Fractions

Fill in the blanks with fractions.

1. Three guavas together weigh 1 kg. If they are roughly of the same size, each guava will roughly weigh .......... kg.
2. A wholesale merchant packed 1 kg of rice in four packets of equal weight. The weight of each packet is .......... kg.
3. Four friends ordered 3 glasses of sugarcane juice and shared it equally among themselves. Each one drank .......... glass of sugarcane juice.
4. The big fish weighs $\dfrac{1}{2}$ kg. The small one weighs $\dfrac{1}{4}$ kg. Together they weigh .......... kg.

Answer

1. Three guavas together weigh 1 kg. If they are roughly of the same size, each guava will roughly weigh $\mathbf{\dfrac{1}{3}}$ kg.

Explanation —

Weight of 3 guavas together = 1 kg

The guavas are roughly of the same size, so 1 kg is shared equally into 3 parts.

$\phantom{=} \text{Weight of each guava} = \dfrac{1}{3} \text{ of 1 kg} \\[1em] = \dfrac{1}{3} \text{ kg}$

2. A wholesale merchant packed 1 kg of rice in four packets of equal weight. The weight of each packet is $\mathbf{\dfrac{1}{4}}$ kg.

Explanation —

Weight of rice packed = 1 kg

This 1 kg is packed into 4 packets of equal weight.

$\phantom{=} \text{Weight of each packet} = \dfrac{1}{4} \text{ of 1 kg} \\[1em] = \dfrac{1}{4} \text{ kg}$

3. Four friends ordered 3 glasses of sugarcane juice and shared it equally among themselves. Each one drank $\mathbf{\dfrac{3}{4}}$ glass of sugarcane juice.

Explanation —

Number of glasses of sugarcane juice = 3

Number of friends sharing them equally = 4

Here 3 glasses are shared equally among 4 friends. So each share is 3 times $\dfrac{1}{4}$ of a glass.

$\phantom{=} \text{Juice each friend drank} = \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} \quad \text{[3 times } \dfrac{1}{4} \text{ glass]} \\[1em] = \dfrac{3}{4} \text{ glass}$

4. The big fish weighs $\dfrac{1}{2}$ kg. The small one weighs $\dfrac{1}{4}$ kg. Together they weigh $\mathbf{\dfrac{3}{4}}$ kg.

Explanation —

Weight of the big fish = $\dfrac{1}{2}$ kg

Weight of the small fish = $\dfrac{1}{4}$ kg

Weight of both fish together = (Weight of big fish) + (Weight of small fish)

Substituting the above values, we get:

$= \left(\dfrac{1}{2} + \dfrac{1}{4}\right) \text{ kg} \\[1em] = \left(\dfrac{1 \times 2}{2 \times 2} + \dfrac{1}{4}\right) \text{ kg} \quad \\[1em] \text{[Making the fractional unit the same]} \\[1em] = \left(\dfrac{2}{4} + \dfrac{1}{4}\right) \text{ kg} \\[1em] = \dfrac{2 + 1}{4} \text{ kg} \\[1em] = \dfrac{3}{4} \text{ kg}$

Arrange these fraction words in order of size from the smallest to the biggest in the empty box below:
One and a half, three quarters, one and a quarter, half, quarter, two and a half.

Answer

First, let us write each fraction word as a fraction:

Since all the fractions are now written using the same fractional unit $\dfrac{1}{4}$, we can compare them by comparing the number of quarters:

$\dfrac{1}{4} \lt \dfrac{2}{4} \lt \dfrac{3}{4} \lt \dfrac{5}{4} \lt \dfrac{6}{4} \lt \dfrac{10}{4}$

So, the fraction words in order of size from the smallest to the biggest are:

Quarter, half, three quarters, one and a quarter, one and a half, two and a half.

∴ The required order is $\mathbf {\dfrac{1}{4} \lt \dfrac{1}{2} \lt \dfrac{3}{4} \lt 1\dfrac{1}{4} \lt 1\dfrac{1}{2} \lt 2\dfrac{1}{2}}$.

By dividing the whole chikki into 6 equal parts in different ways, we get $\dfrac{1}{6}$ chikki pieces of different shapes. Are they of the same size?

Answer

Yes, all the pieces are of the same size.

In both the pictures, the same whole chikki has been cut into 6 equal parts. So each part is 1 out of 6 equal parts, that is, $\dfrac{1}{6}$ of the whole chikki.

In the first picture the chikki is cut by straight cuts into 6 rectangular pieces, and in the second picture it is cut in a different way, so that one of the pieces is a triangle. The shapes of the pieces are different, but each piece is still one of 6 equal shares of the same chikki.

Each piece = $\dfrac{1}{6}$ of the whole chikki

Hence, the pieces have different shapes but they are all of the same size, each being $\dfrac{1}{6}$ of the whole chikki.

The figures below show different fractional units of a whole chikki. How much of a whole chikki is each piece?

Answer

The whole chikki is made up of small equal squares arranged in 4 rows and 6 columns.

Number of small squares in the whole chikki

= 4 × 6

= 24

So, each small square is $\dfrac{1}{24}$ of the whole chikki. Now we count how many such small squares make up each piece.

(a) The piece is a rectangle of 1 column and 2 rows, so it has 2 small squares.

$\phantom{=} \dfrac{2}{24} \\[1em] = \dfrac{1}{12} \quad \text{[Dividing both by 2]}$

∴ Piece (a) is $\mathbf {\dfrac{1}{12}}$ of the whole chikki.

(b) The piece is a triangle which is half of a rectangle of 3 columns and 4 rows.

Number of small squares = $\dfrac{1}{2} \times 3 \times 4$

= 1 × 3 × 2 $\quad$[Dividing 2 and 4 by 2]

= 6

Part of the whole = $\dfrac{6}{24} = \dfrac{1}{4}$

[Dividing both by 6]

∴ Piece (b) is $\mathbf {\dfrac{1}{4}}$ of the whole chikki.

(c) The piece is a triangle which is half of a rectangle of 3 columns and 2 rows.

Number of small squares = $\dfrac{1}{2} \times 3 \times 2$

= 1 × 3 × 1 $\quad$[Dividing 2 and 2 by 2]

= 3

Part of the whole = $\dfrac{3}{24} = \dfrac{1}{8}$

[Dividing both by 3]

∴ Piece (c) is $\mathbf {\dfrac{1}{8}}$ of the whole chikki.

(d) The piece is a rectangle of 1 column and 4 rows, so it has 4 small squares.

