ncert-solutions-class-7-maths-ch-2-fractions-and-decimals

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

NCERT solutions for class 7 maths chapter 2 fractions and decimals topic 2.3.1

1.(a) Find:

$(a)\frac{2}{7}\times 3$

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

Answer:

Given,

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

The product is a proper fraction.

1.(b) Find:

$(b)\frac{9}{7}\times 6$

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

Answer:

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:

$(c)3 \times \frac{1}{8}$

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

Answer:

Given, the product

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

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

$= \frac{3}{8}$

This is a proper fraction.

1.(d) Find:

$(d)\frac{13}{11}\times 6$

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

Answer:

Given the product:

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

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

$= \frac{78}{11}$

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

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

2. Represent pictorially : $2\times \frac{2}{5}=\frac{4}{5}$

Answer:

1.(i) Find:

$(i)5\times 2\frac{3}{7}$

Answer:

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

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

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

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

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

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

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

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

1.(ii) Find :

$(ii)1\frac{4}{9}\times 6$

Answer:

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

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

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

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

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

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

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

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

1. Can you tell, what is $(i) \frac{1}{2}$ of 10? , $(ii) \frac{1}{4}$ of 16? , $(iii) \frac{2}{5}$ of 25?

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

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

1. Find: $\frac{1}{3}\times \frac{4}{5};\frac{2}{3}\times \frac{1}{5}$

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,

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

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

1. Find : $\frac{8}{3}\times \frac{4}{7};\frac{3}{4}\times \frac{2}{3}$

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,

NCERT solutions for class 7 maths chapter 2 fractions and decimals topic 2.3.2

1. Fill in these boxes:

$(i) \frac{1}{2}\times \frac{1}{7}= \frac{1\times 1}{2\times 7}=$ $(ii) \frac{1}{5}\times \frac{1}{7}=$

$(iii) \frac{1}{7}\times \frac{1}{2}=$ $(iv) \frac{1}{7}\times \frac{1}{5}=$

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 class 7 maths chapter 2 fractions and decimals topic 2.4.1

1. Find :

$(i)7\div \frac{2}{5}$ $(ii)6\div \frac{4}{7}$ $(iii)2\div \frac{8}{9}$

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:

$(i)6\div 5\frac{1}{3}$ $(ii)7\div 2\frac{4}{7}$

Answer:

As we know, The division of a number by a / b is equivalent to 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 class 7 maths chapter 2 fractions and decimals topic 2.4.3

3. Find:

$(i)\frac{3}{5}\div \frac{1}{2}$ $(ii)\frac{1}{2}\div \frac{3}{5}$ $(iii)2\frac{1}{2}\div \frac{3}{5}$ $(iv)5\frac{1}{6}\div \frac{9}{2}$

Answer:

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

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

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

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

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

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

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

NCERT solutions forclass 7 chapter 2 fractions and decimals topic 2.6

1. Find:

$(i)2.7\times 4$ $(ii)1.8\times 1.2$ $(iii)2.3\times 4.35$

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.

$(i)2.7\times 4=10.8$

$(ii)1.8\times 1.2=2.16$

$(iii)2.3\times 4.35=10.005$

2. Arrange the products obtained in descending order.

$(i)2.7\times 4$ $(ii)1.8\times 1.2$ $(iii)2.3\times 4.35$

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:

NCERT solutions for class 7 chapter 2 fractions and decimals topic 2.6.1

1. Find:

$(i)0.3\times 10$ $(ii)1.2\times 100$ $(iii)56.3\times 1000$

Answer:

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

2. ( i) 35.7 ÷ 3 = ?;
(ii) 25.5 ÷ 3 = ?

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

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

3. (i) 43.15 ÷ 5 = ?;
(ii) 82.44 ÷ 6 = ?

