## Solve difficult fraction problems quickly by 6 different fraction comparison techniques

Among all types of fraction problem solving questions, comparing fractions is generally most difficult. That's why you would find that *solving more difficult fraction problems would need fraction comparison.*

To make this awkward task of fraction comparison easier, *six comprehensive fraction comparison techniques are presented here. *

The working of the techniques is explained by solving carefully selected examples of difficult fraction problems that involves fraction comparison.

For basic but wide coverage of concepts on fractions, you may refer to our article,

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

### Fraction comparison techniques for solving difficult fraction problems

We will start with the conventional method to compare two fractions—subtracting one from the other.

#### Method 1: Subtraction to compare two fractions

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

Subtracting the second from the first *without giving any heed to which one is larger* we get,

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

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

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

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

Result is positive. So we conclude for sure that $\displaystyle\frac{6}{7}$ is larger. *If the result were negative*, the second fraction would have been the larger. Basic **concept used** is,

Among two fractions with equal denominators, the one with larger numerator will be the larger one.

**Note** that 840 is the LCM of the two denominators. Actually we made both denominators as 840 and suitably changed the numerators, so that the two could be joined together in the numerator for an operation like subtraction. This also is the method for subtraction.

A fallout of this approach is **denominator equalization technique** which is practically same as the subtraction.

Taking up the fraction pair to be compared as $\displaystyle\frac{5}{6}$ and $\displaystyle\frac{97}{120}$, we just multiply numerator 5 by 20, the multiple of 6 to make the denominators equal, and we get first numerator as 100 which is larger than 97. With both denominators 120, this identifies the first fraction $\displaystyle\frac{5}{6}$ as the larger.

If there is no other way to determine which of the two fractions is larger, this is the basic method that we would use. Its obvious **disadvantage is its calculation load involved.**

In each of the methods that follow, we will try to reduce amount of calculations needed to determine the larger or smaller fraction. But remember,

If the pattern of the two fractions supports ease of using the first method, it is generally the fastest way to the solution. The

key patternis, eitherone denominator is a factor of the other denominator, or evaluating their LCM is very easy to carry out.

#### Method 2: Numerator equalization method

In the opposite case, where the **pattern of the two fractions** is such that **one numerator is a factor of the other**, then without doing subtraction we can determine quickly which one is the larger and in this case the **numerator equalization method** will give you the solution quickest.

To show how the method is used, let's compare,

$\displaystyle\frac{12}{17}$ and $\displaystyle\frac{30}{37}$.

Though one numerator is not a factor of the other, we fall back on the LCM 60 of the two and get the two transformed fractions quickly as,

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

We can say now with confidence that the **second fraction with equal numerator and smaller denominator**, is the larger.

The **concept base** here is,

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

For this particular example, the second method will involve lesser amount of calculations and hence time, compared to the conventional method of subtraction.

#### Method 3: Equal denominator numerator difference pattern for fraction comparison

Occasionally, we find problems on fraction comparison, where, examining more closely we discover this pattern of **difference between denominator and numerator for two fractions to be same**. In these cases, we can say in a few seconds that the **fraction with larger numerator is the larger one**.

To show the method in action, let us compare the two fractions,

$\displaystyle\frac{31}{37}$ and $\displaystyle\frac{17}{23}$.

For both the fractions, denominator numerator difference is 6, and so, the first fraction with larger numerator 31 is the larger one. It takes a maximum of ten seconds to determine this.

In the same way,

$\displaystyle\frac{6}{7}$ is larger than $\displaystyle\frac{5}{6}$, or

$\displaystyle\frac{143}{217}$ is larger than $\displaystyle\frac{131}{205}$.

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 larger numerator will be the larger ones.

Because of the simplicity, in a set of say 6 fractions, if we observe that denominator numerator difference is same, we can order the fractions in ascending or descending sequence by the sequence of their numerators only.

Similarly, in a simpler case of comparing two fractions, *if the denominator numerator difference is same, the fraction with the larger numerator will be the larger one.*

For removing your doubts if any, we will explain why this happens. If you are not interested, you may skip this section.

#### Why the fraction comparison method based on equal difference of denominator numerator works

Let us take two generalized fractions,

$ \displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$ for comparison, where $b-a=d-c$ and $a \gt c$.

We will show how in this case, the first fraction will come out as the larger one as expected.

To determine the larger one with certainty we will add and subtract 1 to the subtraction relation between the two,

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

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

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

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

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

This technique of adding 1 and subtracting 1 effectively 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.

**Example 1.**

Order the following three fractions in ascending order,

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

**Solution example 1.**

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

Usually in such cases, we compare the fractions pair by pair. That is the natural way of **ordering or sorting a set of numbers that may be **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, we identify **key pattern** of **all three fractions** to have same denominator numerator difference 1, and order the three in a few seconds by **denominator numerator equal difference technique** *in the order of their numerators*—a simple task,

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

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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