Andhra Pradesh AP Board 5th Class Maths Solutions 8th Lesson Fractions Textbook Exercise Questions and Answers.

## AP State Syllabus 5th Class Maths Solutions Chapter 8 Fractions

I. Observe the following table.

Answer:

What do you observe from the above table?

Answer:

Here, in all fractions the numerator is less than the denominator.

II. If Hema and Gopi got 7 and 9 biscuits then complete this table.

Answer:

III. Observe the fractions : \(\frac{3}{4}, \frac{4}{5}, \frac{6}{7}, \frac{12}{13}, \frac{25}{28}\)

Answer:

What do you say?

Answer:

Here is all fractions have numerator is greater than or equal to denominator. These types of fractions are called improper fractions.

Do this: (TextBook Page No.129)

Question 1.

Write any 5 proper fractions.

Answer:

Proper fractions: \(\frac{3}{4}, \frac{4}{5}, \frac{6}{7}, \frac{12}{13}, \frac{25}{28}\)

Question 2.

Write any 5 1m proper fractions.

Answer:

Improper fractions: \(\frac{7}{6}, \frac{26}{22}, \frac{21}{20}, \frac{28}{25}, \frac{13}{12}\)

Question 3.

Write any 5 mixed fractions.

Answer:

Mixed fractions: \(3 \frac{2}{3}, 7 \frac{1}{2}, 9 \frac{3}{5}, 8 \frac{2}{3}, 6 \frac{5}{7}\)

Question 4.

Convert these fraction into mixed fraction \(\frac{5}{2}, \frac{7}{3}, \frac{9}{4}, \frac{11}{2}\)

Answer:

Conversion of fraction into mixed fraction.

Question 5.

Convert these fractions into improper fraction \(4 \frac{2}{3}, 5 \frac{3}{4}, 6 \frac{2}{5}, 3 \frac{1}{2}\).

Answer:

Conversion of Mixed fractions into improper fractions

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

Do this: (TextBook Page No. 132)

Question 1.

Write any three equivalent fractions to the given fractions.

a) \(\frac{4}{8}\)

b) \(\frac{1}{3}\)

c) \(\frac{3}{7}\)

d) \(\frac{20}{24}\)

Answer:

a) Equivalent fractions to \(\frac{4}{8}=\frac{8}{16}=\frac{12}{24}=\frac{16}{32}\)

b) Equivalent fractions to \(\frac{1}{3}=\frac{3}{9}=\frac{2}{6}=\frac{4}{12}\)

c) Equivalent fractions to \(\frac{3}{7}=\frac{9}{21}=\frac{6}{14}=\frac{12}{28}\)

d) Equivalent fractions to \(\frac{20}{24}=\frac{40}{48}=\frac{60}{72}=\frac{80}{96}\)

Exercise 1:

Question 1.

Simplify the following fractions. (by cancellation method).

(i) \(\frac{105}{15}\)

(ii) \(\frac{200}{20}\)

(iii) \(\frac{7}{10}\)

(iv) \(\frac{666}{66}\)

(v) \(\frac{125}{1000}\)

(vi) \(\frac{120}{200}\)

Answer:

Question 2.

Simplify the following fractions. (by H.C.F. method)

(i) \(\frac{12}{18}\)

(ii) \(\frac{14}{35}\)

(iii) \(\frac{22}{55}\)

(iv) \(\frac{27}{36}\)

(v) \(\frac{128}{164}\)

(vi) \(\frac{210}{427}\)

Answer:

(i) HCF of 12 and 18 is 6.

\(\frac{12 \div 6}{18 \div 6}=\frac{2}{3}\)

So, \(\frac{2}{3}\) is the simplest form of \(\frac{12}{18}\).

(ii) HCF of 14 and 35 is 7.

\(\frac{14 \div 7}{35 \div 7}=\frac{2}{5}\)

So, \(\frac{2}{5}\) is the simplest form of \(\frac{14}{35}\).

