Chapter 7: Fractions

Given: Three guavas together weigh 1 kg.

Step 1 — Identify the total weight and the number of equal parts
• Total weight = $$1$$ kg.
• Number of equal parts (guavas) = $$3$$.

Step 2 — Divide to find the weight of one guava
To share the total weight equally among the 3 guavas, divide the total weight by 3:
$$\text{Weight of one guava} = 1 \div 3 = \dfrac{1}{3}$$ kg.

Step 3 — State the result
Each guava therefore weighs $$\dfrac{1}{3}$$ kg.

Step 1 : Note the total quantity of rice.

Total rice given  = $$1\text{ kg}$$.

Step 2 : Note how many equal packets are made.

Number of packets  = $$4$$.

Step 3 : Use the idea "each share = whole ÷ number of equal parts".

Weight of one packet
= $$\dfrac{\text{total weight}}{\text{number of packets}}$$
= $$\dfrac{1\text{ kg}}{4}$$

Step 4 : Carry out the division.

$$\dfrac{1}{4}=\dfrac14\text{ kg}=0.25\text{ kg}$$

Conclusion : Each packet weighs $$\dfrac14\text{ kg}$$ (or 0.25 kg, i.e. 250 g).

We have to find out what fraction of a glass each friend gets when 3 full glasses of sugarcane juice are shared equally among 4 friends.

Step 1 : Write the total quantity of juice.
Total juice = 3 glasses.

Step 2 : Write the number of equal parts we must make.
Number of friends = 4, so we must divide the 3 glasses into 4 equal parts.

Step 3 : Set up the division.
“Each share” = (total quantity) ÷ (number of shares)
Using fraction notation:
$$\text{Each share} = \dfrac{3\;\text{glasses}}{4}$$

Step 4 : Simplify the fraction if possible.
The numerator and denominator have no common factor other than 1, so the fraction is already in its simplest form.

Therefore, each friend drinks
\[\dfrac{3}{4}\;\text{glass}\]

Step 1 – Write the weights as fractions.
The big fish weighs $$\frac{1}{2}$$ kg and the small fish weighs $$\frac{1}{4}$$ kg.

Step 2 – Set up the addition.
Total weight $$=$$ weight of big fish $$+$$ weight of small fish:
$$\frac{1}{2} + \frac{1}{4}$$.

Step 3 – Make the denominators the same.
The denominators are 2 and 4. Their L.C.M. is 4, so we convert $$\frac{1}{2}$$ to an equivalent fraction with denominator 4.
Multiply numerator and denominator by 2:
$$\frac{1}{2} = \frac{1\times 2}{2\times 2} = \frac{2}{4}.$$

Step 4 – Add the like fractions.
Now add $$\frac{2}{4} + \frac{1}{4}$$ by adding the numerators and keeping the common denominator 4.

Step 5 – Simplify (if needed).
The fraction $$\frac{3}{4}$$ is already in its simplest form.

Conclusion.
Together the two fish weigh $$\frac{3}{4}$$ kg.

Step 1 – Translate each word into a fraction.

• quarter = $$\dfrac{1}{4}$$
• half         = $$\dfrac{1}{2}$$
• three quarters = $$\dfrac{3}{4}$$
• one and a quarter = $$1\dfrac14 = \dfrac{1\times4+1}{4}=\dfrac54$$
• one and a half = $$1\dfrac12 = \dfrac{1\times2+1}{2}=\dfrac32$$
• two and a half = $$2\dfrac12 = \dfrac{2\times2+1}{2}=\dfrac52$$

Step 2 – Rewrite all fractions with a common denominator.
A convenient common denominator is 4 × 2 = 8 (or 4 is enough for most numbers). We choose 4 because many already have it.

Step 3 – Compare the numerators.

• $$\dfrac14$$ has numerator 1
• $$\dfrac24$$ has numerator 2
• $$\dfrac34$$ has numerator 3
• $$\dfrac54$$ has numerator 5
• $$\dfrac64$$ has numerator 6
• $$\dfrac{10}{4}$$ has numerator 10

The larger the numerator (when denominators are equal), the larger the fraction. So the ascending order is

Step 4 – Write the original names in this order.

quarter < half < three quarters < one and a quarter < one and a half < two and a half

Step 1 – Understand the sentence “6 equal parts”
When we say the whole chikki is cut into 6 equal parts, “equal” means that the amount of chikki in every part is the same. In Mathematics that amount is measured by area (if the bar is flat) or by mass/volume (if it is thick). Therefore each part must contain exactly one-sixth of the whole quantity.

Step 2 – Draw a few different ways of cutting
Ask the pupil to draw (or the teacher to show) three pictures of the same rectangular bar:

• Picture A: 6 thin vertical rectangles (like 6 fingers side by side).
• Picture B: 6 thin horizontal rectangles (layers one above another).
• Picture C: 3 vertical cuts and 1 diagonal through every alternate piece, giving 6 triangles. The outline of each part is now triangular.

These cuts give parts whose shapes are obviously different.

Step 3 – Calculate (or reason about) the amount in each part

1. Suppose the full bar is a rectangle of length $$12\,\text{cm}$$ and breadth $$6\,\text{cm}$$.
     • Area of the whole = $$12 \times 6 = 72\,\text{cm}^2$$.
2. Because the bar is divided into 6 equal parts, area of each part = $$\dfrac{72}{6}=12\,\text{cm}^2$$.
3. This $$12\,\text{cm}^2$$ is exactly $$\dfrac16$$ of the whole: $$\dfrac{12}{72}=\dfrac16$$.

The same check can be done with mass: if the bar weighs $$180\,\text{g}$$, every piece obtained in any of the three drawings must weigh $$\dfrac{180}{6}=30\,\text{g}$$.

Step 4 – Size versus shape
“Shape” tells us what the outline looks like (rectangular, triangular, irregular, …). “Size” here means the amount of chikki. Since each part is required to be one-sixth of the whole, the amount (area or mass) of every part is the same. Hence they are all equal in size even though their shapes are different.

Conclusion
Yes, any piece that represents $$\dfrac16$$ of the original chikki contains exactly the same quantity of chikki. Changing the cutting pattern only changes the shape, not the size, of each one-sixth piece.

The given row already lists the first four equivalent fractions of $$\frac{1}{2}$$.

Hence, the next two entries after $$\frac{4}{8}$$ are $$\frac{5}{10}$$ and $$\frac{6}{12}$$.

