Class VI - Mathematics

Other Chapters

**Page 135**

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__Exercise 7.1__

**Q1. Write the fraction representing the shaded portion.
Answer:**

(i) $\frac{2}{4}$

(ii) $\frac{8}{9}$

(iii) $\frac{4}{8}$

(iv) $\frac{1}{4}$

(v) $\frac{3}{7}$

(vi) $\frac{3}{12}$

(vii) $\frac{10}{10}$

(viii) $\frac{4}{9}$

(ix) $\frac{4}{8}$

(x) $\frac{1}{2}$

**Q2. Colour the part according to the given fraction.
Answer:**

**Q3. Identify the error, if any.
Answer:**

Since the given figures are not divided equally, they do not represent the fractions.

**Q4. What fraction of a day is 8 hours?
Answer:**

We know that 1 day = 24 hours

∴ 8 hours will be $\frac{8}{24}$ or $\frac{1}{3}$of a day.

**Q5. What fraction of an hour is 40 minutes?
Answer:**

We know that 1 hour = 60 minutes

∴ 40 minutes will be $\frac{40}{60}$ or $\frac{2}{3}$of an hour.

**Q6. Arya, Abhimanyu, and Vivek shared lunch. Arya has brought two sandwiches, one made of vegetable and one of jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that each person will have an equal share of each sandwich.
(a) How can Arya divide his sandwiches so that each person has an equal share?
(b) What part of a sandwich will each boy receive?
Answer:**

(a) Arya can divide each sandwich in three parts and Arya, Abhimanyu and Vivek can each have one part of each sandwich.

(b) Each boy will recieve $\frac{1}{3}$ part of each sandwich.

**Q7. Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished?
Answer:**

Total dresses to dye = 30

Dresses Dyed = 20

Fraction of dresses Kanchan has finished = $\frac{20}{30}$ = $\frac{2}{3}$

**Q8. Write the natural numbers from 2 to 12. What fraction of them are prime numbers?
Answer:**

Natural numbers from 2 to 12 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

∴ Total natural numbers from 2 to 12 = 11

Prime numbers among Natural numbers from 2 to 12 are 2, 3, 5, 7 and 11.

∴ Total prime numbers from 2 to 12 = 5

Hence, $\frac{5}{11}$ of Natural numbers from 2 to 12 are Prime numbers.

**Q9. Write the natural numbers from 102 to 113. What fraction of them are prime numbers?
Answer:**

Natural numbers from 102 to 113 are 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112 and 113.

∴ Total natural numbers from 102 to 113 = 12

Prime numbers among Natural numbers from 102 to 113 are 103, 107, 109 and 113.

∴ Total prime numbers from 102 to 113 = 4

Hence, $\frac{4}{12}$ or $\frac{1}{3}$rd of Natural numbers from 102 to 113 are Prime numbers.

**Q10. What fraction of these circles have X’s in them?
Answer:**

Total number of circles = 8

Number of circles with X’s = 4

∴ Fraction of circles having X’s = $\frac{4}{8}$ or $\frac{1}{2}$

**Q11. 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?
Answer:**

Number of CDs Kristin bought = 3

Number of CDs received as gift = 5

Total CDs Kristin has now = 3 + 5 = 8

Fraction of Kristin’s total CDs she bought = $\frac{3}{8}$

Fraction of Kristin’s total CDs she received as gifts = $\frac{5}{8}$

**Page 141**

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__Exercise 7.2__

**Q1. Draw number lines and locate the points on them :
(a) $\frac{1}{2},\frac{1}{4},\frac{3}{4},\frac{4}{4}$ (b) $\frac{1}{8},\frac{2}{8},\frac{3}{8},\frac{7}{8}$ (c) $\frac{2}{5},\frac{3}{5},\frac{8}{5},\frac{4}{5}$
Answer:**

**Q2. Express the following as mixed fractions :
(a) $\frac{20}{3}$ (b) $\frac{11}{5}$ (c) $\frac{17}{7}$
(d) $\frac{28}{5}$ (e) $\frac{19}{6}$ (f) $\frac{35}{9}$
Answer:**