(iii) HCF of 22 and 55 is 11.

\(\frac{22 \div 11}{55 \div 11}=\frac{2}{5}\)

So, \(\frac{2}{5}\) is the simplest form of \(\frac{22}{55}\).

(iv) HCF of 27 and 36 is 9.

\(\frac{27 \div 9}{36 \div 9}=\frac{3}{4}\)

So, \(\frac{3}{4}\) is the simplest form of \(\frac{27}{36}\).

(v) HCF of 128 and 164 is 4.

\(\frac{128 \div 4}{164 \div 4}=\frac{32}{41}\)

So, \(\frac{32}{41}\) is the simplest form of \(\frac{128}{164}\).

(vi) HCF of 210 and 427 is 7.

\(\frac{210 \div 7}{427 \div 7}=\frac{30}{61}\)

So, \(\frac{30}{61}\) is the simplest form of \(\frac{210}{427}\).

Question 3.

Convert the following fractions into the simplest form by both the methods.

(i) \(\frac{16}{64}\)

(ii) \(\frac{12}{18}\)

(iii) \(\frac{30}{50}\)

(iv) \(\frac{40}{25}\)

(v) \(\frac{16}{32}\)

(vi) \(\frac{8}{40}\)

Answer:

(i) 1s tMethod: HCF of 16 and 64 is 4.

\(\frac{16 \div 4}{64 \div 4}=\frac{4}{16}=\frac{1}{4}\)

So, \(\frac{1}{4}\) is the simplest form of \(\frac{16}{64}\).

2nd Method =

(ii) 1st Method HCF of 12 and 28 is 4.

\(\frac{12 \div 4}{28 \div 4}=\frac{3}{7}\),

So \(\frac{3}{5}\) is simplest form of \(\frac{12}{28}\) .

2nd Method: =

iii) 1st Method : HCF of 30 and 50 is 10.

\(\frac{30 \div 10}{50 \div 10}=\frac{3}{5}\)

So, \(\frac{3}{5}\) is simplest form of \(\frac{30}{50}\)

2nd method \(\frac{30}{50}=\frac{3}{5}\).

(iv) 1st method: HCF of 40 and 25 is 5

\(\frac{40 \div 5}{25 \div 5}=\frac{8}{5}\)

So \(\frac{8}{5}\) is simplest form of \(\frac{40}{25}\)

2nd method:

(v) 1st method: HCF of 16 and 32 is 16.

\(\frac{16+16}{32+16}=\frac{1}{2}\),

So \(\frac{1}{2}\) is the simplest form of \(\frac{16}{32}\).

2nd method:

vi) 1st method: HCF of 8 and 40 is 8.

\(\frac{8 \div 8}{40 \div 8}=\frac{1}{5}\),

So \(\frac{1}{5}\) is simplest form of \(\frac{8}{48}\).

2nd method:

Question 4.

To get equivalent fractions what should we do a given fraction ?

Answer:

To get equivalent fractions, we multiply / divide both numerator and denominator by the same number.

Question 5.

Write any three equivalent fractions to the given fractions.

(i) \(\frac{5}{8}\)

(ii) \(\frac{32}{64}\)

(iii) \(\frac{3}{7}\)

(iv) \(\frac{125}{255}\)

(v) \(\frac{7}{10}\)

Answer:

(i) Equivalent fractions to \(\frac{5}{8}\) is \(\frac{10}{16}, \frac{15}{24} \text { and } \frac{20}{32}\)

(ii) Equivalent fractions to \(\frac{1}{2}, \frac{2}{4}, \frac{4}{8}, \frac{8}{16}\)

(iii) Equivalent fractions to \(\frac{3}{7}\) is \(\frac{6}{14}, \frac{9}{21}, \frac{12}{28\)

(iv) Equivalent fractions to \(\frac{125}{255}\) is \(\frac{5}{9}, \frac{25}{45} \text { and }\)

(v) Equivalent fractions to \(\frac{7}{10}\) is \(\frac{14}{20}, \frac{21}{30}\) and \(\frac{35}{50}\)

Question 6.

