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

In the simple PDF format provided, you can access NCERT Solutions for Class 7 Maths, specifically addressing Exercise 2.1 within Chapter 2, titled "Fractions and Decimals." This exercise introduces crucial concepts related to fractions, covering aspects such as proper, improper, and mixed fractions, along with their addition and subtraction. These NCERT Solutions are essential tools to comprehensively cover the Class 7 Mathematics syllabus, offering a diverse set of questions that assess students' understanding of these foundational concepts. By utilizing this resource, students can strengthen their grasp of fractions and decimals, ensuring alignment with the NCERT curriculum and the development of strong mathematical skills.

## Access Answers to NCERT Solutions for Class 7 Maths Chapter 2 - Fractions and Decimals

### Exercise 2.1

Question 1 :

Michael finished colouring a picture in (7/12) hour. Vaibhav finished colouring the same picture in (3/4) hour. Who worked longer? By what fraction was it longer?

From the question, it is given that,

Time taken by the Michael to colour the picture is = (7/12)

Time taken by the Vaibhav to colour the picture is = (3/4)

The LCM of 12, 4 = 12

Now, let us change each of the given fraction into an equivalent fraction having 12 as the denominator.

(7/12) = (7/12) × (1/1) = 7/12

(3/4) = (3/4) × (3/3) = 9/12

Clearly, (7/12) < (9/12)

Hence, (7/12) < (3/4)

Thus, Vaibhav worked for longer time.

So, Vaibhav worked longer time by = (3/4) – (7/12)

= (9/12) – (7/12)

= (9 – 7)/12

= (2/12)

= (1/6) of an hour.

Question 2 :

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?

 4/11 9/11 2/11 3/11 5/11 7/11 8/11 1/11 6/11

Sum along the first row = (4/11) + (9/11) + (2/11) = (15/11)

Sum along the second row = (3/11) + (5/11) + (7/11) = (15/11)

Sum along the third row = (8/11) + (1/11) + (6/11) = (15/11)

Sum along the first column = (4/11) + (3/11) + (8/11) = (15/11)

Sum along the second column = (9/11) + (5/11) + (1/11) = (15/11)

Sum along the third column = (2/11) + (7/11) + (6/11) = (15/11)

Sum along the first diagonal = (4/11) + (5/11) + (6/11) = (15/11)

Sum along the second diagonal = (2/11) + (5/11) + (8/11) = (15/11)

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

Question 3 :

A rectangular sheet of paper is 12 ½ cm long and cm wide. Find its perimeter.

From the question, it is given that,

Length = 12 ½ cm = 25/2 cm

cm = 32/3 cm

We know that,

Perimeter of the rectangle = 2 × (length + breadth)

= 2 × [(25/2) + (32/3)]

= 2 × {[(25 × 3) + (32 × 2)]/6}

= 2 × [(75 + 64)/6]

= 2 × [139/6]

= 139/3 cm

Hence, the perimeter of the sheet of paper is
cm

Question 4 :

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

From the fig,

AB = (5/2) cm

AE == 18/5 cm

BE == 11/4 cm

ED = 7/6 cm

(i) We know that,

Perimeter of the triangle = Sum of all sides

Then,

Perimeter of triangle ABE = AB + BE + EA

= (5/2) + (11/4) + (18/5)

The LCM of 2, 4, 5 = 20

Now, let us change each of the given fractions into an equivalent fraction having 20 as the denominator.

= {[(5/2) × (10/10)] + [(11/4) × (5/5)] + [(18/5) × (4/4)]}

= (50/20) + (55/20) + (72/20)

= (50 + 55 + 72)/20

= 177/20

=cm

(ii) Now, we have to find the perimeter of the rectangle,

We know that,

Perimeter of the rectangle = 2 × (length + breadth)

Then,

Perimeter of rectangle BCDE = 2 × (BE + ED)

= 2 × [(11/4) + (7/6)]

The LCM of 4, 6 = 12

Now, let us change each of the given fractions into an equivalent fraction having 20 as the denominator

= 2 × {[(11/4) × (3/3)] + [(7/6) × (2/2)]}

= 2 × [(33/12) + (14/12)]

= 2 × [(33 + 14)/12]

= 2 × (47/12)

= 47/6

=

Finally, we have to find which one is having a greater perimeter.

Perimeter of triangle ABE = (177/20)

Perimeter of rectangle BCDE = (47/6)

The two perimeters are in the form of unlike fractions.

Changing perimeters into like fractions we have,

(177/20) = (177/20) × (3/3) = 531/60

(43/6) = (43/6) × (10/10) = 430/60

Clearly, (531/60) > (430/60)

Hence, (177/20) > (43/6)

∴ Perimeter of Triangle ABE > Perimeter of Rectangle (BCDE)

Question 5 :

Salil wants to put a picture in a frame. The picture is cm wide. To fit in the frame the picture cannot be more than cm wide. How much should the picture be trimmed?

From the question, it is given that,

Picture having a width of == 38/5 cm

Frame having a width of == 73/10 cm

∴ The picture should be trimmed by = [(38/5) – (73/10)]

The LCM of 5, 10 = 10

Now, let us change each of the given fractions into an equivalent fraction having 10 as the denominator.

= [(38/5) × (2/2)] – [(73/10) × (1/1)]

= (76/10) – (73/10)

= (76 – 73)/10

= 3/10 cm

Thus, the picture should be trimmed by (3/10) cm

Question 6 :

Ritu ate (3/5) part of an apple and the remaining apple was eaten by her brother Somu. What part of the apple did Somu eat? Who had the larger share? By how much?

From the question, it is given that,

Part of the apple eaten by Ritu is = (3/5)

Part of the apple eaten by Somu is = 1 – Part of the apple eaten by Ritu

= 1 – (3/5)

The LCM of 1, 5 = 5

Now, let us change each of the given fractions into an equivalent fraction having 10 as the denominator.

