# NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals (EX 2.3) Exercise 2.3

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## NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals (EX 2.3) Exercise 2.3

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### Access NCERT Solutions for Class 7 Maths Chapter 2 – Fractions and Decimals

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### NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.3

NCERT textbooks are a series of textbooks that are written and released by the National Council of Educational Research and Training (NCERT) in India. A list of suggested books has been established by the Central Board of Secondary Education (CBSE), which regulates the use of textbooks in board-affiliated schools. Every subject taught at the educational level is covered in the books from Class 1 to Class 12. NCERT Solutions are available to students to help them understand the concepts presented in the textbooks. The answers in the area of Mathematics are step-by-step presentations that include illustrations, graphs, and figures. NCERT Solutions, for example, NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.3, also includes solutions for in-text examples and sample questions that help teachers assess how well their students have understood the subject matter.

On the contrary, Chapter 2 of the Mathematics subject in Class 7 describes solving the Fractions and Decimals related queries. In this Chapter, students will study the following given topics:

2) Fractions – Proper, Improper, and Mixed Fractions

3) Multiplication of Fractions – by a whole number, by a Fraction

4) Fraction as an operator ‘of’

5) Division of Fractions – division of the whole number by a Fraction and Reciprocal of a Fraction

6) Division of a Fraction by a Whole number and by another Fraction

7) Decimal Numbers – Multiplication

8) Multiplication of Decimal Numbers by 10, 100, and 1000

9) Division of Decimal Numbers – Division by 10, 100, and 1000, by a Whole Number, and by another Decimal Number

In Class 7 Maths Exercise 2.3, there are numerous questions that can be solved quickly with the guidance of NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.3. The questions that are there in Exercise 2.3 are as follows:

1. a) Finding the Fraction of given numbers
2. b) Multiplying Fractions to reduce them to their lowest forms
3. c) Specifying which fraction is greater than the other
4. d) Finding the distance between the saplings
5. e) the simplest form of numbers is asked when an equation is given to provide an outcome after multiplying the equation.

The Class 7 Maths Chapter 2 Exercise 2.3 consists of some vital topics regarding Fractions that can be solved by the solutions of NCERT Class 7 Maths Chapter 2 Exercise 2.3.

### NCERT Solutions for Class 7

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Q.1

$\begin{array}{l}Find:\\ \left(i\right)\text{\hspace{0.17em}}\frac{1}{4}of\text{}\left(a\right)\frac{1}{4}\text{}\text{}\left(b\right)\text{\hspace{0.17em}}\frac{3}{5}\text{}\left(c\right)\text{\hspace{0.17em}}\frac{4}{3}\\ \left(ii\right)\text{\hspace{0.17em}}\frac{1}{7}\text{\hspace{0.17em}}of\left(a\right)\text{\hspace{0.17em}}\frac{2}{9}\text{}\left(b\right)\text{​}\text{\hspace{0.17em}}\frac{6}{5}\text{\hspace{0.17em}}\text{}\left(c\right)\text{​}\frac{3}{10}\end{array}$

Ans.

$\begin{array}{l}\left(i\right)\\ \left(a\right)\text{\hspace{0.17em}}\frac{1}{4}\text{\hspace{0.17em}}\text{of}\frac{1}{4}=\frac{1}{4}×\frac{1}{4}=\overline{)\frac{1}{16}}\\ \left(b\right)\text{\hspace{0.17em}}\frac{1}{4}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{3}{5}=\frac{1}{4}×\frac{3}{5}=\overline{)\frac{3}{20}}\\ \left(c\right)\text{\hspace{0.17em}}\frac{1}{4}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{4}{3}=\frac{1}{\overline{)4}}×\frac{\overline{)4}}{3}=\overline{)\frac{1}{3}}\\ \left(ii\right)\\ \left(a\right)\text{\hspace{0.17em}}\frac{1}{7}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{2}{9}=\frac{1}{7}×\frac{2}{9}=\overline{)\frac{2}{63}}\\ \left(b\right)\text{\hspace{0.17em}}\frac{1}{7}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{6}{5}=\frac{1}{7}×\frac{6}{5}=\overline{)\frac{6}{35}}\\ \left(c\right)\text{\hspace{0.17em}}\frac{1}{7}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{3}{10}=\frac{1}{7}×\frac{3}{10}=\overline{)\frac{3}{70}}\end{array}$

