with the first). By the **rule in numerator equalization method 2** then the mother fraction will be the larger than the new one, and consequently larger than the first one.

Seems complex?

To make it clear let's take this same example of comparing,

$\displaystyle\frac{29}{41}$ and $\displaystyle\frac{31}{42}$.

We will make the denominator of the second to 43 keeping its numerator 31 unchanged to get the new fraction as,

$\displaystyle\frac{31}{43}$.

The denominator numerator difference in this case is 12, same as the first, but with numerator 31 larger. So by the rule in method 3,

$\displaystyle\frac{31}{43} \gt \displaystyle\frac{29}{41}$.

Now we compare,

$\displaystyle\frac{31}{43}$ and $\displaystyle\frac{31}{42}$.

By equal numerator lower denominator larger **rule in method 2**,

$\displaystyle\frac{31}{42} \gt \frac{31}{43} \gt \frac{29}{41}$.

The explanation might seem to be long, but inherently the concept is simple—this method works on **method 3 combined with method 2,** and so **is a hybrid method.**

Next, we will present the most general and so most powerful technique to compare two fractions. This technique is based on the percentage denominator numerator difference concept on which every other fraction comparison is based. In the previous techniques, because of occurrence of special patterns, we could directly work with the patterns instead of comparing by the percentage denominator numerator difference.

#### Method 5. Fraction comparison by percentage denominator numerator difference or numerator closeness measure

To showcase the method, let us compare the two fractions,

$\displaystyle\frac{71}{93}$ and $\displaystyle\frac{5}{7}$.

None of the earlier quick comparison techniques being applicable in this case, we resort to the **most fundamental method of comparing fractions by measuring how close the numerator is to its denominator for each fraction**. We call this measure as **closeness or separation of the numerator to its denominator.**

The **numerator closeness or separation concept** gives us the **rule for determining which fraction is larger**,

The

fraction for which numerator is closer to its denominator, is the larger one,because closer the numerator is to its denominator, closer the fraction is to its maximum value 1, and hence is larger.

More concept details follow which you may skip if not interested.

#### Further on numerator closeness or separation concept

To get the **numerator closeness or separation measure** we just subtract a fraction from 1 as below for a general fraction, $\displaystyle\frac{\text{Numerator}}{\text{Denominator}}$,

$1-\displaystyle\frac{\text{Numerator}}{\text{Denominator}}$

$=\displaystyle\frac{\text{Denominator}-\text{Numerator}}{\text{Denominator}}$.

This new fraction actually is the measure of the **closeness or separation of the fraction from its maximum value of 1** (it is obvious: we just subtracted the fraction from 1). **Lesser this separation is, more is the closeness and larger the fraction will be**.

*We express this fraction separation from 1 in a different way as, the separation of numerator from denominator as a portion of denominator.*

The numerator of this new fraction is the *actual separation (or difference) of numerator and its denominator*, and dividing this actual separation by the denominator gives us **numerator closeness measure** as a **portion of the denominator**. Converting this fraction of numerator closeness measure to its equivalent percentage (by multiplying it by 100) gives us the *percentage denominator numerator difference*.

**Note:** In actual use we do not calculate the percentage, but quickly determine (in most cases mentally) which numerator closeness measure is smaller. This gives us the larger fraction.

**Another note:** We have used just the actual separation or denominator numerator difference in our earlier techniques, but here we are using the most basic concept of numerator closeness measure that is independent of any inherent pattern. **We can apply this concept for any fraction comparison.**

Let us take an example of comparing,

$\displaystyle\frac{4}{5}$ and $\displaystyle\frac{3}{5}$.

For $\displaystyle\frac{3}{5}$, we get the **numerator closeness measure** and then percentage denominator numerator difference by subtracting the fraction from 1 and converting it to a percentage as below,

$1-\displaystyle\frac{3}{5}=\displaystyle\frac{2}{5}=0.4=40$%.

Similarly for the fraction, $\displaystyle\frac{4}{5}$, we get the percentage denominator numerator difference as,

$1-\displaystyle\frac{4}{5}=\displaystyle\frac{1}{5}=20$%.