= [(1/1) × (5/5)] – [(3/5) × (1/1)]

= (5/5) – (3/5)

= (5 – 3)/5

= 2/5

∴ Part of the apple eaten by Somu is (2/5)

So, (3/5) > (2/5) hence, Ritu ate larger size of the apple.

Now, the difference between the 32 shares = (3/5) – (2/5)

= (3 – 2)/5

= 1/5

Thus, Ritu’s share is larger than the share of Somu by (1/5)

Question 7 :

Solve:

(i) 2 – (3/5)

(ii) 4 + (7/8)

(iii) (3/5) + (2/7)

(iv) (9/11) – (4/15)

(v) (7/10) + (2/5) + (3/2)

(vi)

(vii) 812−358

(i) For subtraction of two unlike fractions, first change them to like fractions.

LCM of 1, 5 = 5

Now, let us change each of the given fractions into an equivalent fraction having 5 as the denominator.

= [(2/1) × (5/5)] = (10/5)

= [(3/5) × (1/1)] = (3/5)

Now,

= (10/5) – (3/5)

= [(10 – 3)/5]

= (7/5)

(ii) For addition of two unlike fractions, first change them to like fractions.

LCM of 1, 8 = 8

Now, let us change each of the given fractions into an equivalent fraction having 8 as the denominator.

= [(4/1) × (8/8)] = (32/8)

= [(7/8) × (1/1)] = (7/8)

Now,

= (32/8) + (7/8)

= [(32 + 7)/8]

= (39/8)

=

(iii) For addition of two unlike fractions, first change them to like fractions.

LCM of 5, 7 = 35

Now, let us change each of the given fractions into an equivalent fraction having 35 as the denominator.

= [(3/5) × (7/7)] = (21/35)

= [(2/7) × (5/5)] = (10/35)

Now,

= (21/35) + (10/35)

= [(21 + 10)/35]

= (31/35)

(iv) For subtraction of two unlike fractions, first change them to like fractions.

LCM of 11, 15 = 165

Now, let us change each of the given fractions into an equivalent fraction having 165 as the denominator.

= [(9/11) × (15/15)] = (135/165)

= [(4/15) × (11/11)] = (44/165)

Now,

= (135/165) – (44/165)

= [(135 – 44)/165]

= (91/165)

(v) For addition of two unlike fractions, first change them to like fractions.

LCM of 10, 5, 2 = 10

Now, let us change each of the given fractions into an equivalent fraction having 35 as the denominator.

= [(7/10) × (1/1)] = (7/10)

= [(2/5) × (2/2)] = (4/10)

= [(3/2) × (5/5)] = (15/10)

Now,

= (7/10) + (4/10) + (15/10)

= [(7 + 4 + 15)/10]

= (26/10)

= (13/5)

=

(vi) First convert mixed fraction into improper fraction,

== 8/3

= 3 ½ = 7/2

For addition of two unlike fractions, first change them to like fractions.

LCM of 3, 2 = 6

Now, let us change each of the given fractions into an equivalent fraction having 6 as the denominator.

= [(8/3) × (2/2)] = (16/6)

= [(7/2) × (3/3)] = (21/6)

Now,

= (16/6) + (21/6)

= [(16 + 21)/6]

= (37/6)

=

(vii) First convert mixed fraction into improper fraction,

= 8 ½ = 17/2

== 29/8

For subtraction of two unlike fractions, first change them to like fractions.

LCM of 2, 8 = 8

Now, let us change each of the given fractions into an equivalent fraction having 35 as the denominator.

= [(17/2) × (4/4)] = (68/8)

= [(29/8) × (1/1)] = (29/8)

Now,

= (68/8) – (29/8)

= [(68 – 29)/8]

= (39/8)

=

Question 8 :

Arrange the following in descending order:

(i) 2/9, 2/3, 8/21

(ii) 1/5, 3/7, 7/10

(i) LCM of 9, 3, 21 = 63

Now, let us change each of the given fractions into an equivalent fraction having 63 as the denominator.

[(2/9) × (7/7)] = (14/63) [(2/3) × (21/21)] = (42/63) [(8/21) × (3/3)] = (24/63)

Clearly,

(42/63) > (24/63) > (14/63)

Hence,

(2/3) > (8/21) > (2/9)

Hence, the given fractions in descending order are (2/3), (8/21), (2/9)

(ii) LCM of 5, 7, 10 = 70

Now, let us change each of the given fractions into an equivalent fraction having 70 as the denominator.

[(1/5) × (14/14)] = (14/70) [(3/7) × (10/10)] = (30/70) [(7/10) × (7/7)] = (49/70)

Clearly,

(49/70) > (30/70) > (14/70)

Hence,

(7/10) > (3/7) > (1/5)

Hence, the given fractions in descending order are (7/10), (3/7), (1/5)

### Exercise 2.2

Question 1 :

Multiply and express as a mixed fraction:

(a) 3 ×

(b) 5 × 6 ¾

(c) 7 × 2 ¼

(d) 4 ×

(e) 3 ¼ × 6

(f) × 8

(a) First, convert the given mixed fraction into an improper fraction.

== 26/5

Now,

= 3 × (26/5)

= 78/5

=

(b) First, convert the given mixed fraction into an improper fraction.

= 6 ¾ = 27/4

Now,

= 5 × (27/4)

= 135/4

= 33 ¾

(c) First, convert the given mixed fraction into an improper fraction.

= 2 ¼ = 9/4

Now,

= 7 × (9/4)

= 63/4

= 15 ¾

(d) First, convert the given mixed fraction into an improper fraction.

== 19/3

Now,

= 4 × (19/3)

= 76/3

=

(e) First, convert the given mixed fraction into an improper fraction.