Concept recalled: Two fractions are “equivalent” if they represent the same part of a whole. For any non-zero whole number $$n$$, multiplying both the numerator and the denominator of a fraction by $$n$$ keeps the value of the fraction unchanged:
$$\frac{a}{b}=\frac{a\times n}{b\times n}\;(n\neq 0).$$

The fraction given is $$\frac{1}{4}$$. Let us multiply its numerator and denominator by the whole numbers $$n=1,2,3,4,5,6$$ exactly as was done for $$\frac12$$ in the textbook example.

Every fraction written in the right-most column names the same quantity as $$\frac{1}{4}$$, so the table above is the required “similar table” for $$\frac{1}{4}$$.

Goal. Use only a strip of paper and folding to obtain $$\dfrac{1}{3}$$ and $$\dfrac{1}{6}$$ of the strip.

1. Take the strip. Its whole length is taken as 1 unit.
2. Make $$\tfrac13$$ by an ‘S’-fold (three equal layers).
   1. Gently curl the strip into an ‘S’ (or ‘Z’) shape so that three layers lie one on top of another.
   2. Slide the layers a little until all three sections look the same length and the three short edges line up exactly.
   3. Holding the layers in place, press the two folds firmly to make sharp creases, and then open the strip out flat.
   4. The strip is now divided by two creases into three equal sections. Shade any one of them; the shaded length is $$\dfrac{1}{3}$$ of the whole strip.

   Why not ‘fold an end to the midpoint’? If you first fold the strip in half and then fold one end onto the half-crease, you get a piece of length $$\tfrac14$$, not $$\tfrac13$$ — so that idea does not give thirds. The S-fold above is the natural way to get three equal parts.

3. Why $$\tfrac16$$ can come from $$\tfrac13$$.
   $$\dfrac{1}{6}=\dfrac{1}{2}\times\dfrac{1}{3},$$ so halving each of the three equal parts produces sixths.
4. Make $$\tfrac16$$ with one extra fold.
   1. Fold the strip back into the same three-layer ‘S’-shape from Step 2.
   2. Holding all three layers together, fold the whole packet exactly in half — bring its two short ends together — and press a sharp crease through every layer.
   3. Open the strip out. The single new crease has passed through all three layers, so it appears three times — once inside each third. The strip now has 5 creases dividing it into $$3\times 2 = 6$$ equal parts.
   4. Shade any one of these six parts; its length is $$\dfrac{1}{6}$$ of the whole strip.
5. Check with a ruler (optional). If the strip is $$L\,\text{cm}$$ long, each small section should measure \[\dfrac{L}{6}\,\text{cm},\] confirming that each part is indeed $$\dfrac{1}{6}$$ of the whole.

Conclusion. Yes. The same strip used to make $$\dfrac{1}{3}$$ (via the three-layer ‘S’-fold) can be re-folded into that 3-layer packet and then folded once more in half. The strip now shows 6 equal parts, each of length $$\dfrac{1}{6}$$.

Step 1 — Recall what the denominator tells us
For a fractional unit of the form $$\frac{1}{n}$$ the denominator $$n$$ shows how many equal parts the whole has been cut into. The numerator (here it is 1) shows how many of those parts are being talked about or shaded.

Step 2 — Look carefully at each picture
Because every picture is a complete circle, simply count the number of equal sectors in that circle.

• Picture 1 has 3 equal sectors.
• Picture 2 has 5 equal sectors.
• Picture 3 has 8 equal sectors.
• Picture 4 has 6 equal sectors.

Step 3 — Match the denominator with the correct picture

Therefore, each fractional unit is matched to the picture that shows the same number of equal divisions of the circle.

The complete length represents one whole unit, and it has been cut into three equal parts.

To write the fraction for the blue (shaded) part we follow the basic definition of a fraction.

• The denominator tells how many equal parts the whole is divided into. Here the whole is divided into $$3$$ equal parts, so the denominator is $$3$$.
• The numerator tells how many of those equal parts are taken or shaded. Only one part is coloured blue, so the numerator is $$1$$.

Therefore, the length of the blue line as a fraction of the whole unit is

\[ \frac{1}{3} \]

The question tells us that the whole (the complete line segment) has already been cut into 5 equal parts. That information immediately fixes the denominator of every required fraction:

Total number of equal parts ⇒ $$5$$ ⇒ denominator is $$5$$ for every fraction.

To obtain the numerator we simply count how many of those five parts are coloured blue in each drawing.

1. First box: the blue line covers exactly one of the five parts.
   Numerator = 1, so required fraction is $$\dfrac{1}{5}$$.
2. Second box: the blue line covers two of the five parts.
   Numerator = 2, so required fraction is $$\dfrac{2}{5}$$.
3. Third box: the blue line covers three of the five parts.
   Numerator = 3, so required fraction is $$\dfrac{3}{5}$$.
4. Fourth box: the blue line covers four of the five parts.
   Numerator = 4, so required fraction is $$\dfrac{4}{5}$$.

These four fractions represent the lengths of the blue line segments when the whole is divided into five equal pieces.

When we say that “a unit is divided into 8 equal parts”, we mean that

• the whole (unit) is the denominator, and
• the number of those equal parts that we take, colour or mark is the numerator.

Because the whole is cut into 8 pieces, the denominator in every required fraction is 8.

Each single small part therefore represents the fraction $$\dfrac{1}{8}$$  (one–eighth of the unit).

Let the number of parts chosen be represented by the letter $$n$$. Then the fraction formed is

Writing this fraction for every possible value of $$n$$ from 1 to 8 gives

• 1 part  →  $$\dfrac{1}{8}$$
• 2 parts  →  $$\dfrac{2}{8}$$
• 3 parts  →  $$\dfrac{3}{8}$$
• 4 parts  →  $$\dfrac{4}{8}$$
• 5 parts  →  $$\dfrac{5}{8}$$
• 6 parts  →  $$\dfrac{6}{8}$$
• 7 parts  →  $$\dfrac{7}{8}$$
• 8 parts  →  $$\dfrac{8}{8}=1$$ (the whole unit)

Thus, the appropriate fractions obtained on dividing one whole into eight equal parts are

(You may reduce the fractions like $$\dfrac{2}{8}=\dfrac{1}{4}$$, $$\dfrac{4}{8}=\dfrac{1}{2}$$, etc., if required.)

Goal : show the points that are at distances  $$\frac{1}{10}$$, $$\frac{3}{10}$$ and $$\frac{4}{5}$$ from zero on a single number-line.

