Fractions Class- 7th RS Aggarwal Exe-2C Goyal Brothers ICSE Math Solution . We provide step by step Solutions of lesson-2 Fractions for ICSE Class-7 Foundation RS Aggarwal Mathematics of Goyal Brothers Prakashan . Our Solutions contain all type Questions of Exe-2C to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7 Mathematics.
Fractions Class- 7th RS Aggarwal Exe-2C Goyal Brothers ICSE Math Solution
| Board | ICSE |
| Publications | Goyal brothers Prakashan |
| Subject | Maths |
| Class | 7th |
| Chapter-2 | Fractions |
| Writer | RS Aggrawal |
| Book Name | Foundation |
| Topics | Solution of Exe-2C |
| Academic Session | 2023 – 2024 |
Exercise – 2C
Fractions Class- 7th RS Aggarwal Exe-2C Goyal Brothers ICSE Math Solution
Questions: Simplify :
1. 1(2/3) + (5/6) 0f (24/25)
Solution: 1(2/3) + (5/6) 0f (24/25)
= (5/3) + (5/6) × (24/25)
= (5/3) + (120/150)
= (5/3) × (50/50) + (120/150)
= (250/150) + (120/150)
= (370/150)
= 37/15
= 2(7/15)
2. (1/3) of 4(2/3) ÷ 2(1/3) × 1(1/2)
Solution: (1/3) of 4(2/3) ÷ 2(1/3) × 1(1/2)
= (1/3) × (14/3) ÷ (7/3) × (3/2)
= (1/3) × (14/3) × (3/7) × (3/2)
= (2/2)
= 1
3. 2(1/4) + 1(1/6) – 1(2/3) ÷ 2(2/3) of 3(3/4)
Solution: 2(1/4) + 1(1/6) – 1(2/3) ÷ 2(2/3) of 3(3/4)
= (9/4) + (7/6) – (5/3) ÷ (8/3) × (15/4)
= (9/4) + (7/6) – (5/3) ÷ (120/12)
= (9/4) + (7/6) – (1/6)
= (27 + 14)/12 – (1/6)
= (41/12) – (1/6)
= (41 – 2)/12 = 39/12
= 13/4 = 3(1/4)
4. 1(1/2) × 2(3/4) ÷ 1(4/7) of 2(5/8)
Solution: 1(1/2) × 2(3/4) ÷ 1(4/7) of 2(5/8)
= (3/2) × (11/4) ÷ (11/7) × (21/8)
= (3/2) × (11/4) ÷ (11 × 3)/8
= (3/2) × (11/4) ÷ (33/8)
= (3/2) × (11/4) × (8/33)
= (3/2) × (2/3)
= 1
5. (2(3/4) + 1(5/6)) ÷ 2(1/5) of 3(1/3)
Solution: (2(3/4) + 1(5/6)) ÷ 2(1/5) of 3(1/3)
= (11/4) + (11/6) ÷ (11/5) × (10/3)
= (11 × 3 + 11 × 2)/12 ÷ (11/5) × (10/3)
= (33 + 22)/12 ÷ (22/3)
= (55/12) ÷ (22/3)
= (55/12) × (3/22)
= 5/8
6. (7/15) of (2/3) + (7/12) ÷ (5/6) – (3/5)
Solution: (7/15) of (2/3) + (7/12) ÷ (5/6) – (3/5)
= (7/15) of (8 + 7)/12 ÷ (25 – 18)/30
= (7/15) of (15/12) ÷ (7/30)
= (7/15) × (15/12) ÷ (7/30)
= (7/12) ÷ (7/30)
= (7/12) × (30/7)
= 5/2 = 2(1/2)
7. (22 ÷ 5(1/2) ÷ 2(1/5) of 3(1/3) + 1(5/11)
Solution: (22 ÷ 5(1/2) ÷ 2(1/5) of 3(1/3) + 1(5/11)
= (22/1) ÷ (11/2) ÷ (11/5) of (10/3) + (16/11)
= (22/1) × (2/11) ÷ (11/5) of (10/3) + (16/11)
= (4/1) ÷ (11/5) of (10/3) + (16/11)
= (4/1) ÷ (22/3) + (16/11)
= (4/1) × (3/22) + (16/11)
= (6/11) + (16/11)
= (6 + 16)/11 = (22/11)
= 2
8. 6(1/3) ÷ (2(1/5) + 3(1/2)) of 3(1/3)
Solution: 6(1/3) ÷ (2(1/5) + 3(1/2)) of 3(1/3)
= (19/3) ÷ (11/5) + (7/2) of (10/3)
= (19/3) ÷ (22 + 35)/10 of (10/3)
= (19/3) ÷ (57/10) of (10/3)
= (19/3) ÷ (57/10) × (10/3)
= (19/3) ÷ (19/1)
= (19/3) × (1/19)
= 1/3
— : end of Fractions Class- 7th RS Aggarwal Exe-2C Goyal Brothers ICSE Math Solution:–
Return to- ICSE Class -7 RS Aggarwal Goyal Brothers Math Solutions
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