Fractions

Lesson

If I were to ask you to compare two currencies, say $100$100 American Dollars and $100$100 English Pounds, we might convert each currency to a common currency, say Indian Rupees first.

When it comes to money, we cannot directly compare different currencies.

We might find that $100$100 American Dollars is equivalent to $6818$6818 Indian Rupees, and $100$100 English Pounds is $8422$8422 Indian Rupees. Immediately we realise that $100$100 English pounds is a lot more than $100$100 American dollars.

In the same way, we cannot directly compare fractions with different denominators.

We have to change each fraction to equivalent fractions that have the same denominator.

For example think about the three fractions $\frac{1}{2}$12, $\frac{1}{3}$13 and $\frac{1}{6}$16. They are different fractions just like one dollar and one pound and one rupee are all different currencies. It makes no sense to compare them as they stand because they have different denominators.

But if we convert $\frac{1}{2}$12 into "sixths" by multiplying both the numerator and denominator by $3$3...so that $\frac{1}{2}=\frac{1\times3}{2\times3}=\frac{3}{6}$12=1×32×3=36...

And if we convert $\frac{1}{3}$13 into "sixths" by multiplying both the numerator and denominator by $2$2 so that ...$\frac{1}{3}=\frac{1\times2}{3\times2}=\frac{2}{6}$13=1×23×2=26...

Then we can easily see that $\frac{1}{2}$12 is greater than $\frac{1}{3}$13, because $\frac{3}{6}$36 is greater than $\frac{2}{6}$26.

The idea is to look for a common denominator for both $\frac{1}{2}$12 and $\frac{1}{3}$13 and the trick is to search for the lowest common multiple of the denominators $2$2 and $3$3.

- The multiples of $2$2 start as $2,4,6,8,10,12,14$2,4,6,8,10,12,14,...
- The multiples of $3$3 start as $3,6,9,12,15,18,21$3,6,9,12,15,18,21,...
- So the lowest common multiple is $6$6.

This means when comparing $\frac{1}{2}$12 and $\frac{1}{3}$13 we need to find the equivalent fractions that have a $6$6 in the denominator.

Suppose we ask which fraction is smaller?

$\frac{3}{5}$35 or $\frac{2}{3}$23

We think as follows:

- The multiples of $5$5 begin $5,10,15,20,25,30$5,10,15,20,25,30,...
- The multiples of $3$3 begin $3,6,9,12,15,18,21$3,6,9,12,15,18,21 ...
- Therefore $15$15 is the lowest common multiple we need.

In the fraction $\frac{3}{5}$35 we need to multiply the denominator $5$5 by $3$3 to get to $15$15, and this means that we need to multiply the numerator by $3$3 also to get $9$9. Therefore we replace $\frac{3}{5}$35 with the equivalent fraction $\frac{9}{15}$915.

In a similar way we replace the $\frac{2}{3}$23 with the equivalent fraction $\frac{10}{15}$1015 by multiplying numerator and denominator by $5$5.

Because $\frac{10}{15}$1015 is greater than $\frac{9}{15}$915, we can say that $\frac{2}{3}$23 is the larger fraction.

Mathematically we can write $\frac{2}{3}>\frac{3}{5}$23>35.

Note that this also means that $\frac{3}{5}$35 is closer to $0$0 than $\frac{2}{3}$23 on the number line.

With practise, comparing fractions becomes easy.

Suppose we compare the fractions $\frac{3}{8}$38 and $\frac{1}{4}$14.

The lowest common multiple of $8$8 and $4$4 is $8$8 .

This means that $\frac{1}{4}=\frac{2}{8}$14=28 and so $\frac{3}{8}$38 is larger.

A farmer fenced her estate into four different size pgroups of land. The first three plots represented $\frac{1}{2}$12, $\frac{3}{8}$38 and $\frac{1}{12}$112 of the entire estate. If the entire estate was $96$96 hectares, how many hectares were left for the fourth plot?

One way to answer this question is to find equivalent fractions for the first three plots.

