## Fractions Exercise – 7.1

**Question 1. Write the fraction representing the shaded portion.**

**Solution:**

- \(\frac{2}{4}\)
- \(\frac{8}{9}\)
- \(\frac{4}{8}\)
- \(\frac{1}{4}\)
- \(\frac{3}{7}\)
- \(\frac{3}{12}\)
- \(\frac{10}{10}\)
- \(\frac{4}{9}\)
- \(\frac{4}{8}\)
- \(\frac{1}{2}\)

**Question 2. Color the part according to the given fraction.**

**Solution:**

**Question 3. Identify the error, if any.**

**Solution:**

Shaded parts do not represent the given fractions, because all the figures are not equally divided. For making fractions, it is necessary that the figure is to be divided into equal parts.

**Question** **4. What fraction of a day is 8 hours?**

**Solution:** Since, 1 day = 24 hours

Therefore, the fraction of 8 hours = \(\frac{8}{24}\)

**Question 5. What fraction of an hour is 40 minutes?**

**Solution:** Since, 1 hour = 60 minutes.

Therefore, the fraction of 40 minutes = \(\frac{40}{60}\)

**Question 6. Arya, Abhimanyu, and Vivek shared lunch. Arya has brought two sandwiches, one made of vegetables and one of jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that everyone would have an equal share of each sandwich.**

**How can Arya divide his sandwiches so everyone has an equal share?****What part of a sandwich will each boy receive?**

**Solution:**

- Arya will divide each sandwich into three equal parts and give one part of each sandwich to each one of them.
- Each boy will get part \(\frac{1}{3}\) of a sandwich.

**Question 7. Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished?**

**Solution:** Total number of dresses = 30

Work finished = 20

Fraction of finished work \(=\frac{20}{30}=\frac{2}{3}\)

**Question 8. Write the natural numbers from 2 to 1 2. What fraction of them are prime numbers?**

**Solution:**

Natural numbers from 2 to 12 :

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

Prime numbers from 2 to 12 :

2,3,5,7,11

Hence, fraction of prime numbers \(=\frac{5}{11}\)

**Question 9. Write the natural numbers from 102 to 113. What fraction of them are prime numbers?**

**Solution:**

Natural numbers from 102 to 113 :

102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113

Prime numbers from 102 to 113 :

103, 107, 109, 113

Hence, fraction of prime numbers \(=\frac{4}{12}\)

**Question 10. What fraction of these circles have X’s in them?**

**Solution:**

Total number of circles = 8

And the number of circles having ‘X’ = 4

Hence, the fraction \(=\frac{4}{8}\)

**Question 11. Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts?**

**Solution:**

Total number of CDs = 3 + 5 = 8

Number of CDs purchased = 3

Fraction of CDs purchased \(=\frac{3}{8}\)

Fraction of CDs received as gifts \(=\frac{5}{8}\)

## Fractions Exercise – 7.2

**Question 1. Draw number lines and locate the points on them:**

- \(\frac{1}{2}, \frac{1}{4}, \frac{3}{4}, \frac{4}{4}\)
- \(\frac{1}{8}, \frac{2}{8}, \frac{3}{8}, \frac{7}{8}\)
- \(\frac{2}{5}, \frac{3}{5}, \frac{8}{5}, \frac{4}{5}\)

**Solution:**

**Question 2. Express the following as mixed fractions:**

- \(\frac{20}{3}\)
- \(\frac{11}{5}\)
- \(\frac{17}{7}\)
- \(\frac{28}{5}\)
- \(\frac{19}{6}\)
- \(\frac{35}{9}\)

**Solution:**

∴ \(\frac{20}{3}=6 \frac{2}{3}\)

∴\(\frac{11}{5}=2 \frac{1}{5}\)

∴\( \frac{17}{7}=2 \frac{3}{7}\)

∴\( \frac{28}{5}=5 \frac{3}{5}\)

∴\( \frac{19}{6}=3 \frac{1}{6}\)

∴\( \frac{35}{9}=3 \frac{8}{9}\)

**Question 3. Express the following as improper fractions:**

- \(7 \frac{3}{4}\)
- \(5 \frac{6}{7}\)
- \(2 \frac{5}{6}\)
- \(10 \frac{3}{5}\)
- \(9 \frac{3}{7}\)
- \(8 \frac{4}{9}\)

**Solution:**

- \(7 \frac{3}{4}=\frac{(7 \times 4)+3}{4}=\frac{28+3}{4}=\frac{31}{4}\)
- \(5 \frac{6}{7}=\frac{(5 \times 7)+6}{7}=\frac{35+6}{7}=\frac{41}{7}\)
- \(2 \frac{5}{6}=\frac{(2 \times 6)+5}{6}=\frac{12+5}{6}=\frac{17}{6}\)
- \(10 \frac{3}{5}=\frac{(10 \times 5)+3}{5}=\frac{50+3}{5}=\frac{53}{5}\)
- \(9 \frac{3}{7}=\frac{(9 \times 7)+3}{7}=\frac{63+3}{7}=\frac{66}{7}\)
- \(8 \frac{4}{9}=\frac{(8 \times 9)+4}{9}=\frac{72+4}{9}=\frac{76}{9}\)