= 3 ¼ = 13/4

Now,

= (13/4) × 6

= (13/2) × 3

= 39/2

= 19 ½

(f) First, convert the given mixed fraction into an improper fraction.

== 17/5

Now,

= (17/5) × 8

= 136/5

=

Question 2 :

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

(i) 2 × (1/5) (ii) 2 × ½ (iii) 3 × (2/3) (iv) 3 × ¼

(i) 2 × (1/5) represents the addition of 2 figures, each representing 1 shaded part out of the given 5 equal parts.

∴ 2 × (1/5) is represented by fig (d).

(ii) 2 × ½ represents the addition of 2 figures, each representing 1 shaded part out of the given 2 equal parts.

∴ 2 × ½ is represented by fig (b).

(iii) 3 × (2/3) represents the addition of 3 figures, each representing 2 shaded parts out of the given 3 equal parts.

∴ 3 × (2/3) is represented by fig (a).

(iii) 3 × ¼ represents the addition of 3 figures, each representing 1 shaded part out of the given 4 equal parts.

∴ 3 × ¼ is represented by fig (c).

Question 3 :

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

(i) 3 × (1/5) = (3/5) (ii) 2 × (1/3) = (2/3) (iii) 3 × (3/4) = 2 ¼

(i) 3 × (1/5) represents the addition of 3 figures, each representing 1 shaded part out of the given 5 equal parts and (3/5) represents 3 shaded parts out of 5 equal parts.

∴ 3 × (1/5) = (3/5) is represented by fig (c).

(ii) 2 × (1/3) represents the addition of 2 figures, each representing 1 shaded part out of the given 3 equal parts and (2/3) represents 2 shaded parts out of 3 equal parts.

∴ 2 × (1/3) = (2/3) is represented by fig (a).

(iii) 3 × (3/4) represents the addition of 3 figures, each representing 3 shaded parts out of the given 4 equal parts, and 2 ¼ represents 2 fully and 1 figure having 1 part as shaded out of 4 equal parts.

∴ 3 × (3/4) = 2 ¼ is represented by fig (b).

Question 4 :

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

(i) 7 × (3/5)

(ii) 4 × (1/3)

(iii) 2 × (6/7)

(iv) 5 × (2/9)

(v) (2/3) × 4

(vi) (5/2) × 6

(vii) 11 × (4/7)

(viii) 20 × (4/5)

(ix) 13 × (1/3)

(x) 15 × (3/5)

(i) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (7/1) × (3/5)

= (7 × 3)/ (1 × 5)

= (21/5)

=

(ii) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (4/1) × (1/3)

= (4 × 1)/ (1 × 3)

= (4/3)

=

(iii) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2/1) × (6/7)

= (2 × 6)/ (1 × 7)

= (12/7)

=

(iv) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5/1) × (2/9)

= (5 × 2)/ (1 × 9)

= (10/9)

=

(v) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2/3) × (4/1)

= (2 × 4)/ (3 × 1)

= (8/3)

=

(vi) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5/2) × (6/1)

= (5 × 6)/ (2 × 1)

= (30/2)

= 15

(vii) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (11/1) × (4/7)

= (11 × 4)/ (1 × 7)

= (44/7)

=

(viii) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (20/1) × (4/5)

= (20 × 4)/ (1 × 5)

= (80/5)

= 16

(ix) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (13/1) × (1/3)

= (13 × 1)/ (1 × 3)

= (13/3)

=

(x) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (15/1) × (3/5)

= (15 × 3)/ (1 × 5)

= (45/5)

= 9

Question 5 :

(i) ½ of the circles in box (a) (b) 2/3 of the triangles in box (b)

(iii) 3/5 of the squares in the box (c)

(i) From the question,

We may observe that there are 12 circles in the given box. So, we have to shade ½ of the circles in the box.

∴ 12 × ½ = 12/2

= 6

So we have to shade any 6 circles in the box.

(ii) From the question,

We may observe that there are 9 triangles in the given box. So, we have to shade 2/3 of the triangles in the box.

∴ 9 × (2/3) = 18/3

= 6

So we have to shade any 6 triangles in the box.

(iii) From the question,

We may observe that there are 15 squares in the given box. So, we have to shade 3/5 of the squares in the box.

∴ 15 × (3/5) = 45/5

= 9

So we have to shade any 9 squares in the box.

Question 6 :

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

(i) From the question, it is given that,

Amount of water in the water bottle = 5 litres

Amount of water consumed by Vidya = 2/5 of 5 litres

= (2/5) × 5

= 2 litres

So, the total amount of water drank by Vidya is 2 litres

(ii) From the question, it is given that,

Amount of water in the water bottle = 5 litres

Then,

Amount of water consumed by Pratap = (1 – water consumed by Vidya)

= (1 – (2/5))

= (5-2)/5

= 3/5

∴ The total amount of water consumed by Pratap = 3/5 of 5 litres

= (3/5) × 5

= 3 litres

So, the total amount of water drank by Pratap is 3 litres

Question 7 :

Find:

(a) ½ of (i) 24 (ii) 46

(b) 2/3 of (i) 18 (ii) 27

(c) ¾ of (i) 16 (ii) 36

(d) 4/5 of (i) 20 (ii) 35

(a) (i) 24

We have,

= ½ × 24

= 24/2

= 12

(ii) 46

We have,

= ½ × 46

= 46/2

= 23

(b) (i) 18

We have,

= 2/3 × 18

= 2 × 6

= 12

(ii) 27

We have,

= 2/3 × 27

= 2 × 9

= 18

(c) (i) 16

We have,

= ¾ × 16

= 3 × 4

= 12

(ii) 36

We have

= ¾ × 36

= 3 × 9

= 27

(d) (i) 20

We have,

= 4/5 × 20

= 4 × 4

= 16

(ii) 35

We have,

= 4/5 × 35

= 4 × 7

= 28

Question 8 :

Find:

(a) ½ of (i) 2 ¾ (ii)

(b) 5/8 of (i) (ii)

(a) (i) 2 ¾

First, convert the given mixed fraction into an improper fraction.