Q.2

$\begin{array}{l}\mathrm{Multiply}\mathrm{and}\mathrm{reduce}\mathrm{to}\mathrm{lowest}\mathrm{form}\left(\mathrm{if}\mathrm{possible}\right).\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}}\frac{2}{3}×2\frac{2}{3}\left(\mathrm{ii}\right)\text{\hspace{0.17em}}\frac{2}{7}×\frac{7}{9}\left(\mathrm{iii}\right)\text{\hspace{0.17em}}\frac{3}{8}×\frac{6}{4}\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}}\frac{9}{5}×\frac{3}{5}\left(\mathrm{v}\right)\text{\hspace{0.17em}}\frac{1}{3}×\frac{15}{8}\left(\mathrm{vi}\right)\text{\hspace{0.17em}}\frac{11}{2}×\frac{3}{10}\\ \left(\mathrm{vii}\right)\text{\hspace{0.17em}}\frac{4}{5}×\frac{12}{7}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}}\frac{2}{3}×2\frac{2}{3}=\frac{2}{3}×\frac{8}{3}=\overline{)\frac{16}{9}}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}}\frac{2}{7}×\frac{7}{9}=\frac{2}{\overline{)7}}×\frac{\overline{)7}}{9}=\overline{)\frac{2}{9}}\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}}\frac{3}{8}×\frac{6}{4}=\frac{3}{\overline{)2}×4}×\frac{\overline{)2}×3}{4}=\overline{)\frac{9}{16}}\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}}\frac{9}{5}×\frac{3}{5}=\overline{)\frac{27}{25}}\\ \left(\mathrm{v}\right)\text{\hspace{0.17em}}\frac{1}{3}×\frac{15}{8}=\frac{1}{\overline{)3}}×\frac{\overline{)3}×5}{8}=\overline{)\frac{5}{8}}\\ \left(\mathrm{vi}\right)\text{\hspace{0.17em}}\frac{11}{2}×\frac{3}{10}=\overline{)\frac{33}{20}}\\ \left(\mathrm{vii}\right)\text{\hspace{0.17em}}\frac{4}{5}×\frac{12}{7}=\overline{)\frac{48}{35}}\end{array}$