## Fractions Exercise – 7.3

**Question 1. Write the fractions. Are all these fractions equivalent?**

**Solution:**

- \(\frac{1}{2}, \frac{2}{4}, \frac{3}{6}, \frac{4}{8}\)
- Yes, all of these fractions are equivalent

- \(\frac{4}{12}, \frac{3}{9}, \frac{2}{6}, \frac{1}{3}, \frac{6}{15}\)
- No, all these fractions are not equivalent.

**Question 2. Write the fractions and pair up the equivalent fractions from each row.**

**Solution:**

A. \(\frac{1}{2}\)

B. \(\frac{4}{6}\)

C. \(\frac{3}{9}\)

D. \(\frac{2}{8}\)

E. \(\frac{3}{4}\)

- \(\frac{6}{18}\)
- \(\frac{4}{8}\)
- \(\frac{12}{16}\)
- \(\frac{8}{12}\)
- \(\frac{4}{16}\)

Pairs of equivalent fractions are

(A), (2); (B), (4); (C), (1); (D), (5); (E), (3)

**Question 3. Replace each of the following with the correct number:**

1.

2.

3.

4.

5.

**Solution:**

- \(\frac{2}{7}=\frac{2 \times 4}{7 \times 4}=\frac{8}{28}\)
- \(\frac{5}{8}=\frac{5 \times 2}{8 \times 2}=\frac{10}{16}\)
- \(\frac{3}{5}=\frac{3 \times 4}{5 \times 4}=\frac{12}{20}\)
- \(\frac{45}{60}=\frac{45+3}{60+3}=\frac{15}{20}\)
- \(\frac{18}{24}=\frac{18+6}{24+6}=\frac{3}{4}\)

**Question** **4. Find the equivalent fraction of \(\frac{3}{5}\) having**

**Denominator 20****Numerator 9****Denominator 30****Numerator 27**

**Solution:**

- \(\frac{3}{5}=\frac{3 \times 4}{5 \times 4}=\frac{12}{20}\)
- \(\frac{3}{5}=\frac{3 \times 3}{5 \times 3}=\frac{9}{15}\)
- \(\frac{3}{5}=\frac{3 \times 6}{5 \times 6}=\frac{18}{30}\)
- \(\frac{3}{5}=\frac{3 \times 9}{5 \times 9}=\frac{27}{45}\)

**Question 5. Find the equivalent fraction of \(\frac{36}{48}\) with**

**Numerator 9****Denominator 4**

**Solution:**

- \(\frac{36}{48}=\frac{36/4}{48/4}=\frac{9}{12}\)
- \(\frac{36}{48}=\frac{36+12}{48+12}=\frac{3}{4}\)

**Question 6. Check whether the given fractions are equivalent.**

- \(\frac{5}{9}, \frac{30}{54}\)
- \(\frac{3}{10}, \frac{12}{50}\)
- \(\frac{7}{13}, \frac{5}{11}\)

**Solution:**

⇒ \(\frac{5}{9}, \frac{30}{54}\)

⇒ \(\frac{5}{9}=\frac{5 \times 6}{9 \times 6}=\frac{30}{54}\)

Therefore \(\frac{5}{9} \text { and } \frac{30}{54}\) are equivalent.

⇒ \(\frac{3}{10}, \frac{12}{50}\)

⇒ \(\frac{3}{10}=\frac{3 \times 5}{10 \times 5}=\frac{15}{50} \neq \frac{12}{50}\)

Therefore \(\frac{3}{10} \text { and } \frac{12}{50}\) are not equivalent.

⇒ \(\frac{7}{13}, \frac{5}{11}\)

⇒ \(\frac{7}{13}=\frac{7 \times 11}{13 \times 11}=\frac{77}{143}, \frac{5}{11}=\frac{5 \times 13}{11 \times 13}=\frac{65}{143}\)

∴ \( \frac{77}{143} \neq \frac{65}{143}\)

Therefore,\(\frac{7}{13} \text { and } \frac{5}{11}\) are not equivalent.

