NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

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

Edited By Ramraj Saini | Updated on Feb 07, 2024 04:53 PM IST

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals are discussed here. These NCERT solutions are created by the expert team at Careers360 keeping in mind the latest CBSE syllabus 2023. These solutions are simple and comprehensive and cover step by step solutions to each problem. In the solutions of NCERT Class 7 chapter 2 Fractions and Decimals, we will learn questions related to the multiplication and division of fractions and decimals. The study of fractions includes mixed fractions, proper fraction and improper fraction with their addition and subtraction, equivalent fractions, comparison of fractions, ordering of fractions and representation of fractions on a number line.

NCERT Solutions are the detailed explanation of each and every question of NCERT textbook and this helps to score good marks in the class exams. Some questions related to real-life situations are also explained in the NCERT Solutions for Class 7 . Here you will get solutions to all five exercises of NCERT. Before referring to the solutions, students must complete the NCER Class 7 Maths Syllabus.

## NCERT Solutions for Class 7 Maths Chapter 2 - Important Formulae

fraction = p/q, Where p is the numerator and q is the denominator.

Mixed fraction: = p + q/r, Where p is the whole number, q is the numerator and r is the denominator.

Improper fraction to mixed fraction: (p/q) = (p ÷ q) + ( remaining fraction: p % q)/q.

Mixed Fraction to improper fraction: ( p + q/r ) = (( p×r)+ q)/c

(p/q)(a/b) = (pa)/(qb)

(p/q)/(a/b) = (p/q)(b/a)

p/q + a/b = (pb + aq)/(qb)

p/q - a/b = (pb - aq)/qb

## NCERT Solutions for Class 7 Maths Chapter 2 - Important Points

Important points for maths class 7 chapter 2 are listed below.

Fractions:

• A fraction is written as p/q, where p is the numerator and q is the denominator.
• There are three main types of fractions: proper, improper, and mixed.
• Proper fractions have the numerator (p) smaller than the denominator (q).
• Improper fractions have the numerator (p) greater than or equal to the denominator (q).
• Mixed fractions consist of a whole number and a proper fraction.

Conversions:

• Improper fractions can be converted to mixed fractions.
• To convert an improper fraction to a mixed fraction, divide the numerator by the denominator.
• Mixed fractions can be converted to improper fractions.
• To convert a mixed fraction to an improper fraction, multiply the whole number by the denominator and add the numerator.

Operations:

• Multiplication of fractions involves multiplying the numerators and denominators.
• Division of fractions requires finding the reciprocal of the second fraction and then multiplying.
• Reciprocal of a fraction is the fraction flipped, swapping the numerator and denominator.

• Adding or subtracting fractions with the same denominator is straightforward.
• When denominators differ, find the least common multiple (LCM) and then proceed.

Decimals:

• Decimals represent fractions with denominators of powers of 10.
• Decimal place value determines the value of each digit in a decimal number.
• Tenth place, hundredth place, thousandth place, etc., indicate positions after the decimal point.

Operations with Decimals:

• Multiplying decimals involves ignoring the decimal points, multiplying the numbers, and then placing the decimal point in the product.
• Dividing decimals requires moving the decimal point in the divisor and dividend to make the divisor a whole number. Then divide as usual.

Comparing Decimals:

• Start comparing decimals from the left and move towards the right.
• If digits match up to a certain place value, compare the next digit to determine which decimal is greater.

Free download NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals PDF for CBSE Exam.

## NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals (Intext Questions and Exercise)

NCERT Solutions for chapter 2 maths class 7 Fractions And Decimals Topic 2.3.1

1.(a) Find:

If the product is an improper fraction express it as a mixed fraction.

$\\\Rightarrow \frac{2}{7}\times 3\\=\frac{2\times3}{7}\\=\frac{6}{7}$

The product is a proper fraction.

1.(b) Find:

If the product is an improper fraction express it as a mixed fraction.

Given the product

$\Rightarrow \frac{9}{7}\times 6$

$=\frac{9\times6}{7}$

$=\frac{54}{7}$

This is an improper fraction, so convert it into a mixed fraction.

$\Rightarrow \frac{54}{7}=\frac{49+5}{7}=\frac{49}{7}+\frac{5}{7}=7+\frac{5}{7}=7\frac{5}{7}$ .

1.(c) Find:

If the product is an improper fraction express it as a mixed fraction.

$\Rightarrow 3 \times \frac{1}{8}$

$= \frac{3\times1}{8}$

$= \frac{3}{8}$

This is a proper fraction.

1.(d) Find:

If the product is an improper fraction express it as a mixed fraction.

$\Rightarrow \frac{13}{11}\times 6$

$= \frac{13\times6}{11}$

$= \frac{78}{11}$

This is an improper fraction, so we convert this into a mixed fraction. that is

$\Rightarrow \frac{78}{11}=\frac{77+1}{11}=\frac{77}{11}+\frac{1}{11}=7+\frac{1}{11}=7\frac{1}{11}$ .

1.(i) Find:

Answer: $\Rightarrow 5\times 2\frac{3}{7}=5\times\frac{7\times 2+3}{7}$

$\Rightarrow 5\times 2\frac{3}{7}=5\times\frac{17}{7}$

$\Rightarrow 5\times 2\frac{3}{7}=\frac{5\times17}{7}$

$\Rightarrow 5\times 2\frac{3}{7}=\frac{85}{7}$

$\Rightarrow 5\times 2\frac{3}{7}=\frac{84+1}{7}$

$\Rightarrow 5\times 2\frac{3}{7}=\frac{84}{7}+\frac{1}{7}$

$\Rightarrow 5\times 2\frac{3}{7}=12 +\frac{1}{7}$

$\Rightarrow 5\times 2\frac{3}{7}=12\frac{1}{7}$

Answer: $\Rightarrow 1\frac{4}{9}\times 6=\frac{9\times1+4}{9}\times 6$

$\Rightarrow 1\frac{4}{9}\times 6=\frac{13}{9}\times 6$

$\Rightarrow 1\frac{4}{9}\times 6=\frac{13\times 6}{9}$

$\Rightarrow 1\frac{4}{9}\times 6=\frac{78}{9}$

$\Rightarrow 1\frac{4}{9}\times 6=\frac{72+6}{9}$

$\Rightarrow 1\frac{4}{9}\times 6=\frac{72}{9}+\frac{6}{9}$

$\Rightarrow 1\frac{4}{9}\times 6=8+\frac{2}{3}$

$\Rightarrow 1\frac{4}{9}\times 6=8\frac{2}{3}$ .

Answer: As we know, of means multiply so,

$\frac{1}{2} o\!f\:\:10=\frac{1}{2}\times10=5$

And

$\frac{1}{4} o\!f\:\:16=\frac{1}{4}\times16=4$

And

$\frac{2}{5} o\!f\:\:25=\frac{2}{5}\times25=10$

Answer: As we know in the multiplication of a fraction, the numerator of one fraction is multiplied by the numerator of other fraction, and same for denominator, hence,

$\Rightarrow \frac{1}{3}\times \frac{4}{5}=\frac{4\times1}{5\times3}=\frac{4}{15}$

$\Rightarrow \frac{2}{3}\times \frac{1}{5}=\frac{2\times1}{3\times5}=\frac{2}{15}.$

Answer: As we know in the multiplication of a fraction, the numerator of one fraction is multiplied by the numerator of other fraction, and the same for the denominator, hence,

$\\\Rightarrow \frac{8}{3}\times \frac{4}{7}=\frac{8\times4}{3\times7}=\frac{32}{27}\\\\\Rightarrow \frac{3}{4}\times \frac{2}{3}=\frac{3\times2}{4\times3}=\frac{6}{12}=\frac{1}{2}$

NCERT Solutions for chapter 2 maths class 7 Fractions And Decimals Topic 2.3.2

Answer: $(i) \frac{1}{2}\times \frac{1}{7}= \frac{1\times 1}{2\times 7}=\frac{1}{14}$

$(ii) \frac{1}{5}\times \frac{1}{7}=\frac{1}{35}$

$(iii) \frac{1}{7}\times \frac{1}{2}=\frac{1}{14}$

$(iv) \frac{1}{7}\times \frac{1}{5}=\frac{1}{35}$

NCERT Solutions for chapter 2 maths class 7 Fractions And Decimals Topic 2.4.1

1. Find :

Answer: As we know, The division of a number by a / b is equivalent to multiplication of that number with b / a. So,

