CBSE Class VII (7th) Mathematics -Chapter 2. Fractions and Decimals : Exercise 2.1

Question 1: Solve:

			3
(i)	2	−
			5

			7
(ii)	4	+
			8

	3		2
(iii)		+
	5		7

	9		4
(iv)		−
	11		15

	7		2		3
(v)		+		+
	10		5		2

		2			1
(vi)	2		+	3
		3			2

		1			5
(vii)	8		-	3
		2			8

Answer:

			3		2 × 5		3		10 − 3		7
(i)	2	−		=		−		=		=
			5		5		5		5		5

7

4×8

7

(4×8)+7

39

7

(ii)
4
+

---

=

---

+

---

=

---

=

---

=
4

---

8
8
8
8
8
8

	3		2		3 × 7		2 × 5		21+10		31
(iii)		+		=		+		=		=
	5		7		5 × 7		7 × 5		35		35

	9		4		9 × 15		4 × 11		135 − 44		91
(iv)		−		=		−		=		=
	11		15		11 × 15		15 × 11		165		165

	7		2		3		7		2 × 2		3 × 5		7+ 4 + 15		26		13			3
(v)		+		+		=		+		+		=		=		=		=	2
	10		3		2		10		5 × 2		2 × 5		10		10		5			5

		2			1		8		7		8×2		7×3		16+21		37			1
(vi)	2		+	3		=		+		=		+		=		=		=	6
		3			2		3		2		3×2		2×3		6		6			6

		1			5		17		29		17×4		29		68 − 29		39			7
(vii)	8		−	3		=		-		=		−		=		=		=	4
		2			8		2		8		2×4		8		8		8			8

Question 2: Arrange the following in descending order:

	2	2	8
(i)
	9,	3,	21

	1	3	7
(ii)
	5,	7,	10

Answer:

	2	2	8
(i)
	9,	3,	21

Conveting them to like fraction, we get

2		2×7		14
	=		=
9		9×7		63

2		2×21		42
	=		=
3		3×21		63

8		8×3		24
	=		=
21		21×3		63

Since 42 > 24 >14

	2		8		2
∴		>		>
	3		21		9

	1	3	7
(ii)
	5,	7,	10

Conveting them to like fraction, we get

1		1 × 14		14
	=		=
5		5 × 14		70

3		3×10		30
	=		=
7		7 × 10		70

7		7×7		49
	=		=
10		7 ×7		70

As 49 > 30 >14

	7		3		1
∴		>		>
	10,		7,		5

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?

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

	4		9		2		15
(Along the first row		+		+		=
	11		11		11		11

Answer:

	4		9		2		15
(Along the first row, sum =		+		+		=
	11		11		11		11

	3		5		7		15
(Along the second row, sum =		+		+		=
	11		11		11		11

	8		1		6		15
(Along the third row, sum =		+		+		=
	11		11		11		11

	4		3		8		15
(Along the first column, sum =		+		+		=
	11		11		11		11

	9		5		1		15
(Along the second column, sum =		+		+		=
	11		11		11		11

	2		7		6		15
(Along the third column, sum =		+		+		=
	11		11		11		11

	4		5		6		15
(Along the first diagonal, sum =		+		+		=
	11		11		11		11

	2		5		8		15
(Along the second diagonal, sum =		+		+		=
	11		11		11		11

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 12 ¹/₂ cm long and 10 ²/₃ cm wide. Find its perimeter.

Answer:
Length =12 ¹/₂ cm = 25 / 2
Breadth =10 ²/₃ cm = 32/3cm
Perimeter = 2 × (Length + Breadth)

= 2 ×[

25 32 + 2 3

]

= 2 ×[

(25 × 3) + (32 × 2) 6

]

= 2 ×[

(75 + 64) 6

]

= 2 ×

139 6

=

139 3

=46

1 3

cm

Question5. Find the perimeters of (i) Δ ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

Question6. Salil wants to put a picture in a frame. The picture is 7 3/ 5 cm wide.
To fit in the frame the picture cannot be more than 7 3/10 cm wide. How much should the picture be trimmed?.


Question7: Ritu ate 3/5part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? Who had the larger share? By how much?
Answer:
Part of apple eaten by Ritu =3/5
Part of apple eaten by Somu = 1 − Part of apple eaten by Ritu
=1- 3/5=2/5
Therefore, Somu ate 2/5part of the apple.
Since 3 > 2, Ritu had the larger share.
Difference between the 2 shares =3/5 -2/5=1/5
Therefore, Ritu’s share is larger than the share of Somu by1/5

Question 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?

Answer:
Time taken by Michael =7/12
Time taken by Vaibhav =3/4
Converting these fractions into like fractions, we obtain
3/4=3×3/4×3=9/12
And,7/12
Since 9 > 7,
Vaibhav worked longer.
Difference = 9/12-7/12 =2/12 =1/6hour