**Question** **7. Reduce the following fractions to simplest from:**

- \(\frac{48}{60}\)
- \(\frac{150}{60}\)
- \(\frac{84}{98}\)
- \(\frac{12}{52}\)
- \(\frac{7}{28}\)

**Solution:**

- \(\frac{48}{60}=\frac{2 \times 2 \times 2 \times 2 \times 3}{2 \times 2 \times 3 \times 5}=\frac{4}{5}\)
- \(\frac{150}{60}=\frac{3 \times 5 \times 10}{2 \times 3 \times 10}=\frac{5}{2}\)
- \(\frac{84}{98}=\frac{2 \times 3 \times 14}{7 \times 14}=\frac{6}{7}\)
- \(\frac{12}{52}=\frac{2 \times 2 \times 3}{2 \times 2 \times 13}=\frac{3}{13}\)
- \(\frac{7}{28}=\frac{7}{2 \times 2 \times 7}=\frac{1}{4}\)

**Question** **8. Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her or his pencils.**

**Solution:**

Ramesh: Total pencils = 20

Pencils used = 10

Fraction = \(\frac{10}{20}=\frac{1}{2}\)

Sheelu: Total pencils = 50

Pencils used = 25

Fraction = \(\frac{25}{50}=\frac{1}{2}\)

Jamaal : Total pencils = 80

Pencils used = 40

Fraction = \(\frac{40}{80}=\frac{1}{2}\)

Since all of them used half of their pencils, therefore, each has used up an equal fraction of pencils.

**Question 9. Match the equivalent fractions and write two more for each.**

- \(\frac{250}{400}\)
- \(\frac{180}{200}\)
- \(\frac{660}{990}\)
- \(\frac{180}{360}\)
- \(\frac{220}{550}\)
- \(\text { (a) } \frac{2}{3}\)
- \(\frac{2}{5}\)
- \(\frac{1}{2}\)
- \(\frac{5}{8}\)
- \(\frac{9}{10}\)

**Solution:**

(1) → (d); (2) → (e); (3) → (a); (4) →(c); (5)→(b);

1. \(\frac{250}{400}=\frac{250 \div 50}{400 \div 50}=\frac{5}{8}\)

Also, equivalent fractions of \(\frac{5}{8}\) are

⇒ \(\frac{5}{8}=\frac{5 \times 2}{8 \times 2}=\frac{10}{16}\)

∴ \(\frac{5}{8}=\frac{5 \times 3}{8 \times 3}=\frac{15}{24}\)

2. \(\frac{180}{200}=\frac{180 \div 20}{200 \div 20}=\frac{9}{10}\)

Also,equivalent fractions of \(\frac{9}{10}\) are

⇒ \(\frac{9}{10}=\frac{9 \times 2}{10 \times 2}=\frac{18}{20}\)

∴ \(\frac{9}{10}=\frac{9 \times 3}{10 \times 3}=\frac{27}{30}\)

3. \(\frac{660}{990}=\frac{660 \div 330}{990 \div 330}=\frac{2}{3}\)

Also, equivalent fractions of \(\frac{2}{3}\) are

∴ \(\frac{2 \times 2}{3 \times 2}=\frac{4}{6}, \frac{2 \times 3}{3 \times 3}=\frac{6}{9}\)

4. \(\frac{180}{360}=\frac{180 \div 180}{360 \div 180}=\frac{1}{2}\)

Also,equivalent fractions of \(\frac{1}{2}\) are

⇒ \(\frac{1}{2}=\frac{1 \times 2}{2 \times 2}=\frac{2}{4}\)

∴ \(\frac{1}{2}=\frac{1 \times 3}{2 \times 3}=\frac{3}{6}\)

5. \(\frac{220}{550}=\frac{220 \div 110}{550 \div 110}=\frac{2}{5}\)

Also, equivalent fractions of \(\frac{2}{5}\) are

⇒ \(\frac{2}{5}=\frac{2 \times 2}{5 \times 2}=\frac{4}{10}\)

∴ \(\frac{2}{5}=\frac{2 \times 3}{5 \times 3}=\frac{6}{15}\)

## Fractions Exercise – 7.4

**Question 1. Write the shaded portion as a fraction. Arrange them in ascending and descending order using the correct sign ‘<‘, ‘=’, ’>’ between the fractions:**

**(3) Show \(\frac{2}{6}, \frac{4}{6}, \frac{8}{6} \text { and } \frac{6}{6}\) on the number line. Put appropriate signs between the fractions given.**

**Solution:**

1. \(\frac{3}{8}, \frac{6}{8}, \frac{4}{8}, \frac{1}{8}\)

Ascendung order: \(\frac{1}{8}<\frac{3}{8}<\frac{4}{8}<\frac{6}{8}\)

Descending order: \(\frac{6}{8}>\frac{4}{8}>\frac{3}{8}>\frac{1}{8}\)

2. \(\frac{8}{9}, \frac{4}{9}, \frac{3}{9}, \frac{6}{9}\)

Ascendung order: \(\frac{3}{9}<\frac{4}{9}<\frac{6}{9}<\frac{8}{9}\)

Descending order: \(\frac{8}{9}>\frac{6}{9}>\frac{4}{9}>\frac{3}{9}\)

Number line

**Question 2. Compare the fractions and put an appropriate sign.**

1.

2.

3.

4.

**Solution:**

⇒ \(\frac{3}{6} \text { and } \frac{5}{6}\) are like fractions.