1. Draw the basic number line.
   Draw a long horizontal line and put arrows on both ends to show that it continues indefinitely. Pick a convenient point near the middle, mark it 0.

2. Mark one whole unit.
   Choose a suitable length (for example 10 cm) to represent one unit. Measure that length to the right of 0 and mark the end of it as 1.

3. Divide the unit into ten equal parts.
   Since the smallest denominator we need is 10, use a ruler to split the segment 0 – 1 into 10 equal sub-segments. Label the division points in order:

   0, $$\frac{1}{10}$$, $$\frac{2}{10}$$, $$\frac{3}{10}$$, $$\frac{4}{10}$$, $$\frac{5}{10}=\frac{1}{2}$$, $$\frac{6}{10}$$, $$\frac{7}{10}$$, $$\frac{8}{10}$$, $$\frac{9}{10}$$, 1.

4. Locate $$\frac{1}{10}$$.
   The first tick after 0 is the point whose distance from 0 is $$\frac{1}{10}$$ of the unit. Draw a small vertical line above this point and label it $$\frac{1}{10}$$.

5. Locate $$\frac{3}{10}$$.
   Count three of the small equal parts from 0. The third tick is the required point. Draw a vertical line there and label it $$\frac{3}{10}$$.

6. Locate $$\frac{4}{5}$$.
   Notice that $$\frac{4}{5}=\frac{8}{10}$$. Therefore move eight of the small parts from 0 (or simply read the tick already labelled $$\frac{8}{10}$$). Draw a vertical line above this point and write either $$\frac{4}{5}$$ or $$\frac{8}{10}$$.

   Alternatively, you could first divide 0–1 into five equal parts and count four of them; the point obtained will coincide with the one just described, because

   \[\frac{4}{5}=\frac{8}{10}.\]

Result : the three required lengths are now clearly shown on the number line at their correct positions.

Practical tip : use a sharp pencil and a ruler so that each of the ten sub-segments is truly equal; the accuracy of the fractions depends on it.

Step 1 – Choosing the fractions
To make the plotting simple, let us take all the fractions with the same denominator 5. We choose

• $$\frac{1}{5}$$
• $$\frac{3}{5}$$
• $$\frac{4}{5}$$
• $$\frac{6}{5}$$
• $$\frac{7}{5}$$

The first three are proper fractions (their values lie between 0 and 1). The last two are improper fractions (their values are greater than 1).

Step 2 – Deciding the portion of the number line to draw
Because the largest numerator is 7 and the common denominator is 5, the biggest value is $$\frac{7}{5}=1\frac{2}{5}$$, which lies between 1 and 2. So we need a number line from 0 to a little beyond 2.

Step 3 – Making equal parts
(i) Draw a horizontal line and mark a point as 0.
(ii) Choose a convenient length to the right and mark the point 1. This segment $$0\rightarrow1$$ is the unit length.
(iii) Since the denominator is 5, divide the segment from 0 to 1 into five exactly equal parts. Each small part represents $$\frac{1}{5}$$.
(iv) Extend the line further to the right, mark 2 (one more unit), and again split the segment $$1\rightarrow2$$ into five equal parts. These will locate $$\frac{6}{5},\frac{7}{5},\dots,\frac{10}{5}=2$$.

Step 4 – Locating each chosen fraction

Step 5 – Drawing the final diagram
Draw the number line from 0 to 2. Mark the ten equal divisions (two groups of five). Label the points 0, $$\tfrac{1}{5},\tfrac{2}{5},\tfrac{3}{5},\tfrac{4}{5},1,\tfrac{6}{5},\tfrac{7}{5},\tfrac{8}{5},\tfrac{9}{5},2$$. Finally, circle or colour the five fractions you selected so that they stand out.

Thus the fractions $$\frac{1}{5},\,\frac{3}{5},\,\frac{4}{5},\,\frac{6}{5}\text{ and }\frac{7}{5}$$ are correctly placed on the number line.

Step 1 — Recall what a fraction is
A fraction is written as $$\dfrac{a}{b}$$ where $$a$$ and $$b$$ are whole numbers and $$b \neq 0$$. It represents “a parts out of b equal parts”.
A fraction is between 0 and 1 if its value is greater than 0 but less than 1:

\[0 < \dfrac{a}{b} < 1\]

This inequality is true exactly when the numerator is smaller than the denominator, that is, when $$0 < a < b$$.

Step 2 — Make a short list of such fractions
Writing only a few already shows many possibilities: $$\dfrac12,\;\dfrac13,\;\dfrac23,\;\dfrac14,\;\dfrac34,\;\dfrac15,\;\dfrac25,\;\dfrac35,\;\dfrac45,\ldots$$ They all satisfy $$0<\dfrac{a}{b}<1$$.

Step 3 — Produce still more fractions in a systematic way
Pick any whole number $$n$$ with $$n > 1$$. Now look at the set

\[\dfrac{1}{n},\;\dfrac{2}{n},\;\dfrac{3}{n},\;\ldots,\;\dfrac{n-1}{n}\]

For every fraction in this row the numerator is smaller than the denominator, so each one lies between 0 and 1. How many such fractions did we just obtain? Exactly $$n-1$$ of them.

Step 4 — Argue that the process never ends
Nothing stops us from taking larger and larger values of $$n$$: 2, 3, 10, 100, 1 000, and so on. Every new value of $$n$$ gives us $$n-1$$ new fractions between 0 and 1. Because there is no “largest” natural number, we can keep increasing $$n$$ forever and keep creating new fractions forever.

Step 5 — Draw the conclusion
Since we can generate an unlimited supply of different fractions between 0 and 1, we say

\[\text{The number of fractions between }0\text{ and }1\text{ is infinite.}\]

In everyday words: there are infinitely many fractions between 0 and 1.

Step 1 : Identify the size of one equal part

The distance from the mark $$0$$ to the mark $$1$$ is said to be 1 unit.

This distance is divided into two equal parts. Therefore,

length of each equal part  = $$\dfrac{1}{2}$$ unit.

Step 2 : Find how many such parts make the blue line

• The blue line starts at $$0$$ and stops at the first division mark.
• That first division mark is one part away from $$0$$.
• Hence the blue line covers exactly one part.

blue line  = $$1 \times \dfrac{1}{2}=\dfrac{1}{2}$$ unit. (Already stated in the question.)