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 class 7 chapter 2 fractions and decimals topic 2.7.3

1. Find:

$(i)\frac{7.75}{0.25}$ $(ii)\frac{42.8}{0.02}$ $(iii)\frac{5.6}{1.4}$

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)

(iv)

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

(v)

(vi)

(vii)

2. Arrange the following in descending order:

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

Answer:

(i)

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

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

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

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

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

3. In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square?

(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.

4. A rectangular sheet of paper is $12\tfrac{1}{2}$ cm long and $10\tfrac{2}{3}$ cm wide. Find its perimeter.

Answer:

Given,

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.

<img alt="=2\times\left ( \frac{25}{2}\times\frac{3}{3}+\frac{32}{3}\times\frac{2}{2} \right )" height="45"

src="https://lh5.googleusercontent.com/R6KWawTA47GO7uZo31Jfx8c1uAqZbyAkz1BW7MwAuJyt0HWqsgvWYS_vd5zsXsoicIuBlWHPpA04JVHRMZL00rH3mhh2XPr-OmMOYb07Ct3MdeseXwjWqt1iva4uo0QX9pGnHJw" style="margin-left: 0px; margin-top: 0px;" width="209" />

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

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

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

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

5. Find the perimeters of $(i)\bigtriangleup\! ABE$ (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

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 fraction, 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 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 denominator of both fraction 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.

5. Salil wants to put a picture in a frame. The picture is $7\tfrac{3}{5}$ cm wide. To fit in the frame the picture cannot be more than $7\tfrac{3}{10}$ cm wide. How much should be trimmed?

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.

7. Ritu ate $\frac{3}{5}$ part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? Who had the larger share? By how much?

Answer:

Given,

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 fraction 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.

8. Michael finished colouring a picture in $\frac{7}{12}$ hour. Vaibhav finished colouring the same picture in $\frac{3}{4}$ hour. Who worked longer? By what fraction was it longer?

Answer:

Time taken by Michal in coloring = $\frac{7}{12}$ hours.

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

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

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

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

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

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

1. Which of the drawings (a) to (d) show :

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

Answer:

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

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

ii) b represents two square one part out of two of both square 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)

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

2. Some pictures (a) to (c) are given below. Tell which of them show:

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

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 the 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.

3. Multiply and reduce to lowest form and convert into a mixed fraction:

$(i) 7\times \frac{3}{5}$ $(ii) 4\times \frac{1}{3}$ $(iii) 2\times \frac{6}{7}$ $(iv) 5\times \frac{2}{9}$ $(v) \frac{2}{3}\times 4$

$(vi) \frac{5}{2}\times 6$ $(vii) 11\times \frac{4}{7}$ $(viii) 20\times \frac{4}{5}$ $(ix) 13\times \frac{1}{3}$ $(x) 15\times \frac{3}{5}$

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,

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

4. Shade:

$(i) \frac{1}{2}$ 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:

$(a)\; \frac{1}{2}\;\; of\; (i)24\; \; (ii)46$ $(b)\; \frac{2}{3}\;\; of\; (i)18\; \; (ii)27$

$(c)\; \frac{3}{4}\;\; of\; (i)16\; \; (ii)36$ $(d)\; \frac{4}{5}\;\; of\; (i)20\; \; (ii)35$

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,

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

6. Multiply and express as a mixed fraction :

$(a) 3\times 5\frac{1}{5}$ $(b) 5\times 6\frac{3}{4}$ $(c) 7\times 2\frac{1}{4}$

$(d) 4\times 6\frac{1}{3}$ $(e) 3\frac{1}{4}\times 6$ $(f) 3\frac{2}{5}\times 8$

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:

$(a)\frac{1}{2} \; \; o\! f\; \; (i)\; \; 2\frac{3}{4} \; \; \;(ii)\; 4\frac{2}{9}$ $(b)\frac{5}{8} \; \; o\! f\; \; (i)\; \; 3\frac{5}{6} \; \; \;(ii)\; 9\frac{2}{3}$

Answer:

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

As we know that of is equivalent to multiply,

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

As we know that of is equivalent to multiply,

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

As we know that of is equivalent to multiplication, so

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

As we know that of is equivalent to multiplication, so

8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed $\frac{2}{5}$ of the water. Pratap consumed the remaining water.