$(i)7\div \frac{2}{5}$

$\Rightarrow 7\div \frac{2}{5}=7\times\frac{2}{5}=\frac{14}{5}$

$(ii)6\div \frac{4}{7}$

$\Rightarrow 6\div \frac{4}{7}=6\times\frac{4}{7}=\frac{24}{7}$

$(iii)2\div \frac{8}{9}$

$\Rightarrow 2\div \frac{8}{9}=2\times\frac{8}{9}=\frac{16}{9}$

NCERT Solutions for Class 7 Chapter 2 Fractions And Decimals Topic 2.4.2

2. Find:

Answer: As we know, The division of a number by a / b is equivalent to the multiplication of that number with b / a. So,

$(i)6\div 5\frac{1}{3}$

$\Rightarrow 6\div 5\frac{1}{3}=6\div\frac{16}{3}=6\times\frac{3}{16}=\frac{18}{16}=\frac{9}{8}=1\frac{1}{8}$

$(ii)7\div 2\frac{4}{7}$

$\Rightarrow 7\div 2\frac{4}{7}=7\div\frac{18}{7}=7\times\frac{7}{18}=\frac{49}{18}=2\frac{13}{18}$

NCERT Solutions for chapter 2 maths class 7 Fractions And Decimals Topic 2.4.3

3. Find:

Answer: As we know, The division of a number by a / b is equivalent to the multiplication of that number with b / a. So,

$(i)\frac{3}{5}\div \frac{1}{2}$

$\Rightarrow \frac{3}{5}\div \frac{1}{2}=\frac{3}{5}\times\frac{2}{1}=\frac{6}{5}=1\frac{1}{5}$

$(ii)\frac{1}{2}\div \frac{3}{5}$

$\Rightarrow \frac{1}{2}\div \frac{3}{5}=\frac{1}{2}\times\frac{5}{3}=\frac{5}{6}$

$(iii)2\frac{1}{2}\div \frac{3}{5}$

$2\frac{1}{2}\div \frac{3}{5}=\frac{5}{2}\div \frac{3}{5}=\frac{5}{2}\times\frac{5}{3}=\frac{25}{6}=4\frac{1}{6}$

$(iv)5\frac{1}{6}\div \frac{9}{2}$

$5\frac{1}{6}\div \frac{9}{2}=\frac{31}{6}\div\frac{9}{2}=\frac{31}{6}\times\frac{2}{9}=\frac{62}{54}=1\frac{8}{54}$

NCERT Solutions for Class 7 Chapter 2 Fractions And Decimals Topic 2.6

1. Find:

Answer: As we know, The multiplication of the decimal number is just like the multiplication of normal numbers, we just have to put decimal before a digit such that the number of digits after decimal should remain the same.

In other words,

The number of digits after the decimal should be the same before and after the multiplication.

So.

$(i)2.7\times 4=10.8$

$(ii)1.8\times 1.2=2.16$

$(iii)2.3\times 4.35=10.005$

Answer:$(i)2.7\times 4=10.8$

$(ii)1.8\times 1.2=2.16$

$(iii)2.3\times 4.35=10.005$

The products in descending order are:

$10.8>10.005>2.16$

NCERT Solutions for Class 7 Chapter 2 Fractions And Decimals Topic 2.6.1

1. Find:

Answer: As we know, The multiplication of the decimal number is just like the multiplication of the normal numbers, we just have to put decimal before a digit such that the number of digits after decimal should remain the same.

In other words,

The number of digits after the decimal should be the same before and after the multiplication.

So.

$(i)0.3\times 10=3$

$(ii)1.2\times 100=120$

$(iii)56.3\times 1000=56300$

NCERT Solutions for Class 7 Chapter 2 Fractions And Decimals Topic 2.7.1

1. Find:

(i) 235.4 ÷ 10 (ii) 235.4 ÷100 (iii) 235.4 ÷ 1000

Answer: As we know, while dividing a number by 10, 100 or 1000, the digits of the number and the quotient are same but the decimal point in the quotient shifts to the left by as many places as there are zeros over 1.

So,

(i) 235.4 ÷ 10 = 23.54

(ii) 235.4 ÷100 = 2.354

(iii) 235.4 ÷ 1000 = 0.2354

NCERT Solutions for Class 7 Chapter 2 Fractions And Decimals Topic 2.7.2

Answer: As we know, dividing by a number is equivalent to multiplying by the reciprocal of that number. So

(i) 35.7 ÷ 3

$35.7\div3=\frac{357}{10}\div3=\frac{357}{10}\times\frac{1}{3}=\frac{129}{10}=12.9$

(ii) 25.5 ÷ 3

$25.5\div3=\frac{255}{10}\div3=\frac{255}{10}\times\frac{1}{3}=\frac{85}{10}=8.5$

Answer: As we know, dividing by a number is equivalent to multiplying by the reciprocal of that number. So

(i) 43.15 ÷ 5

$43.15\div5=\frac{4315}{100}\div5=\frac{4315}{100}\times\frac{1}{5}=\frac{863}{100}=8.63$

(ii) 82.44 ÷ 6

$82.44\div6=\frac{8244}{100}\div6=\frac{8244}{100}\times\frac{1}{6}=\frac{2374}{100}=23.74$

4. Find:

(i) 15.5 ÷ 5

(ii) 126.35 ÷ 7

Answer: As we know, dividing by a number is equivalent to multiplying by the reciprocal of that number. So

(i) 15.5 ÷ 5

$15.5\div5=\frac{155}{10}\div5=\frac{155}{10}\times\frac{1}{5}=\frac{31}{10}=3.1$

(ii) 126.35 ÷ 7

$126.35\div7=\frac{12635}{100}\div7=\frac{12635}{100}\times\frac{1}{7}=\frac{1805}{100}=18.05$

NCERT Solutions for C lass 7 Chapter 2 Fractions And Decimals Topic 2.7.3

1. Find:

Answer: As we know, in the dividing of decimal, we first express the decimal in term of fraction and then divides it, So

$(i)\frac{7.75}{0.25}$

$\frac{7.75}{0.25}=\frac{\frac{775}{100}}{\frac{25}{100}}=\frac{775}{25}\times\frac{100}{100}=\frac{31}{1}=31$

$(ii)\frac{42.8}{0.02}$

$\frac{42.8}{0.02}=\frac{\frac{428}{100}}{\frac{2}{100}}=\frac{428}{2}\times\frac{100}{100}=214$

$(iii)\frac{5.6}{1.4}$

$\frac{5.6}{1.4}=\frac{\frac{56}{10}}{\frac{14}{10}}=\frac{56}{14}\times\frac{10}{10}=4$

NCERT Solutions for Class 7 Maths Chapter 2 Fractions And Decimals Exercise 2.1

1. Solve:

(i) $2 - \frac{3}{5}$ (ii) $4 + \frac{7}{8}$ (iii) $\frac{3}{5} + \frac{2}{7}$ (iv) $\frac{9}{11} - \frac{4}{15}$

(v) $\frac{7}{10} + \frac{2}{5} + \frac{3}{2}$ (vi) $2\frac{2}{3} +3 \frac{1}{2}$ (vii) $8\frac{1}{2}-3 \frac{5}{8}$

Answer: As we know we have to make the denominator the same in order to add or subtract the fractions. So,

(i) $2 - \frac{3}{5}=\frac{2}{1}\times\frac{5}{5}-\frac{3}{5}=\frac{10}{5}-\frac{3}{5}=\frac{10-3}{5}=\frac{7}{5}$

(ii)

$4 + \frac{7}{8}=\frac{4}{1}\times\frac{8}{8}+\frac{7}{8}=\frac{32}{8}+\frac{7}{8}=\frac{39}{8}$

(iii)

$\frac{3}{5} + \frac{2}{7}=\frac{3}{5}\times\frac{7}{7}+\frac{2}{7}\times\frac{5}{5}=\frac{21}{35}+\frac{10}{35}=\frac{31}{35}$

(iv)

$\frac{9}{11} - \frac{4}{15}=\frac{15\times 9-11\times 4}{11\times15}=\frac{135-44}{165}=\frac{91}{165}$

(v)

$\frac{7}{10} + \frac{2}{5} + \frac{3}{2}=\frac{7}{10}+\frac{2}{5}\times\frac{2}{2}+\frac{3}{2}\times\frac{5}{5}=\frac{7}{10}+\frac{4}{10}+\frac{15}{10}=\frac{7+4+15}{10}=\frac{26}{10}$

(vi)

$2\frac{2}{3} +3 \frac{1}{2}=\frac{8}{3}+\frac{7}{2}=\frac{16}{6}+\frac{21}{6}=\frac{16+21}{6}=\frac{37}{6}=6\frac{1}{6}$