- The multiples of $2$2 begin $2,4,6,8,10,12,14,16,18,20,24$2,4,6,8,10,12,14,16,18,20,24,...
- The multiples of $8$8 begin $8,16,24$8,16,24,...
- The multiples of $12$12 begin $12,24$12,24,...

This means that $\frac{1}{2}=\frac{12}{24}$12=1224, $\frac{3}{8}=\frac{9}{24}$38=924 and $\frac{1}{12}=\frac{2}{24}$112=224.

How many "twenty-fourths" does this make? It makes $12+9+2=23$12+9+2=23 "twenty-fourths" and so only $\frac{1}{24}$124 of the $96$96 hectares is left. So the fourth plot occupies $\frac{96}{24}=4$9624=4 hectares.

Two boys were asked to measure a broom handle. Peter measures it as $2\frac{7}{12}$2712 metres and Sam measure it as $2\frac{3}{5}$235 metres. Who has the shortest measurement?

We could convert $2\frac{3}{5}$235 to the improper fraction $\frac{13}{5}$135 and similarly convert $2\frac{7}{12}$2712 to $\frac{31}{12}$3112.

Next we need find the lowest common multiple of $12$12 and $5$5 by listing the multiples of each. You might be able to do this in your head. Multiples of $5$5 end in either $0$0 or $5$5, Multiples of $12$12 never end in $5$5. A little bit of thought reveals that $5\times12=60$5×12=60 must be the lowest common multiple.

So $\frac{13}{5}=\frac{13\times12}{5\times12}=\frac{156}{60}$135=13×125×12=15660 and $\frac{31}{12}=\frac{31\times5}{12\times5}=\frac{155}{60}$3112=31×512×5=15560.

How very close these two fractions are! Why there's only a difference of $\frac{1}{60}$160 !

Select the correct number statement.

$\frac{5}{9}<\frac{4}{9}$59<49

A$\frac{5}{9}=\frac{4}{9}$59=49

B$\frac{5}{9}>\frac{4}{9}$59>49

C$\frac{5}{9}<\frac{4}{9}$59<49

A$\frac{5}{9}=\frac{4}{9}$59=49

B$\frac{5}{9}>\frac{4}{9}$59>49

C

Consider the following fractions: $\frac{7}{8}$78, $\frac{1}{9}$19, $\frac{3}{7}$37.

Which of these fractions is closest to $0$0?

$\frac{3}{7}$37

A$\frac{7}{8}$78

B$\frac{1}{9}$19

C$\frac{3}{7}$37

A$\frac{7}{8}$78

B$\frac{1}{9}$19

CWhich of these fractions is closest to $1$1?

$\frac{7}{8}$78

A$\frac{3}{7}$37

B$\frac{1}{9}$19

C$\frac{7}{8}$78

A$\frac{3}{7}$37

B$\frac{1}{9}$19

C

We want to arrange the following in ascending order:

$2\frac{1}{7}$217, $\frac{33}{14}$3314, $\frac{16}{7}$167

First, rewrite all $3$3 fractions as improper fractions with the common denominator of $14$14.

$2\frac{1}{7}=\frac{\editable{}}{14}$217=14

$\frac{33}{14}=\frac{\editable{}}{14}$3314=14

$\frac{16}{7}=\frac{\editable{}}{14}$167=14

Therefore which of the following lists the fractions in ascending order:

$\frac{16}{7}$167, $2\frac{1}{7}$217, $\frac{33}{14}$3314

A$\frac{33}{14}$3314, $\frac{16}{7}$167, $2\frac{1}{7}$217

B$2\frac{1}{7}$217, $\frac{16}{7}$167, $\frac{33}{14}$3314

C$2\frac{1}{7}$217, $\frac{33}{14}$3314, $\frac{16}{7}$167

D$\frac{16}{7}$167, $2\frac{1}{7}$217, $\frac{33}{14}$3314

A$\frac{33}{14}$3314, $\frac{16}{7}$167, $2\frac{1}{7}$217

B$2\frac{1}{7}$217, $\frac{16}{7}$167, $\frac{33}{14}$3314

C$2\frac{1}{7}$217, $\frac{33}{14}$3314, $\frac{16}{7}$167

D