Also, numerator of \(\frac{5}{6}\) is greater than numerator of \(\frac{3}{6}\)

⇒ \(\frac{3}{6}<\frac{5}{6}\)

⇒ \(\frac{1}{7} \text { and } \frac{1}{4}\) are unlike fractions with same numerator. Also, denominator of \(\frac{1}{7}\) greater than denominator of \(\frac{1}{4}\)

⇒ \( \frac{1}{7}<\frac{1}{4}\)

⇒ \(\frac{4}{5} \text { and } \frac{5}{5}\) are unlike fractions with same numerator. Also, denominator of \(\frac{5}{5}\) greater than denominator of \(\frac{4}{5}\)

⇒ \( \frac{4}{5}<\frac{5}{5}\)

⇒ \(\frac{3}{5}\) and \( \frac{3}{7}\) are unlike fractions with same numerator.

Also, denominator of \(\frac{3}{7}\) greater than denominator of \(\frac{3}{5}\)

⇒ \(\frac{3}{5}>\frac{3}{7}\)

**Question** **3. Make five more such pairs and put appropriate signs.**

**Solution:**

1.

2.

3.

4.

5.

**Question 4. Look at the figures and write ‘<‘ or ‘=’ between the given pairs of fractions.**

1.

2.

3.

4.

5.

**Make five more such problems and solve them with your friends. are unlike fractions with**

**Solution:**

**Five more problems:**

1.

2.

3.

4.

5.

**The above problems are solved as follows:**

**Question 5. How quickly can you do this? Fill appropriate sign. (‘<”=”>’)**

**Solution:**

⇒ \(\frac{1}{2} \) and \( \frac{1}{5}\) are unlike fractions with same numerator. Also, denominator of \(\frac{1}{5}\) is greater than denominator of \(\frac{1}{2}\)

⇒ \(\frac{1}{2}>\frac{1}{5}\)

⇒ \(\frac{2}{4}=\frac{2 \div 2}{4 \div 2}=\frac{1}{2}\)

⇒ \(\frac{3}{6}=\frac{3 \div 3}{6 \div 3}=\frac{1}{2}\)

⇒ \(\frac{2}{4}=\frac{3}{6}\)

⇒ \(\frac{3}{5} \) and \( \frac{2}{3}\) are unlike fractions with different numerator.

⇒ \(\frac{3}{5}=\frac{3 \times 3}{5 \times 3}=\frac{9}{15}\)

⇒ \(\frac{2}{3}=\frac{2 \times 5}{3 \times 5}=\frac{10}{15}\)

⇒ \(\frac{9}{15}<\frac{10}{15}\)

⇒ \(\frac{3}{5}<\frac{2}{3}\)

⇒ \(\frac{3}{4} \) and \( \frac{2}{8}\) are unlike fractions with different numerators.

⇒ \(\frac{3}{4}=\frac{3 \times 2}{4 \times 2}=\frac{6}{8}\)

⇒ \(\frac{2}{8}=\frac{2}{8}\)

⇒ \(\quad \frac{6}{8}>\frac{2}{8}\)

⇒ \(\frac{3}{4}>\frac{2}{8}\)

⇒ \(\frac{3}{5} \) and \( \frac{6}{5}\) are like fractions. Also,numerator of \(\frac{6}{5}\) is greater thannumerator of \(\frac{3}{5}\)

⇒ \( \frac{3}{5}<\frac{6}{5}\)

⇒ \(\frac{7}{9} \) and \( \frac{3}{9}\) arelike fractions. Also, numerator of \(\frac{7}{9}\) is greater thannumerator of \(\frac{3}{9}\)

⇒ \( \frac{7}{9}>\frac{3}{9}\)

⇒ \(\frac{1}{4} \) and \( \frac{2}{8}\) are unlike fractions with different numerators

⇒ \(\frac{1}{4}=\frac{1 \times 2}{4 \times 2}=\frac{2}{8}\)

⇒ \(\quad \frac{2}{8}=\frac{2}{8}\)

⇒ \(\frac{1}{4}=\frac{2}{8}\)

⇒ \(\frac{6}{10} \) and \( \frac{4}{5}\) are unlike fractions with different numerator

⇒ \(\frac{6}{10}=\frac{6 \div 2}{10 \div 2}=\frac{3}{5}\)

⇒ \(\quad \frac{3}{5}<\frac{4}{5}\)

⇒ \(\frac{6}{10}<\frac{4}{5}\)

⇒ \(\frac{3}{4} \) and \( \frac{7}{8}\) are unlike fractions with different numerators.

⇒ \(\frac{3}{4}=\frac{3 \times 2}{4 \times 2}=\frac{6}{8}\)

⇒ \( \quad \frac{6}{8}<\frac{7}{8}\)

⇒ \(\frac{3}{4}<\frac{7}{8}\)

⇒ \(\frac{6}{10} \) and \( \frac{3}{5}\) are unlike fractions with different numerators.

