Mathematics NCERT Grade 7, Chapter 2: Fractions and Decimals- We have already studied about fractions and decimals. This chapter deals with the concept of multiplication and division of fractions as well as of decimals.
• A proper fraction is a fraction that represents a part of a whole.
• An improper fraction is a combination of a whole and a proper fraction.
The first half of the chapter deals with the Multiplication of Fractions and the later part deals with the concept of Division of Fractions. Two fractions are multiplied by multiplying their numerators and denominators separately.
Multiplication of fractions is sub-divided into the following topics:
• Multiplication of a fraction by a whole number
• Multiplication of a fraction by a fraction
Before moving onto decimal numbers, the division of fractions is explained. This particular topic contains the following divisions:
• Division of whole number by a fraction
• Division of a fraction by a whole number
• Division of a fraction by another fraction
The same operations are given for decimal numbers.
• Multiplication of decimal numbers
• Multiplication of decimal numbers by 10, 100, and 1000
Reciprocal of a fraction can be studied in this chapter wherein two non-zero numbers whose product with each other is 1 are called reciprocals of each other.
• Division of decimal numbers
• Division of decimal numbers by 10, 100, and 1000
• Division of a decimal number by a whole number
• Division of a decimal number by another decimal number

Understanding of the chapter is made easier as it contains various solved examples and a detailed explanation of every topic.
For quick revision, important points of the chapter are listed in the end.

Solve:

(i) (ii) (iii)

(iv) (v) (vi)

(vii)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

#### Question 2:

Arrange the following in descending order:

(i) (ii)

(i)

Changing them to like fractions, we obtain

Since 42>24>14,

(ii)

Changing them to like fractions, we obtain

As 49 > 30>14,

#### Question 3:

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

 (Along the first row )

Along the first row, sum =

Along the second row, sum =

Along the third row, sum =

Along the first column, sum =

Along the second column, sum =

Along the third column, sum =

Along the first diagonal, sum =

Along the second diagonal, sum =

Since the sum of the numbers in each row, in each column, and along the diagonals is the same, it is a magic square.

#### Question 4:

A rectangular sheet of paper is cm long and cm wide.

Find its perimeter.

Length =

Perimeter = 2 × (Length + Breadth)

#### Question 5:

Find the perimeters of (i) ΔABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

(i) Perimeter of ΔABE = AB + BE + EA

(ii)

Perimeter of rectangle = 2 (Length + Breadth)

Perimeter of ΔABE =

Changing them to like fractions, we obtain

$\frac{177}{20}=\frac{177×3}{20×3}=\frac{531}{60}\phantom{\rule{0ex}{0ex}}\frac{47}{6}=\frac{47×10}{6×10}=\frac{470}{60}$

As 531 > 470,

$⇒\frac{177}{20}>\frac{47}{6}$

Perimeter (ΔABE) > Perimeter (BCDE)

#### Question 6:

Salil wants to put a picture in a frame. The picture is cm wide.

To fit in the frame the picture cannot be more thancm wide. How much should the picture be trimmed?

Width of picture =

Required width =

#### Question 7:

Ritu ate 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?

Part of apple eaten by Ritu =

Part of apple eaten by Somu = 1 − Part of apple eaten by Ritu

=

Therefore, Somu ate part of the apple.

Since 3 > 2, Ritu had the larger share.

Difference between the 2 shares =

Therefore, Ritu’s share is larger than the share of Somu by.

##### Video Solution for fractions and decimals (Page: 32 , Q.No.: 7)

NCERT Solution for Class 7 maths - fractions and decimals 32 , Question 7

#### Question 8:

Michael finished colouring a picture in hour. Vaibhav finished colouring the same picture in hour. Who worked longer? By what fraction was it longer?

Time taken by Michael =

Time taken by Vaibhav =

Converting these fractions into like fractions, we obtain

Since 9 > 7,

Vaibhav worked longer.

Difference = = =

#### Question 1:

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

(i) (ii) (iii) (iv)

(a)

(b)

(c)

(d)

(i) represents addition of 2 figures, each representing 1 shaded part out of 5 equal parts. Hence, is represented by (d).

(ii) represents addition of 2 figures, each representing 1 shaded part out of 2 equal parts. Hence, is represented by (b).

(iii) represents addition of 3 figures, each representing 2 shaded parts out of 3 equal parts. Hence, is represented by (a).

