Chapter 7: Fractions (Fractions)

1 kg shared equally among 3 guavas: each weighs about $\frac{1}{3}$ kg.

$\frac{1}{3}$ kg

$1÷4=\frac{1}{4}$ kg.

$\frac{1}{4}$ kg

Each glass gives each friend $\frac{1}{4}$ glass, so from 3 glasses each one gets $3\times \frac{1}{4}=\frac{3}{4}$ glass.

$\frac{3}{4}$ glass

$\frac{1}{2}=\frac{1}{4}+\frac{1}{4}$, so together they weigh $\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}$ kg.

$\frac{3}{4}$ kg

Quarter $\left(\frac{1}{4}\right)$ < half $\left(\frac{1}{2}\right)$ < three quarters $\left(\frac{3}{4}\right)$ < one and a quarter $\left(1\frac{1}{4}\right)$ < one and a half $\left(1\frac{1}{2}\right)$ < two and a half $\left(2\frac{1}{2}\right)$.

(In Hindi these are paav, aadha, paune, savaa, dedh and dhaai.)

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

Yes. Each piece is one of 6 equal parts of the same whole chikki, so every piece has the same amount (area) of chikki, $\frac{1}{6}$ of the whole, even though the shapes differ.

Yes: each is one-sixth of the same chikki, so they have the same size (area), though different shapes.

Compare each piece with the whole chikki (a rectangle made of 4 × 6 = 24 small squares) and see how many such pieces would make the whole:

(For example, piece b is half of a half: a triangle cut from half the chikki, so $\frac{1}{4}$; piece g is one small square, so $\frac{1}{24}$.)

a. 1/12 b. 1/4 c. 1/8 d. 1/6 e. 1/8 f. 1/6 g. 1/24 h. 1/24

$$2\times \frac{1}{8}=\frac{2}{8}=\frac{1}{4},4\times \frac{1}{8}=\frac{4}{8}=\frac{1}{2},6\times \frac{1}{8}=\frac{6}{8}=\frac{3}{4},8\times \frac{1}{8}=\frac{8}{8}=1$$

2/8 = 1/4, 4/8 = 1/2, 6/8 = 3/4, 8/8 = 1

$$\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=6\text{ times }\frac{1}{2}=3$$

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

6 times 1/2 = 3 (three rotis); 7 times 1/2 = 3½

1/4, 2/4, 3/4, 4/4 = 1, 5/4 = 1¼, … (adding a quarter each time).

Fold the strip into three equal parts (like folding a letter, so the three layers match exactly) and open it: each part is $\frac{1}{3}$. Yes: fold each third in half; the strip now has 6 equal parts, so each part is $\frac{1}{6}$ (half of $\frac{1}{3}$).

Fold into 3 equal parts for 1/3; folding each third in half gives 1/6.

a. One full roti cut into 4 quarters, and one more quarter:

$$\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{5}{4}=1\frac{1}{4}$$

b. Two full rotis (8 quarters) and one more quarter:

$$\underset{9\text{ times}}{\underbrace{\frac{1}{4}+\frac{1}{4}+\cdots +\frac{1}{4}}}=\frac{9}{4}=2\frac{1}{4}$$

a. 5/4 = 1¼ roti b. 9/4 = 2¼ rotis

Count the equal parts of each circle; the shaded part is one of them:

• circle in 3 equal parts → $\frac{1}{3}$
• circle in 5 equal parts → $\frac{1}{5}$
• circle in 6 equal parts → $\frac{1}{6}$
• circle in 8 equal parts → $\frac{1}{8}$

Match by counting the equal parts: 3 parts → 1/3, 5 parts → 1/5, 6 parts → 1/6, 8 parts → 1/8.