Step 3 : Count the parts that make the black line

• Starting again from $$0$$, read each tick of length $$\dfrac{1}{2}$$ unit.
• The black line stretches past $$1$$ and stops at the next tick.
• Let us list the ticks it crosses:
  1. from $$0$$ to $$\tfrac{1}{2}$$  →  first part,
  2. from $$\tfrac{1}{2}$$ to $$1$$  →  second part,
  3. from $$1$$ to $$1\tfrac{1}{2}$$  →  third part.

Altogether it contains three equal parts.

Step 4 : Express this length as a fraction

Each part is $$\dfrac{1}{2}$$ unit long and there are $$3$$ such parts.

\[ \text{length of black line} =3\times \dfrac{1}{2} =\dfrac{3}{2}\;\text{units}. \]

The fraction to be written in the box is therefore $$\dfrac{3}{2}$$.

For every box the complete strip has been marked into equal pieces of the same size. Hence :

• the denominator = total number of equal pieces in the strip,
• the numerator = number of those pieces that have been covered by the thick (black) line.

The textbook shows four identical strips. Each strip is divided into 4 equal parts, so for all the boxes the denominator is $$4$$.

1. First box

   Black line covers 1 part.

   Fraction for its length = $$\dfrac{1}{4}$$.

2. Second box

   Black line covers 2 parts.

   Fraction for its length = $$\dfrac{2}{4}$$; this can also be written in lowest terms as $$\dfrac{1}{2}$$.

3. Third box

   Black line covers 3 parts.

   Fraction for its length = $$\dfrac{3}{4}$$.

4. Fourth box

   Black line covers 4 parts (the whole strip).

   Fraction for its length = $$\dfrac{4}{4}$$, i.e. $$1$$.

Thus the required fractions, in the order in which the strips are printed in the book, are:

$$\dfrac14,\;\dfrac24\;(=\dfrac12),\;\dfrac34,\;\dfrac{4}{4}\;(=1).$$

We have the fraction $$\frac{7}{2}$$.

A whole unit corresponds to a fraction whose numerator and denominator are equal, that is $$\frac{2}{2}=1$$. Hence, to find how many whole units fit into $$\frac{7}{2}$$ we divide the numerator by the denominator:

$$7 \div 2 = 3$$ with remainder $$1$$.

This calculation tells us that

$$7 = 2 \times 3 + 1.$$

Substituting this back into the fraction gives

$$\frac{7}{2}=\frac{2 \times 3 + 1}{2}=\frac{2 \times 3}{2}+\frac{1}{2}=3+\frac{1}{2}.$$

Expressed as a mixed number, we have

\[ \frac{7}{2}=3\frac{1}{2} \]

The whole-number part is $$3$$. Therefore, $$\frac{7}{2}$$ contains 3 complete (whole) units.

Step 1 · Recall what “whole units” means
In an improper fraction like $$\frac{4}{3}$$ or $$\frac{7}{3}$$ the numerator is larger than the denominator. Each complete set of 3 thirds (because the denominator is 3) makes one whole.

Step 2 · Count wholes in $$\frac{4}{3}$$

• Method A: Division
  Divide numerator by denominator: $$4 \div 3 = 1$$ with remainder $$1$$.
  So $$\frac{4}{3} = 1\,\text{whole} + \frac{1}{3}$$.
• Method B: Repeated subtraction
  • Take away one whole (that is $$\frac{3}{3}$$): $$\frac{4}{3} - \frac{3}{3} = \frac{1}{3}$$.
  • We managed to remove $$1$$ whole before we ran out of thirds.

Therefore, there is 1 whole unit in $$\frac{4}{3}$$.

Step 3 · Count wholes in $$\frac{7}{3}$$

• Division again: $$7 \div 3 = 2$$ with remainder $$1$$.
  Hence $$\frac{7}{3} = 2\,\text{wholes} + \frac{1}{3}$$.
• Or repeated subtraction
  • First whole: $$\frac{7}{3} - \frac{3}{3} = \frac{4}{3}$$.
  • Second whole: $$\frac{4}{3} - \frac{3}{3} = \frac{1}{3}$$.
  • Now only $$\frac{1}{3}$$ is left, so we have removed $$2$$ wholes.

Therefore, there are 2 whole units in $$\frac{7}{3}$$.

Step 4 · Summary
Each quotient gives the number of complete wholes:

• $$\frac{4}{3} = 1\,\frac{1}{3} \Rightarrow 1$$ whole.
• $$\frac{7}{3} = 2\,\frac{1}{3} \Rightarrow 2$$ wholes.

We first recall two important words:

• Improper fraction: a fraction whose numerator is greater than or equal to its denominator, e.g. $$\tfrac{9}{4},\;\tfrac{15}{8}$$.
• Mixed number: a whole-number part together with a proper-fraction part, e.g. $$2\tfrac{1}{3},\;5\tfrac{4}{7}$$.

The question asks whether every fraction greater than 1 (every improper fraction whose numerator is strictly larger than the denominator) can always be written as a mixed number.

Step 1 — Take an arbitrary improper fraction. Write it as

Step 2 — Divide $$N$$ by $$D$$ as whole numbers. By the usual division algorithm there exist whole numbers $$q$$ (the quotient) and $$r$$ (the remainder) such that

The key fact is that the remainder $$r$$ is strictly less than the divisor $$D$$.

Step 3 — Divide both sides of the equation by $$D$$.

Because $$0 \le r \lt D$$, the fraction $$\dfrac{r}{D}$$ is a proper fraction, and $$q$$ is a whole number. The right-hand side is therefore the mixed number

Step 4 — Conclude. Our choice of $$N$$ and $$D$$ was completely general, so this construction works for every improper fraction.

• Yes — every fraction greater than 1 can be written as a mixed number.

Step 1 : Write the two fractions.

The two given lengths are $$\frac{1}{2}$$ and $$\frac{3}{6}$$.

Step 2 : Find a common denominator.

The denominators are 2 and 6. Their least common multiple (LCM) is 6.

Step 3 : Convert $$\frac{1}{2}$$ to an equivalent fraction with denominator 6.

Multiply both numerator and denominator by 3 (the same number) so that the denominator becomes 6:

Step 4 : Compare the like fractions.

Now the first length is $$\frac{3}{6}$$ and the second length is also $$\frac{3}{6}$$. Since the numerators as well as the denominators are identical, the two fractions are equal.

Alternate check (cross-multiplication)

We can also compare the fractions without changing the denominators:

Calculate the cross products:

Because both products are equal, the two fractions are equal.

Conclusion

The lengths $$\frac{1}{2}$$ and $$\frac{3}{6}$$ represent exactly the same amount. They are equal.