$\phantom{=} \dfrac{4}{24} = \dfrac{1}{6} \quad \text{[Dividing both by 4]}$

∴ Piece (d) is $\mathbf {\dfrac{1}{6}}$ of the whole chikki.

(e) The piece is an L-shape made of 3 small squares.

$\phantom{=} \dfrac{3}{24} = \dfrac{1}{8} \quad \text{[Dividing both by 3]}$

∴ Piece (e) is $\mathbf {\dfrac{1}{8}}$ of the whole chikki.

(f) The piece is a triangle which is half of a rectangle of 2 columns and 4 rows.

Number of small squares = $\dfrac{1}{2} \times 2 \times 4$

= 1 × 1 × 4 $\quad$ [Dividing 2 and 2 by 2]

= 4

Part of the whole = $\dfrac{4}{24} = \dfrac{1}{6} \quad$ [Dividing both by 4]

∴ Piece (f) is $\mathbf {\dfrac{1}{6}}$ of the whole chikki.

(g) The piece is one small square.

Part of the whole = $\dfrac{1}{24}$

∴ Piece (g) is $\mathbf {\dfrac{1}{24}}$ of the whole chikki.

(h) The piece is a triangle which is half of a rectangle of 1 column and 2 rows.

Number of small squares = $\dfrac{1}{2} \times 1 \times 2$

= 1 × 1 × 1 $\quad$ [Dividing 2 and 2 by 2]

= 1

Part of the whole = $\dfrac{1}{24}$

∴ Piece (h) is $\mathbf {\dfrac{1}{24}}$ of the whole chikki.

Continue this table of $\dfrac{1}{2}$ for 2 more steps.

Answer

The table in the chapter is built by collecting together the fractional unit $\dfrac{1}{2}$, and it ends at 5 times half. The next two steps are 6 times half and 7 times half.

The two new steps can be shown as:

$\dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2}$ = 6 times $\dfrac{1}{2}$

$\dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2}$ = 7 times $\dfrac{1}{2}$

Hence, the next two steps are 6 times half = 3 and 7 times half = $\mathbf {3\dfrac{1}{2}}$.

Can you create a similar table for $\dfrac{1}{4}$?

Answer

Yes, a similar table can be made for $\dfrac{1}{4}$ by collecting together the fractional unit $\dfrac{1}{4}$ (quarter rotis) one at a time.

Hence, a similar table can be made for $\dfrac{1}{4}$, and 4 times $\dfrac{1}{4}$ makes 1 whole.

Make $\dfrac{1}{3}$ using a paper strip. Can you use this to also make $\dfrac{1}{6}$?

Answer

Making $\dfrac{1}{3}$:

Take a paper strip and treat its length as 1 unit. Fold the strip into 3 equal parts and open it out. The two creases divide the strip into 3 equal parts, so the length of each part is $\dfrac{1}{3}$ unit.

Making $\dfrac{1}{6}$ from it:

Yes, the same strip can be used. Now fold the strip once more into two equal parts, that is, fold each $\dfrac{1}{3}$ part into half. Each $\dfrac{1}{3}$ part gets divided into 2 equal parts, so the strip now has 6 equal parts.

Number of equal parts = 3 × 2 = 6

So, length of each part = $\dfrac{1}{6}$ unit

Also, half of $\dfrac{1}{3} = \dfrac{1}{6}$

Hence, folding the strip into 3 equal parts gives $\mathbf {\dfrac{1}{3}}$, and folding it once more into half gives $\mathbf {\dfrac{1}{6}}$.

Draw a picture and write an addition statement as above to show:

(a) 5 times $\dfrac{1}{4}$ of a roti

(b) 9 times $\dfrac{1}{4}$ of a roti

Answer

(a) 5 times $\dfrac{1}{4}$ of a roti

Here 5 quarter pieces of roti are collected together. Four quarter pieces make 1 whole roti and one quarter piece is left over.

The addition statement is:

$\phantom{=} \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} \\[1em] = \dfrac{1 + 1 + 1 + 1 + 1}{4} \\[1em] = \dfrac{5}{4} \\[1em] = 1\dfrac{1}{4} \quad \text{[Converting improper fraction to mixed fraction]}$

∴ 5 times $\mathbf {\dfrac{1}{4}}$ of a roti = $\mathbf {\dfrac{5}{4} = 1\dfrac{1}{4}}$ rotis.

(b) 9 times $\dfrac{1}{4}$ of a roti

Here 9 quarter pieces of roti are collected together. Eight quarter pieces make 2 whole rotis and one quarter piece is left over.

The addition statement is:

$\phantom{=} \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} \\[1em] = \dfrac{1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1}{4} \\[1em] = \dfrac{9}{4} \\[1em] = 2\dfrac{1}{4} \quad \text{[Converting improper fraction to mixed fraction]}$

∴ 9 times $\mathbf {\dfrac{1}{4}}$ of a roti = $\mathbf {\dfrac{9}{4} = 2\dfrac{1}{4}}$ rotis.

Match each fractional unit with the correct picture:

Answer

To match a fractional unit with a picture, we count the number of equal parts into which each circle is divided. If a circle is divided into 8 equal parts, then each shaded part is $\dfrac{1}{8}$, and so on.

Counting the equal parts in the four pictures:

So, the matching is:

Here, the fractional unit is dividing a length of 1 unit into three equal parts. Write the fraction that gives the length of the blue line in the box or in your notebook.

Answer

Here, the length of 1 unit is divided into 3 equal parts.

So, the fractional unit is $\dfrac{1}{3}$.

From the figure, the blue line covers 2 such equal parts.

Length of the blue line = 2 times $\dfrac{1}{3}$

$= 2 \times \dfrac{1}{3} \\[1em] = \dfrac{2 \times 1}{3} \\[1em] = \dfrac{2}{3}$

∴ The length of the blue line = $\mathbf {\dfrac{2}{3}}$ unit.

Here, a unit is divided into 5 equal parts. Write the fraction that gives the length of the blue lines in the respective boxes or in your notebook.

Answer

Here, the length of 1 unit is divided into 5 equal parts.

So, the fractional unit is $\dfrac{1}{5}$.

From the figure, the shorter blue line covers 2 such equal parts.

Length of the shorter blue line = 2 times $\dfrac{1}{5}$

$= 2 \times \dfrac{1}{5} \\[1em] = \dfrac{2 \times 1}{5} \\[1em] = \dfrac{2}{5}$

The longer blue line covers 4 such equal parts.

