Chapter 8: Working with Fractions

Step 1  Write the whole number as a fraction with denominator 1.
$$7 = \frac{7}{1}$$

Step 2  Multiply the two fractions (numerator × numerator and denominator × denominator).
$$\frac{7}{1} \times \frac{3}{5} = \frac{7 \times 3}{1 \times 5} = \frac{21}{5}$$

Step 3  Convert the improper fraction to a mixed fraction.
Divide 21 by 5.
Quotient = 4, Remainder = 1.
Hence $$\frac{21}{5}=4\frac{1}{5}$$.

Step 1  Write the whole number 4 as a fraction.
$$4 = \frac{4}{1}$$

Step 2  Multiply the fractions.
$$\frac{4}{1} \times \frac{1}{3} = \frac{4 \times 1}{1 \times 3} = \frac{4}{3}$$

Step 3  Convert to a mixed fraction.
Divide 4 by 3.
Quotient = 1, Remainder = 1.
Therefore $$\frac{4}{3}=1\frac{1}{3}$$.

Step 1  Write the whole number 6 as a fraction.
$$6 = \frac{6}{1}$$

Step 2  Multiply the fractions.
$$\frac{9}{7} \times \frac{6}{1} = \frac{9 \times 6}{7 \times 1} = \frac{54}{7}$$

Step 3  Convert to a mixed fraction.
Divide 54 by 7.
Quotient = 7, Remainder = 5.
Thus $$\frac{54}{7}=7\frac{5}{7}$$.

Step 1  Express 6 as a fraction.
$$6 = \frac{6}{1}$$

Step 2  Multiply the fractions.
$$\frac{13}{11} \times \frac{6}{1} = \frac{13 \times 6}{11 \times 1} = \frac{78}{11}$$

Step 3  Convert to a mixed fraction.
Divide 78 by 11.
Quotient = 7, Remainder = 1.
Hence $$\frac{78}{11}=7\frac{1}{11}$$.

Part 1 — Compute $$\frac{1}{3}\times\frac{1}{5}$$ using a unit square.

Think of one whole as a unit square.

• First divide the square into 3 equal vertical strips and shade 1 strip. This shaded part represents $$\frac{1}{3}$$ of the whole.
• Next divide the same square into 5 equal horizontal strips and, with a different colour, shade 1 of these horizontal strips. This second shading shows $$\frac{1}{5}$$ of the whole.
• The tiny rectangles that now carry both colours form the overlap region. They represent $$\frac{1}{3}\times\frac{1}{5}$$ of the whole square.
• Because the square is cut into $$3\times5=15$$ equal tiny rectangles, exactly 1 of them is doubly shaded. Therefore the overlapping area is $$\frac{1}{15}$$ of the square.

Algebraically we get the same value:

$$\frac{1}{3}\times\frac{1}{5}=\frac{1\times1}{3\times5}=\frac{1}{15}$$

Hence

Part 2 — Now compute $$\frac{1}{12}\times\frac{1}{18}$$ (the extra product the question asks for).

The same rule applies: multiply the numerators together and the denominators together.

$$\frac{1}{12}\times\frac{1}{18}=\frac{1\times1}{12\times18}=\frac{1}{216}$$

Picture check: if we cut the unit square into 12 equal vertical strips and 18 equal horizontal strips, we obtain $$12\times18=216$$ identical tiny rectangles. The overlap of one vertical strip with one horizontal strip is exactly one of these rectangles, i.e. $$\frac{1}{216}$$ of the whole.

Part 1 — Compute $$\frac{1}{4}\times\frac{1}{3}$$ using a unit square.

• Divide a unit square into 4 equal vertical strips and shade 1 strip → $$\frac{1}{4}$$.
• Now divide the same square into 3 equal horizontal strips and shade 1 of those → $$\frac{1}{3}$$.
• The overlapping region is made of $$4\times3=12$$ tiny rectangles; 1 of them is doubly shaded. So the overlap equals $$\frac{1}{12}$$ of the whole.

By the multiplication rule:

$$\frac{1}{4}\times\frac{1}{3}=\frac{1\times1}{4\times3}=\frac{1}{12}$$

Part 2 — Now compute $$\frac{1}{12}\times\frac{1}{18}$$ (the additional product asked at the end of the question).

Multiply numerators together and denominators together:

$$\frac{1}{12}\times\frac{1}{18}=\frac{1\times1}{12\times18}=\frac{1}{216}$$

Picture check: dividing a unit square into 12 vertical strips and 18 horizontal strips produces $$12\times18=216$$ identical tiny rectangles. The overlap of one vertical strip and one horizontal strip is just one of these rectangles, i.e. $$\frac{1}{216}$$ of the whole square.