The fraction $\displaystyle\frac{4}{5}$ with percentage difference of 20% is **closer to 1**, than the fraction $\displaystyle\frac{3}{5}$ with percentage difference of 40% and **so is larger**. The larger percentage difference for the second fraction shows that its numerator is separated from its denominator to a larger extent and its value becomes that much smaller (smaller numerator means smaller fraction).

But how to evaluate the numerator closeness or separation measure quickly and determine which fraction is larger? Let us explain how.

#### How to evaluate numerator closeness measure quickly and compare for two fractions

**First step** is to *subtract each of the two fractions from 1 getting two new fractions*.

Whichever of these two is smaller, the corresponding original fraction will be the larger, as each of these two **new fractions represents a measure of how far the original fraction is from its maximum value 1**. It makes sense that,

Smaller is the separation of a fraction from maximum value 1, larger the fraction will be.

So our new task is to **compare the two new fractions** created by subtracting each of the original fractions from 1 as below,

$1-\displaystyle\frac{71}{93}=\displaystyle\frac{22}{93}$, and

$1-\displaystyle\frac{5}{7}=\displaystyle\frac{2}{7}$.

We need to determine which of these two fractions is smaller.

It is easy in this case—just divide 93 by 22 to get denominator of the first as 4.2 (with changed numerator 1) and for the second as 3.5 (for a fraction, dividing the larger value of denominator by the smaller value of numerator is easier, and then just compare the evaluated denominators),

$\displaystyle\frac{22}{93}=\frac{1}{\displaystyle\frac{93}{22}}=\frac{1}{4.2}$, and

$\displaystyle\frac{2}{7}=\frac{1}{\displaystyle\frac{7}{2}}=\frac{1}{3.5}$.

Larger denominator makes smaller fraction. So numerator closeness or separation measures are related as,

$\displaystyle\frac{22}{93} \lt \displaystyle\frac{2}{7}$.

And so the corresponding original fractions are related in the reverse manner as,

$\displaystyle\frac{71}{93} \gt \displaystyle\frac{5}{7}$.

We repeat,

Smaller numerator separation belongs to larger fractionbecause, effectively we havemeasured the separation of a fraction from 1, its maximum possible value.

Naturally it follows,

The fraction with smaller percentage denominator numerator difference or numerator separation will be nearer to 1, and so will be the larger one.

This is the **base concept** on which this and other methods rest.

All these methods other than the first involves small amount of numerical calculation. **Our recommendation** is, *if you are not certain of applying a quick method accurately, go ahead and carry out subtraction.*

We will end the session with another approach to fraction comparison taking the last example for highlighting the technique.

#### Method 6. Hybrid technique of fraction comparison, transforming a numerator or denominator to very near to its corresponding element

Compare,

$\displaystyle\frac{71}{93}$ and $\displaystyle\frac{5}{7}$.

We **identify the pattern** that multiplying 5 by 14 **nearly equals** 71. So we decide to compare, $\displaystyle\frac{71}{93}$ and transformed second fraction, $\displaystyle\frac{70}{98}$ using 14 as multiplying factor to both numerator and denominator of the second fraction.

Without any calculation anyone can see that 71 is much closer to 70 than 98 is to 93. Here the underlying mechanism of percentage denominator numerator difference is active at the background.

We can conclude with certainty,

$\displaystyle\frac{71}{93} \gt \displaystyle\frac{5}{7}$.

To be sure we carry out subtraction for you,

$\displaystyle\frac{71}{93}-\displaystyle\frac{5}{7}$

$=\displaystyle\frac{7\times{71}-93\times{5}}{7\times{93}}$

$=\displaystyle\frac{497-465}{7\times{93}}$

—a positive result, as concluded earlier.

Let us do a few exercise problems.

### Exercise problems

#### Exercise problem 1.

Arrange the following three fractions in ascending sequence,

$\displaystyle\frac{29}{45}$, $\displaystyle\frac{67}{88}$, $\displaystyle\frac{59}{80}$.

#### Exercise problem 2.

Arrange the following in descending sequence,

$\displaystyle\frac{5}{6}$, $\displaystyle\frac{29}{36}$, $\displaystyle\frac{35}{43}$.

#### Exercise problem 3.

Arrange the following in ascending order,

$\displaystyle\frac{29}{47}$, $\displaystyle\frac{23}{41}$, $\displaystyle\frac{5}{8}$.