(a) $\frac{20}{3}$

i.e. 6 whole and 2/3 more, or $6\frac{2}{3}$

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

i.e. 2 whole and 1/5 more, or $2\frac{1}{5}$

(c) $\frac{17}{7}$

i.e. 2 whole and 3/7 more, or $2\frac{3}{7}$

(d) $\frac{28}{5}$

i.e. 5 whole and 3/5 more, or $5\frac{3}{5}$

(e) $\frac{19}{6}$

i.e. 3 whole and 1/6 more, or $3\frac{1}{6}$

(f) $\frac{35}{9}$

i.e. 3 whole and 8/9 more, or $3\frac{8}{9}$

**Q3. Express the following as improper fractions :
(a) $7\frac{3}{4}$ (b) $5\frac{6}{7}$ (c) $2\frac{5}{6}$ (d) $10\frac{3}{5}$ (e) $9\frac{3}{7}$ (f) $8\frac{4}{9}$
Answer:**

(a) $7\frac{3}{4}=\frac{\left(7\times 4\right)+3}{4}=\frac{28+3}{4}=\frac{31}{4}$

(b) $5\frac{6}{7}=\frac{\left(5\times 7\right)+6}{7}=\frac{35+6}{7}=\frac{41}{7}$

(c) $2\frac{5}{6}=\frac{\left(2\times 6\right)+5}{6}=\frac{12+5}{6}=\frac{17}{6}$

(d) $10\frac{3}{5}=\frac{\left(10\times 5\right)+3}{5}=\frac{50+3}{5}=\frac{53}{5}$

(e) $9\frac{3}{7}=\frac{\left(9\times 7\right)+3}{7}=\frac{63+3}{7}=\frac{66}{7}$

(f) $8\frac{4}{9}=\frac{\left(8\times 9\right)+4}{9}=\frac{72+4}{9}=\frac{76}{9}$

**Page 146**

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__Exercise 7.3__

**Q1. Write the fractions. Are all these fractions equivalent?
Answer:**

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

All the above fractions are equal.

$\frac{4}{12}$ = $\frac{1}{3}$ $\frac{3}{9}$ = $\frac{1}{3}$ $\frac{2}{6}$ = $\frac{1}{3}$ $\frac{1}{3}$ $\frac{6}{15}$ = $\frac{2}{5}$

All the above fractions are not equal.

**Q2. Write the fractions and pair up the equivalent fractions from each row.
Answer:**

**Q3. Replace ’?’ in each of the following by the correct number :
(a) $\frac{2}{7}=\frac{8}{\mathrm{?}}$
(b) $\frac{5}{8}=\frac{10}{\mathrm{?}}$
(c) $\frac{3}{5}=\frac{\mathrm{?}}{\mathrm{20}}$
(d) $\frac{45}{60}=\frac{15}{\mathrm{?}}$
(e) $\frac{18}{24}=\frac{\mathrm{?}}{\mathrm{4}}$
Answer:**

(a) $\frac{2}{7}=\frac{8}{?}\mathrm{or}\frac{2\times 4}{7\times 4}=\frac{8}{28}$

∴ ’?’ can be replaced by 28

(b) $\frac{5}{8}=\frac{10}{?}\mathrm{or}\frac{5\times 2}{8\times 2}=\frac{10}{16}$

∴ ’?’ can be replaced by 16

(c) $\frac{3}{5}=\frac{?}{20}\mathrm{or}\frac{3\times 4}{5\times 4}=\frac{12}{20}$

∴ ’?’ can be replaced by 12

(d) $\frac{45}{60}=\frac{15}{?}\mathrm{or}\frac{45\xf73}{60\xf73}=\frac{15}{20}$

∴ ’?’ can be replaced by 20

(e) $\frac{18}{24}=\frac{?}{4}\mathrm{or}\frac{18\xf76}{24\xf76}=\frac{3}{4}$

∴ ’?’ can be replaced by 3

**Q4. Find the equivalent fraction of $\frac{3}{5}$ having
(a) denominator 20 (b) numerator 9
**

**(c) denominator 30 (d) numerator 27
Answer:**

**Q5. Find the equivalent fraction of $\frac{36}{48}$ with **

Answer:

**Q6. Check whether the given fractions are equivalent :
(a) $\frac{5}{9},\frac{30}{54}$
(b) $\frac{3}{10},\frac{12}{50}$
(c) $\frac{7}{13},\frac{5}{11}$
Answer:**