Govindamma distributed her 4 acrse of land to her 3 sons, then write the part of land each got in the form of a fraction.

Answer:

Total land = 4 acres

No. of sons = 3

Part of land each got in the form of fraction = \(\frac{4}{3}\) = 1\(\frac{1}{3}\).

Do these : (TextBook Page No.137)

Question 1.

Observe the example and write the correct fractions in the other circles.

Answer:

Question 2.

Find the sum.

(i) \(\frac{2}{10}+\frac{4}{10}\)

(ii) \(\frac{2}{6}+\frac{3}{6}\)

(iii) \(1 \frac{1}{4}+3 \frac{1}{4}\)

(iv) \(2 \frac{1}{5}+3 \frac{1}{5}\)

Answer:

(i) \(\frac{2}{10}+\frac{4}{10}=\frac{2+4}{10}=\frac{6}{10}\)

(ii) \(\frac{2}{6}+\frac{3}{6}=\frac{2+3}{6}=\frac{5}{6}\)

(iii) \(1 \frac{1}{4}+3 \frac{1}{4}=\frac{5}{4}+\frac{13}{4}=\frac{18}{4}\)

(iv) \(2 \frac{1}{5}+3 \frac{1}{5}=\frac{11}{5}+\frac{16}{5}+\frac{27}{5}\)

Question 3.

\(\frac{1}{2}\) kg of a sugar packet, \(\frac{3}{6}\) kg of jaggery are in a bag. Then what is the

total weight of two items in the bag?

Answer:

Weight of sugar packet = \(\frac{1}{2}\) kg

Weight ofjaggery packet = \(\frac{3}{6}\) kg = \(\frac{1}{2}\) kg

Total weight of two items = \(\frac{1}{2}\) + \(\frac{3}{6}\)

\(\frac{3}{6}+\frac{3}{6}=\frac{3+3}{6}=\frac{6}{6}\) = 1.

Question 4.

Sakru paints \(\frac{1}{5}\) th part of a wall on first day. \(\frac{2}{5}\) th part of the wall on second day Then how much part lie painted in both the days?

Answer:

Painting part of wall on first day = \(\frac{1}{5}\)

Painting part of wall on second day = \(\frac{2}{5}\)

Painting part of wall on both the days = \(\frac{1}{5}+\frac{2}{5}=\frac{1+2}{5}=\frac{3}{5}\)

Question 5.

Polamma had some money. She spent \(\frac{3}{6}\)th part of money on books. \(\frac{1}{6}\) th part of money on pens, pencils and erasers. Then how much part money did she spend ¡n total?

Answer:

Money spent on books = \(\frac{3}{6}\)th part

Money spent on pens = \(\frac{1}{6}\)th part

pencils and erasers.

Money spent in total = \(\frac{3}{6}+\frac{1}{6}=\frac{3+1}{6}\) = \(\frac{4}{6}\)th part

Do these: (TextBook Part No. 139)

Question 1.

Complete this.

Answer:

Question 2.

Find the sum.

(i) \(\frac{1}{5}+\frac{3}{4}\)

(ii) \(\frac{3}{4}+\frac{5}{6}\)

(iii) \(1 \frac{2}{3}+2 \frac{5}{6}\)

(iv) \(3 \frac{1}{8}+2 \frac{5}{6}\)

Answer:

Question 3.

Seetamma read \(\frac{1}{5}\) th part of a book on Monday, \(\frac{4}{10}\) part of the book on Tuesday. Then how much part did she complete on two days ?

Answer:

Book read on Monday = \(\frac{1}{5}\) th part

Book read on Tuesday = \(\frac{4}{10}\) th part

Book read on two days = \(\frac{1}{5}\) + \(\frac{4}{10}\)

LCM of 5 and 10 is 10.