1. The line covers 2 of the 3 parts: $\frac{2}{3}$.
2. The lines cover 2 and 4 of the 5 parts: $\frac{2}{5}$ and $\frac{4}{5}$.
3. The marks are $\frac{1}{8},\frac{2}{8},\frac{3}{8},\frac{4}{8},\frac{5}{8},\frac{6}{8},\frac{7}{8}$ and $\frac{8}{8}=1$.

1. 2/3 2. 2/5 and 4/5 3. 1/8, 2/8, 3/8, …, 8/8

Divide the length from 0 to 1 into 10 equal parts. $\frac{1}{10}$ is 1 part and $\frac{3}{10}$ is 3 parts; since $\frac{4}{5}=\frac{8}{10}$, it is 8 parts.

Split 0 to 1 into 10 parts: 1/10 is 1 part, 3/10 is 3 parts and 4/5 = 8/10 is 8 parts.

For example $\frac{1}{4},\frac{1}{2},\frac{3}{4},\frac{5}{4}$ and $\frac{3}{2}$: divide each unit into 4 equal parts.

For example 1/4, 1/2, 3/4, 5/4 and 3/2, marked with each unit split into quarters.

Infinitely many. Between 0 and 1 we have $\frac{1}{2}$, $\frac{1}{3},\frac{2}{3}$, $\frac{1}{4},\frac{3}{4}$, …, $\frac{1}{100}$, … We can always divide the unit into more and more equal parts, and between any two fractions there is always another (e.g. between $\frac{1}{2}$ and $\frac{2}{3}$ lies $\frac{7}{12}$). There is no end to them.

Infinitely many (uncountably many): we can keep dividing the unit into more equal parts.

The blue line is 1 half ($\frac{1}{2}$). The black line covers 3 halves: $\frac{3}{2}$ (one unit and a half).

$\frac{3}{2}$

The black lines end beyond 1, at the 6th, 7th, 8th and 9th marks of fifths: $\frac{6}{5},\frac{7}{5},\frac{8}{5}$ and $\frac{9}{5}$.

6/5, 7/5, 8/5 and 9/5

In the fractions greater than 1, the numerator is larger than the denominator; in those less than 1, the numerator is smaller.

Fractions greater than 1 have a numerator bigger than the denominator.

$\frac{7}{2}=\frac{2}{2}+\frac{2}{2}+\frac{2}{2}+\frac{1}{2}=3+\frac{1}{2}$: 3 whole units.

3 whole units ($\frac{7}{2}=3\frac{1}{2}$)

$\frac{4}{3}=1+\frac{1}{3}$: 1 whole unit. $\frac{7}{3}=2+\frac{1}{3}$: 2 whole units.

1 in 4/3, and 2 in 7/3.

Divide the numerator by the denominator; the quotient is the number of whole units.

a. $8=2\times 3+2$: 2 b. $11=2\times 5+1$: 2 c. $9=2\times 4+1$: 2

a. 2 b. 2 c. 2

Yes, every fraction greater than 1 can be written as a whole number plus a part; but if the numerator is an exact multiple of the denominator, the fractional part is 0 and we just get a whole number, not a mixed number. For example $\frac{8}{4}=2$ and $\frac{9}{3}=3$.

Those whose numerator is a multiple of the denominator (like 8/4 = 2) become whole numbers, not mixed numbers; all others can be written as mixed numbers.

Divide: quotient = whole part, remainder = numerator of the fractional part.

a. $9=4\times 2+1$: $4\frac{1}{2}$ b. $9=1\times 5+4$: $1\frac{4}{5}$ c. $21=1\times 19+2$: $1\frac{2}{19}$

d. $47=5\times 9+2$: $5\frac{2}{9}$ e. $12=1\times 11+1$: $1\frac{1}{11}$ f. $19=3\times 6+1$: $3\frac{1}{6}$

a. 4½ b. 1 4/5 c. 1 2/19 d. 5 2/9 e. 1 1/11 f. 3 1/6

Each whole is (denominator) units, so multiply the whole part by the denominator and add the numerator:

a. $3\frac{1}{4}=\frac{3\times 4+1}{4}=\frac{13}{4}$ b. $\frac{7\times 3+2}{3}=\frac{23}{3}$ c. $\frac{9\times 9+4}{9}=\frac{85}{9}$

d. $\frac{3\times 6+1}{6}=\frac{19}{6}$ e. $\frac{2\times 11+3}{11}=\frac{25}{11}$ f. $\frac{3\times 10+9}{10}=\frac{39}{10}$

a. 13/4 b. 23/3 c. 85/9 d. 19/6 e. 25/11 f. 39/10

1. Yes: on the wall, $\frac{1}{2}$ ends exactly where $\frac{3}{6}$ ends.
2. Yes: $\frac{2}{3}$ and $\frac{4}{6}$ have the same length on the wall ($\frac{2\times 2}{3\times 2}=\frac{4}{6}$).
3. 3 pieces ($\frac{3}{6}=\frac{1}{2}$).
4. 2 pieces ($\frac{2}{6}=\frac{1}{3}$).

1. Yes 2. Yes, they have equal lengths 3. 3 pieces 4. 2 pieces

Yes. Each is half of the whole (the numerator is half the denominator), so on the fraction wall they all have the same length as $\frac{1}{2}$: $\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}$.

Yes: each equals 1/2 (same length on the fraction wall).

$\frac{2}{6}=\frac{1}{3}=\frac{4}{12}$ (also $\frac{3}{9},\frac{6}{18}$, …).

For example 1/3 and 4/12.

$$\frac{4}{6}=\frac{2}{3}=\frac{6}{9}=\frac{8}{12}=\frac{10}{15}=\frac{12}{18}=\frac{14}{21}=\dots$$

2/3, 6/9, 8/12, 10/15, 12/18, …

Cut each roti into 4 quarters and give one quarter of every roti to each child: each child gets 3 quarters, $\frac{3}{4}$ roti.

• Division fact: $3÷4=\frac{3}{4}$
• Addition fact: $3=\frac{3}{4}+\frac{3}{4}+\frac{3}{4}+\frac{3}{4}$
• Multiplication fact: $3=4\times \frac{3}{4}$

Each child gets 3/4 roti; 3 ÷ 4 = 3/4; 3 = 3/4 + 3/4 + 3/4 + 3/4; 3 = 4 × 3/4.

Cut each roti into halves; the 4 halves go to the 4 children: each gets $\frac{1}{2}$ roti.

• Division fact: $2÷4=\frac{2}{4}=\frac{1}{2}$
• Addition fact: $2=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}$
• Multiplication fact: $2=4\times \frac{1}{2}$

Each child gets 1/2 roti; 2 ÷ 4 = 1/2; 2 = ½ + ½ + ½ + ½; 2 = 4 × ½.

Anil gets $2÷5=\frac{2}{5}$ cake.

For 10 children (twice as many) we need twice as many cakes, 4 cakes: $\frac{4}{10}=\frac{2}{5}$. (Putting one group of 2 cakes and 5 children together with another such group gives 4 cakes among 10 children, so $\frac{2}{5}=\frac{4}{10}$.)

2/5 of a cake; 4 cakes for 10 children (4/10 = 2/5).

Equivalent to $\frac{1}{2}$: $\frac{5}{10},\frac{6}{12},\frac{10}{20},\frac{50}{100}$.

The shares are $\frac{2}{3}$, $\frac{4}{6}$ and $\frac{6}{9}$, and they are all equal. In each case, 2 rotis go to every 3 children: 4 rotis among 6 children is just two groups of "2 rotis among 3 children", and 6 among 9 is three such groups. So $\frac{2}{3}=\frac{4}{6}=\frac{6}{9}$.

Relationship: in each fraction the numerator and denominator are the same multiple of 2 and 3 ($\frac{2\times 2}{3\times 2}$, $\frac{2\times 3}{3\times 3}$), so $\frac{2}{3}$ is the simplest form of all of them.