** **

**Q7. Reduce the following fractions to simplest form :
(a) $\frac{48}{60}$
(b) $\frac{150}{60}$
(c) $\frac{84}{98}$
(d) $\frac{12}{52}$
(e) $\frac{7}{28}$
Answer:**

**Q8. 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/his pencils?
Answer:**

**Q9. Match the equivalent fractions and write two more for each.
(i) $\frac{250}{400}$ (a) $\frac{2}{3}$
(ii) $\frac{180}{200}$ (b) $\frac{2}{5}$
(iii) $\frac{660}{990}$ (c) $\frac{1}{2}$
(iv) $\frac{180}{360}$ (d) $\frac{5}{8}$
(v) $\frac{220}{550}$ (e) $\frac{9}{10}$
Answer:**

**Page 152**

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__Exercise 7.4__

**Q1. Write shaded portion as fraction. Arrange them in ascending and descending order using correct sign ‘<’, ‘=’, ‘>’ between the fractions:
(c) Show $\frac{2}{6},\frac{4}{6},\frac{8}{6}\mathrm{and}\frac{6}{6}$ on the number line. Put appropriate signs between the fractions given.
(i) $\frac{5}{6}-\frac{2}{6}$ (ii) $\frac{3}{6}-0$
(iii) $\frac{1}{6}-\frac{6}{6}$ (iv) $\frac{8}{6}-\frac{5}{6}$
Answer:**

(c)

**Q2. Compare the fractions and put an appropriate sign.
(a) $\frac{3}{6}-\frac{5}{6}$ (b) $\frac{1}{7}-\frac{1}{4}$
(c) $\frac{4}{5}-\frac{5}{5}$ (d) $\frac{3}{5}-\frac{3}{7}$
Answer:**

**Q3. Make five more such pairs and put appropriate signs.
Answer:**

**Q4. Look at the figures and write ‘<’ or ‘>’, ‘=’ between the given pairs of fractions
(a) $\frac{1}{6}-\frac{1}{3}$ (b) $\frac{3}{4}-\frac{2}{6}$
(c) $\frac{2}{3}-\frac{2}{4}$ (d) $\frac{6}{6}-\frac{3}{3}$
(e) $\frac{5}{6}-\frac{5}{5}$
Make five more such problems and solve them with your friends.
Answer:
(a) $\frac{1}{6}-\frac{1}{3}$**

**Q5. How quickly can you do this? Fill appropriate sign. (‘<’, ‘=’, ‘>’)
(a) $\frac{1}{2}-\frac{1}{5}$ (b) $\frac{2}{4}-\frac{3}{6}$
(c) $\frac{3}{5}-\frac{2}{3}$
(d) $\frac{3}{4}-\frac{2}{8}$ (e) $\frac{3}{5}-\frac{6}{5}$
(f) $\frac{7}{9}-\frac{3}{9}$
(g) $\frac{1}{4}-\frac{2}{8}$ (h) $\frac{6}{10}-\frac{4}{5}$
(i) $\frac{3}{4}-\frac{7}{8}$
(j) $\frac{6}{10}-\frac{4}{5}$ (k) $\frac{5}{7}-\frac{15}{21}$
Answer:
(a) $\frac{1}{2}-\frac{1}{5}$**

∴ $\frac{1}{2}\frac{1}{5}$

$\frac{2}{4}\mathrm{and}\frac{3}{6}=\frac{1}{2}\mathrm{and}\frac{1}{2}$

∴ $\frac{1}{2}=\frac{1}{2}$

$\frac{3}{5}\mathrm{and}\frac{2}{3}=\left(\frac{3}{5}\times \frac{3}{3}\right)\mathrm{and}\left(\frac{2}{3}\times \frac{5}{5}\right)\times \frac{5}{5}=\frac{9}{15}\mathrm{and}\frac{10}{15}$

Since 10 > 9, therefore $\frac{3}{5}\frac{2}{3}$

To covert the fractions into like fractions we will first divide and multiply $\frac{3}{4}$ by 2.

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

Since 6 > 2, therefore $\frac{3}{4}\frac{2}{8}$

Here the denominators are same so they are like fractions.

Since 6 > 3, therefore $\frac{3}{5}\frac{6}{5}$

Here the denominators are same so they are like fractions.

Since 7 > 3, therefore $\frac{7}{9}\frac{3}{9}$

To covert the fractions into like fractions we will first divide and multiply $\frac{1}{4}$ by 2.