Q.3

$\begin{array}{l}\mathrm{For}\text{}\mathrm{the}\text{}\mathrm{fractions}\text{}\mathrm{given}\text{}\mathrm{below}:\\ \left(\mathrm{a}\right)\text{}\mathrm{Multiply}\text{}\mathrm{and}\text{}\mathrm{reduce}\text{}\mathrm{the}\text{}\mathrm{product}\text{}\mathrm{to}\text{}\mathrm{lowest}\\ \mathrm{form}\text{}\left(\mathrm{if}\text{}\mathrm{possible}\right)\\ \left(\mathrm{b}\right)\text{}\mathrm{Tell}\text{}\mathrm{whether}\text{}\mathrm{the}\text{}\mathrm{fraction}\text{}\mathrm{obtained}\text{}\mathrm{is}\text{}\mathrm{proper}\\ \mathrm{or}\text{}\mathrm{improper}.\\ \left(\mathrm{c}\right)\text{}\mathrm{If}\text{}\mathrm{the}\text{}\mathrm{fraction}\text{}\mathrm{obtained}\text{}\mathrm{is}\text{}\mathrm{improper}\text{}\mathrm{then}\\ \mathrm{convert}\text{}\mathrm{it}\text{}\mathrm{into}\text{}\mathrm{a}\text{}\mathrm{mixed}\text{}\mathrm{fraction}.\\ \text{(i)}\frac{2}{5}×5\frac{1}{4}\\ \text{(ii)}6\frac{2}{5}×\frac{7}{9}\\ \text{(iii)}\frac{3}{2}×5\frac{1}{3}\\ \text{(iv)}\frac{5}{6}×2\frac{3}{7}\\ \left(\mathrm{v}\right)3\frac{2}{5}×\frac{4}{7}\\ \text{(vi)}2\frac{3}{5}×3\\ \text{(vii)}3\frac{4}{7}×\frac{3}{5}\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\\ \left(\text{a}\right)\frac{2}{5}×5\frac{1}{4}=\frac{2}{5}×\frac{21}{4}=\frac{\overline{)2}}{5}×\frac{21}{\overline{)2}×2}=\overline{)\frac{21}{10}}\\ \left(\text{b}\right)\text{It is an improper fraction}.\\ \left(\text{c}\right)\frac{21}{10}=\overline{)2\frac{1}{10}}\\ \left(\mathrm{ii}\right)\\ \left(\mathrm{a}\right)\text{\hspace{0.17em}}6\frac{2}{5}×\frac{7}{9}=\frac{32}{5}×\frac{7}{9}=\overline{)\frac{224}{45}}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}}\text{It is an improper fraction}.\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}\frac{224}{45}=\overline{)4\frac{44}{45}}\\ \left(\mathrm{iii}\right)\\ \left(\mathrm{a}\right)\text{\hspace{0.17em}}\frac{3}{2}×5\frac{1}{3}=\frac{3}{2}×\frac{16}{3}=\frac{\overline{)3}}{\overline{)2}}×\frac{\overline{)2}×8}{\overline{)3}}=\frac{8}{1}=\overline{)8}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}It\hspace{0.17em}is whole number.}\\ \text{(iv)}\\ \text{(a)\hspace{0.17em}}\frac{5}{6}×2\frac{3}{7}\\ =\frac{5}{6}×\frac{17}{7}\\ =\overline{)\frac{85}{42}}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}It is an improper fraction}.\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}\frac{85}{42}=\overline{)2\frac{1}{42}}\\ \left(\mathrm{v}\right)\\ \left(\mathrm{a}\right)\text{\hspace{0.17em}​}3\frac{2}{5}×\frac{4}{7}\\ =\frac{17}{5}×\frac{4}{7}\\ =\frac{68}{35}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}It is an improper fraction}.\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}\frac{68}{35}=\overline{)1\frac{33}{35}}\\ \left(\mathrm{vi}\right)\\ \left(\mathrm{a}\right)\text{\hspace{0.17em}}2\frac{3}{5}×3\\ =\frac{13}{5}×\frac{3}{1}\\ =\frac{39}{5}\\ \left(\text{b}\right)\text{It is an improper fraction}.\\ \left(\text{c}\right)\text{\hspace{0.17em}}\frac{39}{5}=\overline{)7\frac{4}{5}}\\ \left(\mathrm{vii}\right)\\ \left(\mathrm{a}\right)\text{\hspace{0.17em}}3\frac{4}{7}×\frac{3}{5}\\ =\frac{25}{7}×\frac{3}{5}\\ =\frac{5×\overline{)5}}{7}×\frac{3}{\overline{)5}}\\ =\frac{15}{7}\\ \text{(b)\hspace{0.17em}It is an improper fraction}.\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}\frac{15}{7}=\overline{)2\frac{1}{7}}\end{array}$

Q.4

$\begin{array}{l}\mathrm{Which}\mathrm{is}\mathrm{greater}:\\ \left(i\right)\text{\hspace{0.17em}}\frac{2}{7}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\frac{3}{4}\text{\hspace{0.17em}}\mathrm{or}\text{\hspace{0.17em}}\frac{3}{5}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\frac{5}{8}\\ \left(ii\right)\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\frac{6}{7}\text{\hspace{0.17em}}\mathrm{or}\text{\hspace{0.17em}}\frac{2}{3}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\frac{3}{7}\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{\hspace{0.17em}}\\ \frac{2}{7}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{3}{4}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\frac{3}{5}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{5}{8}\\ \frac{2}{7}×\frac{3}{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}or\text{\hspace{0.17em}}\frac{3}{5}×\frac{5}{8}\\ =\frac{\overline{)2}}{7}×\frac{3}{\overline{)2}×2}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\frac{3}{\overline{)5}}×\frac{\overline{)5}}{8}\\ =\frac{3}{14}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\frac{3}{8}\\ \text{Now},\text{the LCM of 14 and 8 is 56}.\text{So},\text{we get}\\ \frac{3}{14}=\frac{3×4}{14×4}=\frac{12}{56}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\frac{3}{8}=\frac{3×7}{8×7}=\frac{21}{56}\\ \text{Since},\text{21}>\text{12},\text{so}\text{\hspace{0.17em}}\frac{3}{8}>\frac{3}{14}\\ \text{Hence},\text{\hspace{0.17em}}\overline{)\frac{3}{5}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{5}{8}>\frac{2}{7}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{3}{4}}\\ \left(\text{ii}\right)\\ \frac{1}{2}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{6}{7}\text{\hspace{0.17em}}or\text{​}\frac{2}{3}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{3}{7}\\ \frac{1}{2}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{6}{7}\text{\hspace{0.17em}}or\text{\hspace{0.17em}}\frac{2}{3}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{3}{7}\\ =\frac{1}{\overline{)2}}×\frac{\overline{)2}×3}{7}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\frac{2}{\overline{)3}}×\frac{\overline{)3}}{7}\\ =\frac{3}{7}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\frac{2}{7}\\ \text{Since},\text{3}>\text{2},\text{so},\frac{3}{7}>\frac{2}{7}\\ \text{Hence},\text{\hspace{0.17em}}\overline{)\frac{1}{2}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{6}{7}>\frac{2}{3}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{3}{7}}\end{array}$