Represent the whole by a square.

• Cut it into 5 equal vertical parts and shade 1 → $$\frac{1}{5}$$.
• Cut the same square into 2 equal horizontal parts and shade 1 → $$\frac{1}{2}$$.
• The square is now split into $$5\times2=10$$ tiny rectangles; 1 of them is doubly shaded, giving $$\frac{1}{10}$$ of the whole.

Algebraic check:

$$\frac{1}{5}\times\frac{1}{2}=\frac{1\times1}{5\times2}=\frac{1}{10}$$

Take a unit square.

• Split it into 6 equal vertical strips and shade 1 → $$\frac{1}{6}$$.
• Split the same square into 5 equal horizontal strips and shade 1 → $$\frac{1}{5}$$.
• There are $$6\times5=30$$ tiny rectangles in all; 1 of them is doubly shaded, i.e. $$\frac{1}{30}$$ of the whole.

Computed directly:

$$\frac{1}{6}\times\frac{1}{5}=\frac{1\times1}{6\times5}=\frac{1}{30}$$

Step 1 — Represent each fraction on one side of a unit square.
Draw a square and call the whole square 1 unit.
• First divide it vertically into 3 equal strips; shade 2 of those 3. The shaded part now shows $$\tfrac{2}{3}$$ of the square.
• Next divide the same square horizontally into 5 equal parts; mark 4 of the 5 horizontal parts along one side. Moving these lines right across the square cuts every vertical strip into 5 small rectangles, so the square is now partitioned into $$3\times5=15$$ equal small rectangles.

Step 2 — Identify the region common to both fractions.
The region that lies in the 2 shaded vertical strips and also in the 4 selected horizontal rows is the overlap. Count those overlapping rectangles:
$$2\;(\text{columns}) \times 4\;(\text{rows}) = 8$$ rectangles are common.

Step 3 — Form the product as a fraction of the whole.
Total tiny rectangles in the unit square = 15.
Overlapping (double–shaded) rectangles = 8.
Therefore
$$\frac{2}{3} \times \frac{4}{5} = \frac{8}{15}.$$ There is no common factor between 8 and 15, so the fraction is already in lowest terms.

Visual model.
Draw a unit square.
• Divide it vertically into 4 equal parts; shade 1 of them to show $$\tfrac{1}{4}$$.
• Now divide the same square horizontally into 3 equal strips and mark 2 of the 3 rows. The square is now cut into $$4\times3=12$$ identical small rectangles.

The overlapping (double–shaded) region lies in 1 vertical strip and 2 horizontal rows:
$$1\times2 = 2$$ rectangles are common out of 12.

Hence
$$\frac{1}{4}\times\frac{2}{3}=\frac{2}{12}=\frac{1}{6}\;\;(\text{dividing numerator and denominator by }2).$$

Unit-square illustration.
• Split the square vertically into 5 equal parts; shade 3 of them to show $$\tfrac{3}{5}$$.
• Now split the square horizontally into 2 equal rows; keep 1 of these rows. Altogether we get $$5\times2=10$$ small rectangles.

The intersection of the 3 shaded columns and the single chosen row contains
$$3\times1 = 3$$ rectangles out of 10.

Therefore
$$\frac{3}{5}\times\frac{1}{2}=\frac{3}{10}.$$ No further simplification is possible.

Step 1 — Fraction model inside a square.
• Divide the square vertically into 6 equal parts; shade 4 parts for $$\tfrac{4}{6}$$.
• Divide it horizontally into 5 equal strips; mark 3 rows for $$\tfrac{3}{5}$$. The whole square is now cut into $$6\times5=30$$ equal rectangles.

Step 2 — Count the overlap.
Common rectangles = $$4\times3 = 12$$ out of 30.

Step 3 — Write and simplify the product.
$$\frac{4}{6}\times\frac{3}{5}=\frac{12}{30}.$$ Both numbers are divisible by 6:
$$\frac{12\div6}{30\div6}=\frac{2}{5}.$$

The tap fills $$\frac{7}{10}$$ of the tank in 1 hour.

Fraction of tank filled in 1 hour = $$\frac{7}{10}$$.

Time given = $$\frac{1}{3}$$ hour.

Fraction filled = $$\left(\frac{1}{3}\right)\times\left(\frac{7}{10}\right)$$

= $$\frac{1\times7}{3\times10}=\frac{7}{30}$$.

Fraction filled in 1 hour = $$\frac{7}{10}$$.

Time given = $$\frac{2}{3}$$ hour.