= 2 ¾ = 11/4

Now,

= ½ × 11/4

By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= ½ × (11/4)

= (1 × 11)/ (2 × 4)

= (11/8)

=

(ii)

First, convert the given mixed fraction into an improper fraction.

== 38/9

Now,

= ½ × (38/9)

By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= ½ × (38/9)

= (1 × 38)/ (2 × 9)

= (38/18)

= 19/9

=

(b)  (i)

First, convert the given mixed fraction into an improper fraction.

== 23/6

Now,

= (5/8) × (23/6)

By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5/8) × (23/6)

= (5 × 23)/ (8 × 6)

= (115/48)

=

(ii)

First, convert the given mixed fraction into an improper fraction.

== 29/3

Now,

= (5/8) × (29/3)

By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5/8) × (29/3)

= (5 × 29)/ (8 × 3)

= (145/24)

=

### Exercise 2.3

Question 1 :

Which is greater?

(i) (2/7) of (3/4) or (3/5) of (5/8)

(ii) (1/2) of (6/7) or (2/3) of (3/7)

(i) We have,

= (2/7) × (3/4) and (3/5) × (5/8)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2/7) × (3/4)

= (2 × 3)/ (7 × 4)

= (1 × 3)/ (7 × 2)

= (3/14) … [i]

And,

= (3/5) × (5/8)

= (3 × 5)/ (5 × 8)

= (3 × 1)/ (1 × 8)

= (3/8) … [ii]

Now, convert [i] and [ii] into like fractions,

LCM of 14 and 8 is 56

Now, let us change each of the given fractions into an equivalent fraction having 56 as the denominator.

[(3/14) × (4/4)] = (12/56) [(3/8) × (7/7)] = (21/56)

Clearly,

(12/56) < (21/56)

Hence,

(3/14) < (3/8)

(ii) We have,

= (1/2) × (6/7) and (2/3) × (3/7)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1/2) × (6/7)

= (1 × 6)/ (2 × 7)

= (1 × 3)/ (1 × 7)

= (3/7) … [i]

And,

= (2/3) × (3/7)

= (2 × 3)/ (3 × 7)

= (2 × 1)/ (1 × 7)

= (2/7) … [ii]

By comparing [i] and [ii],

Clearly,

(3/7) > (2/7)

Question 2 :

Lipika reads a book for 1 ¾ hours every day. She reads the entire book in 6 days. How many hours in all were required by her to read the book?

From the question, it is given that,

Lipika reads the book for = 1 ¾ hours every day = 7/4 hours

Number of days she took to read the entire book = 6 days

∴ Total number of hours required by her to complete the book = (7/4) × 6

= (7/2) × 3

= 21/2

= 10 ½ hours

Hence, the total number of hours required by her to complete the book is 10 ½ hours.

Question 3 :

A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2 ¾ litres of petrol.

From the question, it is given that,

The total number of distance travelled by a car in 1 liter of petrol = 16 km

Then,

Total quantity of petrol = 2 ¾ liter = 11/4 liters

Total number of distance travelled by car in 11/4 liters of petrol = (11/4) × 16

= 11 × 4

= 44 km

∴ Total number of distance travelled by car in 11/4 liters of petrol is 44 km.

Question 4 :

Multiply and reduce to lowest form (if possible):

(i) (2/3) ×

(ii) (2/7) × (7/9)

(iii) (3/8) × (6/4)

(iv) (9/5) × (3/5)

(v) (1/3) × (15/8)

(vi) (11/2) × (3/10)

(vii) (4/5) × (12/7)

(i) First convert the given mixed fraction into improper fraction.

== 8/3

Now,

= (2/3) × (8/3)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2 × 8)/ (3 × 3)

= (16/9)

=

(ii) By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2 × 7)/ (7 × 9)

= (2 × 1)/ (1 × 9)

= (2/9)

(iii) By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (3 × 6)/ (8 × 4)

= (3 × 3)/ (4 × 4)

= (9/16)

(iv) By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (9 × 3)/ (5 × 5)

= (27/25)

=

(v) By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1 × 15)/ (3 × 8)

= (1 × 5)/ (1 × 8)

= (5/8)

(vi) By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (11 × 3)/ (2 × 10)

= (33/20)

=

(vii) By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (4 × 12)/ (5 × 7)

= (48/35)

=

Question 5 :

Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is ¾ m. Find the distance between the first and the last sapling.

From the question, it is given that,

The distance between two adjacent saplings = ¾ m

Number of saplings planted by Saili in a row = 4

Then, number of gap in saplings = ¾ × 4

= 3

∴ The distance between the first and the last saplings = 3 × ¾

= (9/4) m

= 2 ¼ m

Hence, the distance between the first and the last saplings is 2 ¼ m.

Question 6 :

Find:

(i) ¼ of (a) ¼ (b) 3/5 (c) 4/3

(ii) 1/7 of (a) 2/9 (b) 6/5 (c) 3/10

(i) (a) ¼

We have,

= ¼ × ¼

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= ¼ × ¼

= (1 × 1)/ (4 × 4)

= (1/16)

(b) 3/5

We have,

= ¼ × (3/5)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= ¼ × (3/5)

= (1 × 3)/ (4 × 5)

= (3/20)

(c) (4/3)

We have,

= ¼ × (4/3)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= ¼ × (4/3)

= (1 × 4)/ (4 × 3)

= (4/12)

= 1/3

(ii) (a) 2/9

We have,

= (1/7) × (2/9)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1/7) × (2/9)

= (1 × 2)/ (7 × 9)

= (2/63)

(b) 6/5

We have,

= (1/7) × (6/5)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1/7) × (6/5)

= (1 × 6)/ (7 × 5)

= (6/35)

(c) 3/10

We have,

= (1/7) × (3/10)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1/7) × (3/10)

= (1 × 3)/ (7 × 10)

= (3/70)

Question 7 :

Multiply the following fractions:

(i) (2/5) × 5 ¼

(ii) × (7/9)

(iii) (3/2) ×

(iv) (5/6) ×

(v) × (4/7)

(vi) × 3

(vii)  × (3/5)

(i) First convert the given mixed fraction into improper fraction.