= \(\frac{1}{5} \times \frac{2}{2}+\frac{4}{10} \times \frac{1}{1}\)

= \(\frac{2}{10}+\frac{4}{10}=\frac{2+4}{10}=\frac{6}{10}\)

Question 4.

Polayya painted a wall of \(\frac{3}{4}\) th part on 1st day and \(\frac{3}{6}\) th part of the wall on 2nd day. Then how much part he painted the wall in two days ?

Answer:

Painted part of wal on 1 st day = \(\frac{1}{5}\)

Painted part of wall on 2nd day = \(\frac{3}{6}\)

Painted part in two days = \(\frac{1}{5}\) + \(\frac{3}{6}\)

LCM of 5 and 6 is = 30

= \(\frac{1}{5} \times \frac{6}{6}+\frac{3}{6} \times \frac{5}{5}=\frac{21}{30}=\frac{7^{t h}}{10} \text { Part }\)

Try this: (TextBook Part No.139)

Question 1.

Add \(5 \frac{6}{8}+4 \frac{1}{7}\)

Answer:

Do this: (TextBook Page No.141)

Question 1.

Complete this:

Answer:

Question 2.

Do the following.

(i) \(\frac{6}{10}-\frac{1}{10}\)

(ii) \(\frac{3}{15}-\frac{1}{15}\)

(iii) \(1 \frac{3}{15}-1 \frac{1}{15}\)

(iv) \(2 \frac{4}{7}-1 \frac{2}{7}\)

Answer:

(i) \(\frac{6}{10}-\frac{1}{10}\)

= \(\frac{6-1}{10}=\frac{5}{10}\)

(ii) \(\frac{3}{15}-\frac{1}{15}\)

= \(\frac{3-1}{15}-\frac{2}{15}\)

Question 3.

Eswar painted \(\frac{1}{6}\) th part of a wall on first day. Then how much part will remain to complete?

Answer:

Painted part ofwal on Sunday = \(\frac{1}{6}\)th part

Total part = 1

Remaining part to complete = \(\frac{1}{1}-\frac{1}{6}\)

LCM of 1 and 6 is 6.

= \(\frac{1}{1} \times \frac{6}{6}-\frac{1}{6} \times \frac{1}{1}=\frac{6}{6}-\frac{1}{6}\)

= \(\frac{6-1}{6}-\frac{5}{6}\).

Question 4.

Gown completed \(\frac{1}{4}\) part of her homework on Sunday. \(\frac{5}{12}\) part on Sunday morning. How much part did she complete? How much part of home work is left?

Answer:

Work complete on Saturday = \(\frac{1}{4}\)

Work completed on Sunday = \(\frac{5}{12}\)

Work completed both days = \(\frac{1}{4}\) + \(\frac{5}{12}\)

LCM of 4 and 12 is 12.

= \(\frac{1}{4} \times \frac{3}{3}+\frac{5}{12} \times \frac{1}{1}\)

= \(\frac{3}{12}+\frac{5}{12}=\frac{3+5}{12}=\frac{8}{12}\)

Remainig part left = \(\frac{1}{1}-\frac{8}{12}\)

LCM of 1 and 12 is 12

= \(\frac{1}{1} \times \frac{12}{12}-\frac{8}{12} \times \frac{1}{1}\)

= \(\frac{12}{12}-\frac{8}{12}=\frac{12-8}{12}=\frac{4}{12}=\frac{1}{3} \text { th part }\)

Try these: (TextBook Part No.142)

Question 1.

Complete this:

Answer:

Exercise 2:

Question 1.