Objective : To check whether the two given fractions $$\frac{2}{3}$$ and $$\frac{4}{6}$$ are equivalent.

Idea 1 – Reducing to lowest terms

1. Find the greatest common divisor (GCD) of the numerator and the denominator of $$\frac{4}{6}$$.
     • Factors of $$4$$: $$1,\;2,\;4$$
     • Factors of $$6$$: $$1,\;2,\;3,\;6$$
     • The largest common factor is $$2$$, so $$\gcd(4,6)=2$$.
2. Divide both numerator and denominator by this GCD:
   $$\frac{4}{6}=\frac{4\div2}{6\div2}=\frac{2}{3}$$.

Since the simplified form of $$\frac{4}{6}$$ is exactly $$\frac{2}{3}$$, the two fractions represent the same number. Hence they are equivalent.

(Optional cross-multiplication check)
Cross-multiply the numerators and denominators:
$$2\times6=12 \quad \text{and} \quad 3\times4 =12$$.
Because both products are equal, the fractions are equivalent.

Conclusion : $$\frac{2}{3}$$ and $$\frac{4}{6}$$ are equivalent fractions because they have the same simplest form (or give equal products when cross-multiplied).

Step 1 : Understand what is being asked.

We want to know how many pieces, each of length $$\frac{1}{6}$$, are needed to put together a total length of $$\frac{1}{2}$$.

Step 2 : Translate the question into a division problem.

"How many pieces of length $$\frac{1}{6}$$ make $$\frac{1}{2}$$?" means:

Number of pieces = $$\frac{1}{2} \div \frac{1}{6}$$.

Step 3 : Recall how to divide one fraction by another.

• Keep the first fraction as it is.
• Change the division sign to multiplication.
• Flip (take the reciprocal of) the second fraction.

Step 4 : Perform the division.

Change to multiplication and flip the divisor:

Now multiply the numerators and the denominators:

$$\frac{1}{2} \times \frac{6}{1} = \frac{1\times6}{2\times1} = \frac{6}{2}$$

Simplify $$\frac{6}{2}$$:

$$\frac{6}{2} = 3$$

Step 5 : State the result.

Therefore, 3 pieces of length $$\frac{1}{6}$$ will make a total length of $$\frac{1}{2}$$.

Given: Each small piece has length $$\frac{1}{6}$$ unit.
We want to obtain a total length of $$\frac{1}{3}$$ unit by joining such equal pieces.

Let the required number of pieces be n.

When we join these pieces end-to-end, their lengths add. Therefore

$$ n \times \frac{1}{6} = \frac{1}{3}. $$

To find n, we have to divide the total length by the length of one piece:

$$ n = \frac{\frac{1}{3}}{\frac{1}{6}}. $$

Recall that dividing by a fraction is the same as multiplying by its reciprocal:

$$ n = \frac{1}{3} \times \frac{6}{1}. $$

Multiply the numerators and the denominators:

$$ n = \frac{1 \times 6}{3 \times 1} = \frac{6}{3}. $$

Simplify the fraction by dividing numerator and denominator by 3:

$$ n = \frac{6 \div 3}{3 \div 3} = \frac{2}{1} = 2. $$

Therefore, 2 pieces of length $$\frac{1}{6}$$ are needed to make a length of $$\frac{1}{3}$$.

Step 1 – Recall the meaning of “equivalent fractions”

Two or more fractions are called equivalent if they represent the same part of a whole. Mathematically, fractions $$\dfrac{a}{b}$$ and $$\dfrac{c}{d}$$ are equivalent if their cross–products are equal:

\[a \times d = b \times c\]

Another very common method at Class 6 is to reduce every fraction to its lowest (simplest) form. If, after reducing, all the fractions become exactly the same, then the original fractions are equivalent.

Step 2 – Reduce each given fraction to its simplest form

• For $$\dfrac{3}{6}$$, the greatest common divisor (gcd) of 3 and 6 is 3.

  Divide the numerator and the denominator by 3:

  $$\dfrac{3}{6}=\dfrac{3\div3}{6\div3}=\dfrac{1}{2}$$

• For $$\dfrac{4}{8}$$, the gcd of 4 and 8 is 4.

  Divide the numerator and the denominator by 4:

  $$\dfrac{4}{8}=\dfrac{4\div4}{8\div4}=\dfrac{1}{2}$$

• For $$\dfrac{5}{10}$$, the gcd of 5 and 10 is 5.

  Divide the numerator and the denominator by 5:

  $$\dfrac{5}{10}=\dfrac{5\div5}{10\div5}=\dfrac{1}{2}$$

Step 3 – Compare the simplest forms

After simplification, every fraction has become $$\dfrac{1}{2}$$.

Conclusion

Because all three fractions have the same simplest form, they are equivalent fractions. Each of them represents one-half of a whole.

To obtain equivalent fractions we multiply (or divide) the numerator and the denominator of a fraction by the same non–zero whole number.

Given fraction:

$$\frac{2}{6}$$

1. Multiply both terms by 2:

   Numerator: $$2 \times 2 = 4$$
   Denominator: $$6 \times 2 = 12$$

   Hence one equivalent fraction is $$\frac{4}{12}$$.

2. Multiply both terms by 3:

   Numerator: $$2 \times 3 = 6$$
   Denominator: $$6 \times 3 = 18$$

   Thus another equivalent fraction is $$\frac{6}{18}$$.

Therefore, two fractions equivalent to $$\frac{2}{6}$$ are $$\frac{4}{12}$$ and $$\frac{6}{18}$$.

To obtain fractions that are equivalent to $$\frac{4}{6}$$ we either

1. divide numerator and denominator by a common factor, or

2. multiply numerator and denominator by the same non-zero whole number.

Step 1 — Reduce to lowest terms

The greatest common divisor of $$4$$ and $$6$$ is $$2$$, so

\[\frac{4}{6}=\frac{4\div2}{6\div2}=\frac{2}{3}\]

Thus $$\frac{2}{3}$$ is the simplest form of $$\frac{4}{6}$$.

Step 2 — Generate more equivalent fractions

For any whole number $$n\,(n\ge1)$$ we have

$$\frac{4}{6}=\frac{4\times n}{6\times n}$$

Choosing successive natural numbers for $$n$$ gives:

Each fraction in the last column has the same value as $$\frac{4}{6}$$ and therefore as $$\frac{2}{3}$$.