Fraction filled = $$\left(\frac{2}{3}\right)\times\left(\frac{7}{10}\right)$$

= $$\frac{2\times7}{3\times10}=\frac{14}{30}=\frac{7}{15}$$.

Fraction filled in 1 hour = $$\frac{7}{10}$$.

Time given = $$\frac{3}{4}$$ hour.

Fraction filled = $$\left(\frac{3}{4}\right)\times\left(\frac{7}{10}\right)$$

= $$\frac{3\times7}{4\times10}=\frac{21}{40}$$.

Fraction filled in 1 hour = $$\frac{7}{10}$$.

Time given = $$\frac{7}{10}$$ hour.

Fraction filled = $$\left(\frac{7}{10}\right)\times\left(\frac{7}{10}\right)$$

= $$\frac{49}{100}$$.

Let the required time be $$t$$ hours.

Fraction filled in 1 hour = $$\frac{7}{10}$$.

In $$t$$ hours, fraction filled = $$t\times\frac{7}{10}$$.

For the tank to be full, this fraction must be 1.

So, $$t\times\frac{7}{10}=1$$.

Solving for $$t$$:

$$t=1\div\frac{7}{10}=1\times\frac{10}{7}=\frac{10}{7}\text{ hours}$$.

Thus, the tap must run for $$\frac{10}{7}=1\,\frac{3}{7}$$ hours.

(This is about 1 hour 25 minutes 43 seconds.)

Let the whole land be represented by $$1$$.

Land taken by the government = $$\frac{1}{6}$$.

Land left with Somu $$= 1-\frac{1}{6}=\frac{6}{6}-\frac{1}{6}=\frac{5}{6}$$.

Krishna receives half of this leftover land.

Part received by Krishna $$=\frac{1}{2}\times\frac{5}{6}$$.

Multiply the numerators and the denominators:

$$\frac{1}{2}\times\frac{5}{6}=\frac{1\times5}{2\times6}=\frac{5}{12}$$.

Therefore, Krishna got $$\frac{5}{12}$$ of the original land.

Land left after the government’s share $$=\frac{5}{6}$$ (already found in part (a)).

Bora receives one–third of this leftover land.

Part received by Bora $$=\frac{1}{3}\times\frac{5}{6}$$.

$$\frac{1}{3}\times\frac{5}{6}=\frac{1\times5}{3\times6}=\frac{5}{18}$$.

Hence, Bora got $$\frac{5}{18}$$ of the original land.

Total land left with Somu before distributing to her children $$=\frac{5}{6}$$.

Shares already given:

• Krishna: $$\frac{5}{12}$$ (from part (a))
• Bora: $$\frac{5}{18}$$ (from part (b))

Add the children’s shares:

LCM of 12 and 18 is 36.

$$\frac{5}{12}=\frac{5\times3}{12\times3}=\frac{15}{36},\qquad \frac{5}{18}=\frac{5\times2}{18\times2}=\frac{10}{36}$$

Sum $$=\frac{15}{36}+\frac{10}{36}=\frac{25}{36}$$.

Convert $$\frac{5}{6}$$ to denominator 36:

$$\frac{5}{6}=\frac{5\times6}{6\times6}=\frac{30}{36}$$.

Land Somu keeps $$=\frac{30}{36}-\frac{25}{36}=\frac{5}{36}$$.

Thus, Somu kept $$\frac{5}{36}$$ of the original land for herself.

Step 1 – Identify equal parts of the big square
In the figure the big square has been drawn as a 4 × 4 grid. Hence it is cut into $$4\times 4 = 16$$ congruent (equal-sized) small squares.

Step 2 – Count the shaded parts
Exactly 8 of those 16 small squares are shaded.

Step 3 – Form the required fraction
Fraction of the whole square that is shaded is therefore $$\frac{\text{number of shaded equal parts}}{\text{total equal parts}} = \frac{8}{16}.$$ We simplify the fraction by dividing numerator and denominator by the common factor 8:

Thus, the shaded region occupies one-half of the big square.

Step 1 – Identify equal parts of the big square
Again the big square is a 4 × 4 grid, so there are $$4\times 4 = 16$$ identical small squares in all.

Step 2 – Count the shaded parts
This time 12 of the 16 small squares are shaded.

Step 3 – Form the required fraction
Fraction of the big square that is shaded $$=\;\frac{12}{16}.$$ Divide numerator and denominator by their HCF (4):

So three-fourths of the big square is shaded.