Length of the longer blue line = 4 times $\dfrac{1}{5}$

$= 4 \times \dfrac{1}{5} \\[1em] = \dfrac{4 \times 1}{5} \\[1em] = \dfrac{4}{5}$

∴ The shorter blue line is $\mathbf {\dfrac{2}{5}}$ unit long and the longer blue line is $\mathbf {\dfrac{4}{5}}$ unit long.

Now, a unit is divided into 8 equal parts. Write the appropriate fractions in your notebook.

Answer

Here, the length of 1 unit is divided into 8 equal parts.

So, the fractional unit is $\dfrac{1}{8}$.

Each mark on the number line is obtained by collecting together the fractional unit $\dfrac{1}{8}$ again and again.

1 time $\dfrac{1}{8} = \dfrac{1}{8}$, 2 times $\dfrac{1}{8} = \dfrac{2}{8}$, 3 times $\dfrac{1}{8} = \dfrac{3}{8}$, and so on.

So, the fractions marked from 0 to 1 (excluding 0) are:

$\dfrac{1}{8}$, $\dfrac{2}{8}$, $\dfrac{3}{8}$, $\dfrac{4}{8}$, $\dfrac{5}{8}$, $\dfrac{6}{8}$, $\dfrac{7}{8}$ and $\dfrac{8}{8} = 1$

Continuing in the same way, the fractions marked from 1 to 2 (excluding 1) are:

$\dfrac{9}{8}$, $\dfrac{10}{8}$, $\dfrac{11}{8}$, $\dfrac{12}{8}$, $\dfrac{13}{8}$, $\dfrac{14}{8}$, $\dfrac{15}{8}$ and $\dfrac{16}{8} = 2$

∴ The marks from 0 to 1 are $\mathbf {\dfrac{1}{8}, \dfrac{2}{8}, \dfrac{3}{8}, …, \dfrac{8}{8}=1}$, and the marks after 1 up to 2 are $\mathbf {\dfrac{9}{8}, \dfrac{10}{8}, …, \dfrac{16}{8}=2}$.

On a number line, draw lines of lengths $\dfrac{1}{10}$, $\dfrac{3}{10}$, and $\dfrac{4}{5}$.

Answer

To draw all the three lengths on the same number line, the distance between 0 and 1 is divided into 10 equal parts.

So, the length of each part is $\dfrac{1}{10}$ unit.

Length $\dfrac{1}{10}$ = 1 time $\dfrac{1}{10}$, so this line covers 1 part starting from 0.

Length $\dfrac{3}{10}$ = 3 times $\dfrac{1}{10}$, so this line covers 3 parts starting from 0.

To draw the length $\dfrac{4}{5}$ using the same fractional unit, we write:

$= \dfrac{4}{5} \\[1em] = \dfrac{4 \times 2}{5 \times 2} \\[1em] = \dfrac{8}{10}$

So, the line of length $\dfrac{4}{5}$ covers 8 parts starting from 0.

Therefore, draw three separate line segments, all beginning at 0: the first ending at $\dfrac{1}{10}$, the second ending at $\dfrac{3}{10}$, and the third ending at $\dfrac{8}{10} = \dfrac{4}{5}$.

The three required lengths are shown below:

Write five more fractions of your choice and mark them on the number line.

Answer

This is an open-ended question. Five fractions have to be chosen by the student, so the answer will be different for different students.

One possible answer is shown below.

Let the five fractions chosen be $\dfrac{2}{10}$, $\dfrac{4}{10}$, $\dfrac{5}{10}$, $\dfrac{7}{10}$ and $\dfrac{9}{10}$.

Since the fractional unit of each of these fractions is $\dfrac{1}{10}$, the distance between 0 and 1 is divided into 10 equal parts.

Now, starting from 0, we count 2 parts for $\dfrac{2}{10}$, 4 parts for $\dfrac{4}{10}$, 5 parts for $\dfrac{5}{10}$, 7 parts for $\dfrac{7}{10}$ and 9 parts for $\dfrac{9}{10}$, and mark these points on the number line.

Hence, the five chosen fractions are marked on the number line as shown above.

How many fractions lie between 0 and 1? Think, discuss with your classmates, and write your answer.

Answer

Let us divide the length of 1 unit into equal parts, taking a different fractional unit each time.

Dividing 1 unit into 2 equal parts, we get 1 fraction between 0 and 1, that is, $\dfrac{1}{2}$.

Dividing 1 unit into 3 equal parts, we get 2 fractions between 0 and 1, that is, $\dfrac{1}{3}$ and $\dfrac{2}{3}$.

Dividing 1 unit into 4 equal parts, we get 3 fractions between 0 and 1, that is, $\dfrac{1}{4}$, $\dfrac{2}{4}$ and $\dfrac{3}{4}$.

In the same way, on dividing 1 unit into 10 equal parts we get 9 fractions between 0 and 1, on dividing it into 100 equal parts we get 99 fractions between 0 and 1, and so on.

In general, if 1 unit is divided into n equal parts, then the fractions

$\dfrac{1}{n}$, $\dfrac{2}{n}$, $\dfrac{3}{n}$, …, $\dfrac{n-1}{n}$

all lie between 0 and 1.

Since the unit can be divided into as many equal parts as we wish, the value of n can be made larger and larger without ever stopping. So, the fractions lying between 0 and 1 can never be counted completely.

∴ Infinitely many fractions lie between 0 and 1.

What is the length of the blue line and black line shown below? The distance between 0 and 1 is 1 unit long, and it is divided into two equal parts. The length of each part is $\dfrac{1}{2}$. So the blue line is $\dfrac{1}{2}$ units long. Write the fraction that gives the length of the black line in the box.

Answer

Here, the distance between 0 and 1 is 1 unit long and it is divided into 2 equal parts.

So, the fractional unit is $\dfrac{1}{2}$.

The blue line covers 1 such part.

∴ Length of the blue line = 1 time $\dfrac{1}{2} = \dfrac{1}{2}$ unit

From the figure, the black line covers 3 such parts.

Length of the black line = 3 times $\dfrac{1}{2}$

$= 3 \times \dfrac{1}{2} \\[1em] = \dfrac{3 \times 1}{2} \\[1em] = \dfrac{3}{2}$

∴ The blue line is $\mathbf {\dfrac{1}{2}}$ unit long and the black line is $\mathbf {\dfrac{3}{2}}$ units long.

Write the fraction that gives the lengths of the black lines in the respective boxes.

Answer

Here, the length of 1 unit is divided into 5 equal parts.

