NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Mathematics (Maths) Class 7

Class 7 : NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

The document NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev is a part of the Class 7 Course Mathematics (Maths) Class 7.
All you need of Class 7 at this link: Class 7

Exercise 2.1

Q.1. Solve:
(i)NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans:  For subtraction of two unlike fractions, first change them to the like fractions.
LCM of 1, 5 = 5
Now, let us change each of the given fraction 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)  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans:  For the addition of two unlike fractions, first change them to the like fractions.
LCM of 1, 8 = 8
Now, let us change each of the given fraction 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)  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans: For the addition of two unlike fractions, first change them to the like fractions.
LCM of 5, 7 = 35
Now, let us change each of the given fraction 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)  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

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

LCM of 11, 15 = 165
Now, let us change each of the given fraction 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)  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans:  For the addition of two unlike fractions, first change them to the like fractions.
LCM of 10, 5, 2 = 10
Now, let us change each of the given fraction 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)  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans: First convert mixed fraction into improper fraction,
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev= 8/3
= 3 ½ = 7/2
For the addition of two unlike fractions, first, change them to the like fractions.
LCM of 3, 2 = 6
Now, let us change each of the given fraction 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)  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans:  First convert mixed fraction into improper fraction,
= 8 ½ = 17/2
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev = 29/8

For the Subtraction of two, unlike fractions, first, change them to the like fractions.
LCM of 2, 8 = 8
Now, let us change each of the given fraction 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)


Q.2. Arrange the following in descending order:
NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
(ii)NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Ans.
(i) NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
[Converting into like fractions]
NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
[Arranging in descending order]
NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Therefore,NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

(ii)  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
[Converting into like fractions]

⇒  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev 
[Arranging in descending order]

⇒  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev   

Therefore,NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Q.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?

NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans. Sum of first row NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Sum of second row  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Sum of third row  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Sum of first column  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Sum of second column  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Sum of third column  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Sum of first diagonal [left to right) NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Sum of second diagonal [left to right]  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Since the sum of fractions in each row, in each column and along the diagonals are same, therefore it is a magic square.

Q.4. A rectangular sheet of paper is NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev cm wide. Find its perimeter. 
Ans. 
Given: The sheet of paper is in rectangular form.
Length of sheet =NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev  cm and Breadth of sheetNCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev cm
Perimeter of rectangle = 2 (length + breadth)

NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Thus, the perimeter of the rectangular sheet is NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Q.5. Find the perimeter of 
(i) ΔABE 
(ii) the rectangle BCDE in this figure. 
Whose perimeter is greater?

NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans.
(i) In ΔABE NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
The perimeter of ΔABE = AB + BE + AE

NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Thus, the perimeter of ΔABE isNCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRevcm

(ii) In rectangle BCDE,NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Perimeter of rectangle = 2 [length + breadth]

NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Thus, the perimeter of rectangle BCDE isNCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Comparing the perimeter of triangle and that of rectangle,

NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Therefore, the perimeter of triangle ABE is greater than that of rectangle BCDE.

Q.6. Salil wants to put a picture in a frame. The picture is NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev cm wide. To fit in the frame the picture cannot be more thanNCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev cm wide. How much should the picture be trimmed?
Ans.
Given: The width of the pictureNCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
and the width of picture frame NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Therefore, the picture should be trimmed
NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Thus, the picture should be trimmed by  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Q.7. Ritu ate 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? 
Ans: The part of an apple eaten by Ritu = 3/5
The part of an apple eaten by Somu NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Comparing the parts of apple eaten by both Ritu and Somu  NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
The larger share will be more b NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Thus, Ritu's part is 1/5 more than Somu’s part.

Q.8. 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? 
Ans: Time is taken by Michael to colour the picture = 7/2 hour
Time is taken by Vaibhav to colour the picture = 3/4 hour
Converting both fractions in like fractions,NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Here, NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Thus, Vaibhav worked longer time.
Vaibhav worked longer time by NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev
Thus, Vaibhav took 1/6 hour more than Michael.


Exercise 2.2 

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

NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans:

(i) 2 × (1/5) represents the addition of 2 figures, each represents 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 represents 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 represents 2 shaded part 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 represents 1 shaded part out of the

given 4 equal parts.
∴ 3 × ¼ is represented by fig (c).