### Exercise problem solutions

#### Solution exercise problem 1.

We need to arrange the following in ascending sequence with smallest first and largest last,

$\displaystyle\frac{29}{45}$, $\displaystyle\frac{67}{88}$, $\displaystyle\frac{59}{80}$.

There is no easy way. So difference analysis begins. We detect equal denominator numerator difference between 2nd and the third, and so make the **first conclusion:** with larger numerator and equal denominator numerator difference of 12, second fraction is larger than the third,

$\displaystyle\frac{59}{80} \lt \displaystyle\frac{67}{88}$.

Next, doubling 29 we get 58 very near to the third numerator. So we decide to compare first with third after doubling the numerator and denominator of the first.

Comparing $\displaystyle\frac{58}{90}$ with $\displaystyle\frac{59}{80}$, we observe the numerators to be very close, but the first denominator is much larger than the second and so we make the second conclusion easily,

$\displaystyle\frac{29}{45} \lt \displaystyle\frac{59}{80}$.

That settles the **final conclusion of the ascending sequence**,

$\displaystyle\frac{29}{45} \lt \displaystyle\frac{59}{80} \lt \displaystyle\frac{67}{88}$.

No tortuous calculations, just observing patterns, creating patterns by simple methods and using the rules we learned.

#### Solution to exercise problem 2.

We need to arrange the following three fractions in descending sequence,

$\displaystyle\frac{5}{6}$, $\displaystyle\frac{29}{36}$, $\displaystyle\frac{35}{43}$.

Numerator 5 is a factor of third numerator 35, and so after equalization of numerators of first and the third, the denominator 42 of the first is smaller, and so the first fraction is larger than the third,

$\displaystyle\frac{5}{6} \gt \displaystyle\frac{35}{43}$.

Again we observe that the denominator 6 of the first is a factor of the denominator 36 of the second. So after equalizing the denominators, the transformed numerator 30 of the first is larger than numerator 29 of the second. So first fraction is also greater than the second fraction,

$\displaystyle\frac{5}{6} \gt \displaystyle\frac{29}{36}$.

We have to compare the second with the third now.

Analyzing the differences of the two we decide to go forward with percentage difference technique. Accordingly, **subtracting the two fractions from 1** we get,

$\displaystyle\frac{7}{36}$, and $\displaystyle\frac{8}{43}$,

Or, $\displaystyle\frac{1}{\displaystyle\frac{36}{7}}$, and $\displaystyle\frac{1}{\displaystyle\frac{43}{8}}$.

Denominators of the two,

$5.1 \lt 5.3$.

So, $\displaystyle\frac{7}{36} \gt \displaystyle\frac{8}{43}$.

For comparison of the original fractions then, the result will be reverse, that is,

$\displaystyle\frac{29}{36} \lt \displaystyle\frac{35}{43}$.

Finally then,

$\displaystyle\frac{5}{6} \gt \displaystyle\frac{35}{43} \gt \displaystyle\frac{29}{36}$.

This was a bit more involved than the first problem, but then it is like unraveling a small mystery.

#### Solution to exercise problem 3.

We need to arrange the following in ascending order,

$\displaystyle\frac{29}{47}$, $\displaystyle\frac{23}{41}$, $\displaystyle\frac{5}{8}$.

The first two have equal differences, and so we can conclude immediately,

$\displaystyle\frac{23}{41} \lt \displaystyle\frac{29}{47}$.

Next we **create the pattern of equal differences** by multiplying the third fraction by factor 6 and get by equal difference rule,

$\displaystyle\frac{29}{47} \lt \displaystyle\frac{30}{48}$.

**Finally** thus,

$\displaystyle\frac{23}{41} \lt \displaystyle\frac{29}{47} \lt \displaystyle\frac{5}{8}$.

### End note

Fraction comparison is a tricky process. It involves **identification of patterns in the fractions, occasional creation of patterns and applying the most suitable quick method**. If not, go for subtraction that would take more time but will be accurate.

Of course, in a competitive test if you can apply a quick method and save valuable seconds that might mean the difference between success and failure. This is the reason why in all our solutions we try to search for the quickest path to the solution.

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