= 5 ¼ = 21/4

Now,

= (2/5) × (21/4)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2 × 21)/ (5 × 4)

= (1 × 21)/ (5 × 2)

= (21/10)

=

(ii) First convert the given mixed fraction into improper fraction.

== 32/5

Now,

= (32/5) × (7/9)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (32 × 7)/ (5 × 9)

= (224/45)

=

(iii) First convert the given mixed fraction into improper fraction.

== 16/3

Now,

= (3/2) × (16/3)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (3 × 16)/ (2 × 3)

= (1 × 8)/ (1 × 1)

= 8

(iv) First convert the given mixed fraction into improper fraction.

== 17/7

Now,

= (5/6) × (17/7)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5 × 17)/ (6 × 7)

= (85/42)

=

(v) First convert the given mixed fraction into improper fraction.

== 17/5

Now,

= (17/5) × (4/7)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (17 × 4)/ (5 × 7)

= (68/35)

=

(vi) First convert the given mixed fraction into improper fraction.

== 13/5

Now,

= (13/5) × (3/1)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (13 × 3)/ (5 × 1)

= (39/5)

=

(vii) First convert the given mixed fraction into improper fraction.

== 25/7

Now,

= (25/7) × (3/5)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (25 × 3)/ (7 × 5)

= (5 × 3)/ (7 × 1)

= (15/7)

=

Question 8 :

(a) (i) provide the number in the box [ ], such that (2/3) × [ ] = (10/30)

(ii) The simplest form of the number obtained in [ ] is

(b) (i) provide the number in the box [ ], such that (3/5) × [ ] = (24/75)

(ii) The simplest form of the number obtained in [ ] is

(a) (i)

Let the required number be x,

Then,

= (2/3) × (x) = (10/30)

By cross multiplication,

= x = (10/30) × (3/2)

= x = (10 × 3) / (30 × 2)

= x = (5 × 1) / (10 × 1)

= x = 5/10

∴ The required number in the box is (5/20)

(ii) The number in the box is 5/10

Then,

The simplest form of 5/10 is ½

(b) (i)

Let the required number be x,

Then,

= (3/5) × (x) = (24/75)

By cross multiplication,

= x = (24/75) × (5/3)

= x = (24 × 5) / (75 × 3)

= x = (8 × 1) / (15 × 1)

= x = 8/15

∴ The required number in the box is (8/15)

(ii) The number in the box is 8/15

Then,

The simplest form of 8/15 is 8/15

### Exercise 2.4

Question 1 :

Find.

(i) 12 ÷ ¾

(ii) 14 ÷ (5/6)

(iii) 8 ÷ (7/3)

(iv) 4 ÷ (8/3)

(v) 3 ÷

(vi) 5 ÷

(i) We have

= 12 × reciprocal of ¾

= 12 × (4/3)

= 4 × 4

= 16

(ii) We have

= 14 × reciprocal of (5/6)

= 14 × (6/5)

= 84/5

(iii) We have

= 8 × reciprocal of (7/3)

= 8 × (3/7)

(iv) We have

= 4 × reciprocal of (8/3)

= 4 × (3/8)

= 1 × (3/2)

= 3/2

(v) While dividing a whole number by a mixed fraction, first convert the mixed fraction into an improper fraction.

We have

== 7/3

Then,

= 3 ÷ (7/3)

= 3 × reciprocal of (7/3)

= 3 × (3/7)

= 9/7

(vi) While dividing a whole number by a mixed fraction, first convert the mixed fraction into an improper fraction.

We have

== 25/7

Then,

= 5 ÷ (25/7)

= 5 × reciprocal of (25/7)

= 5 × (7/25)

= 1 × (7/5)

= 7/5

Question 2 :

Find.

(i) (2/5) ÷ (½)

(ii) (4/9) ÷ (2/3)

(iii) (3/7) ÷ (8/7)

(iv) ÷ (3/5)

(v) 3 ½ ÷ (8/3)

(vi) (2/5) ÷ 1 ½

(vii) ÷

(viii) ÷

(i) We have

= (2/5) × reciprocal of ½

= (2/5) × (2/1)

= (2 × 2) / (5 × 1)

= 4/5

(ii) We have

= (4/9) × reciprocal of (2/3)

= (4/9) × (3/2)

= (4 × 3) / (9 × 2)

= (2 × 1) / (3 × 1)

= 2/3

(iii) We have

= (3/7) × reciprocal of (8/7)

= (3/7) × (7/8)

= (3 × 7) / (7 × 8)

= (3 × 1) / (1 × 8)

= 3/8

(iv) First, convert the mixed fraction into an improper fraction.

We have,

== 7/3

Then,

= (7/3) × reciprocal of (3/5)

= (7/3) × (5/3)

= (7 × 5) / (3 × 3)

= 35/9

(v) First, convert the mixed fraction into an improper fraction.

We have

= 3 ½ = 7/2

Then,

= (7/2) × reciprocal of (8/3)

= (7/2) × (3/8)

= (7 × 3) / (2 × 8)

= 21/16

(vi) First, convert the mixed fraction into an improper fraction.

We have

= 1 ½ = 3/2

Then,

= (2/5) × reciprocal of (3/2)

= (2/5) × (2/3)

= (2 × 2) / (5 × 3)

= 4/15

(vii) First, convert the mixed fraction into an improper fraction.