5/10, 6/12, …; the shares 2/3, 4/6, 6/9 are equal because each case is made of groups of 2 rotis for 3 children.

a. Twice the friends need twice the juice: 10 glasses; $\frac{5}{4}=\frac{10}{8}$.

b. 12 kg is 3 times 4 kg, so 3 times the bags: 9 bags; $\frac{4}{3}=\frac{12}{9}$.

c. For example 14 rotis among 10 children: $\frac{7}{5}=\frac{14}{10}$ (also 21 among 15, 28 among 20, …).

a. 10 b. 9 c. e.g. 14 rotis among 10 children

Each child's share increases, because there is more to share among the same number of children. So, with the same denominator, the fraction with the larger numerator is larger: $\frac{1}{5}<\frac{2}{5}$ and $\frac{3}{7}<\frac{4}{7}$. For $\frac{1}{2}$ and $\frac{5}{8}$: $\frac{1}{2}=\frac{4}{8}$, which is 4 units among 8 children, and $\frac{5}{8}$ is 5 units among 8 children, so $\frac{1}{2}<\frac{5}{8}$.

The share increases: same number of children, more to share. So 1/5 < 2/5, 3/7 < 4/7, and 1/2 = 4/8 < 5/8.

1. $\frac{3}{4}=\frac{30}{40}$ and $\frac{7}{10}=\frac{28}{40}$, so Group 1 gets more ($\frac{3}{4}>\frac{7}{10}$).

2. $\frac{4}{7}$ and $\frac{5}{7}$: same number of children, more juice in Group 2, so $\frac{5}{7}>\frac{4}{7}$.

The second pair was easier to compare, because the number of children (denominator) was the same, so we only had to compare the numerators.

1. Group 1 (3/4 > 7/10) 2. Group 2 (5/7 > 4/7). The second was easier because the denominators were the same.

Use a common multiple of the denominators:

a. 35/10, 6/10 b. 16/6, 5/6 c. 15/20, 12/20 d. 30/35, 56/35 e. 9/4, 10/4 f. 9/90, 20/90 g. 32/12, 33/12 h. 39/18, 2/18

Divide the numerator and denominator by their highest common factor:

a. $\frac{17}{51}=\frac{17÷17}{51÷17}=\frac{1}{3}$ b. $\frac{64}{144}=\frac{64÷16}{144÷16}=\frac{4}{9}$

c. $\frac{126}{147}=\frac{126÷21}{147÷21}=\frac{6}{7}$ d. $\frac{525}{112}=\frac{525÷7}{112÷7}=\frac{75}{16}$

a. 1/3 b. 4/9 c. 6/7 d. 75/16

a. $\frac{8}{3}=\frac{16}{6}$, $\frac{5}{2}=\frac{15}{6}$: $\frac{8}{3}>\frac{5}{2}$

b. $\frac{4}{9}=\frac{28}{63}$, $\frac{3}{7}=\frac{27}{63}$: $\frac{4}{9}>\frac{3}{7}$

c. $\frac{7}{10}=\frac{49}{70}$, $\frac{9}{14}=\frac{45}{70}$: $\frac{7}{10}>\frac{9}{14}$

d. Same denominator, 12 > 8: $\frac{12}{5}>\frac{8}{5}$

e. $\frac{5}{2}=\frac{10}{4}$: $\frac{9}{4}<\frac{5}{2}$

a. 8/3 > 5/2 b. 4/9 > 3/7 c. 7/10 > 9/14 d. 12/5 > 8/5 e. 9/4 < 5/2

a. With denominator 30: $\frac{21}{30},\frac{22}{30},\frac{12}{30}$, so $\frac{2}{5}<\frac{7}{10}<\frac{11}{15}$.

b. With denominator 24: $\frac{19}{24},\frac{20}{24},\frac{14}{24}$, so $\frac{7}{12}<\frac{19}{24}<\frac{5}{6}$.