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

Answer:

Given

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 :

<img alt="=\left (1-\frac{2}{5} \right )\:\:o\!f 5\:liter=\frac{3}{5}\times5=3\:liter" height="45"

src="https://lh5.googleusercontent.com/9jWswPkkyGbC-jLdTzPYsuAZI8PGxbD1yDb7q2Rl4Ya0x1OGPrEWHPcofv4JC630jQD4Pr7AOlkwKI5zVXFX-aJscox5VT4PVktawrZmE28fSGlN4dtnyK3bYoxOcmbp5Ww19jQ" style="margin-left: 0px; margin-top: 0px;" width="304" />

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

NCERT solutions for class 7 maths chapter 2 fractions and decimals exercise 2.3

1. Find:

Answer:

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

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

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

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

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

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

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

2. Multiply and reduce to lowest form (if possible) :

Answer:

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

3. Multiply the following fractions:

Answer:

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

So,

4. Which is greater:

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

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

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,

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

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

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

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

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,

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

5. Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is $\frac{3}{4}m$ . Find the distance between the first and the last sapling.

Answer:

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

distance between the first and the last sapling :

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

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

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

6. Lipika reads a book for $1\frac{3}{4}$ hours everyday. She reads the entire book in 6 days.How many hours in all were required by her to read the book?

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.

7. A car runs 16 km using 1 litre of petrol. How much distance will it cover using $2\frac{3}{4}$ litres of petrol.

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.

8. (a) (i) Provide the number in the box , such that $\frac{2}{3}\times$ = $\frac{10}{30}$ .

(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 class 7 maths chapter 2 fractions and decimals exercise 2.4

1. Find:

$(i) 12\div \frac{3}{4}$ $(ii) 14\div \frac{5}{6}$ $(iii) 8\div \frac{7}{3}$ $(iv) 4\div \frac{8}{3}$ $(v) 3\div 2\frac{1}{3}$ $(vi) 5\div 3\frac{4}{7}$

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

2. Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.

$(i)\frac{3}{7}$ $(ii)\frac{5}{8}$ $(iii)\frac{9}{7}$ $(iv)\frac{6}{5}$ $(v)\frac{12}{7}$ $(vi)\frac{1}{8}$ $(vii)\frac{1}{11}$

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:

$(i)\; \frac{7}{3}\div 2$ $(ii)\; \frac{4}{9}\div 5$ $(iii)\; \frac{6}{13}\div 7$ $(iv)\; 4\frac{1}{3}\div 3$

$(v)\; 3\frac{1}{2}\div 4$ $(vi)\; 4\frac{3}{7}\div 7$

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:

$(i)\; \frac{2}{5}\div \frac{1}{2}$ $(ii)\; \frac{4}{9}\div \frac{2}{3}$ $(iii)\; \frac{3}{7}\div \frac{8}{7}$ $(iv)\;2 \frac{1}{3}\div \frac{3}{5}$ $(v)\;3 \frac{1}{2}\div \frac{8}{3}$ $(vi)\;\frac{2}{5}\div 1\frac{1}{2}$ $(vii)\;3\frac{1}{5}\div 1\frac{2}{3}$ $(viii)\;2\frac{1}{5}\div 1\frac{1}{5}$

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,

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

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

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

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

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

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

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

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

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

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

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

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

NCERT solutions for class 7 maths chapter 2 fractions and decimals exercise 2.5

1 . Which is greater?

(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.

