New Learning Composite Mathematics Class 6 SK Gupta Anubhuti Gangal Fractions Chapter 5J Solution
(1) 1 4/5
= 9/5
(2) Anil spends on supplies 2/5 of his budget
Anil spends on advertising 2/12 of his budget
Anil spends on taxes 2/15 of his budget
LCM of 5, 12 and 25 = 300
2×60/5×60 = 120/300
2×25/12×35 = 100/300
2×12/25×12 = 24/300
Ans/ Anil spend more on advertising.
(3) Sumer ran in 2 days = (1/3 + 3 1/4) km
= 1/3 + 13/4
= 4+39/12
= 43/12
= 3 7/12
Ans. On two days Sumer ran 3 7/12 km.
(4) Tails appear = (100 – 55) times
= 45
In fraction = 45/100
Ans. Total tails are 9/20
(5) Crows was standing = 1 – (3/4 + 1/8)
= 1 – (6 + 1/8)
= 1 – 7/8
= 8 – 7 / 8
= 1/8
Ans. fraction of crowd standing is 1/8
6) sales occured in morning = 3/8
Sales occured in afternoon 7/10
Sales occured in evening = 3/16
LCM of 8, 16, 19 = 80
3×10/8×10 = 30/80
7×8/10×8 = 56/80
3×5/16×5 = 15/80
Ans. During afternoon most of her sales occured.
(7) Fraction remains = 1 – (1/5 + 1/3 + 1/4)
= 1 – 12+20+15/60
= 1 – 47/60
= 13/60
∴ Pages remains = 240 x 13/60
= 52
Ans. Fraction remains to be read is 13/60 pages remains to be read are 52.
(8) Shekhar jogs = (1/4) + 2 1/8 + 1 1/20) km
= (5/4 + 17/8 + 21/20) km
= 50 + 85 + 42 / 40
= 177/40
∴ (5 – 177/40) km
= 200-177/40
= 23/40 km
= 575 m
ans. One round is 23/40 km or 575 m less than 5 km.
(9) Sudhir walked 3/5 km
∴ Prem walked = (3/5 – 7/20) km
= 12 – 7 / 20
= 5/20
= 1/4 km
Rohan walked = (1/4 + 5/8)
= 2+5/8
= 7/8
Ans – Rohan walked 7/8 km.
10) Jouney remains = {1 – (1/8 + 2/5 + 3/10)}
= {1 – 5+16+12/40}
= {1 – 33/40}
= 40-33/40
= 7/40 km
Whole distance is 80 km
∴ I ride before breakfast = (80 x 1/8) km
= 10 km
∴ I ride between breakfast and lunch = (80 x 2/5) km
= 32 km
∴ I ride between lunch and tea = (80 x 3/10) km
= 24 km
Ans – Journey remains 7/40 km
I ride 10 km before breakfast, 32 km between breakfast and lunch, 24 km between lunch and tea.