Do the following:

a) \(\frac{3}{4}+\frac{7}{4}\)

b) \(1 \frac{1}{2}\)

c) \(\frac{8}{3}+\frac{2}{5}\)

d) \(\frac{6}{3}+\frac{7}{4}\)

e) \(\frac{3}{5}+\frac{9}{11}\)

f) \(\frac{10}{10}+\frac{5}{20}\)

g) \(\frac{9}{10}+\frac{4}{15}\)

h) \(\frac{5}{20}+\frac{13}{30}\)

Answer:

Question 2.

Do the following.

a) \(\frac{3}{7}-\frac{1}{7}\)

b) 6 – \(\frac{1}{3}\)

c) \(\frac{3}{8}-\frac{3}{16}\)

d) \(\frac{8}{7}-\frac{5}{8}\)

e) \(\frac{8}{7}-\frac{5}{8}\)

f) \(\frac{13}{15}-\frac{7}{20}\)

g) \(\frac{63}{40}-\frac{9}{10}\)

h) \(\frac{7}{15}-\frac{3}{10}\)

Answer:

a) \(\frac{3}{7}-\frac{1}{7}\)

= \(\frac{3-1}{7}-\frac{2}{7}\)

Question 3.

Find the difference between 5\(\frac{1}{3}\) and 2\(\frac{4}{7}\)

Answer:

\(\frac{15+1}{3}-\frac{14+4}{7}=\frac{16}{3}-\frac{18}{7}\)

LCM of 3 and 7 = 21

\(\frac{16}{3} \times \frac{7}{7}-\frac{18}{7} \times \frac{3}{3}\)

= \(\frac{112}{21}-\frac{54}{21}=\frac{58}{21}\)

Question 4.

Seetha purchased 1\(\frac{1}{2}\) litre of sunflower oil, \(\frac{3}{4}\) litre of groundnut oil. How much of oil she purchased in total?

Answer:

Purchased quantity of sunflower oil = 1\(\frac{1}{2}\) litr.

Purchased quantity of ground nut oil = \(\frac{3}{4}\) litr.

Purchased quantity of total oil = 1 \(\frac{1}{2}\) + \(\frac{3}{4}\)

= \(\frac{3}{2}\) + \(\frac{3}{4}\)

LCM of 2, 4 is 4

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

= \(\frac{6}{4}+\frac{3}{4}=\frac{6+3}{4}=\frac{9}{4}\)

Question 5.

Vimala purchased 1 \(\frac{3}{4}\) m of cotton cloth for skirt, \(\frac{3}{4}\) m of cloth for blouse. How much cloth is purchased by her?

Answer:

Purchased cloth for skirt = 1 \(\frac{3}{4}\) m

Purchased cloth for blouse = \(\frac{3}{4}\) m

Purchased cloth Total = 1\(\frac{3}{4}\) + \(\frac{3}{4}\)

= \(\frac{7}{4}\) + \(\frac{3}{4}\) = \(\frac{10}{4}\) m.

Question 6.

A water tank is filled with \(\frac{9}{10}\)th part of water, but \(\frac{3}{5}\) th part of water is consumed in a day. Then find the remaining part of water in the tank?

Answer:

Total quantity of water infilled in tank = \(\frac{9}{10}\) th part

Consumed quantity of water is = \(\frac{3}{5}\) th part

Remaining part of water in the tank = \(\frac{9}{10}-\frac{3}{5}=\frac{9-6}{10}=\frac{3}{10} \text { th part }\)

Do these: (TextBook Page No.145)

Question 1.

Read 485.267

Answer:

Four hundred and eighty five (point) two six seven.

Question 2.

Write the place value of all digits in 293.819

Answer:

Given numbers = 293.819

place valu e of 2 = 200

place value of 9 = 90

place value of 5 = 5

place value of 8 = \(\frac{1}{80}\)

place value of 1 = \(\frac{1}{900}\)

place value of 9 = \(\frac{9}{1000}\)

Question 3.

Write any 5 examples for decimal fractions.