Q.5

$\mathrm{Saili}\text{}\mathrm{plants}\text{}4\text{}\mathrm{saplings},\text{}\mathrm{in}\text{}\mathrm{a}\text{}\mathrm{row},\text{}\mathrm{in}\text{}\mathrm{her}\text{}\mathrm{garden}.\text{}\mathrm{The}\text{}\mathrm{distance}\text{}\mathrm{between}\text{}\mathrm{two}\text{}\mathrm{adjacent}\mathrm{}\mathrm{saplings}\mathrm{}\mathrm{is}\text{}\frac{3}{4}\text{}m.\phantom{\rule{0ex}{0ex}}\text{\hspace{0.17em}}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{distance}\text{}\mathrm{between}\text{}\mathrm{the}\text{}\mathrm{first}\text{}\mathrm{and}\text{}\mathrm{the}\text{}\mathrm{last}\text{}\mathrm{sapling}.$

Ans.

$\begin{array}{l}\text{Here},\text{the distance between the two saplings is given as}\text{\hspace{0.17em}}\frac{3}{4}\text{\hspace{0.17em}}\text{m}\\ \text{So},\text{the distance between first and the last sapling is}\\ \frac{3}{4}\text{\hspace{0.17em}}\text{m}×\text{3=}\frac{9}{4}\text{m}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\overline{)\text{2}\frac{1}{4}\text{m}}\end{array}$

Q.6

$\begin{array}{l}\mathrm{Lipika}\mathrm{}\mathrm{reads}\mathrm{}\mathrm{a}\mathrm{}\mathrm{book}\mathrm{}\mathrm{for}\mathrm{ }1\frac{3}{4}\mathrm{ }\mathrm{hours}\mathrm{}\mathrm{every}\mathrm{}\mathrm{day}.\mathrm{}\mathrm{She}\\ \mathrm{reads}\mathrm{}\mathrm{the}\mathrm{}\mathrm{entire}\mathrm{}\mathrm{book}\mathrm{}\mathrm{in}\mathrm{}6\mathrm{}\mathrm{days}.\mathrm{}\mathrm{How}\mathrm{}\mathrm{many}\mathrm{}\mathrm{hours}\\ \mathrm{in}\mathrm{}\mathrm{all}\mathrm{}\mathrm{were}\mathrm{}\mathrm{required}\mathrm{}\mathrm{by}\mathrm{}\mathrm{her}\mathrm{}\mathrm{to}\mathrm{}\mathrm{read}\mathrm{}\mathrm{the}\mathrm{}\mathrm{book}?\end{array}$

Ans.

$\begin{array}{l}\text{Lipika reads}\text{\hspace{0.17em}}1\frac{3}{4}=\frac{7}{4}\text{hours a day}.\\ \text{So},\text{Number of hours required by her to read the book}\\ \text{in 6 days is}\text{​}\frac{7}{4}\text{hour}×6=\frac{7}{2×\overline{)2}}\text{hour}×\left(\overline{)2}×3\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{21}{2}\text{hour}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\overline{)\text{10}\frac{1}{2}\text{\hspace{0.17em}}\text{hour}}\end{array}$

Q.7

$\begin{array}{l}\mathrm{A}\mathrm{}\mathrm{car}\mathrm{}\mathrm{runs}\mathrm{}16\mathrm{}\mathrm{km}\mathrm{}\mathrm{using}\mathrm{}1\mathrm{}\mathrm{litre}\mathrm{}\mathrm{of}\mathrm{}\mathrm{petrol}.\mathrm{}\mathrm{How}\\ \mathrm{much}\mathrm{}\mathrm{distance}\mathrm{}\mathrm{will}\mathrm{}\mathrm{it}\mathrm{}\mathrm{cover}\mathrm{}\mathrm{using}\mathrm{ }2\frac{3}{4}\mathrm{litres}\mathrm{}\mathrm{of}\mathrm{}\mathrm{petrol}?\end{array}$

Ans.