Result

So we can write, for example,

$$\frac{4}{6}=\frac{2}{3}=\frac{8}{12}=\frac{12}{18}=\frac{16}{24}=\frac{20}{30}=\cdots$$

Since $$n$$ can be any natural number, there are infinitely many such equivalent fractions.

Step 1 – Understand the situation
There are 3 whole rotis that must be shared equally among 4 children.

Step 2 – Cutting the rotis (describe the picture)
To make four equal shares, cut every roti into 4 equal pieces (quarters). In your notebook draw:

• Three big circles to represent the three rotis.
• In each circle draw two diameters at right angles so that each circle is divided into 4 identical sectors.
• Shade one sector of every roti for Child A, another sector of every roti for Child B, and so on, until the four children each have one sector from each roti.

Step 3 – Fraction each child gets
Each child’s share = one quarter from Roti 1 + one quarter from Roti 2 + one quarter from Roti 3:
$$\tfrac14 + \tfrac14 + \tfrac14 = \tfrac34.$$
So every child receives $$\tfrac34$$ of a whole roti.

Step 4 – Write the arithmetic facts

Conclusion
Dividing 3 rotis equally among 4 children gives each child $$\tfrac34$$ of a roti. The picture with three circles cut into quarters, and the division, addition and multiplication statements, all describe the same sharing.

Step 1 — Understand the sharing.

We have 2 whole rotis and 4 children. “Sharing equally” means every child must receive the same amount.

Step 2 — Draw the picture.

• Draw two identical circles to represent the two rotis.
• Draw two diameters at right angles in each circle so that every roti is cut into 4 equal wedges (quarters).
• Colour the wedges for the four children A, B, C, D so that each child gets 2 wedges — one from each roti:
  • Child A → shade any 2 wedges, one from each roti.
  • Child B → shade another 2 wedges, one from each roti.
  • Child C → shade another 2 wedges, one from each roti.
  • Child D → shade the remaining 2 wedges, one from each roti.
  So each child gets 2 of the 8 equal wedges.

Step 3 — Find each child's fraction.

Total number of equal wedges = 4 wedges/roti $$\times$$ 2 rotis = 8 wedges.

Wedges per child = 8 $$\div$$ 4 = 2 wedges.

So one child's share is

Step 4 — Division fact.

Two whole rotis are shared by 4 children:

Step 5 — Repeated-addition fact.

Adding the shares of all four children gives back the original quantity:

Step 6 — Multiplication facts.

Repeated addition of the same fraction is multiplication, so

Conclusion. When 2 rotis are divided equally among 4 children, every child gets $$\dfrac{1}{2}$$ of a roti.

Step 1 : Understand the situation
There are $$2$$ whole cakes, and they have to be shared equally among $$5$$ children (Anil and four others).

Step 2 : Write the division that represents the sharing
To find the amount of cake each child gets, divide the total number of cakes by the number of children:

$$\text{cake for one child} = \dfrac{\text{total cakes}}{\text{number of children}} = \dfrac{2}{5}.$$

Step 3 : Interpret the fraction
The fraction $$\dfrac{2}{5}$$ means “two fifths.” One whole cake is cut into $$5$$ equal parts, and each part is called “one fifth.” Here we have $$2$$ such fifths.

Step 4 : Conclude
Each child, including Anil, receives two fifths of a cake.

Key result:

\[ \dfrac{2}{5}\;\text{cake} \]

Step 1 — Same number of children, more units to share.

Suppose $$n$$ children sit together and share identical bars of chocolate.

• Case A: they share 1 bar. Each child's share is $$\dfrac{1}{n}$$.
• Case B: they share 2 bars. Each child's share is $$\dfrac{2}{n}$$.

With the same $$n$$ children but more bars to share, every child clearly gets more. So

The same idea works for any common denominator: when the denominator is the same, the fraction with the larger numerator is the larger fraction.

Step 2 — Apply this to the given pairs with equal denominators.

• $$\dfrac{1}{5}$$ and $$\dfrac{2}{5}$$ have the same denominator 5 (the same five children). Two bars give each child more than one bar does, so

• $$\dfrac{3}{7}$$ and $$\dfrac{4}{7}$$ have the same denominator 7 (the same seven children). Four bars give each child more than three bars do, so

Step 3 — A pair with different denominators: $$\dfrac{1}{2}$$ and $$\dfrac{5}{8}$$.

First make the denominators the same. Convert $$\dfrac{1}{2}$$ to eighths:

Both fractions now share the denominator 8 (the same eight equal pieces). Since $$5 \gt 4$$,

Conclusion. Keeping the number of children (denominator) fixed and increasing the number of units shared (numerator) increases each child's share. When the denominators differ, first rewrite the fractions with a common denominator and then compare the numerators in the same way.

Step 1 : Recall the three ways in which fractions can be presented
When we are asked to arrange or compare several fractions, normally three different situations arise.

• All the fractions already have the same denominator (for example, $$\frac{1}{8},\,\frac{3}{8},\,\frac{7}{8}$$).
• All the fractions already have the same numerator (for example, $$\frac{3}{2},\,\frac{3}{5},\,\frac{3}{11}$$).
• The fractions have different numerators and different denominators (for example, $$\frac{2}{5},\,\frac{3}{8},\,\frac{5}{6}$$).

Step 2 : Examine how much work is needed in each situation

• Fractions with the same denominator
  Because the pieces are already of the same size, we decide the order simply by looking at the numerators. No extra calculation is required. For instance
  $$\frac{1}{8}<\frac{3}{8}<\frac{7}{8}$$
• Fractions with the same numerator
  The number of equal pieces is the same in every fraction. A larger denominator means each piece is smaller, so the fraction is smaller. Thus we also finish in one direct observation. For example
  $$\frac{3}{11}<\frac{3}{7}<\frac{3}{4}<\frac{3}{2}$$
• Fractions with different numerators and denominators
  Here we must first convert every fraction to an equivalent fraction with a common denominator (or change each to a decimal). Only afterwards can we start comparing. This takes two or three extra steps.

Step 3 : Conclude which groups were easiest and why
The groups that were easiest are the first two:

• Fractions that already have the same denominator.
• Fractions that already have the same numerator.

They are easier because no conversion to like fractions is necessary. A single look at the numerators (when denominators are equal) or at the denominators (when numerators are equal) is sufficient to tell which fraction is larger or smaller.

Goal: Add $$\tfrac{4}{7}$$ and $$\tfrac{6}{7}$$ with the help of a number line and confirm that the result matches the usual fraction-addition rule.