(vii)

$8\frac{1}{2}-3 \frac{5}{8}=\frac{17}{2}-\frac{29}{8}=\frac{68}{8}-\frac{29}{8}=\frac{68-29}{8}=\frac{39}{9}=4\frac{1}{3}$

(i) $\frac{2}{9}, \frac{2}{3}, \frac{8}{21}$ (ii) $\frac{1}{5}, \frac{3}{7}, \frac{7}{10}$

$\frac{2}{9}=\frac{2}{3\times3}\times\frac{7}{7}=\frac{14}{3\times3\times7}$

$\frac{2}{3}=\frac{2}{3}\times\frac{7}{7}\times\frac{3}{3}=\frac{42}{3\times3\times7}$

$\frac{8}{21}=\frac{8}{3\times7}\times\frac{3}{3}=\frac{24}{3\times3\times7}$

As 14 < 24 < 42

$\frac{2}{3}>\frac{8}{21}>\frac{2}{9}$

(ii) $\frac{1}{5}=\frac{1}{5}\times\frac{14}{14}=\frac{14}{70}$

$\frac{3}{7}=\frac{3}{7}\times\frac{10}{10}=\frac{30}{70}$

$\frac{7}{10}=\frac{7}{10}\times\frac{7}{7}=\frac{49}{70}$

As 14 < 30 < 49

$\frac{7}{10}>\frac{3}{7}>\frac{1}{5}$

(Along the first row $\frac{4}{11} + \frac{9}{11} + \frac{2}{11} = \frac{15}{11}$ ).

Answer: As, we can see that the sum of every row, column or diagonal is 15 / 11 . so yes this is a magic square.

Length of the rectangle :

$l=12\tfrac{1}{2}cm=\frac{12\times 2+1}{2}=\frac{25}{2}cm$

Width of the rectangle :

$b=10\tfrac{2}{3}cm=\frac{10\times 3+2}{3}=\frac{32}{3}cm$

Now, As we know,

Perimeter of the rectangle = 2 x ( length + width )

So,

The perimeter of the given rectangle :

$=2\times(l+b)$

$=2\times\left ( \frac{25}{2}+\frac{32}{3} \right )$

Now let's make the denominator of both fractions equal.

$=2\times\left ( \frac{25}{2}\times\frac{3}{3}+\frac{32}{3}\times\frac{2}{2} \right )$

$=2\times\left ( \frac{75}{6}+\frac{64}{6} \right )$

$=2\times\left ( \frac{139}{6} \right )$

$=\frac{139}{3}cm$

Hence, the perimeter of the rectangle is $\frac{139}{3}cm$ .

Answer: The perimeter of $\bigtriangleup\! ABE^{}$ = AB + BE + AE

$=\frac{5}{2}+2\frac{3}{4}+3\frac{3}{5}$

$=\frac{5}{2}+\frac{4\times2+3}{4}+\frac{3\times5+3}{5}$

$=\frac{5}{2}+\frac{11}{4}+\frac{18}{5}$

Noe, The LCM of 2,4 and 5 is 20. So let's make the denominator of all fractions, 20.

So,

The perimeter of $\bigtriangleup\! ABE^{}$ :

$=\frac{5}{2}\times\frac{10}{10}+\frac{11}{4}\times\frac{5}{5}+\frac{18}{5}\times\frac{4}{4}$

$=\frac{50}{20}+\frac{55}{20}+\frac{72}{20}$

$=\frac{50+55+72}{20}$

$=\frac{177}{20}cm$

Now,

The perimeter of rectangle BCDE = 2 x ( BE + ED )

$=2\times\left (2\frac{3}{4}+\frac{7}{6} \right )$

$=2\times\left (\frac{4\times2+3}{4}+\frac{7}{6} \right )$

$=2\times\left (\frac{11}{4}+\frac{7}{6} \right )$

The LCM of 4 and 6 is 12. So let's make the denominator of both fractions equal to 12.

$=2\times\left (\frac{11}{4}\times\frac{3}{3}+\frac{7}{6}\times\frac{2}{2} \right )$

$=2\times\left (\frac{33}{12}+\frac{14}{12} \right )$

$=2\times\left (\frac{33+14}{12} \right )$

$=2\times\left (\frac{47}{12} \right )$

$=\frac{47}{6}cm$

Hence The perimeter of the Triangle is $177/20 \:cm$ and the perimeter of the rectangle is $47/6 \: cm$ .

Now, we have

$\frac{177}{20} \:\: and\:\:\frac{47}{6}$

LCM of 20 and 6 is 60, so let's make the denominator of both fractions equal to 60.

So,

$\frac{177}{20} =\frac{177}{20}\times\frac{3}{3}=\frac{531}{60}$

And

$\frac{47}{6}=\frac{47}{6}\times\frac{10}{10}=\frac{470}{60}$

Now, Since 531 > 470

$\Rightarrow \frac{177}{20}>\frac{47}{6}.$

$\Rightarrow$ Area of Triangle > Area of Rectangle.

Answer: Given, the width of the picture = $7\tfrac{3}{5}$ cm.

The maximum width of the picture which can fit in the frame = $7\tfrac{3}{10}$ cm.

Hence the length Salil should trim :

$\Rightarrow 7\tfrac{3}{5}-7\tfrac{3}{10}$

$\Rightarrow \frac{7\times5+3}{5}-\frac{10\times7+3}{10}$

$\Rightarrow \frac{38}{5}-\frac{73}{10}$

Now LCM of 5 and 10 is 10. So, let's make the denominator of both fractions equal to 10. So,

$\Rightarrow \frac{38}{5}\times\frac{2}{2}-\frac{73}{10}\times\frac{1}{1}$

$\Rightarrow \frac{76}{10}-\frac{73}{10}$

$\Rightarrow \frac{76-73}{10}$

$\Rightarrow \frac{3}{10}$

Hence Salil should cut 3/10 cm of the picture in order to fit it in the frame.

Part of Apple eaten by Ritu = $\frac{3}{5}$ .

Part of Apple eaten by Somu:

$\Rightarrow 1-\frac{3}{5}$

$\Rightarrow \frac{1}{1}-\frac{3}{5}$

LCM of 1 and 5 is 5. So making the denominator of both fractions equal to 5, we get

$\Rightarrow \frac{1}{1}\times\frac{5}{5}-\frac{3}{5}$

$\Rightarrow \frac{5}{5}-\frac{3}{5}$

$\Rightarrow \frac{5-3}{5}$

$\Rightarrow \frac{2}{5}$

Hence, The Part of Apple eaten by Somu is 2/5.

Now, As

$\frac{3}{5}> \frac{2}{5}$

$\frac{3}{5}- \frac{2}{5}=\frac{3-2}{5}=\frac{1}{5}$

Hence, Ritu had a larger share of the apple by $\frac{1}{5}$ part.

Answer: Time taken by Michal in colouring = $\frac{7}{12}$ hours.

Time taken by Vaibhav in colouring = $\frac{3}{4}$ hours.

The LCM of 12 and 4 is 12. So making the denominator of both fractions equal to 12, we get,

$\frac{7}{12}\:\:and\:\:\frac{3}{4}\times\frac{3}{3}$

$\Rightarrow \frac{7}{12}\:\:and\:\:\frac{9}{12}$

As

$\frac{9}{12}>\frac{7}{12}$

For calculating by how much, we do Subtraction

$\frac{9}{12}-\frac{7}{12}=\frac{9-7}{12}=\frac{2}{12}=\frac{1}{6}$

Vaibhav worked longer by the fraction $\frac{1}{6}$ .

NCERT Solutions for Class 7 Maths Chapter 2 Fractions And Decimals Exercise 2.2

$(i)2\times \frac{1}{5}$ $(ii)2\times \frac{1}{2}$ $(iii)3\times \frac{2}{3}$ $(iv)3\times \frac{1}{4}$

(d) represents two circles with 1 part shaded out of 5 parts So, it represents

$2\times \frac{1}{5}$ .

ii) b represents two squares one part out of two of both squares are shaded, So it represents

$2\times \frac{1}{2}$

(iii)

(a) represents 3 circles 2 parts out of three of all circles are shaded. So it represents

$3\times \frac{2}{3}$

(iv)

(c) represents 3 squares with one part out of four-part shaded in each square hence it represents

$3\times \frac{1}{4}$ .

Answer:$(i) 3\times \frac{1}{5}= \frac{3}{5}$

As in option (c), In the Left-hand side, there are three figures in which one part out of three-part is shaded and in the right-hand side, three out of five portions are shaded.

