Fractions Class- 7th RS Aggarwal Exe-2C Goyal Brothers ICSE Math Solution

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

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

Thanks

Please share with yours friends if you find it helpful. 

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.