Answer:

(i) \(\frac{4756}{100}\) = 47.56

(ii) \(\frac{87685}{1000}\) = 87.685

(iii) \(\frac{763407}{1000}\) = 763.407

(iv) \(\frac{86734}{10000}\) = 8.6734

(v) \(\frac{96302}{10}\)= 9630.2

Exercise 3:

Question 1.

Fill in the blanks:

a) In improper fraction, numerator is ………………… than the denominator.

Answer:

greater

b) \(\frac{6}{6}\) is ………………… fraction (which type?)

Answer:

improper

c) 3\(\frac{1}{2}\) is ………………… fraction (which type?)

Answer:

mixed

d) \(\frac{9}{6}\) ………………… is fraction (which type?)

Answer:

improper

e) \(\frac{2}{5}\) is ………………… fraction (which type?)

Answer:

proper

f) A function having whole number and proper fraction is called ………………… fraction.

Answer:

(mixed)

Question 2.

Convert \(\frac{9}{6}\) into mixed fraction.

Answer:

Mixed fraction of \(\frac{9}{6}=1 \frac{3}{6}\).

Question 3.

Convert 2\(\frac{1}{5}\) into an improper fraction.

Answer:

Improper fraction of \(2 \frac{1}{5}=\frac{2 \times 5+1}{5}=\frac{11}{5}\).

Question 4.

Write any 5 equivalent fractions to \(\frac{2}{3}\).

Answer:

Equivalent fractions of \(\frac{2}{3}\) is \(\frac{4}{6}, \frac{6}{9}, \frac{8}{12}, \frac{10}{15}\) and \(\frac{16}{18}\).

Question 5.

Write simplest form of fraction for \(\frac{25}{75}\).

Answer:

Cancilation method: =

∴ Simplest form of \(\frac{25}{75}\) is \(\frac{1}{3}\).

Question 6.

Write two equivalent fractions to \(\frac{64}{36}\).

Answer:

Equivalent fractions to \(\frac{64}{36}\)

= \(\frac{64 \div 2}{36 \div 2}=\frac{32}{18}\)

= \(\frac{64 \div 4}{36 \div 4}=\frac{16}{9}\).

Question 7.

Classify the following as like and unlike frations.

\(\frac{3}{5}, \frac{2}{7}, \frac{8}{5}, \frac{9}{5}, \frac{8}{4}, \frac{1}{5}\)

Answer:

Like fractions: \(\frac{1}{5}, \frac{3}{5}, \frac{8}{5}, \frac{9}{5}\)

Unlike fractions = \(\frac{2}{7}, \frac{8}{4}\).

Question 8.

Fill in the blanks.

a) \(\frac{15}{20}=\frac{3}{\square}\)

b) \(\frac{2}{5}=\frac{\square}{50}\)

c) \(\frac{3}{5}=\frac{\square}{30}\)

Answer:

a) \(\frac{15}{20}=\frac{3}{4}\)

b) \(\frac{2}{5}=\frac{20}{50}\)

c) \(\frac{3}{5}=\frac{18}{30}\)

Question 9.

Fill in the blanks with = or ≠ (≠ denotes not equal to)

a) \(\frac{1}{2}\) ____ \(\frac{8}{16}\)

b) \(\frac{9}{15}\) ____ \(\frac{27}{30}\)

c) \(\frac{6}{13}\) ____ \(\frac{12}{39}\)

Answer:

a) \(\frac{1}{2}\) = \(\frac{8}{16}\)

b) \(\frac{9}{15}\) ≠ \(\frac{27}{30}\)

c) \(\frac{6}{13}\) ≠ \(\frac{12}{39}\)

Question 10.

Fill in the blanks with equivalent fractions.

a) \(\frac{1}{2}\) —- \(\frac{8}{16}\) ____, ____, ____

Answer:

a) \(\frac{1}{2}\) —- \(\frac{8}{16} \frac{2}{4}, \frac{3}{6}, \frac{5}{10}\).