We have

== 16/5

== 5/3

Then,

= (16/5) × reciprocal of (5/3)

= (16/5) × (3/5)

= (16 × 3) / (5 × 5)

= 48/25

(viii) First, convert the mixed fraction into an improper fraction.

We have

== 11/5

== 6/5

Then,

= (11/5) × reciprocal of (6/5)

= (11/5) × (5/6)

= (11 × 5) / (5 × 6)

= (11 × 1) / (1 × 6)

= 11/6

Question 3 :

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

(i) 3/7

(ii) 5/8

(iii) 9/7

(iv) 6/5

(v) 12/7

(vi) 1/8

(vii) 1/11

(i) Reciprocal of (3/7) is (7/3) [∵ ((3/7) × (7/3)) = 1]

So, it is an improper fraction.

An improper fraction is a fraction in which the numerator is greater than its denominator.

(ii) Reciprocal of (5/8) is (8/5) [∵ ((5/8) × (8/5)) = 1]

So, it is an improper fraction.

An improper fraction is a fraction in which the numerator is greater than its denominator.

(iii) Reciprocal of (9/7) is (7/9) [∵ ((9/7) × (7/9)) = 1]

So, it is a proper fraction.

A proper fraction is a fraction in which the denominator is greater than the numerator of the fraction.

(iv) Reciprocal of (6/5) is (5/6) [∵ ((6/5) × (5/6)) = 1]

So, it is a proper fraction.

A proper fraction is a fraction in which the denominator is greater than the numerator of the fraction.

(v) Reciprocal of (12/7) is (7/12) [∵ ((12/7) × (7/12)) = 1]

So, it is a proper fraction.

A proper fraction is a fraction in which the denominator is greater than the numerator of the fraction.

(vi) Reciprocal of (1/8) is (8/1) or 8 [∵ ((1/8) × (8/1)) = 1]

So, it is a whole number.

Whole numbers are collections of all positive integers, including 0.

(vii) Reciprocal of (1/11) is (11/1) or 11 [∵ ((1/11) × (11/1)) = 1]

So, it is a whole number.

Whole numbers are collections of all positive integers, including 0.

Question 4 :

Find:

(i) (7/3) ÷ 2

(ii) (4/9) ÷ 5

(iii) (6/13) ÷ 7

(iv) ÷ 3

(v) 3 ½ ÷ 4

(vi)  ÷ 7

(i) We have

= (7/3) × reciprocal of 2

= (7/3) × (1/2)

= (7 × 1) / (3 × 2)

= 7/6

=

(ii) We have

= (4/9) × reciprocal of 5

= (4/9) × (1/5)

= (4 × 1) / (9 × 5)

= 4/45

(iii) We have

= (6/13) × reciprocal of 7

= (6/13) × (1/7)

= (6 × 1) / (13 × 7)

= 6/91

(iv) First, convert the mixed fraction into an improper fraction.

We have,

== 13/3

Then,

= (13/3) × reciprocal of 3

= (13/3) × (1/3)

= (13 × 1) / (3 × 3)

= 13/9

(v) First, convert the mixed fraction into an improper fraction.

We have

= 3 ½ = 7/2

Then,

= (7/2) × reciprocal of 4

= (7/2) × (1/4)

= (7 × 1) / (2 × 4)

= 7/8

(vi) First, convert the mixed fraction into an improper fraction.

We have

== 31/7

Then,

= (31/7) × reciprocal of 7

= (31/7) × (1/7)

= (31 × 1) / (7 × 7)

= 31/49

### Exercise 2.5

Question 1 :

(i) Express 5 cm in meter and kilometer

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

(i) We know that,

= 1 meter = 100 cm

Then,

= 1 cm = (1/100) m

= 5 cm = (5/100)

= 0.05 m

Now,

= 1 km = 1000 m

Then,

= 1 m = (1/1000) km

= 0.05 m = (0.05/1000)

= 0. 00005 km

(ii) We know that,

= 1 cm = 10 mm

Then,

= 1 mm = (1/10) cm

= 35 mm = (35/10) cm

= 3.5 cm

And,

= 1 meter = 100 cm

Then,

= 1 cm = (1/100) m

= 3.5 cm = (3.5/100) m

= (35/1000) m

= 0.035 m

Now,

= 1 km = 1000 m

Then,

= 1 m = (1/1000) km

= 0.035 m = (0.035/1000)

= 0. 000035 km

Question 2 :

Write the following decimal numbers in the expanded form:

(i) 20.03

(ii) 2.03

(iii) 200.03

(iv) 2.034

(i) We have,

20.03 = (2 × 10) + (0 × 1) + (0 × (1/10)) + (3 × (1/100))

(ii) We have,

2.03 = (2 × 1) + (0 × (1/10)) + (3 × (1/100))

(iii) We have,

200.03 = (2 × 100) + (0 × 10) + (0 × 1) + (0 × (1/10)) + (3 × (1/100))

(iv) We have,

2.034 = (2 × 1) + (0 × (1/10)) + (3 × (1/100)) + (4 × (1/1000))

Question 3 :

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?

From the question, it is given that,

Distance travelled by Dinesh = AB + BC

= 7.5 + 12.7

= 20.2 km

∴Dinesh travelled 20.2 km

Distance travelled by Ayub = AD + DC

= 9.3 + 11.8

= 21.1 km

∴Ayub travelled 21.1km

Clearly, Ayub travelled more distance by = (21.1 – 20.2)

= 0.9 km

∴Ayub travelled 0.9 km more than Dinesh.

Question 4 :

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?

From the question, it is given that,

Fruits bought by Shyama = 5 kg 300 g

= 5 kg + (300/1000) kg

= 5 kg + 0.3 kg

= 5.3 kg

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

= (4 + (800/1000)) + (4 + (150/1000))

= (4 + 0.8) kg + (4 + .150) kg

= 4.8 kg + 4.150kg

= 8.950 kg

So, Sarala bought more fruits.