⇒ \(\frac{6}{10}=\frac{6 \div 2}{10 \div 2}=\frac{3}{5}\)

⇒ \(\quad \frac{3}{5}=\frac{3}{5}\)

⇒ \(\frac{6}{10}=\frac{3}{5}\)

⇒ \(\frac{5}{7} \) and \( \frac{15}{21}\) are unlike fractions with different numerators.

⇒ \(\frac{5}{7}=\frac{5 \times 3}{7 \times 3}=\frac{15}{21}\)

⇒ \(\quad \frac{15}{21}=\frac{15}{21}\)

∴ \(\frac{5}{7}=\frac{15}{21}\)

**Question 6. The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.**

- \(\frac{2}{12}\)
- \(\frac{3}{15}\)
- \(\frac{8}{50}\)
- \(\frac{16}{100}\)
- \(\frac{10}{60}\)
- \(\frac{15}{75}\)
- \(\frac{12}{60}\)
- \(\frac{16}{96}\)
- \(\frac{12}{75}\)
- \(\frac{12}{72}\)
- \(\frac{3}{18}\)
- \(\frac{4}{25}\)

**Solution:**

- \(\frac{2}{12}=\frac{2 \div 2}{12 \div 2}=\frac{1}{6}\)
- \(\frac{3}{15}=\frac{3 \div 3}{15 \div 3}=\frac{1}{5}\)
- \(\frac{8}{50}=\frac{8 \div 2}{50 \div 2}=\frac{4}{25}\)
- \(\frac{16}{100}=\frac{16 \div 4}{100 \div 4}=\frac{4}{25}\)
- \(\frac{10}{60}=\frac{10 \div 10}{60 \div 10}=\frac{1}{6}\)
- \(\frac{15}{75}=\frac{15 \div 15}{75 \div 15}=\frac{1}{5}\)
- \(\frac{12}{60}=\frac{12 \div 12}{60 \div 12}=\frac{1}{5}\)
- \(\frac{16}{96}=\frac{16 \div 16}{96 \div 16}=\frac{1}{6}\)
- \(\frac{12}{75}=\frac{12 \div 3}{75 \div 3}=\frac{4}{25}\)
- \(\frac{12}{72}=\frac{12 \div 12}{72 \div 12}=\frac{1}{6}\)
- \(\frac{3}{18}=\frac{3 \div 3}{18 \div 3}=\frac{1}{6}\)
- \(\frac{4}{25}=\frac{4}{25}\)

**Equivalent groups :**

1. group :\(\frac{1}{5}[(b),(f),(g)]\)

2. group:\(\frac{1}{6}[(\mathrm{a}),(\mathrm{e}),(\mathrm{h}),(\mathrm{j}),(\mathrm{k})]\)

3.group:\(\frac{4}{25}[(\mathrm{c}),(\mathrm{d}),(\mathrm{i}),(\mathrm{l})]\)

**Question 7. Find answers to the following. Write and indicate how you solved them.**

Is \(\frac{5}{9} \text { equal to } \frac{4}{5} \text { ? }\)

Is \(\frac{9}{16} \text { equal to } \frac{5}{9} ?\)

Is \(\frac{4}{5} \text { equal to } \frac{16}{20} ?\)

Is \(\frac{1}{15} \text { equal to } \frac{4}{30} \text { ? }\)

**Solution:**

⇒ \(\frac{5}{9} \text { and } \frac{4}{5} \Rightarrow \frac{5 \times 5}{9 \times 5}=\frac{25}{45} \text { and } \frac{4 \times 9}{5 \times 9}=\frac{36}{45}\)

[L.C.M. of 9 and 5 is 45]

Since, \(\frac{25}{45} \neq \frac{36}{45}\)

Therefore, \(\frac{5}{9} \neq \frac{4}{5}\)

⇒ \(\frac{9}{16} \text { and } \frac{5}{9}\)

⇒ \(\Rightarrow \frac{9 \times 9}{16 \times 9}=\frac{81}{144} \text { and } \frac{5 \times 16}{9 \times 16}=\frac{80}{144}\)

[L.C.M. of 16 and 9 is 144]

Since, \(\frac{81}{144} \neq \frac{80}{144}\)

Therefore, \(\frac{9}{16} \neq \frac{5}{9}\)

⇒ \(\frac{4}{5} \text { and } \frac{16}{20}\)

⇒ \(\frac{4 \times 4}{5 \times 4}=\frac{16}{20} \text { and } \frac{16 \times 1}{20 \times 1}=\frac{16}{20}\)

[L.C.M. of 5 and 20 is 20]

Since, \(\frac{16}{20}=\frac{16}{20}\)

Therefore, \(\frac{4}{5}=\frac{16}{20}\)

⇒ \(\frac{1}{15} \text { and } \frac{4}{30}\)

⇒ \(\frac{1 \times 2}{15 \times 2}=\frac{2}{30} \text { and } \frac{4 \times 1}{30 \times 1}=\frac{4}{30}\)