So, the fractional unit is $\dfrac{1}{5}$.

Counting the number of such equal parts covered by each black line, starting from the shortest line at the bottom, we get:

The shortest black line covers 6 parts.

∴ Its length = 6 times $\dfrac{1}{5} = \dfrac{6}{5}$ units

The next black line covers 7 parts.

∴ Its length = 7 times $\dfrac{1}{5} = \dfrac{7}{5}$ units

The next black line covers 8 parts.

∴ Its length = 8 times $\dfrac{1}{5} = \dfrac{8}{5}$ units

The longest black line at the top covers 9 parts.

∴ Its length = 9 times $\dfrac{1}{5} = \dfrac{9}{5}$ units

∴ The lengths of the black lines are $\mathbf {\dfrac{6}{5}, \dfrac{7}{5}, \dfrac{8}{5}}$ and $\mathbf {\dfrac{9}{5}}$ units, and these are the fractions to be written in the boxes at the respective marks.

How many whole units are there in $\dfrac{7}{2}$?

Answer

We know that $\dfrac{7}{2}$ means 7 times $\dfrac{1}{2}$.

Also, 2 times $\dfrac{1}{2} = \dfrac{2}{2} = 1$ whole unit.

So, we collect the halves in groups of 2:

$\phantom{=} \dfrac{7}{2} \\[1em] = \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} \\[1em] = \Big(\dfrac{2}{2}\Big) + \Big(\dfrac{2}{2}\Big) + \Big(\dfrac{2}{2}\Big) + \dfrac{1}{2} \\[1em] = 1 + 1 + 1 + \dfrac{1}{2} \\[1em] = 3 + \dfrac{1}{2}$

So, $\dfrac{7}{2}$ contains 3 whole units and one more half unit.

∴ There are 3 whole units in $\mathbf {\dfrac{7}{2}}$.

How many whole units are there in $\dfrac{4}{3}$ and in $\dfrac{7}{3}$?

Answer

We know that 3 times $\dfrac{1}{3} = \dfrac{3}{3} = 1$ whole unit.

For $\dfrac{4}{3}$:

$\phantom{=} \dfrac{4}{3} \\[1em] = \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3} \\[1em] = \Big(\dfrac{3}{3}\Big) + \dfrac{1}{3} \\[1em] = 1 + \dfrac{1}{3}$

So, $\dfrac{4}{3}$ contains 1 whole unit and one more third.

For $\dfrac{7}{3}$:

$\phantom{=} \dfrac{7}{3} \\[1em] = \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3} \\[1em] = \Big(\dfrac{3}{3}\Big) + \Big(\dfrac{3}{3}\Big) + \dfrac{1}{3} \\[1em] = 1 + 1 + \dfrac{1}{3} \\[1em] = 2 + \dfrac{1}{3}$

So, $\dfrac{7}{3}$ contains 2 whole units and one more third.

∴ There is 1 whole unit in $\mathbf {\dfrac{4}{3}}$ and there are 2 whole units in $\mathbf {\dfrac{7}{3}}$.

Figure out the number of whole units in each of the following fractions:

(a) $\dfrac{8}{3}$

(b) $\dfrac{11}{5}$

(c) $\dfrac{9}{4}$

Answer

(a) $\dfrac{8}{3}$

Here, the fractional unit is $\dfrac{1}{3}$ and 3 times $\dfrac{1}{3} = \dfrac{3}{3} = 1$ whole unit.

$\phantom{=} \dfrac{8}{3} \\[1em] = \dfrac{3}{3} + \dfrac{3}{3} + \dfrac{2}{3} \\[1em] = 1 + 1 + \dfrac{2}{3} \\[1em] = 2 + \dfrac{2}{3}$

∴ There are 2 whole units in $\mathbf {\dfrac{8}{3}}$.

(b) $\dfrac{11}{5}$

Here, the fractional unit is $\dfrac{1}{5}$ and 5 times $\dfrac{1}{5} = \dfrac{5}{5} = 1$ whole unit.

$\phantom{=} \dfrac{11}{5} \\[1em] = \dfrac{5}{5} + \dfrac{5}{5} + \dfrac{1}{5} \\[1em] = 1 + 1 + \dfrac{1}{5} \\[1em] = 2 + \dfrac{1}{5}$

∴ There are 2 whole units in $\mathbf {\dfrac{11}{5}}$.

(c) $\dfrac{9}{4}$

Here, the fractional unit is $\dfrac{1}{4}$ and 4 times $\dfrac{1}{4} = \dfrac{4}{4} = 1$ whole unit.

$\phantom{=} \dfrac{9}{4} \\[1em] = \dfrac{4}{4} + \dfrac{4}{4} + \dfrac{1}{4} \\[1em] = 1 + 1 + \dfrac{1}{4} \\[1em] = 2 + \dfrac{1}{4}$

∴ There are 2 whole units in $\mathbf {\dfrac{9}{4}}$.

Can all fractions greater than 1 be written as such mixed numbers?

Answer

No, all fractions greater than 1 cannot be written as mixed numbers.

A mixed number has a whole part and a fractional part which is less than 1. So, a fraction greater than 1 can be written as a mixed number only when something is left over after taking out the complete whole units from it.

If the numerator is a multiple of the denominator, then nothing is left over. For example:

$\phantom{=} \dfrac{8}{4} \\[1em] = \dfrac{4}{4} + \dfrac{4}{4} \\[1em] = 1 + 1 \\[1em] = 2 \quad \text{[No fractional part is left]}$

Here $\dfrac{8}{4}$ is exactly 2 whole units. As there is no fractional part, it cannot be written as a mixed number.

∴ Only those fractions greater than 1 whose numerator is not a multiple of the denominator can be written as mixed numbers.

