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

## 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.
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Exercise 2.1

Q.1. Solve:
(i)

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)

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)

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)

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)

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)

Ans: First convert mixed fraction into improper fraction,
= = 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)

Ans:  First convert mixed fraction into improper fraction,
= 8 ½ = 17/2
=  = 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:

(ii)
Ans.
(i)
[Converting into like fractions]

[Arranging in descending order]

Therefore,

(ii)
[Converting into like fractions]

⇒
[Arranging in descending order]

⇒

Therefore,

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?

Ans. Sum of first row
Sum of second row
Sum of third row
Sum of first column
Sum of second column
Sum of third column
Sum of first diagonal [left to right)
Sum of second diagonal [left to right]
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  cm wide. Find its perimeter.
Ans.
Given: The sheet of paper is in rectangular form.
Length of sheet =  cm and Breadth of sheet cm
Perimeter of rectangle = 2 (length + breadth)

Thus, the perimeter of the rectangular sheet is

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

Ans.
(i) In ΔABE
The perimeter of ΔABE = AB + BE + AE

=

Thus, the perimeter of ΔABE iscm

(ii) In rectangle BCDE,
Perimeter of rectangle = 2 [length + breadth]

Thus, the perimeter of rectangle BCDE is
Comparing the perimeter of triangle and that of rectangle,

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  cm wide. To fit in the frame the picture cannot be more than cm wide. How much should the picture be trimmed?
Ans.
Given: The width of the picture
and the width of picture frame
Therefore, the picture should be trimmed

Thus, the picture should be trimmed by

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
Comparing the parts of apple eaten by both Ritu and Somu
The larger share will be more b
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,
Here,
Thus, Vaibhav worked longer time.
Vaibhav worked longer time by
Thus, Vaibhav took 1/6 hour more than Michael.

Exercise 2.2

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

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:

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)

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)
=

(ii)

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)
=

(iii)

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)
=

(iv)

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)
=

(v)

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)
=

(vi)

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)

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)
=

(viii)

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)

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)
=

(x)

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

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.

(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.

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:

Ans: First convert the given mixed fraction into improper fraction.
=  = 26/5
Now,
= 3 × (26/5)
= 78/5
=

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

Ans:

First convert the given mixed fraction into improper fraction.

= 2 ¼ = 9/4

Now,

= 7 × (9/4)

= 63/4

= 15 ¾

Ans:  First convert the given mixed fraction into improper fraction.
= = 19/3
Now,
= 4 × (19/3)
= 76/3
=

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

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

=   = 17/5

Now,

= (17/5) × 8

= 136/5
=

Q.7. Find:

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)
=

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

Ans:

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

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

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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## Mathematics (Maths) Class 7

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