$\begin{array}{c}\text{In 1 litre of petrol},\text{car runs 16 km}.\\ \text{So},\text{in}\text{\hspace{0.17em}}2\frac{3}{4}\text{litres of petrol},\text{distance covered by car is}\text{\hspace{0.17em}}\\ 2\frac{3}{4}×16\text{\hspace{0.17em}}\text{km}\\ =\frac{11}{\overline{)4}}×\left(\overline{)4}×4\right)\text{\hspace{0.17em}}\text{km}\\ =\overline{)\text{44}\text{\hspace{0.17em}}\text{km}}\end{array}$

Q.8

$\begin{array}{l}\left(\mathrm{a}\right)\\ \left(\mathrm{i}\right)\text{​}\mathrm{Provide}\mathrm{the}\mathrm{number}\mathrm{in}\mathrm{the}\mathrm{box}\text{\hspace{0.17em}​}\overline{)}\text{\hspace{0.17em}},\mathrm{such}\mathrm{that}\text{\hspace{0.17em}}\frac{2}{3}\text{}×\overline{)}=\frac{10}{30}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}}\mathrm{The}\mathrm{simplest}\mathrm{form}\mathrm{of}\mathrm{the}\mathrm{number}\mathrm{obtained}\mathrm{in}\text{\hspace{0.17em}}\overline{)}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}____\\ \left(\mathrm{b}\right)\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}}\mathrm{Provide}\mathrm{the}\mathrm{number}\mathrm{in}\mathrm{the}\mathrm{box}\text{\hspace{0.17em}}\overline{)},\mathrm{such}\mathrm{that}\frac{3}{5}×\overline{)}=\frac{24}{75}\\ \left(\mathrm{ii}\right)\\ \mathrm{The}\mathrm{simplest}\mathrm{form}\mathrm{of}\mathrm{the}\mathrm{number}\mathrm{obtained}\mathrm{in}\text{\hspace{0.17em}}\overline{)}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}____\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)\\ \left(\text{i}\right)\text{\hspace{0.17em}}\frac{\text{2}}{\text{3}}×\overline{)}\text{=}\frac{\text{10}}{\text{30}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\text{2×5}}{\text{3×10}}=\frac{\text{2}}{\text{3}}\text{×}\frac{\text{5}}{\text{10}}\\ \text{So,}\text{\hspace{0.17em}}\frac{\text{2}}{\text{3}}×\frac{\text{5}}{\text{10}}=\frac{\text{10}}{\text{30}}\\ \text{Therefore, the number in}\text{\hspace{0.17em}}\overline{)}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\frac{\text{5}}{\text{10}}\text{.}\\ \text{(ii)}\\ \text{The simplest form of}\text{\hspace{0.17em}}\frac{\text{5}}{\text{10}}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\frac{\text{5}}{\text{10}}=\frac{}{}=\overline{)\frac{\text{1}}{\text{2}}}\\ \text{(b)}\\ \left(\text{i}\right)\text{\hspace{0.17em}}\frac{\text{3}}{\text{5}}×\overline{)}\text{=}\frac{\text{24}}{\text{75}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\text{3×8}}{\text{5×15}}=\frac{\text{3}}{\text{5}}\text{×}\frac{\text{8}}{\text{15}}\\ \text{So,}\text{\hspace{0.17em}}\frac{\text{3}}{\text{5}}\text{×}\frac{\text{8}}{\text{15}}=\frac{\text{24}}{\text{75}}\\ \text{Therefore, the number in}\text{\hspace{0.17em}}\overline{)}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\frac{\text{8}}{\text{15}}\text{.}\\ \text{(ii)}\\ \text{The simplest form of}\text{\hspace{0.17em}}\frac{\text{8}}{\text{15}}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\overline{)\frac{\text{8}}{\text{15}}}\end{array}$