Hence this represents

$3\times \frac{1}{5}= \frac{3}{5}$ .

$(ii) 2\times \frac{1}{3}= \frac{2}{3}$

Option (a) represents the equation pictorially.

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

Option (b) represents this equation pictorially.

Answer: $(i) 7\times \frac{3}{5}$

On Multiplying, we get

$\Rightarrow \frac{7\times3}{5}=\frac{21}{5}=\frac{20+1}{5}=\frac{20}{5}+\frac{1}{5}=4+\frac{1}{5}=4\frac{1}{5}$

$(ii) 4\times \frac{1}{3}$

On Multiplying, we get

$\Rightarrow \frac{4\times1}{3}=\frac{4}{3}=\frac{3+1}{3}=\frac{3}{3}+\frac{1}{3}=1+\frac{1}{3}=1\frac{1}{3}$

$(iii) 2\times \frac{6}{7}$

On Multiplying, we get

$\Rightarrow \frac{2\times6}{7}=\frac{12}{7}=\frac{7+5}{7}=\frac{7}{7}+\frac{5}{7}=1+\frac{5}{7}=1\frac{5}{7}$

$(iv) 5\times \frac{2}{9}$

On Multiplying, we get

$\Rightarrow \frac{5\times2}{9}=\frac{10}{9}=\frac{9+1}{9}=\frac{9}{9}+\frac{1}{9}=1+\frac{1}{9}=1\frac{1}{9}$

$(v) \frac{2}{3}\times 4$

On Multiplying, we get

$\Rightarrow \frac{2\times4}{3}=\frac{8}{3}=\frac{6+2}{3}=\frac{6}{3}+\frac{2}{3}=2+\frac{2}{3}=2\frac{2}{3}$

$(vi) \frac{5}{2}\times 6$

On Multiplying, we get

$\Rightarrow \frac{5\times6}{2}=\frac{30}{2}=15$

$(vii) 11\times \frac{4}{7}$

On Multiplying, we get

$\Rightarrow \frac{11\times4}{7}=\frac{44}{7}$

Converting this into a mixed fraction, we get

$\Rightarrow \frac{44}{7}=\frac{42+2}{7}=\frac{42}{7}+\frac{2}{7}=6+\frac{2}{7}=6\frac{2}{7}.$

$(viii) 20\times \frac{4}{5}$

On Multiplying, we get

$\Rightarrow \frac{20\times4}{5}=\frac{80}{5}=16$

$(ix) 13\times \frac{1}{3}$

On multiplying, we get

$\Rightarrow \frac{13\times1}{3}=\frac{13}{3}$

Converting this into a mixed fraction,

$\Rightarrow\frac{13}{3}=\frac{12+1}{3}=\frac{12}{3}+\frac{1}{3}=4+\frac{1}{3}=4\frac{1}{3}$ .

$(x) 15\times \frac{3}{5}$

On multiplying, we get,

$\Rightarrow \frac{15\times3}{5}=\frac{45}{5}=9$ .

$(i) \frac{1}{2}$ of the circles in box (a) $(ii) \frac{2}{3}$ of the triangles in box (b)

$(iii) \frac{3}{5}$ of the squares in box (c).

Answer: 1) In figure a there are 12 circles: half of 12 = 6

2) In figure b there are 9 triangles: 2/3 of 9 = 6

3) In figure c there are 15 triangles: 3/5 of 15 = 9

5. Find:

Answer: $(a)(i)\; \frac{1}{2}\;\; of\; 24$

On Multiplying we get,

$\; \frac{1}{2}\;\; of\; 24=\frac{1}{2}\times24=\frac{1\times24}{2}=12$

$(a)\;ii) \frac{1}{2}\;\; of\;46$

On multiplying, we get

$\Rightarrow \frac{1}{2}\;\; of\;46=\frac{1}{2}\times46=\frac{1\times46}{2}=23.$

$(b)(i)\; \frac{2}{3}\;\; of\;18$

On Multiplying, we get

$\Rightarrow \frac{2}{3}\;\; of\;18=\frac{2}{3}\times18=\frac{2\times18}{3}=\frac{36}{3}=12.$

$(b)(ii)\; \frac{2}{3}\;\; of\;27$

On multiplying, we get

$\Rightarrow \frac{2}{3}\:\:of\:\:27=\frac{2}{3}\times27=\frac{2\times27}{3}=\frac{54}{3}=18$

$(c)(i)\; \frac{3}{4}\;\; of\; 16$

On multiplying, we get

$\Rightarrow \frac{3}{4}\:\:of\:\:16=\frac{3}{4}\times16=\frac{3\times16}{4}=\frac{48}{4}=12.$

$(c)(ii)\; \frac{3}{4}\;\; of\; 36$

$\Rightarrow \frac{3}{4}\:\:of\:\:36=\frac{3}{4}\times36=\frac{3\times36}{4}=\frac{108}{4}=27$

$(d)(i)\; \frac{4}{5}\;\; of\; 20\;$

On multiplying, we get

$\frac{4}{5}\;\; of\; 20\;=\frac{4}{5}\times20=\frac{4\times20}{5}=\frac{80}{5}=16$

$(d)(ii)\; \frac{4}{5}\;\; of\; 35\;$

On Multiplying, we get,

$\; \frac{4}{5}\;\; of\; 35\;=\frac{4}{5}\times35=\frac{4\times35}{5}=\frac{140}{4}=28$

Answer:$(a) 3\times 5\frac{1}{5}$

On Multiplying, we get

$\Rightarrow 3\times\frac{5\times5+1}{5}=3\times\frac{26}{5}=\frac{3\times26}{5}=\frac{78}{5}$

Converting This into Mixed Fraction,

$\Rightarrow \frac{78}{5}=\frac{75+3}{5}=\frac{75}{5}+\frac{3}{5}=15+\frac{3}{5}=15\frac{3}{5}$

$(b) 5\times 6\frac{3}{4}$

On Multiplying, we get

$\Rightarrow 5\times\frac{6\times4+3}{4}=5\times\frac{27}{4}=\frac{5\times27}{4}=\frac{135}{4}$

Converting This into Mixed Fraction,

$\Rightarrow \frac{135}{4}=\frac{132+3}{4}=\frac{132}{4}+\frac{3}{4}=33+\frac{3}{4}=33\frac{3}{4}$

$(c) 7\times 2\frac{1}{4}$

On multiplying, we get

$7\times 2\frac{1}{4}=7\times \frac{4\times2+1}{4}=7\times\frac{9}{4}=\frac{7\times9}{4}=\frac{63}{4}$

Converting it into a mixed fraction, we get

$\frac{63}{4}=\frac{60+3}{4}=\frac{60}{4}+\frac{3}{4}=15+\frac{3}{4}=15\frac{3}{4}$

$(d) 4\times 6\frac{1}{3}$

On multiplying, we get

$4\times 6\frac{1}{3}=4\times\frac{3\times6+1}{3}=4\times\frac{19}{3}=\frac{4\times 19}{3}=\frac{76}{3}$

Converting it into a mixed fraction,

$\frac{76}{3}=\frac{75+1}{3}=\frac{75}{3}+\frac{1}{3}=25+\frac{1}{3}=25\frac{1}{3}$

$(e) 3\frac{1}{4}\times 6$

Multiplying them, we get

$3\frac{1}{4}\times 6=\frac{4\times3+1}{4}\times6=\frac{13}{4}\times6=\frac{13\times6}{4}=\frac{78}{4}$

Now, converting the result fraction we got to mixed fraction,

$\frac{78}{4}=\frac{76+2}{4}=\frac{76}{4}+\frac{2}{4}=19+\frac{1}{2}=19\frac{1}{2}$

$(f) 3\frac{2}{5}\times 8$

On multiplying, we get

$3\frac{2}{5}\times 8=\frac{5\times3+2}{5}\times8=\frac{17}{5}\times8=\frac{17\times8}{5}=\frac{136}{5}$

Converting this into a mixed fraction, we get

$\frac{136}{5}=\frac{135+1}{5}=\frac{135}{5}+\frac{1}{5}=27+\frac{1}{5}=27\frac{1}{5}$

7. Find:

Answer: $(a) (i)\frac{1}{2} \; \; o\! f\; \;\; \; 2\frac{3}{4} \;$

As we know that of is equivalent to multiply,

$\frac{1}{2} \; \; o\! f\; \;\; \; 2\frac{3}{4} \;=\frac{1}{2}\times2\frac{3}{4}=\frac{1}{2}\times\frac{11}{4}=\frac{11}{8}=1\frac{3}{8}$