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

Alternatively, we can also simplify $\frac{2}{8}$

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

Since both like fractions are same, therefore $\frac{1}{4}=\frac{2}{8}$

To convert the fractions into like fractions, we will simplify $\frac{6}{10}$

$\frac{6}{10}=\frac{3}{5}$

Since 4 > 3, therefore $\frac{6}{10}\frac{4}{5}$

To covert the fractions into like fractions we will first divide and multiply $\frac{3}{4}$ by 2.

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

Since 7 > 6, therefore $\frac{3}{4}\frac{7}{8}$

To convert the fractions into like fractions, we will simplify $\frac{6}{10}$

$\frac{6}{10}=\frac{3}{5}$

Since 4 > 3, therefore $\frac{6}{10}\frac{4}{5}$

To convert the fractions into like fractions, we will simplify $\frac{15}{21}$

$\frac{15}{21}=\frac{5}{7}$

Since both like fractions are same, therefore $\frac{5}{7}=\frac{15}{21}$

**Q6. The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.
(a) $\frac{2}{12}$ (b) $\frac{3}{15}$
(c) $\frac{8}{50}$ (d) $\frac{16}{100}$
(e) $\frac{10}{60}$ (f) $\frac{15}{75}$
(g) $\frac{12}{60}$ (h) $\frac{16}{96}$
(i) $\frac{12}{75}$ (j) $\frac{12}{72}$
(k) $\frac{3}{18}$ (l) $\frac{4}{25}$
Answer:**

(a) $\frac{2}{12}=\frac{1}{6}$ (b) $\frac{3}{15}=\frac{1}{5}$ (c) $\frac{8}{50}=\frac{4}{25}$ (d) $\frac{16}{100}=\frac{4}{25}$

(e) $\frac{10}{60}=\frac{1}{6}$ (f) $\frac{15}{75}=\frac{1}{5}$ (g) $\frac{12}{60}=\frac{1}{5}$ (h) $\frac{16}{96}=\frac{1}{6}$

(i) $\frac{12}{75}=\frac{4}{25}$ (j) $\frac{12}{72}=\frac{1}{6}$ (k) $\frac{3}{18}=\frac{1}{6}$ (l) $\frac{4}{25}$ can't be reduced further.

The three groups of equivalent fractions in the given fractions are

(i) $\frac{1}{5}$ - (b), (f) and (g)

(ii) $\frac{1}{6}$ - (a), (e), (h), (j) and (k)

(iii) $\frac{4}{25}$ - (c), (d), (i) and (l)

**Q7. Find answers to the following. Write and indicate how you solved them.
(a) Is $\frac{5}{9}$ equal to $\frac{4}{5}$? (b) Is $\frac{9}{16}$ equal to $\frac{5}{9}$?
(c) Is $\frac{4}{5}$ equal to $\frac{16}{20}$? (d) Is $\frac{1}{15}$ equal to $\frac{4}{30}$?
Answer:
(a) Is $\frac{5}{9}$ equal to $\frac{4}{5}$**

L.C.M. of 9 and 5 is 45, therefore

$\Rightarrow \left(\frac{5}{9}\times \frac{5}{5}\right)\mathrm{and}\left(\frac{4}{5}\times \frac{9}{9}\right)\Rightarrow \frac{25}{45}\mathrm{and}\frac{36}{45}$

Since 36 ≠ 25, hence $\frac{5}{9}\mathrm{and}\frac{4}{5}$ are not equal.

L.C.M. of 16 and 9 is 144, therefore

$\Rightarrow \left(\frac{9}{16}\times \frac{9}{9}\right)\mathrm{and}\left(\frac{5}{9}\times \frac{16}{16}\right)\Rightarrow \frac{81}{144}\mathrm{and}\frac{80}{144}$

Since 81 ≠ 80, hence $\frac{9}{16}\mathrm{and}\frac{5}{9}$ are not equal.

L.C.M. of 5 and 20 is 20, therefore

$\Rightarrow \left(\frac{4}{5}\times \frac{4}{4}\right)\mathrm{and}\frac{16}{20}\Rightarrow \frac{16}{20}\mathrm{and}\frac{16}{20}$

Since 16 = 16, hence $\frac{4}{5}\mathrm{and}\frac{16}{20}$ are equal.