Write the following fractions as mixed fractions (e.g., $\dfrac{9}{2} = 4\dfrac{1}{2}$):

(a) $\dfrac{9}{2}$

(b) $\dfrac{9}{5}$

(c) $\dfrac{21}{19}$

(d) $\dfrac{47}{9}$

(e) $\dfrac{12}{11}$

(f) $\dfrac{19}{6}$

Answer

(a) $\dfrac{9}{2}$

We have:

$\phantom{=} \dfrac{9}{2} \\[1em] = \dfrac{8 + 1}{2} \quad \text{[Since 9 = 8 + 1]} \\[1em] = \dfrac{8}{2} + \dfrac{1}{2} \\[1em] = 4 + \dfrac{1}{2} \quad \text{[Taking out the whole units]} \\[1em] = 4\dfrac{1}{2}$

∴ $\mathbf {\dfrac{9}{2} = 4\dfrac{1}{2}}$

(b) $\dfrac{9}{5}$

We have:

$\phantom{=} \dfrac{9}{5} \\[1em] = \dfrac{5 + 4}{5} \quad \text{[Since 9 = 5 + 4]} \\[1em] = \dfrac{5}{5} + \dfrac{4}{5} \\[1em] = 1 + \dfrac{4}{5} \quad \text{[Taking out the whole units]} \\[1em] = 1\dfrac{4}{5}$

∴ $\mathbf {\dfrac{9}{5} = 1\dfrac{4}{5}}$

(c) $\dfrac{21}{19}$

We have:

$\phantom{=} \dfrac{21}{19} \\[1em] = \dfrac{19 + 2}{19} \quad \text{[Since 21 = 19 + 2]} \\[1em] = \dfrac{19}{19} + \dfrac{2}{19} \\[1em] = 1 + \dfrac{2}{19} \quad \text{[Taking out the whole units]} \\[1em] = 1\dfrac{2}{19}$

∴ $\mathbf {\dfrac{21}{19} = 1\dfrac{2}{19}}$

(d) $\dfrac{47}{9}$

We have:

$\phantom{=} \dfrac{47}{9} \\[1em] = \dfrac{45 + 2}{9} \quad \text{[Since 47 = 45 + 2]} \\[1em] = \dfrac{45}{9} + \dfrac{2}{9} \\[1em] = 5 + \dfrac{2}{9} \quad \text{[Taking out the whole units]} \\[1em] = 5\dfrac{2}{9}$

∴ $\mathbf {\dfrac{47}{9} = 5\dfrac{2}{9}}$

(e) $\dfrac{12}{11}$

We have:

$\phantom{=} \dfrac{12}{11} \\[1em] = \dfrac{11 + 1}{11} \quad \text{[Since 12 = 11 + 1]} \\[1em] = \dfrac{11}{11} + \dfrac{1}{11} \\[1em] = 1 + \dfrac{1}{11} \quad \text{[Taking out the whole units]} \\[1em] = 1\dfrac{1}{11}$

∴ $\mathbf {\dfrac{12}{11} = 1\dfrac{1}{11}}$

(f) $\dfrac{19}{6}$

We have:

$\phantom{=} \dfrac{19}{6} \\[1em] = \dfrac{18 + 1}{6} \quad \text{[Since 19 = 18 + 1]} \\[1em] = \dfrac{18}{6} + \dfrac{1}{6} \\[1em] = 3 + \dfrac{1}{6} \quad \text{[Taking out the whole units]} \\[1em] = 3\dfrac{1}{6}$

∴ $\mathbf {\dfrac{19}{6} = 3\dfrac{1}{6}}$

Write the following mixed numbers as fractions:

(a) $3\dfrac{1}{4}$

(b) $7\dfrac{2}{3}$

(c) $9\dfrac{4}{9}$

(d) $3\dfrac{1}{6}$

(e) $2\dfrac{3}{11}$

(f) $3\dfrac{9}{10}$

Answer

(a) $3\dfrac{1}{4}$

We have:

$\phantom{=} 3\dfrac{1}{4} \\[1em] = 3 + \dfrac{1}{4} \\[1em] = \dfrac{3 \times 4}{4} + \dfrac{1}{4} \\[1em] \text{[Writing 3 as a fraction with denominator 4]} \\[1em] = \dfrac{12 + 1}{4} \\[1em] = \dfrac{13}{4}$

∴ $\mathbf {3\dfrac{1}{4} = \dfrac{13}{4}}$

(b) $7\dfrac{2}{3}$

We have:

$\phantom{=} 7\dfrac{2}{3} \\[1em] = 7 + \dfrac{2}{3} \\[1em] = \dfrac{7 \times 3}{3} + \dfrac{2}{3} \\[1em] \text{[Writing 7 as a fraction with denominator 3]} \\[1em] = \dfrac{21 + 2}{3} \\[1em] = \dfrac{23}{3}$

∴ $\mathbf {7\dfrac{2}{3} = \dfrac{23}{3}}$

(c) $9\dfrac{4}{9}$

We have:

$\phantom{=} 9\dfrac{4}{9} \\[1em] = 9 + \dfrac{4}{9} \\[1em] = \dfrac{9 \times 9}{9} + \dfrac{4}{9} \\[1em] \text{[Writing 9 as a fraction with denominator 9]} \\[1em] = \dfrac{81 + 4}{9} \\[1em] = \dfrac{85}{9}$

∴ $\mathbf{9\dfrac{4}{9} = \dfrac{85}{9}}$

(d) $3\dfrac{1}{6}$

We have:

$\phantom{=} 3\dfrac{1}{6} \\[1em] = 3 + \dfrac{1}{6} \\[1em] = \dfrac{3 \times 6}{6} + \dfrac{1}{6} \\[1em] \text{[Writing 3 as a fraction with denominator 6]} \\[1em] = \dfrac{18 + 1}{6} \\[1em] = \dfrac{19}{6}$

∴ $\mathbf {3\dfrac{1}{6} = \dfrac{19}{6}}$

(e) $2\dfrac{3}{11}$

We have:

$\phantom{=} 2\dfrac{3}{11} \\[1em] = 2 + \dfrac{3}{11} \\[1em] = \dfrac{2 \times 11}{11} + \dfrac{3}{11} \\[1em] \text{[Writing 2 as a fraction with denominator 11]} \\[1em] = \dfrac{22 + 3}{11} \\[1em] = \dfrac{25}{11}$

∴ $\mathbf {2\dfrac{3}{11} = \dfrac{25}{11}}$

(f) $3\dfrac{9}{10}$

We have:

$\phantom{=} 3\dfrac{9}{10} \\[1em] = 3 + \dfrac{9}{10} \\[1em] = \dfrac{3 \times 10}{10} + \dfrac{9}{10} \\[1em] \text{[Writing 3 as a fraction with denominator 10]} \\[1em] = \dfrac{30 + 9}{10} \\[1em] = \dfrac{39}{10}$

∴ $\mathbf {3\dfrac{9}{10} = \dfrac{39}{10}}$

Check whether $\dfrac{1}{3}$ and $\dfrac{2}{6}$ are equivalent fractions or not, using paper strips.

Answer

Take two paper strips of the same length. Each strip is taken to be 1 unit long.