[L.C.M. of 15 and 30 is 30]

Since, \(\frac{2}{30} \neq \frac{4}{30}\)

∴ \(\frac{1}{15} \neq \frac{4}{30}\)

**Question 8. Ila read 25 pages of a book containing 100 pages. Lalita read \(\frac{2}{5}\) of the same book. Who reads less?**

**Solution:**

Ila read 25 pages out of 100 pages. Fraction of reading the pages

⇒ \(=\frac{25}{100}=\frac{1}{4}\) of book

⇒ \(\frac{1}{4} \text { and } \frac{2}{5}\frac{1 \times 5}{4 \times 5}=\frac{5}{20} \text { and } \frac{2 \times 4}{5 \times 4}=\frac{8}{20}\)

[L.C.M. of 5 and 4 is 20]

⇒ \(\frac{5}{20}<\frac{8}{20}\)

∴ \(\quad \frac{1}{4}<\frac{2}{5}\)

Therefore, Ila read less.

**Question 9. Rafiq exercised for \(\frac{3}{6}\) of an hour, while Rohit exercised for \(\frac{3}{4}\) of an hour. Who exercised for a longer time?**

**Solution:**

Rafiq exercised \(\frac{3}{6}\) of an hour.

\(\) Rohit exercised \(\frac{3}{4}\) of an hour.

Since, \(\frac{3}{4}>\frac{3}{6}\)

Therefore, Rohit exercised for a longer time.

**Question 10. In a class A of 25 students, 20 passed with 60% or more marks; in another class B of 30 students, 24 passed with 60% or more marks. In which class was a greater fraction of students getting 60% or more marks?**

**Solution:**

In class A, 20 passed with 60% or more marks out of 25

Fraction overpassed students \(=\frac{20}{25}=\frac{4}{5}\)

In class B, 24 passed with 60% or more marks out of 30

Fraction of passed students \(=\frac{24}{30}=\frac{4}{5}\)

Hence, both classes have the same fraction of students getting 60% or more marks

## Fractions Exercise – 7.5

**Question 1. Write these fractions appropriately as additions or subtractions:**

**Solution:**

**Question 2. Solve:**

⇒ \(\frac{1}{18}+\frac{1}{18}\)

⇒ \(\frac{8}{15}+\frac{3}{15}\)

⇒ \(\frac{7}{7}-\frac{5}{7}\)

⇒ \(\frac{1}{22}+\frac{21}{22}\)

⇒ \(\frac{12}{15}-\frac{7}{15}\)

⇒ \(\frac{5}{8}+\frac{3}{8}\)

⇒ \(1-\frac{2}{3}\left(1=\frac{3}{3}\right)\)

⇒ \(\frac{1}{4}+\frac{0}{4}\)

∴ \(3-\frac{12}{5}\)

**Solution:**

⇒ \(\frac{1}{18}+\frac{1}{18}=\frac{1+1}{18}=\frac{2}{18}=\frac{1}{9}\)

⇒ \(\frac{8}{15}+\frac{3}{15}=\frac{8+3}{15}=\frac{11}{15}\)

⇒ \(\frac{7}{7}-\frac{5}{7}=\frac{7-5}{7}=\frac{2}{7}\)

⇒ \(\frac{1}{22}+\frac{21}{22}=\frac{1+21}{22}=\frac{22}{22}=1\)

⇒ \(\frac{12}{15}-\frac{7}{15}=\frac{12-7}{15}=\frac{5}{15}=\frac{1}{3}\)

⇒ \(\frac{5}{8}+\frac{3}{8}=\frac{5+3}{8}=\frac{8}{8}=1\)

⇒ \(1-\frac{2}{3}=\frac{3}{3}-\frac{2}{3}=\frac{3-2}{3}=\frac{1}{3}\)

⇒ \(\frac{1}{4}+\frac{0}{4}=\frac{1+0}{4}=\frac{1}{4}\)

∴ \(3-\frac{12}{5}=\frac{15}{5}-\frac{12}{5}=\frac{15-12}{5}=\frac{3}{5}\)

**Question 3. Shubham painted \(\frac{2}{3}\) of the wall space in his room. His sister Madhavi helped and painted \(\frac{1}{3}\) of the wall space. How much did they paint together?**

**Solution:**

Fraction of wall painted by Shubham = \(\frac{2}{3}\)

Fraction of wall painted by Madhavi = \(\frac{1}{3}\)

Total painting by both of them \(=\frac{2}{3}+\frac{1}{3}=\frac{2+1}{3}=\frac{3}{3}=1\)

Therefore, they painted a complete wall.