1. Draw a suitable number line.
   Mark the whole numbers 0, 1 and 2, because the result may be bigger than 1. Between every pair of whole numbers, divide the segment into 7 equal parts (since the denominator is 7). Label the points $$0,\;\tfrac17,\;\tfrac27,\;\tfrac37,\;\tfrac47,\;\tfrac57,\;\tfrac67,\;1=\tfrac77,$$
   then $$1\tfrac17=\tfrac87,\;1\tfrac27=\tfrac97,\;1\tfrac37=\tfrac{10}{7},\;\dots$$

2. Show the first jump of $$\tfrac47$$.
   Start at 0 and move right four divisions (each division is $$\tfrac17$$). You land at the point labelled $$\tfrac47$$.

3. Show the second jump of $$\tfrac67$$.
   From $$\tfrac47$$, move six more equal divisions to the right. Count:

   • 1st step: $$\tfrac57$$
   • 2nd step: $$\tfrac67$$
   • 3rd step: $$\tfrac77 = 1$$ (you have reached 1 but keep counting)
   • 4th step: $$\tfrac87 = 1\tfrac17$$
   • 5th step: $$\tfrac97 = 1\tfrac27$$
   • 6th step: $$\tfrac{10}{7} = 1\tfrac37$$

   The arrow finally stops at $$\tfrac{10}{7}$$.

4. Read the answer from the number line.
   The endpoint is $$\tfrac{10}{7}$$, which is the mixed number $$1\tfrac37$$.

5. Check by the standard rule.
   Because the denominators are the same (7), add the numerators: \[\frac47 + \frac67 = \frac{4+6}{7} = \frac{10}{7} = 1\tfrac37.\] The number-line result exactly matches the usual calculation.

Conclusion: Using a number line or using the fractional addition rule gives the same answer, $$\tfrac{10}{7}$$ (that is, $$1\tfrac37$$).

Rahim has two quantities of paint:

• Yellow paint: $$\frac{2}{3}$$ litre
• Blue paint: $$\frac{3}{4}$$ litre

To find the total volume of green paint, we must add the two fractions.

Step 1 – Express both fractions with a common denominator

The denominators are 3 and 4. Their least common multiple (LCM) is 12.

For the yellow paint:

$$\frac{2}{3}=\frac{2\times4}{3\times4}=\frac{8}{12}$$

For the blue paint:

$$\frac{3}{4}=\frac{3\times3}{4\times3}=\frac{9}{12}$$

Step 2 – Add the fractions

Now that the two fractions have the same denominator, add the numerators:

$$\frac{8}{12}+\frac{9}{12}=\frac{8+9}{12}=\frac{17}{12}$$

Step 3 – Convert to a mixed fraction (optional)

Since 12 goes into 17 once with 5 left over,

$$\frac{17}{12}=1\frac{5}{12}$$

Therefore, the total volume of green paint Rahim makes is

\[\frac{17}{12}\text{ litres } = 1\frac{5}{12}\text{ litres}\]

Given data

• Geeta’s lace length  = $$\frac{2}{5}$$ m
• Shamim’s lace length = $$\frac{3}{4}$$ m
• Perimeter of the table­-cloth  = $$1$$ m

Step 1  — Express both fractions with a common denominator

The denominators are $$5$$ and $$4$$. Their least common multiple (LCM) is $$20$$.

• Convert $$\frac{2}{5}$$ to twentieths:
  $$\frac{2}{5}=\frac{2\times4}{5\times4}=\frac{8}{20}$$
• Convert $$\frac{3}{4}$$ to twentieths:
  $$\frac{3}{4}=\frac{3\times5}{4\times5}=\frac{15}{20}$$

Step 2  — Add the two lengths

Add the numerators (the denominators are now the same):

$$\frac{8}{20}+\frac{15}{20}=\frac{8+15}{20}=\frac{23}{20}$$

The total length of lace bought is therefore

Step 3  — Convert to a mixed fraction

$$\frac{23}{20}=1+\frac{3}{20}=1\frac{3}{20}\text{ m}$$

Step 4  — Compare with the required length

The border needs $$1$$ metre. The lace available is $$1\frac{3}{20}$$ m.

Because $$1\frac{3}{20}\text{ m}>1\text{ m}$$, the lace is sufficient. In fact, the extra lace left over is

$$1\frac{3}{20}-1=\frac{3}{20}\text{ metre}.$$

Conclusion

• Total lace bought  = $$1\frac{3}{20}$$ m.
• The lace will cover the whole 1 m border, and $$\frac{3}{20}$$ m of lace will remain.

Concept recap: To subtract two fractions, they must have the same denominator. Then we only subtract the numerators and keep that common denominator.

Step 1 – Check the denominators
Both fractions are $$\frac{5}{8}$$ and $$\frac{3}{8}$$. Their denominators are the same (8), so they are like fractions. No change is needed.

Step 2 – Subtract the numerators
Keep the denominator 8 and subtract the numerators 5 and 3:

$$\frac{5}{8} - \frac{3}{8} = \frac{5 - 3}{8}$$

Compute the difference in the numerator:

$$5 - 3 = 2$$

Therefore:

$$\frac{5 - 3}{8} = \frac{2}{8}$$

Step 3 – Simplify the fraction (if possible)
The numerator 2 and the denominator 8 have a common factor of 2. Divide both by 2:

$$\frac{2 \div 2}{8 \div 2} = \frac{1}{4}$$

Final simplified result:

\[ \frac{5}{8} - \frac{3}{8} = \frac{2}{8} = \frac{1}{4} \]

We need to evaluate the difference $$\frac{7}{9} - \frac{5}{9}$$.

Step 1: Check the denominators.
Both fractions have the same denominator $$9$$, so they already refer to parts of equal size.

Step 2: Subtract the numerators while keeping the denominator unchanged:

$$7 - 5 = 2$$

Hence the fraction becomes

\[\frac{7}{9} - \frac{5}{9} = \frac{2}{9}\]

Step 3: Simplify if possible.
The numerator $$2$$ and the denominator $$9$$ have no common factor other than $$1$$, so $$\frac{2}{9}$$ is already in simplest form.

Therefore, the value of $$\frac{7}{9} - \frac{5}{9}$$ is $$\frac{2}{9}$$.

We need to subtract the two fractions $$\frac{10}{27}$$ and $$\frac{1}{27}$$.