$(a)(ii)\frac{1}{2} \; \; o\! f\;\; 4\frac{2}{9}$

As we know that of is equivalent to multiply,

$\frac{1}{2} \; \; o\! f\;\; 4\frac{2}{9}=\frac{1}{2}\times4\frac{2}{9}=\frac{1}{2}\times\frac{38}{9}=\frac{38}{18}=\frac{19}{9}=2\frac{1}{9}$

$(b)(i)\frac{5}{8} \; \; o\! f\; \; 3\frac{5}{6} \; \;$

As we know that of is equivalent to multiplication, so

$\frac{5}{8} \; \; o\! f\; \; 3\frac{5}{6} \; =\frac{5}{8}\times3\frac{5}{6}=\frac{5}{8}\times\frac{23}{6}=\frac{115}{48}=2\frac{19}{48}$

$(b)(ii)\frac{5}{8} \; \; o\! f\; \;\; 9\frac{2}{3}$

As we know that of is equivalent to multiplication, so

$\frac{5}{8} \; \; o\! f\; \;\; 9\frac{2}{3}=\frac{5}{8}\times \frac{29}{3}=\frac{145}{24}=6\frac{1}{24}$

(i) How much water did Vidya drink?
(ii) What fraction of the total quantity of water did Pratap drink?

Total water = 5 litre.

i) The amount of water vidya consumed :

$=\frac{2}{5}\:\:o\!f 5\:liter=\frac{2}{5}\times5=2\:liter$

Hence vidya consumed 2 liters of water from the bottle.

ii) The amount of water Pratap consumed :

$=\left (1-\frac{2}{5} \right )\:\:o\!f 5\:liter=\frac{3}{5}\times5=3\:liter$

Hence, Pratap consumed 3 liters of water from the bottle.

NCERT S olutions for fractions and decimals class 7 Exercise 2.3

1. Find:

Answer: As we know, the term "of " means multiplication. So,

$(i)(a)\frac{1}{4}\; o\! f\;\; \frac{1}{4}=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}$

$(i)(b)\frac{1}{4}\; o\! f\;\; \frac{3}{5}=\frac{1}{4}\times\frac{3}{5}=\frac{3}{20}$

$(i)(c)\frac{1}{4}\; o\! f\;\; \frac{4}{3}=\frac{1}{4}\times\frac{4}{3}=\frac{4}{12}=\frac{1}{3}$

$(ii) (a)\frac{1}{7}\; o\! f\;\; \frac{2}{9}=\frac{1}{7}\times\frac{2}{9}=\frac{2}{63}$

$(ii) (b)\frac{1}{7}\; o\! f\;\; \frac{6}{5}=\frac{1}{7}\times\frac{6}{5}=\frac{6}{35}$

$(ii) (c)\frac{1}{7}\; o\! f\;\; \frac{3}{10}=\frac{1}{7}\times\frac{3}{10}=\frac{3}{70}$

Answer: As we know, in the multiplication of fraction, the numerator gets multiplied with numerator and denominator gets multiplied by the denominator. So,

$\\(i) \; \frac{2}{3}\times 2\frac{2}{3}=\frac{2}{3}\times\frac{8}{3}=\frac{2\times8}{3\times3}=\frac{16}{9}=1\frac{7}{9} \; \; \;\\\\ (ii) \frac{2}{7}\times \frac{7}{9}=\frac{2\times7}{7\times9}=\frac{2}{9}\; \; \; \; \;\\\\ (iii)\frac{3}{8}\times \frac{6}{4}=\frac{3\times6}{8\times4}=\frac{18}{32}=\frac{9}{16}\; \; \; \;$

$\\(iv)\; \; \frac{9}{5}\times \frac{3}{5}=\frac{9\times3}{5\times5}=\frac{27}{25}=1\frac{2}{25} \\\\ (v)\frac{1}{3}\times \frac{15}{8}=\frac{1\times15}{3\times8}=\frac{15}{24}=\frac{5}{8}\; \; \; \\\\ (vi)\; \; \frac{11}{2}\times \frac{3}{10}=\frac{11\times3}{2\times10}=\frac{33}{20}=1\frac{13}{20} \; \; \; \\\\ (vii)\; \; \frac{4}{5}\times \frac{12}{7}=\frac{4\times12}{5\times7}=\frac{48}{35}=1\frac{13}{35}$ .

Answer: As we know in the multiplication of fractions, the numerator is multiplied with numerator and denominator is multiplied by the denominator.

So,

$\\(i) \; \frac{2}{5}\times 5\frac{1}{4}=\frac{2}{5}\times\frac{21}{4}=\frac{2\times21}{5\times4}=\frac{42}{20}=\frac{21}{10}=2\frac{1}{10}\; \; \; \;\\ \\(ii)\; 6\frac{2}{5}\times \frac{7}{9}=\frac{32}{5}\times\frac{7}{9}=\frac{32\times7}{5\times9}=\frac{224}{45}=4\frac{44}{45}\; \; \; \; \; \\\\(iii) \frac{3}{2}\times 5\frac{1}{3}=\frac{3}{2}\times\frac{16}{3}=\frac{3\times16}{2\times3}=\frac{48}{6}=8$

$\\(iv) \; \frac{5}{6}\times 2\frac{3}{7}=\frac{5}{6}\times\frac{17}{7}=\frac{5\times17}{6\times7}=\frac{85}{42}=2\frac{1}{42}\; \; \; \; \\\\(v)\; 3\frac{2}{5}\times \frac{4}{7}=\frac{17}{5}\times\frac{4}{7}=\frac{17\times4}{5\times7}=\frac{68}{35}=1\frac{33}{35}\; \; \; \; \; \\\\(vi) 2\frac{3}{5}\times 3=\frac{13}{5}\times3=\frac{13\times3}{5}=\frac{39}{5}=7\frac{4}{5}\; \; \; \;\\\\ (vii)3\frac{4}{7}\times \frac{3}{5}=\frac{25}{7}\times\frac{3}{5}=\frac{25\times3}{7\times5}=\frac{75}{35}=\frac{15}{7}=2\frac{1}{7}$

Answer: $(i)\; \frac{2}{7}\; o\! f\; \frac{3}{4}\; \; or\; \; \frac{3}{5}\; \; of \; \frac{5}{8}$

$\Rightarrow \; \frac{2}{7} \times\ \frac{3}{4}\; \; or\; \; \frac{3}{5}\; \times \frac{5}{8}$

$\Rightarrow \frac{2\times3}{7\times4}\:\:or\:\:\frac{3\times5}{5\times8}$

$\Rightarrow \frac{3}{14}\:\:or\:\:\frac{3}{8}$

Now, As we Know, When the numerator of two fractions is the same the fraction with lesser denominator is the bigger fraction. So,

$\Rightarrow \frac{3}{14}\:\:<\:\:\frac{3}{8}$

Thus,

$\; \frac{2}{7}\; o\! f\; \frac{3}{4}\; \; <\; \; \frac{3}{5}\; \; of \; \frac{5}{8}$ .

$(ii)\; \frac{1}{2}\; o\! f\; \frac{6}{7}\; \; or\; \; \frac{2}{3}\; \; of \; \frac{3}{7}$

$\Rightarrow \; \frac{1}{2}\; \times\; \frac{6}{7}\; \; or\; \; \frac{2}{3}\; \; \times \; \frac{3}{7}$

$\Rightarrow \frac{1\times6}{2\times7}\:\:or\:\:\frac{2\times3}{3\times7}$

$\Rightarrow \frac{3}{7}\:\:or\:\:\frac{2}{7}$

As we know, When the denominator of two fractions are the same, the fraction with the bigger numerator is the bigger fraction, so,

$\Rightarrow \frac{3}{7}\:\:>\:\:\frac{2}{7}$

Thus,

$\; \frac{1}{2}\; \times\; \frac{6}{7}\; \; >\; \; \frac{2}{3}\; \; \times \; \frac{3}{7}$ .

Answer: distance between two adjacent saplings = $\frac{3}{4}m$

distance between the first and the last sapling :

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

$=\frac{3\times3}{4}$

$=\frac{9}{4}$ .

Answer: Number of time spent in one day = $1\frac{3}{4}$ hour

The number of time spent in 6 days :

$=6\times1\frac{3}{4}=6\times\frac{7}{4}=\frac{6\times7}{4}=\frac{42}{4}=\frac{21}{2}=10\frac{1}{2}\:hours$

Hence $10\frac{1}{2}\:hours$ are required by Lipika to complete the book.