2. Express as rupees using decimals :

(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,

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

<img alt="ii) 7 \:Rupees ,7 \:paise=7+\frac{7}{100}=7+0.07=7.07\:Rupees" height="39"

src="https://lh6.googleusercontent.com/z__Ep8XlfLu_LRLBR12pSBA6ATKczfEfBRPsE78UhfRx46KXlf9mY4aU_2tM1ZPGsHAItQ9GmNjSMwdj0wH3mzXJOWqbT6gIZT70P5z1PiS-G1Hv4GDRUTlUFV4LTV-bLcw74Kw" style="margin-left: 0px; margin-top: 0px;" width="441" />

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

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

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

3.(i) Express 5 cm in metre and kilometre

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

Answer:

As we know,

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

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

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

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

4. Express in kg:

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

Answer:

As we know,

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

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

5. Write the following decimal numbers in the expanded form:
(i) 20.03 (ii) 2.03 (iii) 200.03 (iv) 2.034

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

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

6. Write the place value of 2 in the following decimal numbers:
(i) 2.56 (ii) 21.37 (iii) 10.25 (iv) 9.42 (v) 63.352.

Answer:

(i) 2.56

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.

7. Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much?

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.

8. Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?

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

9. How much less is 28 km than 42.6 km?

Answer:

Difference = 42.6 km - 28 km

= 14.6 km

Hence 28 is 14.6 km less than 42.6.

NCERT solutions for 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:

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

2. Find the area of rectangle whose length is 5.7cm and breadth is 3 cm.

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 .

Hence Area of the rectangle is 17.1 .

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:

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

4. A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?

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:

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 class 7 chapter 2 fractions and decimals 2.7

1. Find:

$(i)\; 0.4\div 2$ $(ii)\; 0.35\div 5$ $(iii)\; 2.48\div 4$ $(iv)\; 65.4\div 6$

$(v)\; 651.2\div 4$ $(vi)\; 14.49\div 7$ $(vii)\; 3.96\div 4$ $(viii)\; 0.80\div 5$

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

$(i)\; 0.4\div 2$

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

$(ii)\; 0.35\div 5$

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

$(iii)\; 2.48\div 4$

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

$(iv)\; 65.4\div 6$

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

$(v)\; 651.2\div 4$

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

$(vi)\; 14.49\div 7$

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

$(vii)\; 3.96\div 4$

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

$(viii)\; 0.80\div 5$

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

2. Find:

$(i)\; 4.8\div 10$ $(ii)\; 52.5\div 10$ $(iii)\; 0.7\div 10$ $(iv)\; 33.1\div 10$

$(v)\; 272.23\div 10$ $(vi)\; 0.56\div 10$ $(vii)\; 3.97\div 10$

Answer:

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

$(i)\; 4.8\div 10$

$\; 4.8\div 10=0.48$

$(ii)\; 52.5\div 10$

$\; 52.5\div 10=5.25$

$(iii)\; 0.7\div 10$

$\; 0.7\div 10=0.07$

$(iv)\; 33.1\div 10$

$\; 33.1\div 10=3.31$

$(v)\; 272.23\div 10$

$\; 272.23\div 10=27.223$

$(vi)\; 0.56\div 10$

$\; 0.56\div 10=0.056$

$(vii)\; 3.97\div 10$

$\; 3.97\div 10=0.397$

3. Find:

$(i)\; 2.7\div 100$ $(ii)\; 0.3\div 100$ $(iii)\; 0.78\div 100$ $(iv)\; 432.6\div 100$

$(v)\; 23.6\div 100$ $(vi)\; 98.53\div 100$

Answer:

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

$(i)\; 2.7\div 100$

$\; 2.7\div 100=0.027$

$(ii)\; 0.3\div 100$

$\; 0.3\div 100=0.003$

$(iii)\; 0.78\div 100$

$\; 0.78\div 100=0.0078$

$(iv)\; 432.6\div 100$

$\; 432.6\div 100=4.326$

$(v)\; 23.6\div 100$

$\; 23.6\div 100=0.236$

$(vi)\; 98.53\div 100$

$\; 98.53\div 100=0.9853$

4. Find:

$(i)\; 7.9\div 1000$ $(ii)\; 26.3\div 1000$ $(iii)\; 38.53\div 1000$ $(iv)\; 128.9\div 1000$

$(v)\; 0.5\div 1000$

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:

$(i)\; 7\div 3.5$ $(ii)\; 36\div 0.2$ $(iii)\; 3.25\div 0.5$ $(iv)\; 30.94\div 0.7$

$(v)\; 0.5\div 0.25$ $(vi)\; 7.75\div 0.25$ $(vii)\; 76.5\div 0.15$ $(viii)\; 37.8\div 1.4$

$(ix)\; 2.73\div 1.3$

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.