a. 2/5 < 7/10 < 11/15 b. 7/12 < 19/24 < 5/6

a. With denominator 32: $\frac{50}{32},\frac{28}{32},\frac{104}{32},\frac{17}{32}$, so $\frac{13}{4}>\frac{25}{16}>\frac{7}{8}>\frac{17}{32}$.

b. With denominator 60: $\frac{45}{60},\frac{144}{60},\frac{35}{60},\frac{75}{60}$, so $\frac{12}{5}>\frac{5}{4}>\frac{3}{4}>\frac{7}{12}$.

a. 13/4 > 25/16 > 7/8 > 17/32 b. 12/5 > 5/4 > 3/4 > 7/12

They ate $\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$, so $1-\frac{3}{4}=\frac{1}{4}$ of the chikki is left.

$\frac{1}{4}$ of the chikki

Divide each unit into 7 equal parts. Start at 0, jump 4 parts to reach $\frac{4}{7}$, then 6 more parts: we land 10 parts from 0, which is 3 parts beyond 1. So $\frac{4}{7}+\frac{6}{7}=\frac{10}{7}=1\frac{3}{7}$, the same answer as with strips.

Yes: 4 sevenths and then 6 more sevenths reach 10/7 = 1 3/7.

(Parts e and k are the same sum in the textbook; f and g show that the order of adding does not matter.)

a. 13/7 b. 13/12 c. 3/2 d. 20/21 e. 77/60 f. 22/15 g. 22/15 h. 49/40 i. 23/4 j. 62/21 k. 77/60 l. 199/105 m. 83/12

$$\frac{2}{3}+\frac{3}{4}=\frac{8}{12}+\frac{9}{12}=\frac{17}{12}=1\frac{5}{12}\text{litres}$$

$1\frac{5}{12}$ litres ($\frac{17}{12}$ L)

$$\frac{2}{5}+\frac{3}{4}=\frac{8}{20}+\frac{15}{20}=\frac{23}{20}=1\frac{3}{20}\text{m}$$

Yes, it is sufficient: $1\frac{3}{20}$ m is more than the 1 m needed ($\frac{3}{20}$ m is left over).

$1\frac{3}{20}$ m; yes, it is more than the 1 m needed.

1. $\frac{5-3}{8}=\frac{2}{8}=\frac{1}{4}$ 2. $\frac{7-5}{9}=\frac{2}{9}$ 3. $\frac{10-1}{27}=\frac{9}{27}=\frac{1}{3}$

1. 1/4 2. 2/9 3. 1/3

a. $\frac{5}{15}=\frac{1}{3}$ b. $\frac{6}{15}-\frac{4}{15}=\frac{2}{15}$ c. $\frac{15}{18}-\frac{8}{18}=\frac{7}{18}$ d. $\frac{4}{6}-\frac{3}{6}=\frac{1}{6}$

a. 1/3 b. 2/15 c. 7/18 d. 1/6

"A from B" means B − A.

a. $\frac{10}{3}-\frac{13}{4}=\frac{40}{12}-\frac{39}{12}=\frac{1}{12}$

b. $\frac{23}{3}-\frac{18}{5}=\frac{115}{15}-\frac{54}{15}=\frac{61}{15}=4\frac{1}{15}$

c. $\frac{45}{7}-\frac{29}{7}=\frac{16}{7}=2\frac{2}{7}$

a. 1/12 b. 61/15 c. 16/7

a. $\frac{7}{10}-\frac{1}{2}=\frac{7}{10}-\frac{5}{10}=\frac{2}{10}=\frac{1}{5}$ km

b. $\frac{10}{3}=\frac{40}{12}$ and $\frac{13}{4}=\frac{39}{12}$. Namit takes less time, by $\frac{40}{12}-\frac{39}{12}=\frac{1}{12}$ minute (5 seconds).

a. 1/5 km b. Namit, by 1/12 minute.