Question 5 :

How much less is 28 km than 42.6 km?

Now, we have to find the difference of 42.6 km and 28 km

42.6

-28.0

14.6

∴ 14.6 km less is 28 km than 42.6 km.

Question 6 :

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) By comparing whole number, 0 = 0

By comparing the tenths place digit, 5 > 0

∴ 0.5 > 0.05

(ii) By comparing whole number, 0 = 0

By comparing the tenths place digit, 7 > 5

∴ 0.7 > 0.5

(iii) By comparing whole number, 7 > 0

∴ 7 > 0.7

(iv) By comparing whole number, 1 = 1

By comparing the tenths place digit, 3 < 4

∴ 1.37 < 1.49

(v) By comparing whole number, 2 = 2

By comparing the tenths place digit, 0 < 3

∴ 2.03 < 2.30

(vi) By comparing whole number, 0 = 0

By comparing the tenths place digit, 8 = 8

By comparing the hundredths place digit, 0 < 8

∴ 0.8 < 0.88

Question 7 :

Express as rupees as decimals:

(i) 7 paise

(ii) 7 rupees 7 paise

(iii) 77 rupees 77 paise

(iv) 50 paise

(v) 235 paise

(i) We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 7 paise = ₹ (7/100)

= ₹ 0.07

(ii) We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 7 rupees 7 paise = ₹ 7 + ₹ (7/100)

= ₹ 7 + ₹ 0.07

= ₹ 7.07

(iii) We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 77 rupees 77 paise = ₹ 77 + ₹ (77/100)

= ₹ 77 + ₹ 0.77

= ₹ 77.77

(iv) We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 50 paise = ₹ (50/100)

= ₹ 0.50

(v) We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 235 paise = ₹ (235/100)

= ₹ 2.35

Question 8 :

Express in kg:

(i) 200 g

(ii) 3470 g

(iii) 4 kg 8 g

(i) We know that,

= 1 kg = 1000 g

Then,

= 1 g = (1/1000) kg

= 200 g = (200/1000) kg

= (2/10)

= 0.2 kg

(ii) We know that,

= 1 kg = 1000 g

Then,

= 1 g = (1/1000) kg

= 3470 g = (3470/1000) kg

= (3470/100)

= 3.470 kg

(iii) We know that,

= 1 kg = 1000 g

Then,

= 1 g = (1/1000) kg

= 4 kg 8 g = 4 kg + (8/1000) kg

= 4 kg + 0.008

= 4.008 kg

Question 9 :

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) From the question, we observe that,

The place value of 2 in 2.56 is ones

(ii) From the question, we observe that,

The place value of 2 in 21.37 is tens

(iii) From the question, we observe that,

The place value of 2 in 10.25 is tenths.

(iv) From the question, we observe that,

The place value of 2 in 9.42 is hundredth.

(v) From the question, we observe that,

The place value of 2 in 63.352 is thousandth.

### Exercise 2.6

Question 1 :

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

(i) On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 1.3 × 10 = 13

(ii) On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 36.8 × 10 = 368

(iii) On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 153.7 × 10 = 1537

(iv) On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 168.07 × 10 = 1680.7

(v) On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 31.1 × 100 = 3110

(vi) On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 156.1 × 100 = 15610

(vii) On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 3.62 × 100 = 362

(viii) On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 43.07 × 100 = 4307

(ix) On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 0.5 × 10 = 5

(x) On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 0.08 × 10 = 0.8

(xi) On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 0.9 × 100 = 90

(xii) On multiplying a decimal by 1000, the decimal point is shifted to the right by three places.

We have,

= 0.03 × 1000 = 30

Question 2 :

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

From the question,

Length of the rectangle = 5.7 cm

Breadth of the rectangle = 3 cm

Then,

Area of the rectangle = length × Breadth

= 5.7 × 3

= 17.1 cm2

Question 3 :

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?

It is given that,

Distance covered by two-wheeler in 1 litre of petrol = 55.3 km

Then,

Distance covered by two-wheeler in 10 litres of petrol = (10 × 55.3)

= 553 km

∴ The two-wheeler will cover a distance of 553 km in 10 litres of petrol.

Question 4 :

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) We have,

= (2/10) × 6

= (12/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 1.2

(ii) We have,

= (8) × (46/10)

= (368/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 36.8

(iii) We have,

= (271/100) × 5

= (1355/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 13.55

(iv) We have,

= (201/10) × 4

= (804/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 80.4

(v) We have,

= (5/100) × 7

= (35/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.35

(vi) We have,

= (21102/100) × 4

= (84408/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 844.08

(vii) We have,

= (2) × (86/100)

= (172/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 1.72

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) We have,

= (25/10) × (3/10)

= (75/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.75

(ii) We have,

= (1/10) × (517/10)

= (517/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 5.17

(iii) We have,

= (2/10) × (3168/10)

= (6336/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 63.36

(iv) We have,

= (13/10) × (31/10)

= (403/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 4.03

(v) We have,

= (5/10) × (5/100)

= (25/1000)

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

Then,

= 0.025

(vi) We have,

= (112/10) × (15/100)

= (1680/1000)

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

Then,

= 1.680

(vii) We have,

= (107/100) × (2/100)

= (214/10000)

On dividing a decimal by 10000, the decimal point is shifted to the left by four places.

Then,

= 0.0214

(viii) We have,

= (1005/100) × (105/100)

= (105525/10000)

On dividing a decimal by 10000, the decimal point is shifted to the left by four places.

Then,

= 10.5525

(ix) We have,

= (10101/100) × (1/100)

= (10101/10000)

On dividing a decimal by 10000, the decimal point is shifted to the left by four places.

Then,

= 1.0101

(x) We have,

= (10001/100) × (11/10)

= (110011/1000)

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

Then,

= 110.011

### Exercise 2.7

Question 1 :

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?