$(i)\; 7\div 3.5$

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

$(ii)\; 36\div 0.2$

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

$(iii)\; 3.25\div 0.5$

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

$(iv)\; 30.94\div 0.7$

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

$(v)\; 0.5\div 0.25$

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

$(vi)\; 7.75\div 0.25$

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

$(vii)\; 76.5\div 0.15$

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

$(viii)\; 37.8\div 1.4$

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

$(ix)\; 2.73\div 1.3$

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

6. A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it cover in one litre of petrol?

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.

Responsive Table

Chapter No.Chapter Name

Chapter 1	NCERT solutions for class 7 maths chapter 1 Integers
Chapter 2	NCERT solutions for class 7 maths chapter 2 Fractions and Decimals
Chapter 3	NCERT solutions for class 7 maths chapter 3 Data Handling
Chapter 4	NCERT solutions for class 7 maths chapter 4 Simple Equations
Chapter 5	NCERT solutions for class 7 maths chapter 5 Lines and Angles
Chapter 6	NCERT solutions for class 7 maths chapter 6 The Triangle and its Properties
Chapter 7	NCERT solutions for class 7 maths chapter 7 Congruence of Triangles
Chapter 8	NCERT solutions for class 7 maths chapter 8 comparing quantities
Chapter 9	NCERT solutions for class 7 maths chapter 9 Rational Numbers
Chapter 10	NCERT solutions for class 7 maths chapter 10 Practical Geometry
Chapter 11	NCERT solutions for class 7 maths chapter 11 Perimeter and Area
Chapter 12	NCERT solutions for class 7 maths chapter 12 Algebraic Expressions
Chapter 13	NCERT solutions for class 7 maths chapter 13 Exponents and Powers
Chapter 14	NCERT solutions for class 7 maths chapter 14 Symmetry
Chapter 15	NCERT solutions for class 7 maths chapter 15 visualising-solid-shapes

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NCERT solutions for class 7 maths chapter 2 fractions and decimals topic 2.3.1

NCERT solutions for class 7 maths chapter 2 fractions and decimals topic 2.3.2

NCERT solutions for class 7 maths chapter 2 fractions and decimals topic 2.4.1

NCERT solutions for class 7 chapter 2 fractions and decimals topic 2.4.2

NCERT solutions for class 7 maths chapter 2 fractions and decimals topic 2.4.3

NCERT solutions forclass 7 chapter 2 fractions and decimals topic 2.6

NCERT solutions for class 7 chapter 2 fractions and decimals topic 2.6.1

NCERT solutions for class 7 chapter 2 fractions and decimals topic 2.7.1

NCERT solutions for class 7 chapter 2 fractions and decimals topic 2.7.2

NCERT solutions for class 7 chapter 2 fractions and decimals topic 2.7.3

NCERT solutions for class 7 maths chapter 2 fractions and decimals exercise 2.1

NCERT solutions for class 7 maths chapter 2 fractions and decimals exercise 2.2

NCERT solutions for class 7 maths chapter 2 fractions and decimals exercise 2.3

NCERT solutions for class 7 maths chapter 2 fractions and decimals exercise 2.4

NCERT solutions for class 7 maths chapter 2 fractions and decimals exercise 2.5

NCERT solutions for class 7 chapter 2 fractions and decimals exercise 2.6

NCERT solutions for class 7 chapter 2 fractions and decimals 2.7

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