Answer: Distance covered in 1 liter petrol = 16 km

The distance that will be covered in $2\frac{3}{4}$ liter petrol.

$=16\times2\frac{3}{4}$

$=16\times\frac{11}{4}$

$=\frac{11\times16}{4}$

$=11\times4$

$=44\:km$

Hence 44 km can be covered using $2\frac{3}{4}$ liter petrol.

(ii) The simplest form of the number obtained in is _________.

(b) (i) Provide the number in the box , such that $\frac{3}{5}\times$ = $\frac{24}{75}$ .

(ii) The simplest form of the number obtained in is _________.

Answer: As we Know, That in the multiplication of fraction, the numerator of both fraction are multiplied to give numerator of new fraction, and denominator of both fractions is multiplied to give the denominator of answer fraction. So,

$\frac{2}{3}\times$ = $\frac{10}{30}$ .

We have to multiply 5 with 2 in the numerator to get numerator equal to 10 and 10 with 3 in the denominator to get 30. So,

$\Rightarrow \frac{2}{3}\times\frac{5}{10}=\frac{10}{30}$

The Simplest Form :

$\Rightarrow \frac{5}{10}=\frac{1}{2}$ .

NCERT Solutions for fractions and decimals class 7 Exercise 2.4

1. Find:

Answer: As we know, The division of a number by a / b is equivalent to multiplication of that number with b / a. So,

$(i) 12\div \frac{3}{4}$

$\Rightarrow 12\div \frac{3}{4}=12\times\frac{4}{3}=\frac{48}{3}=16$

$(ii) 14\div \frac{5}{6}$

$\Rightarrow 14\div \frac{5}{6}=14\times\frac{6}{5}=\frac{84}{5}=16\frac{4}{5}$

$(iii) 8\div \frac{7}{3}$

$\Rightarrow 8\div \frac{7}{3}=8\times\frac{3}{7}=\frac{24}{7}=3\frac{3}{7}$

$(iv) 4\div \frac{8}{3}$

$\Rightarrow 4\div \frac{8}{3}=4\times\frac{3}{8}=\frac{12}{8}=\frac{3}{2}=1\frac{1}{2}$

$(v) 3\div 2\frac{1}{3}$

$\Rightarrow 3\div 2\frac{1}{3}=3\div\frac{7}{3}=3\times\frac{3}{7}=\frac{9}{7}=1\frac{2}{7}$

$(vi) 5\div 3\frac{4}{7}$

$\Rightarrow 5\div 3\frac{4}{7}=5\div\frac{25}{7}=5\times\frac{7}{25}=\frac{35}{25}=\frac{7}{5}=1\frac{2}{5}$

Answer: As we know, in the reciprocal of any fraction, the numerator and denominator get exchanged. Basically, we flip the number upside down. So

$i)Reciprocal\:\: o\!f \: \frac{3}{7}=\frac{7}{3}$

As numerator is greater than the denominator, it is an improper fraction.

$ii)Reciprocal\:\: o\!f \: \frac{5}{8}=\frac{8}{5}$

As numerator is greater than the denominator, it is an improper fraction.

$iii)Reciprocal\:\: o\!f \: \frac{9}{7}=\frac{7}{9}$

As the Denominator is greater than Numerator, it is a proper fraction.

$iv)Reciprocal\:\: o\!f \: \frac{6}{5}=\frac{5}{6}$

As the Denominator is greater than Numerator, it is a proper fraction.

$v)Reciprocal\:\: o\!f \: \frac{12}{7}=\frac{7}{12}$

As the Denominator is greater than Numerator, it is a proper fraction.

$vi)Reciprocal\:\: o\!f \: \frac{1}{8}=\frac{8}{1}=8$

It is an integer and hence a whole Number.

$vii)Reciprocal\:\: o\!f \: \frac{1}{11}=\frac{11}{1}=11$

It is an integer and hence a whole Number.

3. Find:

Answer: As we know, The division of a number by a / b is equivalent to multiplication of that number with b / a which is also called the reciprocal of a / b. So,

$(i)\; \frac{7}{3}\div 2$

$\Rightarrow \; \frac{7}{3}\div 2=\frac{7}{3}\times\frac{1}{2}=\frac{7}{6}=1\frac{1}{6}$

$(ii)\; \frac{4}{9}\div 5$

$\Rightarrow \; \frac{4}{9}\div 5=\frac{4}{9}\times\frac{1}{5}=\frac{4}{45}$

$(iii)\; \frac{6}{13}\div 7$

$\Rightarrow \; \frac{6}{13}\div 7=\frac{6}{13}\times\frac{1}{7}=\frac{6}{91}$

$(iv)\; 4\frac{1}{3}\div 3$

$\Rightarrow \; 4\frac{1}{3}\div 3=\frac{13}{3}\div3=\frac{13}{3}\times\frac{1}{3}=\frac{13}{9}=1\frac{4}{9}$

$(v)\; 3\frac{1}{2}\div 4$

$\Rightarrow \; 3\frac{1}{2}\div 4=\frac{7}{2}\div4=\frac{7}{2}\times\frac{1}{4}=\frac{7}{8}$

$(vi)\; 4\frac{3}{7}\div 7$

$\Rightarrow \; 4\frac{3}{7}\div 7=\frac{31}{7}\div7=\frac{31}{7}\times\frac{1}{7}=\frac{31}{49}$

4. Find:

Answer: As we know, The division of a number by a / b is equivalent to multiplication of that number with b / a which is also called the reciprocal of a / b. So,

$(i)\; \frac{2}{5}\div \frac{1}{2}$

$\Rightarrow \; \frac{2}{5}\div \frac{1}{2}=\frac{2}{5}\times\frac{2}{1}=\frac{4}{5}$

$(ii)\; \frac{4}{9}\div \frac{2}{3}$

$\Rightarrow \; \frac{4}{9}\div \frac{2}{3}=\frac{4}{9}\times\frac{3}{2}=\frac{12}{18}=\frac{2}{3}$

$(iii)\; \frac{3}{7}\div \frac{8}{7}$

$\Rightarrow \; \frac{3}{7}\div \frac{8}{7}=\frac{3}{7}\times\frac{7}{8}=\frac{3}{8}$

$(iv)\;2 \frac{1}{3}\div \frac{3}{5}$

$\Rightarrow \;2 \frac{1}{3}\div \frac{3}{5}=\frac{7}{3}\div\frac{3}{5}=\frac{7}{3}\times\frac{5}{3}=\frac{35}{9}=3\frac{8}{9}$

$(v)\;3 \frac{1}{2}\div \frac{8}{3}$

$\;3 \frac{1}{2}\div \frac{8}{3}=\frac{7}{2}\div\frac{8}{3}=\frac{7}{2}\times\frac{3}{8}=\frac{21}{16}=1\frac{5}{16}$

$(vi)\;\frac{2}{5}\div 1\frac{1}{2}$

$\Rightarrow \;\frac{2}{5}\div 1\frac{1}{2}=\frac{2}{5}\div\frac{3}{2}=\frac{2}{5}\times\frac{2}{3}=\frac{4}{15}$

$(vii)\;3\frac{1}{5}\div 1\frac{2}{3}$

$\Rightarrow \;3\frac{1}{5}\div 1\frac{2}{3}=\frac{16}{5}\div\frac{5}{3}=\frac{16}{5}\times\frac{3}{5}=\frac{48}{25}=1\frac{23}{25}$

$(viii)\;2\frac{1}{5}\div 1\frac{1}{5}$

$\Rightarrow \;2\frac{1}{5}\div 1\frac{1}{5}=\frac{11}{5}\div\frac{6}{5}=\frac{11}{5}\times\frac{5}{6}=\frac{11}{6}=1\frac{5}{6}$

NCERT Solutions for fractions and decimals class 7 Exercise 2.5

(i) 0.5 or 0.05 (ii) 0.7 or 0.5 (iii) 7 or 0.7
(iv) 1.37 or 1.49 (v) 2.03 or 2.30 (vi) 0.8 or 0.88.

Answer: As we know, Any decimal is equivalent to the number without decimal divided by 10 x (number of integers after the decimal). in other words,

$0.5=\frac{5}{10}$ $\:\:and\:\:0.05=\frac{5}{100}$

So,

(i) 0.5 > 0.05

(ii) 0.7 > 0.5

(iii) 7 > 0.7

(iv) 1.37 < 1.49

(v) 2.03 < 2.30

(vi) 0.8 < 0.88.

(i) 7 paise (ii) 7 rupees 7 paise (iii) 77 rupees 77 paise
(iv) 50 paise (v) 235 paise.