Fold the first strip into 3 equal parts and shade 1 part. The shaded length is $\dfrac{1}{3}$ unit.

Fold the second strip into 6 equal parts and shade 2 parts. The shaded length is $\dfrac{2}{6}$ unit.

Now place the second strip just below the first strip so that their left ends match.

The two shaded lengths are seen to be exactly equal.

We can also check this by using fractional units:

$\phantom{=} \dfrac{2}{6} \\[1em] = \dfrac{2 \div 2}{6 \div 2} \\[1em] \text{[Dividing numerator and denominator by 2]} \\[1em] = \dfrac{1}{3}$

Hence, $\mathbf {\dfrac{1}{3}}$ and $\mathbf {\dfrac{2}{6}}$ are equivalent fractions.

Answer the following questions after looking at the fraction wall:

(i) Are the lengths $\dfrac{1}{2}$ and $\dfrac{3}{6}$ equal?

(ii) Are $\dfrac{2}{3}$ and $\dfrac{4}{6}$ equivalent fractions? Why?

(iii) How many pieces of length $\dfrac{1}{6}$ will make a length of $\dfrac{1}{2}$?

(iv) How many pieces of length $\dfrac{1}{6}$ will make a length of $\dfrac{1}{3}$?

Answer

(i) From the fraction wall, the strip showing $\dfrac{1}{2}$ and the 3 pieces of the strip showing $\dfrac{1}{6}$ end at the same point. So, the two lengths are equal.

$\phantom{=} \dfrac{3}{6} \\[1em] = \dfrac{3 \div 3}{6 \div 3} \\[1em] \text{[Dividing numerator and denominator by 3]} \\[1em] = \dfrac{1}{2}$

Hence, the lengths $\mathbf {\dfrac{1}{2}}$ and $\mathbf {\dfrac{3}{6}}$ are equal.

(ii) In the fraction wall, the length of 2 pieces of $\dfrac{1}{3}$ is exactly the same as the length of 4 pieces of $\dfrac{1}{6}$.

$\phantom{=} \dfrac{4}{6} \\[1em] = \dfrac{4 \div 2}{6 \div 2} \\[1em] \text{[Dividing numerator and denominator by 2]} \\[1em] = \dfrac{2}{3}$

Both fractions denote the same length, though they are expressed in different fractional units.

Hence, $\mathbf {\dfrac{2}{3}}$ and $\mathbf {\dfrac{4}{6}}$ are equivalent fractions.

(iii) We need the number of pieces of length $\dfrac{1}{6}$ that together make a length of $\dfrac{1}{2}$.

$\phantom{=} \dfrac{1}{2} \\[1em] = \dfrac{1 \times 3}{2 \times 3} \\[1em] \text{[Making the fractional unit one-sixth]} \\[1em] = \dfrac{3}{6} \\[1em] = 3 \times \dfrac{1}{6}$

Hence, 3 pieces of length $\mathbf {\dfrac{1}{6}}$ will make a length of $\mathbf {\dfrac{1}{2}}$.

(iv) We need the number of pieces of length $\dfrac{1}{6}$ that together make a length of $\dfrac{1}{3}$.

$\phantom{=} \dfrac{1}{3} \\[1em] = \dfrac{1 \times 2}{3 \times 2} \\[1em] \text{[Making the fractional unit one-sixth]} \\[1em] = \dfrac{2}{6} \\[1em] = 2 \times \dfrac{1}{6}$

Hence, 2 pieces of length $\mathbf {\dfrac{1}{6}}$ will make a length of $\mathbf {\dfrac{1}{3}}$.

Are $\dfrac{3}{6}$, $\dfrac{4}{8}$, $\dfrac{5}{10}$ equivalent fractions? Why?

Answer

Let us write each of the given fractions in its simplest form:

$\phantom{=} \dfrac{3}{6} = \dfrac{3 \div 3}{6 \div 3} = \dfrac{1}{2} \\[1em] \phantom{=} \dfrac{4}{8} = \dfrac{4 \div 4}{8 \div 4} = \dfrac{1}{2} \\[1em] \phantom{=} \dfrac{5}{10} = \dfrac{5 \div 5}{10 \div 5} = \dfrac{1}{2}$

Each of the three fractions denotes the same length, namely $\dfrac{1}{2}$ of a unit. This can also be seen in the fraction wall, where 3 pieces of $\dfrac{1}{6}$, 4 pieces of $\dfrac{1}{8}$ and 5 pieces of $\dfrac{1}{10}$ all end at the same point as $\dfrac{1}{2}$.

Hence, $\mathbf {\dfrac{3}{6}, \dfrac{4}{8}}$ and $\mathbf {\dfrac{5}{10}}$ are equivalent fractions, as they all represent the same length $\mathbf {\dfrac{1}{2}}$.

Write two equivalent fractions for $\dfrac{2}{6}$.

Answer

Equivalent fractions are obtained by multiplying or dividing the numerator and the denominator by the same number.

Dividing the numerator and the denominator by 2, we get:

$\phantom{=} \dfrac{2}{6} \\[1em] = \dfrac{2 \div 2}{6 \div 2} \\[1em] = \dfrac{1}{3}$

Multiplying the numerator and the denominator by 3, we get:

$\phantom{=} \dfrac{2}{6} \\[1em] = \dfrac{2 \times 3}{6 \times 3} \\[1em] = \dfrac{6}{18}$

Hence, two equivalent fractions for $\mathbf {\dfrac{2}{6}}$ are $\mathbf {\dfrac{1}{3}}$ and $\mathbf {\dfrac{6}{18}}$.

Note: Other equivalent fractions such as $\dfrac{3}{9}$, $\dfrac{4}{12}$, $\dfrac{5}{15}$, … are also correct answers.

$\dfrac{4}{6} = \dfrac{\square}{\square} = \dfrac{\square}{\square} = \dfrac{\square}{\square} =$ ............ (Write as many as you can)

Answer

Dividing the numerator and the denominator of $\dfrac{4}{6}$ by 2, we get:

$\dfrac{4}{6} = \dfrac{4 \div 2}{6 \div 2} = \dfrac{2}{3}$

Now, multiplying the numerator and the denominator of $\dfrac{2}{3}$ by 3, 4, 5, … in turn, we get:

$\phantom{=} \dfrac{2}{3} = \dfrac{2 \times 3}{3 \times 3} = \dfrac{6}{9} \\[1em] \phantom{=} \dfrac{2}{3} = \dfrac{2 \times 4}{3 \times 4} = \dfrac{8}{12} \\[1em] \phantom{=} \dfrac{2}{3} = \dfrac{2 \times 5}{3 \times 5} = \dfrac{10}{15}$

Hence, $\mathbf {\dfrac{4}{6} = \dfrac{2}{3} = \dfrac{6}{9} = \dfrac{8}{12} = \dfrac{10}{15} =}$ ............