(iv) represents addition of 3 figures, each representing 1 shaded part out of 4 equal parts. Hence, is represented by (c).

#### Question 2:

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

(i) (ii) (iii)

(a)

(b)

(c)

(i) represents the addition of 3 figures, each representing 1 shaded part out of 5 equal parts and represents 3 shaded parts out of 5 equal parts. Hence, is represented by (c).

(ii) represents the addition of 2 figures, each representing 1 shaded part out of 3 equal parts and represents 2 shaded parts out of 3 equal parts. Hence, is represented by (a).

(iii) represents the addition of 3 figures, each representing 3 shaded parts out of 4 equal parts and represents 2 fully shaded figures and one figure having 1 part as shaded out of 4 equal parts. Hence, is represented by (b)

#### Question 3:

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

(i) (ii) (iii) (iv)

(v) (vi) (vii) (viii)

(ix) (x)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

#### Question 4:

(i) of the circles in box (a) (ii) of the triangles in box (b)

(iii) of the squares in box (c)

 (a) (b) (c)

(i) It can be observed that there are 12 circles in the given box. We have to shade of the circles in it. As, therefore, we will shade any 6 circles of it.

(ii) It can be observed that there are 9 triangles in the given box. We have to shade of the triangles in it. As, therefore, we will shade any 6 triangles of it.

(iii) It can be observed that there are 15 squares in the given box. We have to shade of the squares in it. As, therefore, we will shade any 9 squares of it.

#### Question 5:

Find:

(a) of (i) 24 (ii) 46

(b) of (i) 18 (ii) 27

(c) of (i) 16 (ii) 36

(d) of (i) 20 (ii) 35

(a) (i)

(ii)

(b) (i)

(ii)

(c) (i)

(ii)

(d) (i)

(ii)

#### Question 6:

Multiply and express as a mixed fraction:

(a) (b)

(c) (d)

(e) (f)

(a)

(b)

(c)

(d)

(e)

(f)

#### Question 7:

Find (a) of (i) (ii) (b) of (i) (ii)

(a) (i)

(ii)

(b) (i)

(ii)

#### Question 8:

Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed 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?

(i) Water consumed by Vidya = of 5 litres

(ii) Water consumed by Pratap = of the total water

##### Video Solution for fractions and decimals (Page: 37 , Q.No.: 8)

NCERT Solution for Class 7 maths - fractions and decimals 37 , Question 8

#### Question 1:

Find:

(i) of (a) (b) (c)

(ii) of (a) (b) (c)

(i) (a)

(b)

(c)

(ii) (a)

(b)

(c)

#### Question 2:

Multiply and reduce to lowest form (if possible):

(i) (ii) (iii)

(iv) (v) (vi)

(vii)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

#### Question 3:

Multiply the following fractions:

(i) (ii) (iii)

(iv) (v) (vi)

(vii)

(i)

This is an improper fraction and it can be written as a mixed fraction as .

(ii)

This is an improper fraction and it can be written as a mixed fraction as .

(iii)

This is a whole number.

(iv)

This is an improper fraction and it can be written as a mixed fraction as .

(v)

This is an improper fraction and it can be written as a mixed fraction as .

(vi)

This is an improper fraction and it can be written as a mixed fraction as .

(vii)

This is an improper fraction and it can be written as a mixed fraction as .

#### Question 4:

Which is greater:

(i) of orof

(ii) of orof

(i)

Converting these fractions into like fractions,

Therefore, of is greater.

(ii)

Therefore, of is greater.

#### Question 5:

Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is m. Find the distance between the first and the last sapling.

From the figure, it can be observed that gaps between 1st and last sapling = 3

Length of 1 gap =

Therefore, distance between I and IV sapling =

##### Video Solution for fractions and decimals (Page: 42 , Q.No.: 5)

NCERT Solution for Class 7 maths - fractions and decimals 42 , Question 5

#### Question 6:

Lipika reads a book for hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book?

Number of hours Lipika reads the book per day =

Number of days = 6

Total number of hours required by her to read the book =

#### Question 7:

A car runs 16 km using 1 litre of petrol. How much distance will it cover using litres of petrol.

Number of kms a car can run per litre petrol = 16 km

Quantity of petrol =

Number of kms a car can run for litre petrol = = 44 km

It will cover 44 km distance by using litres of petrol.