**Question 4. Fill in the missing fractions.**

**Solution:**

⇒ \(\frac{7}{10}-\frac{4}{10}=\frac{7-4}{10}=\frac{3}{10}\)

⇒ \(\frac{8}{21}-\frac{3}{21}=\frac{8-3}{21}=\frac{5}{21}\)

⇒ \(\frac{6}{6}-\frac{3}{6}=\frac{6-3}{6}=\frac{3}{6}\)

∴\(\frac{7}{27}+\frac{5}{27}=\frac{7+5}{27}=\frac{12}{27}\)

**Question 5. Javed was given \(\frac{5}{7}\) of a basket of oranges. What fraction of oranges was left in the basket?**

**Solution:**

Consider the total number of oranges to be the whole portion or 1.

Fraction of oranges left \(=1-\frac{5}{7}\)

⇒ \(=\frac{7}{7}-\frac{5}{7}=\frac{7-5}{7}=\frac{2}{7}\)

Thus, \(\frac{2}{7}\) oranges was left in the basket.

## Fractions Exercise – 7.6

**Question** **1. Solve:**

⇒ \(\frac{2}{3}+\frac{1}{7}\)

⇒ \(\frac{3}{10}+\frac{7}{15}\)

⇒ \(\frac{4}{9}+\frac{2}{7}\)

⇒ \(\frac{5}{7}+\frac{1}{3}\)

⇒ \(\frac{2}{5}+\frac{1}{6}\)

⇒ \(\frac{4}{5}+\frac{2}{3}\)

⇒ \(\frac{3}{4}-\frac{1}{3}\)

⇒ \(\frac{5}{6}-\frac{1}{3}\)

⇒ \(\frac{2}{3}+\frac{3}{4}+\frac{1}{2}\)

⇒ \(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)

⇒ \(1 \frac{1}{3}+3 \frac{2}{3}\)

⇒ \(4 \frac{2}{3}+3 \frac{1}{4}\)

⇒ \(\frac{16}{5}-\frac{7}{5}\)

∴ \(\frac{4}{3}-\frac{1}{2}\)

**Solution:**

L.C.M. of 3 and 7 is 21

⇒ \(\frac{2}{3}+\frac{1}{7}=\frac{2 \times 7+1 \times 3}{21}=\frac{14+3}{21}=\frac{17}{21}\)

L.C.M. of 10 and 15 is 30

⇒ \(\frac{3}{10}+\frac{7}{15}=\frac{3 \times 3+7 \times 2}{30}=\frac{9+14}{30}=\frac{23}{30}\)

L.C.M. of 9 and 7 is 63

⇒ \(\frac{4}{9}+\frac{2}{7}=\frac{4 \times 7+2 \times 9}{63}=\frac{28+18}{63}=\frac{46}{63}\)

L.C.M. of 7 and 3 is 21

⇒ \(\frac{5}{7}+\frac{1}{3}=\frac{5 \times 3+7 \times 1}{21}=\frac{15+7}{21}=\frac{22}{21}=1 \frac{1}{21}\)

L.C.M. of 5 and 6 is 30

⇒ \(\frac{2}{5}+\frac{1}{6}=\frac{2 \times 6+5 \times 1}{30}=\frac{12+5}{30}=\frac{17}{30}\)

L.C.M. of 5 and 3 is 15

⇒ \(\frac{4}{5}+\frac{2}{3}=\frac{4 \times 3+2 \times 5}{15}=\frac{12+10}{15}=\frac{22}{15}=1 \frac{7}{15}\)

L.C.M. of 4 and 3 is 12

⇒ \(\frac{3}{4}-\frac{1}{3}=\frac{3 \times 3-4 \times 1}{12}=\frac{9-4}{12}=\frac{5}{12}\)

L.C.M. of 6 and 3 is 6

⇒ \(\frac{5}{6}-\frac{1}{3}=\frac{5 \times 1-2 \times 1}{6}=\frac{5-2}{6}=\frac{3}{6}=\frac{1}{2}\)

L.C.M. of 2, 3, and 6 is 6

⇒ \(\frac{2}{3}+\frac{3}{4}+\frac{1}{2}=\frac{2 \times 4+3 \times 3+1 \times 6}{12}\)

⇒ \(=\frac{8+9+6}{12}=\frac{23}{12}=1 \frac{11}{12}\)

L.C.M. of 3 and 3 is 3

⇒ \(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{1 \times 3+1 \times 2+1 \times 1}{6}\)

⇒ \(=\frac{3+2+1}{6}=\frac{6}{6}=1\)

L.C.M. of 3 and 4 is 12

⇒ \(\frac{14}{3}+\frac{13}{4}=\frac{14 \times 4+13 \times 3}{12}\)

⇒ \(=\frac{56+39}{12}=\frac{95}{12}=7 \frac{11}{12}\)

L.C.M. of 5 and 5 is 5

⇒ \(\frac{16}{5}-\frac{7}{5}=\frac{16-7}{5}=\frac{9}{5}=1 \frac{4}{5}\)

L.C.M. of 3 and 2 is 6

⇒ \(\frac{4}{3}-\frac{1}{2}=\frac{4 \times 2-1 \times 3}{6}=\frac{8-3}{6}=\frac{5}{6}\)

