New Learning Composite Mathematics Class 6 SK Gupta Anubhuti Gangal Fractions Chapter 5I Solution
(1) (a) 5 2/3 + 3 3/4
= 17/3 + 15/4
= 68 + 45 / 12 [LCD of 3, 4 = 12]
= 113/12
= 9 5/12
(b) 3 1/5 + 9 7/10
= 16/5 + 97/10
= 32 + 97 / 10 [LCD of 5, 10 = 10]
= 129/10
= 12 9/10
(c) 19 7/10 + 12 5/8
= 197/10 + 101/8
= 788+505/40 [LCD of 10, 8 = 40]
= 1293/40
= 32 13/’40
(d) 2 15/22 + 3 20/33
= 59/22 + 119/33
= 177+238/66 [LCD of 22, 33 = 66]
= 415/66
= 6 19/66
(2) (a) 7 5/6 – 4 1/3
= 47/6 – 13/3
= 47 – 26 / 6
= 21/6
= 7/2
= 3 1/2
(b) 13 3/5 – 7 7/15
= 68/5 – 112/15 [LCD of 5, 15 = 15]
= 204-112/15
= 92/15
= 6 2/15
(c) 3 3/4 – 1 2/5
= 15/4 – 17/5
= 75-28/20 [LCD of 4, 5 = 20]
= 47/20
= 2 7/20
(d) 2 3/4 – 1 5/21
= 31/14 – 26/21 [LCD of 14, 21 = 42]
= 93 – 52/42
= 41/42
(3) 7 1/3 + 4 5/8 + 2 1/6
= 22/3 + 37/8 + 13/6
= 176 + 111 + 52 / 24 [LCD of 3, 8 , 6 = 24]
= 339/24
= 14 1/8
(4) 2 1/2 + 3 3/4 – 1 7/8 + 5 2/3
= 5/2 + 15/4 – 15/8 + 17/3
= 60 + 90 – 45 + 136/24 [LCD of 2, 4, 8, 3 = 24]
= 241/24
= 10 1/24
(5) 4 1/8 – (1 1/4 + 2 1/2)
= 33/8 – (5/4 + 5/2) [LCD of 4, 2 = 4]
= 33/8 – (5 + 10 / 4)
= 33/8 – 15/4
= 33 – 30 / 8 [LCD of 8, 4 = 8]
= 3 / 8
(6) 6 1/5 – (4 3/10 – 1 1/2)
= 31/5 – (43/10 – 3/2)
= 31/5 – (43-15/10) [LCD of 10, 2 = 10]
= 31/5 – 28/10
= 62 – 28 / 10 [LCD of 5, 10 = 10]
= 34/10
= 3 2/5