Answer: As we know, there are 100 paise in 1 rupee. i.e.

$\:1\:paise=\frac{1}{100}Rupees$

So,

$i) 7 \:paise=\frac{7}{100}=0.07\:Rupees$

$ii) 7 \:Rupees ,7 \:paise=7+\frac{7}{100}=7+0.07=7.07\:Rupees$

$iii) 77 \:Rupees ,77 \:paise=77+\frac{77}{100}=7+0.77=7.77\:Rupees$

$iv)50 \:paise=\frac{50}{100}=0.5\:Rupees$

$v)235 \:paise=\frac{235}{100}=\frac{200+35}{100}=\frac{200}{100}+\frac{35}{100}=2+0.35=2.35\:Rupees$

(ii) Express 35 mm in cm, m and km

1 centimeter = 10 millimeter

1 meter = 100 centimeter.

And

1 kilometer = 1000 meter

So,

i) 5 cm

$5 \:cm=\frac{5}{100}m=0.05m$

$5 \:cm=\frac{5}{100}m=0.05m=\frac{0.005}{1000}=0.000005\:km$

ii) 35 mm

$35\:mm=\frac{35}{10}cm=3.5cm$

$35\:mm=\frac{35}{10}cm=3.5cm=\frac{3.5}{100}m=0.035m$

$35\:mm=\frac{35}{10}cm=3.5cm=\frac{3.5}{100}m=0.035m=\frac{0.035}{1000}km=0.000035km$

(i) 200 g (ii) 3470 g (iii) 4 kg 8 g

1 kg = 1000 g.

$1\:g=\frac{1}{1000}kg$

So,

i) 200 g

$200g=\frac{200}{1000}kg=0.2kg$

(ii) 3470 g

$3470g=\frac{3470}{1000}kg=3.47kg$

(iii) 4 kg 8 g

$4kg,\:\:8g=4kg+\frac{8}{1000}kg=4kg+0.008kg=4.0008kg$

Answer: Decimal in their expanded form are

(i) 20.03

$20.03=2\times10+0\times1+\frac{1}{10}\times0+\frac{1}{100}\times3$

(ii) 2.03

$2.03=2\times1+\frac{1}{10}\times0+\frac{1}{100}\times3$

(iii) 200.03

$200.03=2\times100+0\times10+0\times1+\frac{1}{10}\times0+\frac{1}{100}\times3$

(iv) 2.034

$2.034=2\times1+\frac{1}{10}\times0+\frac{1}{100}\times3+4\times\frac{1}{1000}$

2 is in one's position.

(ii) 21.37

2 is in ten's position

(iii) 10.25

2 is in one-tenths position

(iv) 9.42

2 is in one-hundredths position.

(v) 63.352.

2 is in one-thousandth's position.

Answer: Total distance travelled by Dinesh = AB + BC

= 7.5km + 12.7km

= 20.2 km

Total distance travelled by Ayub = AD + DC

= 9.3km + 11.8 km

= 21.1 km

Hence Ayub travelled More distance than Ayub as 21.1 > 20.2

The difference between path travelled by them = 21.1km - 20.2 km

= 0.9 km.

Hence Ayub travelled 0.9 km more than Dinesh.

Answer: For comparing two quantities, we should make their unit the same first. So,

Fruits bought by Shyama in kg = 5 kg + 300 g + 3 kg + 250 g

= 8 kg + 550 g

= 8 kg + 0.55 kg

= 8.55 kg.

Fruits bought by Sarala in kg = 4 kg + 800 g + 4 kg + 150 g

= 8 kg + 950 kg

= 8 kg + 0.95 kg

= 8.95 kg.

Hence Sarala bought more fruits as 8.95 > 8.55

The difference between the amount of Fruits they bought = 8.95 kg - 8.55 kg

= 0.4 kg

Answer: Difference = 42.6 km - 28 km

= 14.6 km

Hence 28 is 14.6 km less than 42.6.

NCERT Solutions for maths class 7 chapter 2 Fractions And Decimals Exercise 2.6

1. Find:

(i) 0.2 × 6 (ii) 8 × 4.6 (iii) 2.71 × 5 (iv) 20.1 × 4

(v) 0.05 × 7 (vi) 211.02 × 4 (vii) 2 × 0.86

Answer: As we know, The multiplication of the decimal number is just like multiplication of normal number, we just have to put decimal before a digit such that the number of digits after decimal should remain same.

In other words,

The number of digits after the decimal should be the same before and after the multiplication.

So.

(i) 0.2 × 6 = 1.2

(ii) 8 × 4.6 = 36.8

(iii) 2.71 × 5 = 13.55

(iv) 20.1 × 4 = 80.4

(v) 0.05 × 7 = 0.35

(vi) 211.02 × 4 = 844.08

(vii) 2 × 0.86 = 1.72

Answer: Given, Length of rectangle = 5.7 cm.

Width of rectangle = 3 cm

Area of the rectangle = Length x width

= 5.7 x 3

= 17.1 $cm^2$ .

Hence Area of the rectangle is 17.1 $cm^2$ .

3. Find:

(i) 1.3 × 10 (ii) 36.8 × 10 (iii) 153.7 × 10 (iv) 168.07 × 10

(v) 31.1 × 100 (vi) 156.1 × 100 (vii) 3.62 × 100 (viii) 43.07 × 100

(ix) 0.5 × 10 (x) 0.08 × 10 (xi) 0.9 × 100 (xii) 0.03 × 1000

Answer: As we know, The multiplication of the decimal number is just like multiplication of normal number, we just have to put decimal before a digit such that the number of digits after decimal should remain same.

In other words,

The number of digits after the decimal should be the same before and after the multiplication.

So.

(i) 1.3 × 10 = 13

(ii) 36.8 × 10 = 368

(iii) 153.7 × 10 = 1537

(iv) 168.07 × 10 = 1680.7

(v) 31.1 × 100 = 3110

(vi) 156.1 × 100 =15610

(vii) 3.62 × 100 = 362

(viii) 43.07 × 100 = 4307

(ix) 0.5 × 10 = 5

(x) 0.08 × 10 = 0.8

(xi) 0.9 × 100 = 90

(xii) 0.03 × 1000 =30

Answer: Distance covered by two-wheeler in 1 liter of petrol = 55.3 km.

Distance two-wheeler will cover in 10 liters of petrol = 10 x 55.3 km

= 553 km

Hence two-wheeler will cover a distance of 553 km in 10 liters of petrol.

5. Find:

(i) 2.5 × 0.3 (ii) 0.1 × 51.7 (iii) 0.2 × 316.8 (iv) 1.3 × 3.1 (v) 0.5 × 0.05 (vi) 11.2 × 0.15 (vii) 1.07 × 0.02

(viii) 10.05 × 1.05 (ix) 101.01 × 0.01 (x) 100.01 × 1.1

Answer: As we know, The multiplication of the decimal number is just like multiplication of normal number, we just have to put decimal before a digit such that the number of digits after decimal should remain same.

In other words,

The number of digits after the decimal should be the same before and after the multiplication.

So.

(i) 2.5 × 0.3 = 0.75

(ii) 0.1 × 51.7 = 5.17

(iii) 0.2 × 316.8 = 63.36

(iv) 1.3 × 3.1 = 4.03

(v) 0.5 × 0.05 = 0.025

(vi) 11.2 × 0.15 = 1.680

(vii) 1.07 × 0.02 = 0.0214

(viii) 10.05 × 1.05 = 10.5525

(ix) 101.01 × 0.01 = 1.0101

(x) 100.01 × 1.1 = 110.11

NCERT Solutions for maths class 7 chapter 2 Fractions And Decimals 2.7

1. Find:

Answer: As we know, while dividing decimal, we first express the decimal in term of fraction and then divides it,

And

Dividing by a number is equivalent to multiplying by the reciprocal of that number. So

So

$(i)\; 0.4\div 2$

$\; 0.4\div 2=\frac{4}{10}\div2=\frac{4}{10}\times\frac{1}{2}=\frac{2}{10}=\frac{1}{5}$

$(ii)\; 0.35\div 5$

$\; 0.35\div 5=\frac{35}{100}\div5=\frac{35}{100}\times\frac{1}{5}=\frac{7}{100}=0.07$

$(iii)\; 2.48\div 4$

$\; 2.48\div 4=\frac{248}{100}\div4=\frac{248}{100}\times\frac{1}{4}=\frac{62}{100}=0.62$

$(iv)\; 65.4\div 6$

$\; 65.4\div 6=\frac{654}{100}\div6=\frac{654}{100}\times\frac{1}{6}=\frac{109}{100}=1.09$

$(v)\; 651.2\div 4$

$\; 651.2\div 4=\frac{6512}{10}\div4=\frac{6512}{10}\times\frac{1}{4}=\frac{1628}{10}=162.8$

$(vi)\; 14.49\div 7$

$\; 14.49\div 7=\frac{1449}{100}\div7=\frac{1449}{100}\times\frac{1}{7}=\frac{207}{100}=2.07$

$(vii)\; 3.96\div 4$

$\; 3.96\div 4=\frac{396}{100}\div4=\frac{396}{100}\times\frac{1}{4}=\frac{99}{100}=0.99$

$(viii)\; 0.80\div 5$

$\; 0.80\div 5=\frac{80}{100}\div5=\frac{80}{100}\times\frac{1}{5}=\frac{16}{100}=0.16$

2. Find:

Answer: As we know, When we divide a decimal number by 10, the decimal point gets shifted by one digit in the left.