#### Question 8:

(a) (i) Provide the number in the box , such that .

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

(b) (i) Provide the number in the box , such that ?

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

(a) (i) As ,

Therefore, the number in the box , such that is

.

(ii) The simplest form of is .

(b) (i) As ,

Therefore, the number in the box , such that is

.

(ii) As cannot be further simplified, therefore, its simplest form is

Find:

(i) (ii) (iii)

(iv) (v) (vi)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

#### Question 2:

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

(i) (ii) (iii)

(iv) (v) (vi)

(vii)

A proper fraction is the fraction which has its denominator greater than its numerator while improper fraction is the fraction which has its numerator greater than its denominator. Whole numbers are a collection of all positive integers including 0.

(i)

Reciprocal =

Therefore, it is an improper fraction.

(ii)

Reciprocal =

Therefore, it is an improper fraction.

(iii)

Reciprocal =

Therefore, it is a proper fraction.

(iv)

Reciprocal =

Therefore, it is a proper fraction.

(v)

Reciprocal =

Therefore, it is a proper fraction.

(vi)

Reciprocal =

Therefore, it is a whole number.

(vii)

Reciprocal =

Therefore, it is a whole number.

Find:

(i) (ii) (iii)

(iv) (v) (vi)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Find:

(i) (ii) (iii)

(iv) (v) (vi)

(vii) (viii)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

#### Question 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

(i) 0.5 or 0.05

Converting these decimal numbers into equivalent fractions,

It can be observed that both fractions have the same denominator.

As 50 > 5,

Therefore, 0.5 > 0.05

(ii) 0.7 or 0.5

Converting these decimal numbers into equivalent fractions,

It can be observed that both fractions have the same denominator.

As 7 > 5,

Therefore, 0.7 >0.5

(iii) 7 or 0.7

Converting these decimal numbers into equivalent fractions,

It can be observed that both fractions have the same denominator.

As 70 > 7,

Therefore, 7 > 0.7

(iv) 1.37 or 1.49

Converting these decimal numbers into equivalent fractions,

It can be observed that both fractions have the same denominator.

As 137 < 149,

Therefore, 1.37 < 1.49

(v) 2.03 or 2.30

Converting these decimal numbers into equivalent fractions,

It can be observed that both fractions have the same denominator.

As 203 < 230,

Therefore, 2.03 < 2.30

(vi) 0.8 or 0.88

Converting these decimal numbers into equivalent fractions,

It can be observed that both fractions have the same denominator.

As 80 < 88,

Therefore, 0.8 < 0.88

#### Question 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

There are 100 paise in 1 rupee. Therefore, if we want to convert paise into rupees, then we have to divide paise by 100.

(i) 7 paise =

(ii) 7 Rs 7 paise =

= Rs 7.07

(iii) 77 Rs 77 paise = Rs 77.77

(iv) 50 paise

(v) 235 paise

#### Question 3:

(i) Express 5 cm in metre and kilometre

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

(i) 5 cm

(ii) 35 mm

#### Question 4:

Express in kg:

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

(i) 200 g

(ii) 3470 g

(iii) 4 kg 8 g = 4.008 kg

#### Question 5:

Write the following decimal numbers in the expanded form:

(i) 20.03 (ii) 2.03 (iii) 200.03

(iv) 2.034

(i) 20.03

(ii) 2.03

(iii) 200.03

(iv) 2.034

#### Question 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

(i) 2.56

Ones

(ii) 21.37

Tens

(iii) 10.25

Tenths

(iv) 9.42

Hundredths

(v) 63.352

Thousandths

#### Question 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?

Distance travelled by Dinesh = AB + BC = (7.5 + 12.7) km

Therefore, Dinesh travelled 20.2 km.

Distance travelled by Ayub = AD + DC = (9.3 + 11.8) km

Therefore, Ayub travelled 21.1 km.

Hence, Ayub travelled more distance.

Difference = (21.1 − 20.2) km

Therefore, Ayub travelled 0.9 km more than Dinesh.

##### Video Solution for fractions and decimals (Page: 48 , Q.No.: 7)

NCERT Solution for Class 7 maths - fractions and decimals 48 , Question 7

#### Question 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?

Total fruits bought by Shyama = 5 kg 300 g + 3 kg 250 g

= 8 kg 550 g

=

= 8.550 kg

Total fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g

= 8 kg 950 g

=

= 8.950 kg

∴ Sarala bought more fruits.