Note: This list never ends. We can keep multiplying the numerator and the denominator of $\dfrac{2}{3}$ by the same number to get as many equivalent fractions as we want.

Three rotis are shared equally by four children. Show the division in the picture and write a fraction for how much each child gets. Also, write the corresponding division facts, addition facts and multiplication facts.

Fraction of roti each child gets is ............... .

Division fact:

Addition fact:

Multiplication fact:

Compare your picture and answers with your classmates!

Answer

Given:

Number of rotis = 3

Number of children = 4

Divide each roti into 4 equal parts. Each part is $\dfrac{1}{4}$ roti.

From the 3 rotis, we get 12 such parts in all. Distributing them equally, each child gets 3 parts, that is, 3 pieces of $\dfrac{1}{4}$ roti.

$\phantom{=} \text{Share of each child} = 3 \times \dfrac{1}{4} \text{ roti} \\[1em] = \dfrac{3}{4} \text{ roti}$

Fraction of roti each child gets is $\dfrac{3}{4}$.

Division fact: $3 \div 4 = \dfrac{3}{4}$

Addition fact: $3 = \dfrac{3}{4} + \dfrac{3}{4} + \dfrac{3}{4} + \dfrac{3}{4}$

Multiplication fact: $3 = 4 \times \dfrac{3}{4}$

Hence, each child gets $\mathbf {\dfrac{3}{4}}$ roti.

Draw a picture to show how much each child gets when 2 rotis are shared equally by 4 children. Also, write the corresponding division facts, addition facts, and multiplication facts.

Answer

Given:

Number of rotis = 2

Number of children = 4

Divide each roti into 2 equal halves. Each half is $\dfrac{1}{2}$ roti.

From the 2 rotis, we get 4 halves in all. Distributing these 4 equal halves among 4 children, each child gets 1 half.

$\phantom{=} \text{Share of each child} = 2 \div 4 \text{ roti} \\[1em] = \dfrac{2}{4} \text{ roti} \\[1em] = \dfrac{1}{2} \text{ roti} \\[1em]$

Fraction of roti each child gets is $\dfrac{1}{2}$.

Division fact: $2 \div 4 = \dfrac{2}{4} = \dfrac{1}{2}$

Addition fact: $2 = \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2}$

Multiplication fact: $2 = 4 \times \dfrac{1}{2}$

Hence, each child gets $\mathbf {\dfrac{1}{2}}$ roti.

Anil was in a group where 2 cakes were divided equally among 5 children. How much cake would Anil get?

Answer

Given:

Number of cakes = 2

Number of children = 5

Divide each cake into 5 equal parts. Each part is $\dfrac{1}{5}$ cake.

From the 2 cakes, we get 10 such parts in all. Distributing them equally, each child gets 2 parts, that is, 2 pieces of $\dfrac{1}{5}$ cake.

$\phantom{=} \text{Share of each child} = 2 \div 5 \\[1em] = 2 \times \dfrac{1}{5} \text{ cake} \\[1em] = \dfrac{2}{5} \text{ cake}$

Hence, Anil would get $\mathbf {\dfrac{2}{5}}$ of a cake.

Find some more fractions equivalent to $\dfrac{1}{2}$. Write them in the boxes here:

$\square$ $\square$ $\square$ $\square$

Answer

To get a fraction equivalent to a given fraction, we multiply both its numerator and its denominator by the same number.

We have:

$\phantom{=} \dfrac{1}{2} = \dfrac{1 \times 4}{2 \times 4} = \dfrac{4}{8} \\[1em] \phantom{=} \dfrac{1}{2} = \dfrac{1 \times 5}{2 \times 5} = \dfrac{5}{10} \\[1em] \phantom{=} \dfrac{1}{2} = \dfrac{1 \times 6}{2 \times 6} = \dfrac{6}{12} \\[1em] \phantom{=} \dfrac{1}{2} = \dfrac{1 \times 7}{2 \times 7} = \dfrac{7}{14}$

So, the four boxes can be filled as shown below:

$\boxed{\dfrac{4}{8}}$ $\boxed{\dfrac{5}{10}}$ $\boxed{\dfrac{6}{12}}$ $\boxed{\dfrac{7}{14}}$

∴ Four more fractions equivalent to $\mathbf {\dfrac{1}{2}}$ are $\mathbf {\dfrac{4}{8}, \dfrac{5}{10}, \dfrac{6}{12}}$ and $\mathbf {\dfrac{7}{14}}$.

Equally divide the rotis in the situations shown below and write down the share of each child. Are the shares in each of these cases the same? Why?

(i) Do you notice anything about the relationship between the numerator and denominator in each of these fractions?

Answer

When a whole number of rotis is shared equally among a number of children, the share of each child is given by:

Share of each child = $\dfrac{\text{Number of rotis}}{\text{Number of children}}$

Dividing the rotis equally in each of the three situations shown, we get:

Now, let us express the second and the third share in the simplest form:

$\phantom{=} \dfrac{4}{6} = \dfrac{4 \div 2}{6 \div 2} = \dfrac{2}{3} \\[1em] \phantom{=} \dfrac{6}{9} = \dfrac{6 \div 3}{9 \div 3} = \dfrac{2}{3}$

⇒ $\dfrac{2}{3} = \dfrac{4}{6} = \dfrac{6}{9}$

∴ The share of each child is the same in all the three situations.

The shares are the same because in each situation the number of rotis and the number of children have both been multiplied by the same number. In the second situation twice as many rotis are shared among twice as many children, and in the third situation three times as many rotis are shared among three times as many children. So the quantity of roti received by one child does not change.

Hence, the share of each child is $\mathbf {\dfrac{2}{3}}$ roti in every case, and the three shares are the same because $\mathbf {\dfrac{2}{3}, \dfrac{4}{6}}$ and $\mathbf {\dfrac{6}{9}}$ are equivalent fractions.