L.C.M. of 15 and 30 is 30, therefore

$\Rightarrow \left(\frac{1}{15}\times \frac{2}{2}\right)\mathrm{and}\frac{4}{30}\Rightarrow \frac{2}{30}\mathrm{and}\frac{4}{30}$

Since 2 ≠ 4, hence $\frac{1}{15}\mathrm{and}\frac{4}{30}$ are not equal.

**Q8. Ila read 25 pages of a book containing 100 pages. Lalita read $\frac{2}{5}$ of the same book. Who read less?
Answer:**

Total number of pages = 100

Pages Ila read = 25

∴ Fraction of book Ila read = $\frac{25}{100}=\frac{1}{4}$ ...(i)

Fraction of book Lalita read = $\frac{2}{5}$ ...(ii)

To convert the fractions to like fraction, we multiply (i) by 5 and (ii) by 4.

$\frac{1}{4}\times \frac{5}{5}=\frac{5}{20}$

$\frac{2}{5}\times \frac{4}{4}=\frac{8}{20}$

Since $\frac{8}{2}\frac{5}{20}$ so $\frac{2}{5}\frac{1}{4}$

∴ between Ila and Lalita, Ila read less number of pages in the book.

**Q9. 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?
Answer:**

Time Rafiq exercised = $\frac{3}{6}$ of an hour ...(i)

Time Rohit exercised = $\frac{3}{4}$ of an hour ...(ii)

To convert the fractions to like fractions, we multiply (i) by 2 and (ii) by 3.

$\frac{3}{6}\times \frac{2}{2}=\frac{6}{12}$

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

Since $\frac{9}{12}\frac{6}{12}$ so $\frac{3}{4}\frac{3}{6}$

∴ between Rohit and Rafiq, Rohit exercised for a longer time.

**Q10. In a class A of 25 students, 20 passed in first class; in another class B of 30 students, 24 passed in first class. In which class was a greater fraction of students getting first class?
Answer:**

Total students in class A = 25

Fraction of students who passed in first class in class A = $\frac{20}{25}=\frac{4}{5}$

Total students in class B = 30

Fraction of students who passed in first class in class B = $\frac{24}{30}=\frac{4}{5}$

∴ Class A and B both have the same fraction of students getting first class.

** **

**Page 157**

** **

__Exercise 7.5__

**Q1. Write these fractions appropriately as additions or subtractions :
Answer:**

**Q2. Solve :
(a) $\frac{1}{18}+\frac{1}{18}$ (b) $\frac{8}{15}+\frac{3}{15}$
(c) $\frac{7}{7}-\frac{5}{7}$ (d) $\frac{1}{22}+\frac{21}{22}$
(e) $\frac{12}{15}-\frac{7}{15}$
(f) $\frac{5}{8}+\frac{3}{8}$ (g) $1-\frac{2}{3}\left(1=\frac{3}{3}\right)$ (h) $\frac{1}{4}+\frac{0}{4}$
(i) $3-\frac{12}{5}$
Answer:
(a) $\frac{1}{18}+\frac{1}{18}$**

$=\frac{1+1}{18}=\frac{2}{18}=\frac{{\displaystyle 1}}{9}$

$=\frac{8+3}{15}=\frac{11}{15}$

$=\frac{7-5}{7}=\frac{2}{7}$

$=\frac{1+21}{22}=\frac{22}{22}=1$

$=\frac{12-7}{15}=\frac{5}{15}=\frac{1}{3}$

$=\frac{5+3}{8}=\frac{8}{8}=1$

$\frac{3}{3}-\frac{2}{3}=\frac{3-2}{3}=\frac{1}{3}$

$=\frac{1+0}{4}=\frac{1}{4}$

$\left(\frac{3}{1}\times \frac{5}{5}\right)-\frac{12}{5}=\frac{15}{5}-\frac{12}{5}=\frac{15-12}{5}=\frac{3}{5}$

**Q3. 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?
Answer:**

Wall space painted by Shubham = $\frac{2}{3}$

Wall space painted by Madhavi = $\frac{1}{3}$

Total wall space painted by both of them = Wall space painted by Shubham + Wall space painted by Madhavi

Total wall space painted by both of them $=\frac{2}{3}+\frac{1}{3}$

$=\frac{2+1}{3}=\frac{3}{3}=1$

∴ together they both painted 1 wall.