Write the whole number as a fraction: $$3 = \frac{3}{1}$$

To divide by a fraction, multiply by its reciprocal:

$$\frac{3}{1} \div \frac{7}{9} = \frac{3}{1}\times\frac{9}{7}$$

Multiply numerators and denominators: $$= \frac{3\times 9}{1\times 7}=\frac{27}{7}$$

The answer can stay as an improper fraction or be written as a mixed number:

$$\frac{27}{7}=3\,\frac{6}{7}$$

Simplify the dividend first: $$\frac{14}{4}=\frac{7}{2}$$

Write the divisor as a fraction: $$2=\frac{2}{1}$$

Change division to multiplication by the reciprocal:

$$\frac{7}{2}\div \frac{2}{1}=\frac{7}{2}\times\frac{1}{2}=\frac{7}{4}$$

Convert to a mixed number if desired: $$\frac{7}{4}=1\,\frac{3}{4}$$

Both fractions are the same. Dividing a number by itself gives 1, but we show the reciprocal step:

$$\frac{2}{3}\div\frac{2}{3}=\frac{2}{3}\times\frac{3}{2}=1$$

Simplify both fractions first: $$\frac{14}{6}=\frac{7}{3}$$

Now divide: $$\frac{7}{3}\div\frac{7}{3}=\frac{7}{3}\times\frac{3}{7}=1$$

Keep the first fraction, multiply by the reciprocal of the second:

$$\frac{4}{3}\div\frac{3}{4}=\frac{4}{3}\times\frac{4}{3}=\frac{16}{9}$$

As a mixed number: $$\frac{16}{9}=1\,\frac{7}{9}$$

Multiply by the reciprocal of $$\frac{1}{7}$$:

$$\frac{7}{4}\div\frac{1}{7}=\frac{7}{4}\times\frac{7}{1}=\frac{49}{4}$$

Mixed number: $$\frac{49}{4}=12\,\frac{1}{4}$$

Simplify the dividend: $$\frac{8}{2}=4=\frac{4}{1}$$

Now divide by $$\frac{4}{15}$$:

$$\frac{4}{1}\div\frac{4}{15}=\frac{4}{1}\times\frac{15}{4}=\frac{60}{4}=15$$

Multiply by the reciprocal of $$\frac{1}{9}$$:

$$\frac{1}{5}\div\frac{1}{9}=\frac{1}{5}\times\frac{9}{1}=\frac{9}{5}=1\,\frac{4}{5}$$

Use the reciprocal of $$\frac{11}{12}$$:

$$\frac{1}{6}\div\frac{11}{12}=\frac{1}{6}\times\frac{12}{11}=\frac{12}{66}$$

Simplify: divide numerator and denominator by 6

$$=\frac{2}{11}$$

Convert the mixed numbers to improper fractions:

$$3\,\frac{2}{3}=\frac{3\times3+2}{3}=\frac{11}{3}$$

$$1\,\frac{3}{8}=\frac{1\times8+3}{8}=\frac{11}{8}$$

Divide by multiplying with the reciprocal of $$\frac{11}{8}$$:

$$\frac{11}{3}\div\frac{11}{8}=\frac{11}{3}\times\frac{8}{11}=\frac{8}{3}$$

As a mixed number: $$\frac{8}{3}=2\,\frac{2}{3}$$

Total lace available = $$8\text{ m}$$.
Lace required for one bag = $$\frac{1}{4}\text{ m}$$.

Number of bags = (total length) ÷ (length per bag).

Correct choice: option (iii) $$8 \div \frac{1}{4}$$.

Calculation:
Write $$8 = \frac{8}{1}$$. Dividing by a fraction is the same as multiplying by its reciprocal, so

$$\frac{8}{1} \div \frac{1}{4} = \frac{8}{1} \times \frac{4}{1} = \frac{32}{1} = 32$$.

Maria decorated 32 bags.

Total ribbon used = $$\frac{1}{2}\text{ m}$$ for 8 badges.

Ribbon per badge = (total length) ÷ (number of badges).

Correct choice: option (iv) $$\frac{1}{2} \div 8$$.

Calculation:
$$\frac{1}{2} \div 8 = \frac{1}{2} \times \frac{1}{8} = \frac{1}{16}$$.

Each badge needs $$\frac{1}{16}\text{ m}$$ of ribbon.

Flour available = $$5\text{ kg}$$.
Flour needed for one loaf = $$\frac{1}{6}\text{ kg}$$.

Number of loaves = (total flour) ÷ (flour per loaf).

Correct choice: option (iii) $$5 \div \frac{1}{6}$$.

Calculation:
Write $$5 = \frac{5}{1}$$.
$$\frac{5}{1} \div \frac{1}{6} = \frac{5}{1} \times 6 = 30$$.

The baker can make 30 loaves.