Question 11.

a) \(\frac{6}{5}+\frac{1}{5}\) = ———-

b) \(\frac{5}{7}+\frac{2}{14}\) = ———-

c) \(\frac{15}{32}+\frac{3}{8}\) = ———-

d) \(\frac{11}{16}+1 \frac{1}{8}\) = ———-

Answer:

Question 12.

Kavitha studied \(\frac{1}{2}\) part of a book on 1st day, \(\frac{1}{3}\) part on 2nd day. then how much part she studied in both the days ?

Answer:

On I st day completed book part = \(\frac{1}{2}\)

On 2nd day completed book part = \(\frac{1}{3}\)

On both days completed book part = \(\frac{1}{2}\) + \(\frac{1}{3}\)

LCM of 2 and 3 is 6.

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

= \(\frac{3}{6}+\frac{2}{6}\)

= \(\frac{3+2}{6}=\frac{5}{6}\).

Question 13.

Koushik went to school km by walk. He went on bicycle with his friend for the remaining distance \(\frac{3}{4}\) km. Then find the distance to school from his house.

Answer:

Distance covered by walk = \(\frac{1}{4}\) km

Distance covered by bicycle = \(\frac{3}{4}\) km

Total distance to school from his house = \(\frac{1}{4}\) + \(\frac{3}{4}\)

= \(\frac{1+3}{4}=\frac{4}{4}\) km

Question 14.

a) \(\frac{8}{10}-\frac{2}{10}\) = ………………

b) \(\frac{1}{3}-\frac{1}{9}\) = ………………

c) \(\frac{15}{32}-\frac{3}{8}\) = ………………

d) \(6 \frac{1}{16}-1 \frac{1}{8}\) = ………………

Answer:

Question 15.

\(\frac{2}{3}\)rd nd part of students in a school is boys. Find the part of girls.

Answer:

Part of boys in a school = \(\frac{2}{3}\)rd

Part of girls inaschool= 1 – \(\frac{2}{3}\).

= \(\frac{3-2}{3}=\frac{1}{3}\)rd.

Question 16.

Subtract \(\frac{21}{4}\) from the total of \(\frac{7}{2}\) and \(\frac{8}{3}\).

Answer:

Total of \(\frac{7}{2}\) and \(\frac{8}{3}\) = \(\frac{7}{2} \times \frac{3}{3}+\frac{8}{3} \times \frac{2}{2}\)

= \(\frac{21}{6}+\frac{16}{6}=\frac{21+16}{6}=\frac{37}{6}\)

Subtract of \(\frac{37}{6}-\frac{21}{4}\) =

LCM of 6 and 4 is 12 = \(\frac{37}{6} \times \frac{2}{2}-\frac{21}{4} \times \frac{3}{3}=\frac{74-63}{12}=\frac{11}{12}\)

Question 17.

Govind studied \(\frac{2}{5}\) th part of a book on 1st day, \(\frac{1}{7}\) th part on 2nd day. Then how much part is yet to be completed?

Answer:

Completed part of book on 1st day = \(\frac{2}{5}\)th

Completed part of book on 2nd day = \(\frac{1}{7}\)th

Completed part of book on two days = \(\frac{2}{5}\) + \(\frac{1}{7}\)

LCM of 5 and 7 is 35.

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

= \(\frac{14}{35}+\frac{5}{35}=\frac{14+5}{35}=\frac{19}{35}\)

remaining part of book yet to be completed = 1 – \(\frac{19}{35}\)

= \(\frac{1 \times 35}{35}-\frac{19}{35} \times \frac{1}{1}\)

= \(\frac{35}{35}-\frac{19}{35}=\frac{35-19}{35}=\frac{14}{35}\)

Question 18.

Write in words 189.257

Answer:

One hundred and eight nine point two five seven.

Question 19.

Write the place value of 6 ¡n 489.167

Answer:

6 is in \(\frac{1}{100}\)th place.