**Question 2. Sarita bought \(\frac{2}{5}\) metre of ribbon and Lalita \(\frac{3}{4}\) metre ofribbon. What is the total length of the ribbon they bought?**

Ribbon bought by Sarita = \(\frac{2}{5}\)m

And Ribbon bought by Lalita =\(\frac{3}{4}\)

TinTotal length, of ribbon = \(\frac{2}{5}+\frac{3}{4}=\frac{2 \times 4+5 \times 3}{20}\)

[L.C.M. of 5 and 4 is 20]

⇒ \(=\frac{8+15}{20}=\frac{23}{20}=1 \frac{3}{20} \mathrm{~m}\)

Therefore, they bought\(1 \frac{3}{20} \mathrm{~m}\) of ribbon.

**Question 3. Naina was given \(1 \frac{1}{2}\) piece of cake and Najma was given \(1 \frac{1}{3}\) a piece of cake. Please find the total amount of cake that was given to both of them.**

**Solution:**

Cake taken by Naina = \(1 \frac{1}{2}\) Piece.

And cake taken by Najma =\(1 \frac{1}{3}\) Piece.

Total cake taken = \(1 \frac{1}{2}+1 \frac{1}{3}=\frac{3}{2}+\frac{4}{3}\)

⇒ \(=\frac{3 \times 3+4 \times 2}{6}\)

[ L.C.M. of 2 and 3 is 6]

⇒ \(=\frac{9+8}{6}=\frac{17}{6}=2 \frac{5}{6}\)

Therefore, the total amount of cake given to both of them = \(2 \frac{5}{6}.\)

**Question 4. Fill in the boxes:**

**Solution:**

⇒ \(\frac{1}{4}+\frac{5}{8}=\frac{2+5}{8}=\frac{7}{8}\)

⇒ \(\frac{1}{2}+\frac{1}{5}=\frac{5+2}{10}=\frac{7}{10}\)

⇒ \(\frac{1}{2}-\frac{1}{6}=\frac{3-1}{6}=\frac{2}{6}=\frac{1}{3}\)

**Question** **5. Complete the addition-subtraction box**

**Solution:**

**Question 6. A piece of wire \(\frac{7}{8}\) meter long broke into two pieces. One piece was \(\frac{1}{4}\) metre long. How long is the other piece?**

**Solution:**

Total length of wire = \(\frac{7}{8}\)

Length of first part = \(\frac{7}{8}\)

Remaining part = \(\frac{7}{8}-\frac{1}{4}=\frac{7 \times 1-2 \times 1}{8}\)

[L.C.M. of 8 and 4 is 8]

⇒ \(=\frac{7-2}{8}=\frac{5}{8} \text { metre }\)

Therefore, the length of the remaining part is \(\frac{5}{8}\) meter.

**Question 7. Nandini’s house is \(\frac{9}{10}\) km from her school. She walked some distance and then took a bus for \(\frac{1}{2}\) km to reach the school. How far did she walk?**

**Solution:**

The total distance between school and

house = \(\frac{9}{10}\)km

Distance covered by bus = \(\frac{1}{2} \mathrm{~km}\)

Remaining Distance = \(\frac{9}{10}-\frac{1}{2}=\frac{9 \times 1-1 \times 5}{10}\)

[ L.C.M. of 10 and 2 is 10]

⇒ \(=\frac{9-5}{10}=\frac{4}{10}=\frac{2}{5} \mathrm{~km}\)

Therefore, distance covered by walking is \(\frac{2}{5} \mathrm{~km}\)

**Question** **8. Asha and Samuel have bookshelves of the 5 same size partly filled with books. Asha’s shelf is \(\frac{5}{6}\)th full and Samuel’s shelf is \(\frac{2}{5}\)th full. Whose bookshelf is full? By what fraction?**

**Solution:**

[ L.C.M. of 6 and 5 is 30]

⇒ \(\frac{25}{30}>\frac{12}{30} \Rightarrow \frac{5}{6}>\frac{2}{5}\)

Asha’s bookshelf is more covered than Samuel’s.

Difference = \(\frac{25}{30}-\frac{12}{30}=\frac{13}{30}\)

**Question** **9. Jaidev takes \(2 \frac{1}{5}\) minutes to walk across the school ground. Rahul takes \(\frac{7}{4}\) minutes to do the same. Who takes less time and by what fraction?**

**Solution**:

Time taken by Jaidev = \(2 \frac{1}{5}\)minutes

=\(\frac{11}{5}\)minutes

Time taken by Rahul = \(\frac{7}{4} \text { minutes }\)

Difference = \(=\frac{11}{5}-\frac{7}{4}=\frac{11 \times 4-7 \times 5}{20}\)

[ L.C.M. of 5 and 4 is 20]

⇒ \(=\frac{44-35}{20}=\frac{9}{20} \text { minutes }\)

Thus, Rahul takes less time, which is \(\frac{9}{20}\) minutes.