From the question, it is given that,

Total distance covered by the vehicle in 2.4 litres of petrol = 43.2 km

Then,

Distance covered in 1 litre of petrol = 43.2 ÷ 2.4

= (432/10) ÷ (24/10)

= (432/10) × (10/24)

= (432 × 10)/ (10 × 24)

= (36 × 1)/ (1 × 2)

= (18 × 1)/ (1 × 1)

= 18 km

∴ The total distance covered in 1 litre of petrol is 18 km.

Question 2 :

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) We have,

= (4/10) ÷ 2

Then,

= (4/10) × (1/2)

= (2/10) × (1/1)

= (2/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 0.2

(ii) We have,

= (35/100) ÷ 5

Then,

= (35/100) × (1/5)

= (7/100) × (1/1)

= (7/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.07

(iii) We have,

= (248/100) ÷ 4

Then,

= (248/100) × (1/4)

= (62/100) × (1/1)

= (62/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.62

(iv) We have,

= (654/10) ÷ 6

Then,

= (654/10) × (1/6)

= (109/10) × (1/1)

= (109/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 10.9

(v) We have,

= (6512/10) ÷ 4

Then,

= (6512/10) × (1/4)

= (1628/10) × (1/1)

= (1628/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 162.8

(vi) We have,

= (1449/100) ÷ 7

Then,

= (1449/100) × (1/7)

= (207/100) × (1/1)

= (207/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 2.07

(vii) We have,

= (396/100) ÷ 4

Then,

= (396/100) × (1/4)

= (99/100) × (1/1)

= (99/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.99

(viii) We have,

= (80/100) ÷ 5

Then,

= (80/100) × (1/5)

= (16/100) × (1/1)

= (16/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.16

Question 3 :

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

(i) On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 4.8 ÷ 10

= (4.8/10)

= 0.48

(ii) On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 52.5 ÷ 10

= (52.5/10)

= 5.25

(iii) On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 0.7 ÷ 10

= (0.7/10)

= 0.07

(iv) On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 33.1 ÷ 10

= (33.1/10)

= 3.31

(v) On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 272.23 ÷ 10

= (272.23/10)

= 27.223

(vi) On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 0.56 ÷ 10

= (0.56/10)

= 0.056

(vii) On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 3.97 ÷ 10

= (3.97/10)

= 0.397

Question 4 :

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

(i) On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 2.7 ÷ 100

= (2.7/100)

= 0.027

(ii) On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 0.3 ÷ 100

= (0.3/100)

= 0.003

(iii) On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 0.78 ÷ 100

= (0.78/100)

= 0.0078

(iv) On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 432.6 ÷ 100

= (432.6/100)

= 4.326

(v) On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 23.6 ÷ 100

= (23.6/100)

= 0.236

(vi) On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 98.53 ÷ 100

= (98.53/100)

= 0.9853

Question 5 :

Find:

(i) 7.9 ÷ 1000

(ii) 26.3 ÷ 1000

(iii) 38.53 ÷ 1000

(iv) 128.9 ÷ 1000

(v) 0.5 ÷ 1000

(i) On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 7.9 ÷ 1000

= (7.9/1000)

= 0.0079

(ii) On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 26.3 ÷ 1000

= (26.3/1000)

= 0.0263

(iii) On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 38.53 ÷ 1000

= (38.53/1000)

= 0.03853

(iv) On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 128.9 ÷ 1000

= (128.9/1000)

= 0.1289

(v) On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 0.5 ÷ 1000

= (0.5/1000)

= 0.0005

Question 6 :

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) We have,

= 7 ÷ (35/10)

= 7 × (10/35)

= 1 × (10/5)

= 2

(ii) We have,

= 36 ÷ (2/10)

= 36 × (10/2)

= 18 × 10

= 180

(iii) We have,

= (325/100) ÷ (5/10)

= (325/100) × (10/5)

= (325 × 10)/ (100 × 5)

= (65 × 1)/ (10 × 1)

= 65/10

= 6.5

(iv) We have,

= (3094/100) ÷ (7/10)

= (3094/100) × (10/7)

= (3094 × 10)/ (100 × 7)

= (442 × 1)/ (10 × 1)

= 442/10

= 44.2

(v) We have,

= (5/10) ÷ (25/100)

= (5/10) × (100/25)

= (5 × 100)/ (10 × 25)

= (1 × 10)/ (1 × 5)

= 10/5

= 2

(vi) We have,

= (775/100) ÷ (25/100)

= (775/100) × (100/25)

= (775 × 100)/ (100 × 25)

= (155 × 1)/ (1 × 5)

= (31 × 1)/ (1 × 1)

= 31

(vii) We have,

= (765/10) ÷ (15/100)

= (765/10) × (100/15)

= (765 × 100)/ (10 × 15)

= (51 × 10)/ (1 × 1)

= 510

(viii) We have,

= (378/10) ÷ (14/10)

= (378/10) × (10/14)

= (378 × 10)/ (10 × 14)

= (27 × 1)/ (1 × 1)

= 27

(ix) We have,

= (273/100) ÷ (13/10)

= (273/100) × (10/13)

= (273 × 10)/ (100 × 13)

= (21 × 1)/ (10 × 1)

= 21/10

= 2.1

The NCERT solution for Class 7 Chapter 2: Fractions and Decimals is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.

Yes, the NCERT solution for Class 7 Chapter 2: Fractions and Decimals is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. Fractions and Decimalsally, they can solve the practice questions and exercises that allow them to get exam-ready in no time.

You can get all the NCERT solutions for Class 7 Maths Chapter 2 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.

Yes, students must practice all the questions provided in the NCERT solution for Class 7 Maths Chapter 2: Fractions and Decimals as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.

Students can utilize the NCERT solution for Class 7 Maths Chapter 2 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.

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