**Q4. Fill in the missing fractions.
(a) $\frac{7}{10}-?=\frac{3}{10}$ (b) $?-\frac{3}{21}=\frac{5}{21}$
(c) $?-\frac{3}{6}=\frac{3}{6}$ (d) $?-\frac{5}{27}=\frac{12}{27}$
Answer:
(a) $\frac{7}{10}-?=\frac{3}{10}$**

$?=\frac{7}{10}-\frac{3}{10}=\frac{7-3}{10}=\frac{4}{10}\mathrm{or}\frac{2}{5}$

$?=\frac{5}{21}+\frac{3}{21}=\frac{5+3}{21}=\frac{8}{21}$

$?=\frac{3}{6}+\frac{3}{6}=\frac{3+3}{6}=\frac{6}{6}=1$

$?=\frac{12}{27}+\frac{5}{27}=\frac{12+5}{27}=\frac{17}{27}$

**Q5. Javed was given $\frac{5}{7}$ of a basket of oranges. What fraction of oranges was left in the basket?
Answer:**

Oranges given to Javed = $\frac{5}{7}$

Oranges left in the basket = 1 - Oranges given to Javed

$1-\frac{5}{7}=\left(\frac{{\displaystyle 1}}{{\displaystyle 1}}\times \frac{7}{7}\right)-\frac{5}{7}=\frac{7}{7}-\frac{5}{7}=\frac{7-5}{7}=\frac{2}{7}$

∴ $\frac{2}{7}$ oranges are left in the basket.

**Page 160**

** **

__Exercise 7.6__

**Q1. Solve :
(a) $\frac{2}{3}+\frac{1}{7}$ (b) $\frac{3}{10}+\frac{7}{15}$
(c) $\frac{4}{9}+\frac{2}{7}$ (d) $\frac{5}{7}+\frac{1}{3}$
(e) $\frac{2}{5}+\frac{1}{6}$
(f) $\frac{4}{5}+\frac{2}{3}$ (g) $\frac{3}{4}+\frac{1}{3}$
(h) $\frac{5}{6}+\frac{1}{3}$ (i) $\frac{2}{3}+\frac{3}{4}+\frac{1}{2}$
(j) $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$
(k) $1\frac{1}{3}+3\frac{2}{3}$ (l) $4\frac{2}{3}+3\frac{1}{4}$
(m) $\frac{16}{5}+\frac{7}{5}$ (n) $\frac{4}{3}+\frac{1}{2}$
Answer:**

(a) $\frac{2}{3}+\frac{1}{7}=\left(\frac{2}{3}\times \frac{7}{7}\right)+\left(\frac{1}{7}\times \frac{3}{3}\right)=\frac{14}{21}+\frac{3}{21}=\frac{14+3}{21}=\frac{17}{21}$

(b) $\frac{3}{10}+\frac{7}{15}=\left(\frac{3}{10}\times \frac{3}{3}\right)+\left(\frac{7}{15}\times \frac{2}{2}\right)=\frac{9}{30}+\frac{14}{30}=\frac{9+14}{30}=\frac{23}{30}$

(c) $\frac{4}{9}+\frac{2}{7}=\left(\frac{4}{9}\times \frac{7}{7}\right)+\left(\frac{2}{7}\times \frac{9}{9}\right)=\frac{28}{63}+\frac{18}{63}=\frac{28+18}{63}=\frac{46}{63}$

(d) $\frac{5}{7}+\frac{1}{3}=\left(\frac{5}{7}\times \frac{3}{3}\right)+\left(\frac{1}{3}\times \frac{7}{7}\right)=\frac{15}{21}+\frac{7}{21}=\frac{15+7}{21}=\frac{22}{21}$

(e) $\frac{2}{5}+\frac{1}{6}=\left(\frac{2}{5}\times \frac{6}{6}\right)+\left(\frac{1}{6}\times \frac{5}{5}\right)=\frac{12}{30}+\frac{5}{30}=\frac{12+5}{30}=\frac{17}{30}$

(f) $\frac{4}{5}+\frac{2}{3}=\left(\frac{4}{5}\times \frac{3}{3}\right)+\left(\frac{2}{3}\times \frac{5}{5}\right)=\frac{12}{15}+\frac{10}{15}=\frac{12+10}{15}=\frac{22}{15}$