#### Question 9:

How much less is 28 km than 42.6 km?

Therefore, 28 km is 14.6 km less than 42.6 km.

#### Question 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

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

#### Question 2:

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

Length = 5.7 cm

= 5.7 × 3 = 17.1 cm2

#### Question 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

We know that when a decimal number is multiplied by 10, 100, 1000, the decimal point in the product is shifted to the right by as many places as there are zeroes. Therefore, these products can be calculated as

(i) 1.3 × 10 = 13

(ii) 36.8 × 10 = 368

(iii) 153.7 × 10 = 1537

(vi) 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

(xiii) 0.03 × 1000 = 30

#### Question 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?

Distance covered in 1 litre of petrol = 55.3 km

Distance covered in 10litre of petrol = 10 × 55.3 = 553 km

Therefore, it will cover 553 km distance in 10 litre petrol.

#### Question 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

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

#### Question 1:

Find:

(i) 0.4 ÷ 2 (ii) 0.35 ÷ 5 (iii) 2.48 ÷ 4

(iv) 65.4 ÷ 6 (v) 651.2 ÷ 4 (vi) 14.49 ÷ 7

(vii) 3.96 ÷ 4 (viii) 0.80 ÷ 5

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

#### Question 2:

Find:

(i) 4.8 ÷ 10 (ii) 52.5 ÷ 10 (iii) 0.7 ÷ 10

(iv) 33.1 ÷ 10 (v) 272.23 ÷ 10 (vi) 0.56 ÷ 10

(vii) 3.97 ÷ 10

We know that when a decimal number is divided by a multiple of 10 only (i.e., 10, 100, 1000, etc.), the decimal point will be shifted to the left by as many places as there are zeroes. Since here we are dividing by 10, the decimal will shift to the left by 1 place.

(i) 4.8 ÷ 10 = 0 .48

(ii) 52.5 ÷ 10 = 5.25

(iii) 0.7 ÷ 10 = 0.07

(iv) 33.1 ÷ 10 = 3.31

(v) 272.23 ÷ 10 = 27.223

(vi) 0.56 ÷ 10 = 0.056

(vii) 3.97 ÷ 10 = 0.397

#### Question 3:

Find:

(i) 2.7 ÷ 100 (ii) 0.3 ÷ 100 (iii) 0.78 ÷ 100

(iv) 432.6 ÷ 100 (v) 23.6 ÷ 100 (vi) 98.53 ÷ 100

We know that when a decimal number is divided by a multiple of 10 only (i.e., 10, 100, 1000, etc.), the decimal point will be shifted to the left by as many places as there are zeroes. Since here we are dividing by 100, the decimal will shift to the left by 2 places.

(i) 2.7 ÷ 100 = 0.027

(ii) 0.3 ÷ 100 = 0.003

(iii) 0.78 ÷ 100 = 0.0078

(iv) 432.6 ÷ 100 = 4.326

(v) 23.6 ÷ 100 = 0.236

(vi) 98.53 ÷ 100 = 0.9853

#### Question 4:

Find:

(i) 7.9 ÷ 1000 (ii) 26.3 ÷ 1000 (iii) 38.53 ÷ 1000

(iv) 128.9 ÷ 1000 (v) 0.5 ÷ 1000

We know that when a decimal number is divided by a multiple of 10 only (i.e., 10, 100, 1000, etc.), the decimal point will be shifted to the left by as many places as there are zeroes. Since here we are dividing by 1000, the decimal will shift to the left by 3 places.

(i) 7.9 ÷ 1000 = 0.0079

(ii) 26.3 ÷ 1000 = 0.0263

(iii) 38.53 ÷ 1000 = 0.03853

(iv) 128.9 ÷ 1000 = 0.1289

(v) 0.5 ÷ 1000 = 0.0005

#### Question 5:

Find:

(i) 7 ÷ 3.5 (ii) 36 ÷ 0.2 (iii) 3.25 ÷ 0.5

(iv) 30.94 ÷ 0.7 (v) 0.5 ÷ 0.25 (vi) 7.75 ÷ 0.25

(vii) 76.5 ÷ 0.15 (viii) 37.8 ÷ 1.4 (ix) 2.73 ÷ 1.3

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

#### Question 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?