(i) In each of these fractions, the numerator and the denominator of $\dfrac{2}{3}$ have been multiplied by the same number:

$\phantom{=} \dfrac{2}{3} = \dfrac{2 \times 1}{3 \times 1} \\[1em] \phantom{=} \dfrac{4}{6} = \dfrac{2 \times 2}{3 \times 2} \\[1em] \phantom{=} \dfrac{6}{9} = \dfrac{2 \times 3}{3 \times 3}$

Hence, multiplying (or dividing) the numerator and the denominator of a fraction by the same non-zero number gives an equivalent fraction, and $\mathbf {\dfrac{2}{3}}$ is the simplest form of both $\mathbf {\dfrac{4}{6}}$ and $\mathbf {\dfrac{6}{9}}$.

Find the missing numbers:

5 glasses of juice shared equally among 4 friends is the same as .......... glasses of juice shared equally among 8 friends.

So, $\dfrac{5}{4} = \dfrac{\square}{8}$.

Answer

The number of friends has increased from 4 to 8, that is, it has become 2 times. So, the number of glasses of juice must also become 2 times for each friend to get the same share.

We have:

$\phantom{=} \dfrac{5}{4} = \dfrac{5 \times 2}{4 \times 2} \\[1em] = \dfrac{10}{8}$

∴ 5 glasses of juice shared equally among 4 friends is the same as 10 glasses of juice shared equally among 8 friends, that is, $\mathbf {\dfrac{5}{4} = \dfrac{10}{8}}$.

Find the missing numbers:

4 kg of potatoes divided equally in 3 bags is the same as 12 kg of potatoes divided equally in .......... bags.

So, $\dfrac{4}{3} = \dfrac{12}{\square}$

Answer

The quantity of potatoes has increased from 4 kg to 12 kg, that is, it has become 3 times. So, the number of bags must also become 3 times for each bag to hold the same quantity.

We have:

$\phantom{=} \dfrac{4}{3} = \dfrac{4 \times 3}{3 \times 3} \\[1em] = \dfrac{12}{9}$

∴ 4 kg of potatoes divided equally in 3 bags is the same as 12 kg of potatoes divided equally in 9 bags, that is, $\mathbf {\dfrac{4}{3} = \dfrac{12}{9}}$.

Find the missing numbers:

7 rotis divided among 5 children is the same as .......... rotis divided among .......... children.

So, $\dfrac{7}{5} = \dfrac{\square}{\square}$.

Answer

Here we may choose any multiplier we like, as long as the number of rotis and the number of children are both multiplied by the same number. Taking the multiplier as 2, we get:

$\phantom{=} \dfrac{7}{5} = \dfrac{7 \times 2}{5 \times 2} \\[1em] = \dfrac{14}{10}$

Some other correct answers are $\dfrac{7}{5} = \dfrac{21}{15}$ and $\dfrac{7}{5} = \dfrac{28}{20}$.

∴ 7 rotis divided among 5 children is the same as 14 rotis divided among 10 children, that is, $\mathbf {\dfrac{7}{5} = \dfrac{14}{10}}$.

In which group will each child get more chikki?
1 chikki divided between 2 children or 5 chikkis divided among 8 children.

Answer

Given:

Share of each child in the first group = $\dfrac{1}{2}$ chikki

Share of each child in the second group = $\dfrac{5}{8}$ chikki

To compare $\dfrac{1}{2}$ and $\dfrac{5}{8}$, we make their fractional units the same.

Since 8 is a multiple of 2, we take 8 as the common denominator.

$\phantom{=} \dfrac{1}{2} = \dfrac{1 \times 4}{2 \times 4} = \dfrac{4}{8} \\[1em] \text{[Multiplying numerator and denominator by 4]} \\[1em] \phantom{=} \dfrac{5}{8} = \dfrac{5}{8} \quad \text{[Already has denominator 8]}$

Both fractions now use the same fractional unit $\dfrac{1}{8}$, so we compare their numerators:

4 < 5

⇒ $\dfrac{4}{8} \lt \dfrac{5}{8}$

∴ $\dfrac{1}{2} \lt \dfrac{5}{8}$

Hence, each child in the second group, where 5 chikkis are divided among 8 children, will get more chikki.

In which group will each child get more?
1 chikki divided between 2 children or 4 chikkis divided among 7 children.

Answer

Given:

Share of each child in the first group = $\dfrac{1}{2}$ chikki

Share of each child in the second group = $\dfrac{4}{7}$ chikki

First, we write $\dfrac{1}{2}$ as an equivalent fraction whose numerator is 4:

$\phantom{=} \dfrac{1}{2} = \dfrac{1 \times 4}{2 \times 4} = \dfrac{4}{8} \\[1em] \text{[Multiplying numerator and denominator by 4]}$

Now we compare $\dfrac{4}{7}$ and $\dfrac{4}{8}$.

In both of these, 4 chikkis are being shared, but in the first case they are shared among 7 children and in the second case among 8 children. If the number of units shared is the same but the number of children is more, then each share is less.

⇒ $\dfrac{4}{7} \gt \dfrac{4}{8}$

∴ $\dfrac{4}{7} \gt \dfrac{1}{2} \hspace{1cm}$ [Since $\dfrac{4}{8} = \dfrac{1}{2}$]

Hence, each child in the second group, where 4 chikkis are divided among 7 children, will get more chikki.

Suppose the number of children is kept the same, but the number of units that are being shared is increased? What can you say about each child's share now? Why? Discuss how your reasoning explains $\dfrac{1}{5} \lt \dfrac{2}{5}$, $\dfrac{3}{7} \lt \dfrac{4}{7}$, and $\dfrac{1}{2} \lt \dfrac{5}{8}$.

Answer

If the number of children is kept the same and the number of units being shared is increased, then each child's share becomes larger.

This is because the same number of children now share a greater quantity, so every child receives more. In terms of fractions, when the denominator (the number of children) remains unchanged and the numerator (the number of units) increases, each child simply receives a greater number of the same fractional unit.

Let us apply this reasoning to the given pairs.

$\dfrac{1}{5}$ and $\dfrac{2}{5}$: In both cases the units are shared among 5 children, so the fractional unit is $\dfrac{1}{5}$ in both. Since 2 units of $\dfrac{1}{5}$ are more than 1 unit of $\dfrac{1}{5}$,

∴ $\dfrac{1}{5} \lt \dfrac{2}{5}$

$\dfrac{3}{7}$ and $\dfrac{4}{7}$: In both cases the units are shared among 7 children, so the fractional unit is $\dfrac{1}{7}$ in both. Since 4 units of $\dfrac{1}{7}$ are more than 3 units of $\dfrac{1}{7}$,

∴ $\dfrac{3}{7} \lt \dfrac{4}{7}$