(g) $\frac{3}{4}+\frac{1}{3}=\left(\frac{3}{4}\times \frac{3}{3}\right)+\left(\frac{1}{3}\times \frac{4}{4}\right)=\frac{9}{12}+\frac{4}{12}=\frac{9+4}{12}=\frac{13}{12}$

(h) $\frac{5}{6}+\frac{1}{3}=\frac{5}{6}+\left(\frac{1}{3}\times \frac{2}{2}\right)=\frac{5}{6}+\frac{3}{6}=\frac{5+3}{6}=\frac{8}{6}=\frac{4}{3}$

(i) $\frac{2}{3}+\frac{3}{4}+\frac{1}{2}=\left(\frac{2}{3}\times \frac{4}{4}\right)+\left(\frac{3}{4}\times \frac{3}{3}\right)+\left(\frac{1}{2}\times \frac{6}{6}\right)=\frac{8}{12}+\frac{9}{12}+\frac{6}{12}=\frac{8+9+6}{12}=\frac{23}{12}$

(j) $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\left(\frac{1}{2}\times \frac{3}{3}\right)+\left(\frac{1}{3}\times \frac{2}{2}\right)+\frac{1}{6}=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{3+2+1}{6}=\frac{6}{6}=1$

(k) $1\frac{1}{3}+3\frac{2}{3}=\frac{4}{3}+\frac{11}{3}=\frac{4+11}{3}=\frac{15}{3}=5$

(l) $4\frac{2}{3}+3\frac{1}{4}=\frac{14}{3}+\frac{13}{4}=\left(\frac{14}{3}\times \frac{4}{4}\right)+\left(\frac{13}{4}\times \frac{3}{3}\right)=\frac{56}{12}+\frac{39}{12}=\frac{56+39}{12}=\frac{95}{12}\mathrm{or}7\frac{11}{12}$

(m) $\frac{16}{5}+\frac{7}{5}=\frac{16+7}{5}=\frac{23}{5}$

(n) $\frac{4}{3}+\frac{1}{2}=\left(\frac{4}{3}\times \frac{2}{2}\right)+\left(\frac{1}{2}\times \frac{3}{3}\right)=\frac{8}{6}+\frac{3}{6}=\frac{8+3}{6}=\frac{11}{6}$

**Q2. Sarita bought $\frac{2}{5}$ metre of ribbon and Lalita $\frac{3}{4}$ metre of ribbon. What is the total length of the ribbon they bought?
Answer:**

Length of Sarita's ribbon = $\frac{2}{5}$ metre

Length of Lalita's ribbon = $\frac{3}{4}$ metre

Total length of ribbon = Length of Sarita's ribbon + Length of Lalita's ribbon

$=\frac{2}{5}+\frac{3}{4}=\left(\frac{2}{5}\times \frac{4}{4}\right)+\left(\frac{3}{4}\times \frac{5}{5}\right)=\frac{8}{20}+\frac{15}{20}=\frac{8+15}{20}=\frac{23}{20}$ metre

∴ the total length of ribbon they brought is $\frac{23}{20}$ metre.

**Q3. Naina was given $1\frac{1}{2}$ piece of cake and Najma was given $1\frac{1}{3}$ piece of cake. Find the total amount of cake was given to both of them.
Answer:**

Piece of cake given to Naina = $1\frac{1}{2}=\frac{3}{2}$

Piece of cake given to Najma = $1\frac{1}{3}=\frac{4}{3}$

Total amount of cake given to both of them = Piece of cake given to Naina + Piece of cake given to Najma

$\frac{3}{2}+\frac{4}{3}=\left(\frac{3}{2}\times \frac{3}{3}\right)+\left(\frac{4}{3}\times \frac{2}{2}\right)=\frac{9}{6}+\frac{8}{6}=\frac{17}{6}$

∴ the total amount of cake given to both of them was $\frac{17}{6}\mathrm{or}2\frac{5}{6}$.