Since both fractions already have the same denominator (27), keep that denominator and subtract their numerators:

$$\frac{10}{27}-\frac{1}{27}=\frac{10-1}{27}$$

Work out the subtraction in the numerator:

$$\frac{10-1}{27}=\frac{9}{27}$$

Next, reduce $$\frac{9}{27}$$ to lowest terms. The greatest common divisor of 9 and 27 is 9. Divide both numerator and denominator by 9:

$$\frac{9}{27}=\frac{9\div9}{27\div9}=\frac{1}{3}$$

Therefore,

\[ \frac{10}{27}-\frac{1}{27}=\frac{1}{3} \]

Let the three different fractional units be

We must find whole numbers $$a \lt b \lt c$$ satisfying equation (1):

Step 1 — Bound the smallest denominator $$a$$.

• Each fraction is positive and their sum is exactly 1, so none of them can be 1 by itself. Hence $$a \ge 2$$.
• Because $$a \lt b \lt c$$, we have $$\dfrac{1}{a} \gt \dfrac{1}{b} \gt \dfrac{1}{c}$$, so $$\dfrac{1}{a}$$ is the largest of the three. Therefore

which gives $$a \le 3$$. So $$a$$ is either 2 or 3.

Step 2 — Show $$a=3$$ is impossible.

Suppose $$a=3$$. Then $$b \gt 3$$ forces $$b \ge 4$$, and $$c \gt b \ge 4$$ forces $$c \ge 5$$. Hence

But equation (1) with $$a=3$$ requires

Now $$\dfrac{9}{20}=\dfrac{27}{60}$$ and $$\dfrac{27}{60} \lt \dfrac{40}{60}$$, so the largest possible value of $$\dfrac{1}{b}+\dfrac{1}{c}$$ is already smaller than what is required. Hence $$a=3$$ gives no solution.

Step 3 — Solve the only remaining case: $$a=2$$.

With $$a=2$$, equation (1) becomes

Since $$b \gt a=2$$, we have $$b \ge 3$$. Try consecutive values of $$b$$:

1. $$b=3$$: $$\dfrac{1}{c}=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}\Rightarrow c=6$$. Since $$2 \lt 3 \lt 6$$, this set works.
2. $$b=4$$: $$\dfrac{1}{c}=\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}\Rightarrow c=4$$. But then $$b=c=4$$, so the three fractions would not all be different. Reject.
3. $$b \ge 5$$: Then $$\dfrac{1}{b}\le\dfrac{1}{5}$$, so by (2)
   $$\dfrac{1}{c}=\dfrac{1}{2}-\dfrac{1}{b}\ge\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10} \gt \dfrac{1}{5}\ge\dfrac{1}{b}.$$ This gives $$\dfrac{1}{c} \gt \dfrac{1}{b}$$, i.e. $$c \lt b$$, contradicting $$b \lt c$$. So no $$b \ge 5$$ works.

Hence the only admissible triple (with $$a \lt b \lt c$$) is

Step 4 — Verify the sum.

The L.C.M. of 2, 3, 6 is 6, so

Conclusion

The only way to write 1 as a sum of three different unit fractions is

Every other solution is just a rearrangement of these three.

Step 1 – Begin with the biggest possible unit fraction.
Every unit fraction is at most $$1\!/2$$ (because $$1\!/1=1$$ is already the whole sum).
Therefore the first term must be $$\tfrac12$$; otherwise the remaining three fractions could never reach the total 1.

Step 2 – Choose the second fraction.
If we now picked any unit fraction larger than $$\tfrac13$$, the two together would exceed 1. Hence there are only two sensible choices for the second fraction:

• Case A : take $$\tfrac13$$, or
• Case B : take $$\tfrac14$$.

We treat the two cases separately.

Case A : second fraction $$=\tfrac13$$

After adding $$\tfrac12+\tfrac13$$ we have used $$\tfrac56$$ of the whole. The part still left is

$$ 1-\frac12-\frac13 = \frac16. $$

So the last two fractions must satisfy

$$ \frac1a+\frac1b = \frac16, \qquad a,b > 3,\; a\neq b. $$

Multiply by $$6ab$$ to clear the denominators:

$$ 6b + 6a = ab. $$

Bring everything to one side and complete the rectangle:

$$ ab - 6a - 6b = 0 \;\Longrightarrow\; ab - 6a - 6b + 36 = 36 \;\Longrightarrow\; (a-6)(b-6)=36. $$

The product 36 has only a few factor-pairs. Remember also that $$a>b>6$$ (otherwise we repeat a denominator or get a negative fraction). Listing the pairs gives

Each pair immediately produces a complete solution:

• $$\tfrac12+\tfrac13+\tfrac17+\tfrac1{42}=1$$
• $$\tfrac12+\tfrac13+\tfrac18+\tfrac1{24}=1$$
• $$\tfrac12+\tfrac13+\tfrac19+\tfrac1{18}=1$$
• $$\tfrac12+\tfrac13+\tfrac1{10}+\tfrac1{15}=1$$

Case B : second fraction $$=\tfrac14$$

This time the two fractions $$\tfrac12+\tfrac14$$ make $$\tfrac34$$, so the balance to be filled is

$$ 1-\frac12-\frac14 = \frac14. $$

We want

$$ \frac1a+\frac1b = \frac14, \qquad a,b > 4,\; a\neq b. $$

Clear the denominators (multiply by $$4ab$$):

$$ 4b+4a = ab \;\Longrightarrow\; ab-4a-4b=0 \;\Longrightarrow\; (a-4)(b-4)=16. $$

Now 16 = $$1\times16$$ or $$2\times8$$. Converting these gives:

So we obtain two further solutions:

• $$\tfrac12+\tfrac14+\tfrac15+\tfrac1{20}=1$$
• $$\tfrac12+\tfrac14+\tfrac16+\tfrac1{12}=1$$

Step 3 – Collect the results.
Adding the four solutions from Case A and the two from Case B we get exactly six different ways to write 1 as a sum of four distinct unit fractions:

1. $$\displaystyle \frac12+\frac13+\frac17+\frac1{42}=1$$
2. $$\displaystyle \frac12+\frac13+\frac18+\frac1{24}=1$$
3. $$\displaystyle \frac12+\frac13+\frac19+\frac1{18}=1$$
4. $$\displaystyle \frac12+\frac13+\frac1{10}+\frac1{15}=1$$
5. $$\displaystyle \frac12+\frac14+\frac15+\frac1{20}=1$$
6. $$\displaystyle \frac12+\frac14+\frac16+\frac1{12}=1$$

No other combination is possible, so the exercise is complete.

Answer found!