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

NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

 Ans:

(i) 3 × (1/5) represents the addition of 3 figures, each represents 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 represents 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 represents 3 shaded part 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)

Q.3. Multiply and reduce to lowest form and convert into a mixed fraction: 
(i) NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (7/1) × (3/5)
= (7 × 3)/ (1 × 5)
= (21/5)
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev


(ii) NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (4/1) × (1/3)
= (4 × 1)/ (1 × 3)
= (4/3)
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev


(iii) NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (2/1) × (6/7)
= (2 × 6)/ (1 × 7)
= (12/7)
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev


(iv) NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/1) × (2/9)
= (5 × 2)/ (1 × 9)
= (10/9)
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev


(v) NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (2/3) × (4/1)
= (2 × 4)/ (3 × 1)
= (8/3)
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev


(vi) NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/2) × (6/1)
= (5 × 6)/ (2 × 1)
= (30/2)
= 15


(vii) NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (11/1) × (4/7)
= (11 × 4)/ (1 × 7)

= (44/7)
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev


(viii) NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (20/1) × (4/5)
= (20 × 4)/ (1 × 5)
= (80/5)
= 16


(ix) NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans:  By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (13/1) × (1/3)
= (13 × 1)/ (1 × 3)
= (13/3)
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev


(x) NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans: By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (15/1) × (3/5)
= (15 × 3)/ (1 × 5)
= (45/5)
= 9

Q.4. Shade:

NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans:

(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.
NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

(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.

NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

(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. 

NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev


Q.5. Find: 
(a) 1/2 of (i) 24 (ii) 46

Ans: 

(i) 24
We have,
= ½ × 24
= 24/2
= 12

(ii) 46
We have,
= ½ × 46
= 46/2
= 23

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

Ans: 

(i) 18
We have,
= 2/3 × 18
= 2 × 6
= 12

(ii) 27
We have,
= 2/3 × 27
= 2 × 9
= 18

(c) 3/4 of  (i) 16 (ii) 36

Ans: 

(i) 16
We have,
= ¾ × 16
= 3 × 4
= 12

(ii) 36
We have
= ¾ × 36
= 3 × 9
= 27

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

Ans: 

(i) 20
We have,
= 4/5 × 20
= 4 × 4
= 16

(ii) 35
We have,
= 4/5 × 35
= 4 × 7
= 28

Q.6. Multiply and express as a mixed fraction: 
NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans: First convert the given mixed fraction into improper fraction.
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev = 26/5
Now,
= 3 × (26/5)
= 78/5
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev


NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans:  First convert the given mixed fraction into improper fraction.
= 6 ¾ = 27/4
Now,
= 5 × (27/4)
= 135/4
= 33 ¾ 


NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans: 

First convert the given mixed fraction into improper fraction.

= 2 ¼ = 9/4

Now,

= 7 × (9/4)

= 63/4

= 15 ¾ 


NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans:  First convert the given mixed fraction into improper fraction.
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev= 19/3
Now,
= 4 × (19/3)
= 76/3
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans:  First convert the given mixed fraction into improper fraction.
= 3 ¼ = 13/4
Now,
= (13/4) × 6
= (13/2) × 3

= 39/2
= 19 ½ 


NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans: First convert the given mixed fraction into improper fraction.

= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev  = 17/5

Now,

= (17/5) × 8

= 136/5
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev


Q.7. Find:
NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans: 

(i) First convert the given mixed fraction into an improper fraction.
= 2 ¾ = 11/4
Now,
= ½ × 11/4
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= ½ × (11/4)
= (1 × 11)/ (2 × 4)
= (11/8)
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

(ii) First convert the given mixed fraction into improper fraction.
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev = 38/9

Now,
= ½ × (38/9)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= ½ × (38/9)
= (1 × 38)/ (2 × 9)
= (38/18)
= 19/9
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev


NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

Ans: 

(i) First convert the given mixed fraction into an improper fraction.
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev = 23/6
Now,
= (5/8) × (23/6)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/8) × (23/6)
= (5 × 23)/ (8 × 6)
= (115/48)
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev

(ii) First convert the given mixed fraction into improper fraction.
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev = 29/3

Now,
= (5/8) × (29/3)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/8) × (29/3)
= (5 × 29)/ (8 × 3)
= (145/24)
= NCERT Solutions(Part - 1) - Fractions and Decimals Class 7 Notes | EduRev


Q.8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of 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? 
Ans.
(i) From the question, it is given that,
Amount of water in the water bottle = 5 liters
Amount of water consumed by Vidya = 2/5 of 5 liters
= (2/5) × 5
= 2 liters
So, the total amount of water drank by Vidya is 2 liters

(ii) From the question, it is given that,
Amount of water in the water bottle = 5 liters
Then,
Amount of water consumed by Pratap = (1 – water consumed by Vidya)
= (1 – (2/5))
= (5-2)/5
= 3/5
∴ Total amount of water consumed by Pratap = 3/5 of 5 liters
 = (3/5) × 5
 = 3 liters
So, the total amount of water drank by Pratap is 3 liters 

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