So

$(i)\; 4.8\div 10$

$\; 4.8\div 10=0.48$

$(ii)\; 52.5\div 10$

$\; 52.5\div 10=5.25$

$(iii)\; 0.7\div 10$

$\; 0.7\div 10=0.07$

$(iv)\; 33.1\div 10$

$\; 33.1\div 10=3.31$

$(v)\; 272.23\div 10$

$\; 272.23\div 10=27.223$

$(vi)\; 0.56\div 10$

$\; 0.56\div 10=0.056$

$(vii)\; 3.97\div 10$

$\; 3.97\div 10=0.397$

3. Find:

Answer: As we know while dividing a decimal number by 100, the decimal point gets shifted to left by two digits.

So

$(i)\; 2.7\div 100$

$\; 2.7\div 100=0.027$

$(ii)\; 0.3\div 100$

$\; 0.3\div 100=0.003$

$(iii)\; 0.78\div 100$

$\; 0.78\div 100=0.0078$

$(iv)\; 432.6\div 100$

$\; 432.6\div 100=4.326$

$(v)\; 23.6\div 100$

$\; 23.6\div 100=0.236$

$(vi)\; 98.53\div 100$

$\; 98.53\div 100=0.9853$

4. Find:

Answer: As we know, while dividing a decimal number by 1000 we shift the decimal to left by 3 digits, So

$(i)\; 7.9\div 1000$

$\; 7.9\div 1000=0.0079$

$(ii)\; 26.3\div 1000$

$\; 26.3\div 1000=0.0263$

$(iii)\; 38.53\div 1000$

$\; 38.53\div 1000=0.03853$

$(iv)\; 128.9\div 1000$

$\; 128.9\div 1000=0.1289$

$(v)\; 0.5\div 1000$

5. Find:

Answer: As we know, while dividing decimal, we first express the decimal in term of fraction and then divides it,

And

Dividing by a number is equivalent to multiplying by the reciprocal of that number.

So

$(i)\; 7\div 3.5$

$\; 7\div 3.5=7\div \frac{35}{10}=7\times\frac{10}{35}=\frac{70}{35}=2$

$(ii)\; 36\div 0.2$

$\; 36\div 0.2=36\div\frac{2}{10}=36\times\frac{10}{2}=180$

$(iii)\; 3.25\div 0.5$

$\; 3.25\div 0.5=\frac{325}{100}\div\frac{5}{10}=\frac{325}{100}\times\frac{10}{5}=\frac{65}{10}=6.5$

$(iv)\; 30.94\div 0.7$

$\; 30.94\div 0.7=\frac{3094}{100}\div\frac{7}{10}=\frac{3094}{100}\times\frac{10}{7}=\frac{442}{10}=44.2$

$(v)\; 0.5\div 0.25$

$\; 0.5\div 0.25=\frac{5}{10}\div\frac{25}{100}=\frac{5}{10}\times\frac{100}{25}=\frac{10}{5}=2$

$(vi)\; 7.75\div 0.25$

$\; 7.75\div 0.25=\frac{775}{100}\div\frac{25}{100}=\frac{775}{100}\times\frac{100}{25}=31$

$(vii)\; 76.5\div 0.15$

$\; 76.5\div 0.15=\frac{765}{10}\div\frac{15}{100}=\frac{765}{10}\times\frac{100}{15}=510$

$(viii)\; 37.8\div 1.4$

$\; 37.8\div 1.4=\frac{378}{10}\div\frac{14}{10}=\frac{378}{10}\times\frac{10}{14}=27$

$(ix)\; 2.73\div 1.3$

$\; 2.73\div 1.3=\frac{273}{100}\div\frac{13}{10}=\frac{273}{100}\times\frac{10}{13}=\frac{21}{10}=2.1$

Answer: Distance travelled by vehicle in 2.4 litres of petrol = 43.2 km

Distance travelled by vehicle in 1 litre of petrol:

$=\frac{43.2}{2.4}=\frac{432}{24}\times\frac{10}{10}=18$

Hence Distance travelled by vehicle in 1 litre is 18 km.

## Fractions and Decimals Class 7 Maths Chapter 2-Topics

Here students can find all the topics which are discussed in this chapter, fractions and decimals class 7.

• How Well Have You Learnt About Fractions?
• Multiplication Of Fractions
• Division Of Fractions
• How Well Have You Learnt About Decimal Numbers
• Multiplication Of Decimal Numbers
• Division Of Decimal Numbers

### NCERT Solutions for Class 7 Maths Chapter Wise

 Chapter No. Chapter Name Chapter 1 Integers Chapter 2 Fractions and Decimals Chapter 3 Data Handling Chapter 4 Simple Equations Chapter 5 Lines and Angles Chapter 6 The Triangle and its Properties Chapter 7 Congruence of Triangles Chapter 8 Comparing quantities Chapter 9 Rational Numbers Chapter 10 Practical Geometry Chapter 11 Perimeter and Area Chapter 12 Algebraic Expressions Chapter 13 Exponents and Powers Chapter 14 Symmetry Chapter 15 Visualising Solid Shapes

### NCERT Solutions for Class 7 Subject Wise

 NCERT Solutions for Class 7 Maths NCERT Solutions for Class 7 Science

### Importance of NCERT Solutions for Class 7 Chapter 2 Fractions and Decimals

• Now homework of the chapter becomes an easy breezy thing with CBSE NCERT solutions for Class 7 chapter 2 Fractions and Decimals in hand.
• Solutions of NCERT Class 7 chapter 2 Fractions and Decimals in hand helps students in self-evaluation
• The NCERT solutions for Class 7 Maths chapter 2 Fractions and Decimals are helpful in preparation for the exam. A similar type of question can be expected for the class exams.

Also Check NCERT Books and NCERT Syllabus here:

1. Do I need to acquire knowledge of all the topics covered in the NCERT Solutions for Class 7 Maths Chapter 2?

Absolutely, it is crucial to thoroughly grasp all the topics presented in the NCERT Solutions for Class 7 Maths Chapter 2 in order to achieve high marks in the Class 7 board exams. These solutions are meticulously crafted by subject matter experts who have compiled model questions encompassing all the exercise questions from the textbook. Emphasis is placed on elucidating the solutions in a manner that is easily comprehensible for students, thereby aiding their understanding.

2. Can I rely solely on the NCERT Solutions for Class 7 Maths Chapter 2 to confidently tackle all the questions in the board exam?

Indeed, the NCERT Solutions for Class 7 Maths Chapter 2 are sufficient to solve all the questions that appear in the board exam. Thoroughly practicing this chapter enables students to grasp the concepts flawlessly. These questions have been meticulously designed in accordance with the NCERT syllabus and guidelines, ensuring that students can achieve good marks in their final exams. To study both online and offline, students can download class 7 maths chapter 2 pdf.

3. List the topics included in the NCERT Solutions for class 7 chapter 2 maths.

The main topics covered in the NCERT Solutions for class 7 chapter 2 maths include:

1. Addition and subtraction of fractions
2. Multiplication of fractions
3. Multiplication of a fraction by a whole number
4. Multiplication of a fraction by a fraction
5. Division of fractions
6. Division of a whole number by a fraction
7. Reciprocal of a fraction
8. Division of a fraction by a whole number
9. Division of a fraction by another fraction
10. Multiplication of decimal numbers
11. Multiplication of decimal numbers by 10, 100, and 1000
12. Division of decimal numbers
13. Division of decimals by 10, 100, and 1000
14. Division of a decimal number by a whole number
15. Division of a decimal number by another decimal number.

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