**Q4. Fill in the blank :
(a) $\_-\frac{5}{8}=\frac{1}{4}$ (b) $\_-\frac{1}{5}=\frac{1}{2}$
(c) $\frac{1}{2}-\_=\frac{1}{6}$
Answer:**

(a) $\_-\frac{5}{8}=\frac{1}{4}$

$?=\frac{1}{4}+\frac{5}{8}=\left(\frac{1}{4}\times \frac{2}{2}\right)+\frac{5}{8}=\frac{2}{8}+\frac{5}{8}=\frac{2+5}{8}=\frac{7}{8}$

(b) $\_-\frac{1}{5}=\frac{1}{2}$

$?=\frac{1}{2}+\frac{1}{5}=\left(\frac{1}{2}\times \frac{5}{5}\right)+\left(\frac{1}{5}\times \frac{2}{2}\right)=\frac{5}{10}+\frac{2}{10}=\frac{5+2}{10}=\frac{7}{10}$

(c) $\frac{1}{2}-\_=\frac{1}{6}$

$?=\frac{1}{2}-\frac{1}{6}=\left(\frac{1}{2}\times \frac{3}{3}\right)-\frac{1}{6}=\frac{3}{6}-\frac{1}{6}=\frac{3-1}{6}=\frac{2}{6}=\frac{1}{3}$

**Q5. Complete the addition-subtraction box.
Answer:**

**Q6. A piece of wire $\frac{7}{8}$ metre long broke into two pieces. One piece was $\frac{1}{4}$ metre long. How long is the other piece?
Answer:**

Given: Length of wire before breaking = $\frac{7}{8}$ metre

After breaking into two pieces, length of one piece = $\frac{1}{4}$ metre

Total length of wire = Length of one piece + Length of second piece

Length of second piece = Total length of wire - Length of one piece

$\frac{7}{8}-\left(\frac{1}{4}\times \frac{2}{2}\right)=\frac{7}{8}-\frac{2}{8}=\frac{7-2}{8}=\frac{5}{8}$

∴ the other piece of wire is $\frac{5}{8}$ metre long.

**Q7. 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?
Answer:**

Given: Total distance between Nandini's house and school = $\frac{9}{10}$ km

Distance travelled by Nandini by bus = $\frac{1}{2}$ km

Total distance = Distance travelled by Nandini walking + Distance travelled by Nandini by bus

Distance travelled by Nandini walking = Total Distance - Distance travelled by Nandini by bus

$\frac{9}{10}-\left(\frac{1}{2}\times \frac{5}{5}\right)=\frac{9}{10}-\frac{5}{10}=\frac{9-5}{10}=\frac{4}{10}\mathrm{or}\frac{2}{5}$

∴ Nandini walked $\frac{2}{5}$ km.

**Q8. Asha and Samuel have bookshelves of the 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 more full? By what fraction?
Answer:**

Asha's bookshelf is filled = $\frac{5}{6}$... (i)

Samuel's bookshelf is filled = $\frac{2}{5}$...(ii)

Since the fractions are not like fractions, we convert these to like fractions by multiplying and dividing (i) with 5 and (ii)) with 6.

$\frac{5}{6}\times \frac{5}{5}=\frac{25}{30}$

$\frac{2}{5}\times \frac{6}{6}=\frac{12}{30}$

We know that 25 > 12 so we can say that $\frac{5}{6}$ > $\frac{2}{5}$

The difference between them

$\frac{5}{6}-\frac{2}{5}=\frac{25}{30}-\frac{12}{30}=\frac{25-12}{30}=\frac{13}{30}$

∴ Asha's bookshelf is $\frac{13}{30}$ more full than Samuel's bookshelf.

**Q9. 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?
Answer:**

Time taken by Jaidev to walk across the school ground = $2\frac{1}{5}=\frac{11}{5}$... (i)

Time taken by Rahul to walk across the school ground = $\frac{7}{4}$...(ii)

Since the fractions are not like fractions, we convert these to like fractions by multiplying and dividing (i) with 4 and (ii)) with 5.

$\frac{11}{5}\times \frac{4}{4}=\frac{44}{20}$

$\frac{7}{4}\times \frac{5}{5}=\frac{35}{20}$

We know that 44 > 35 so we can say that $\frac{11}{5}$ > $\frac{7}{4}$

The difference between the time taken by Jaidev and Rahul

$\frac{11}{5}-\frac{7}{4}=\frac{44}{20}-\frac{35}{20}=\frac{44-35}{20}=\frac{9}{20}$

∴ Rahul takes $\frac{9}{20}$mins less time than Jaidev to walk across the school ground.

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