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## Chapter 2 Fractions and Decimals Exercise 2.1

Question 1: Solve:
(i) 2 – (3/5)

= 2/1 - 3/5
LCM of 1, 5 = 5
= 2/1 × 5/5 = 10/5
= 3/5 × 1/1 = 3/5
= 10/5 - 3/5
= 10 - 3/5
= 7/5

(ii) 4 + (7/8)

= 4/1 + 7/8
LCM of 1, 8 = 8
= 4/1 × 8/8 = 32/8
= 7/8 × 1/1 = 7/8
= 32/8 + 7/8
= 32 + 7/8
= 39/8 or 4 7/8

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

LCM of 5, 7 = 35
= 3/5 × 7/7 = 21/35
= 2/7 × 5/5 = 10/35
= 21/35 + 10/35
= 21 + 10/35
= 31/35

(iv) (9/11) – (4/15)

LCM of 11, 15 = 165
= 9/11 × 15/15 = 135/165
= 4/15 × 11/11 = 44/165
= 135/165 - 44/165
= 135 - 44/165
= 91/165

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

LCM of 2, 5, 10 = 10
= 7/10 × 1/1 = 7/10
= 2/5 × 2/2 = 4/10
= 3/2 × 5/5 = 15/10
= 7/10 + 4/10 + 15/10
= 7 + 4 + 15/10
= 26/10
= 13/5 or 2 3/5

(vi) 2 2/3 + 3 1/2

= 2 2/3 = 8/3
= 3 1/2 = 7/2
LCM of 2, 3 = 6
= 8/3 × 2/2 = 16/6
= 7/2 × 3/3 = 21/6
= 16/6 + 21/6
= 16 + 21/6
= 37/6 or 6 1/6

(vii) 8 1/2 - 3 5/8

= 8 1/2 = 17/2
= 3 5/8 = 29/8
LCM of 2, 8 = 8
= 17/2 × 4/4 = 68/8
= 29/8 × 1/1 = 29/8
= 68/8 - 29/8
= 68 - 29/8
= 39/8 or 4 7/8

Question 2: Arrange the following in descending order:
(i) 2/9, 2/3, 8/21

LCM of 3, 9, 21 = 63
= 2/9 × 7/7 = 14/63
= 2/3 × 21/21 = 42/63
= 8/21 × 3/3 = 24/63
= 42/63 > 24/63 > 14/63
= 2/3 > 8/21 > 2/9

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

LCM of 5, 7, 10 = 70
= 1/5 × 14/14 = 14/70
= 3/7 × 10/10 = 30/70
= 7/10 × 7/7 = 49/70
= 49/70 > 30/70 > 14/70
= 7/10 > 3/7 > 1/5

Question 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?

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, this 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.

Length = 12 1/2 cm (25/2 cm)
Breadth = 10 2/3 cm (32/3 cm)
Perimeter of the rectangle = 2 × (length + breadth)
= 2 × (25/2 + 32/3)
= 2 × 139/6
= 278/6
= 139/3 cm
Therefore, the perimeter of the sheet of paper is 139/3 cm or 46 1/3 cm.

Question 5: Find the perimeters of
(i) Triangle ABE
(ii) the rectangle BCDE in this figure. Whose perimeter is greater?

(i)
AB = 5/2 cm
AE = 3 3/5 (18/5 cm)
BE = 2 3/4 (11/4 cm)
Perimeter of the triangle = Sum of all sides
= AB + BE + EA
= 5/2 + 11/4 + 18/5
The LCM of 2, 4, 5 = 20
= 5/2 × 10/10 = 50/20
= 11/4 × 5/5 = 55/20
= 18/5 × 4/4 = 72/20
= 50/20 + 55/20 + 72/20
= 177/20
= 8 17/20 cm

(ii)
ED = 7/6 cm
BE = 2 3/4 cm (11/4 cm)
Perimeter of the rectangle = 2 × (length + breadth)
= 2 × (7/6 + 11/4)
= 2 × 47/12
= 94/12
= 7 5/6 cm
Comparing 8 17/20 [] 7 5/6
Therefore,   ⃤  ABE has the more perimeter.

Question 6: 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?

Picture having a width of = 7 3/5 (38/5 cm)
Frame having a width of = 7 3/10 (73/10 cm)
The picture should be trimmed by = 38/5 - 73/10
The LCM of 5, 10 = 10
= 38/5 × 2/2 = 76/10
= 73/10 × 1/1 = 73/10
= 76/10 - 73/10
= 76 - 73/10
= 3/10
Therefore, the picture should be trimmed by 3/10 cm.

Question 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?

Part of apple eaten by Ritu is = 3/5
Part of apple eaten by Somu is = ?
= 1 - 3/5
= 1/1 - 3/5
The LCM of 1, 5 = 5
= 1/1 × 5/5 = 5/5
= 3/5 × 1/1 = 3/5
= 5/5 - 3/5
= 5 - 3/5
= 2/5
Therefore, part of apple eaten by Somu is 2/5.
Comparing 3/5 [] 2/5
= 3/5 > 2/5
Therefore, Ritu ate more apple.
= 3/5 - 2/5
= 3 - 2/5
= 1/5
Therefore, Ritu ate more apple by 1/5.

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?

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
= 7/12 × 1/1 = 7/12
= 3/4 × 3/3 = 9/12
Comparing 7/12 [] 9/12
= 7/12 < 9/12
Therefore, Vaibhav worked longer.
= 9/12 - 7/12
= 9 - 7/12
= 2/12
= 1/6
Therefore, Vaibhav worked longer by 1/6 hour.