Monday, 17 October 2011

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

Question 1: Solve:
3
(i)2
5
7
(ii)4+
8


32
(iii)
+
57
94
(iv)

1115
723
(v)
+
+
1052
21
(vi)2
+3
32
15
(vii)8
-3
28




Answer:

32 × 5310 − 37
(i)2
=

=
=
55555


74×87(4×8)+7397



(ii)
4
+

=

+

=

=

=
4




8
8
8
8
8
8


323 × 72 × 521+1031
(iii)
+
=
+
=
=
575 × 77 × 53535
949 × 154 × 11135 −

44
91
(iv)

=

=
=
111511 × 1515 × 11165165
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


21878×27×316+21371
(vi)2
+3
=
+
=
+
=
=
=6
32323×22×3666

15172917×42968 − 29397
(vii)8
3
=
-
=

=
=
=4
28282×48888


Question 2: Arrange the following in descending order:

228
(i)


9,3,21

137
(ii)


5,7,10

Answer:
228
(i)


9,3,21
Conveting them to like fraction, we get
22×714

=
=
99×763


22×2142

==
33×2163

88×324

=
=
2121×363
Since 42 > 24 >14
282

>
>
3219
137
(ii)


5,7,10
Conveting them to like fraction, we get
11 × 1414

=
=
55 × 1470
33×1030

=
=
77 × 1070
77×749

=
=
107 ×770
As 49 > 30 >14
731

>
>
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


49215
(Along the first row 
+
+

11111111


Answer:
49215
(Along the first row, sum =
+
+
=
11111111
35715
(Along the second row, sum =
+
+
=
11111111
81615
(Along the third row, sum =
+
+
=
11111111
43815
(Along the first column, sum =
+
+
=
11111111
95115
(Along the second column, sum =
+
+
=
11111111
27615
(Along the third column, sum =
+
+
=
11111111
45615
(Along the first diagonal, sum =
+
+
=
11111111
25815
(Along the second diagonal, sum =
+
+
=
11111111
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 ×[
2532

+
23
]
= 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

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