ML Aggarwal Integers Exe-1.5 Class 7 ICSE Maths Solutions

ML Aggarwal Integers Exe-1.5 Class 7 ICSE Maths Solutions. We Provide Step by Step Answer of  Exe-1.5 Questions for Integers as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-7.

ML Aggarwal Integers Exe-1.5 Class 7 Maths Solutions

Board ICSE
Publications Avichal Publishig Company (APC)
Subject Maths
Class 7th
Chapter-1 Integers
Writer ML Aggarwal
Book Name Understanding
Topics Solution of Exe-1.5 Questions
Edition 2023-2024

Integers Exe-1.5

ML Aggarwal Class 7 ICSE Maths Solutions

Page-23

Question 1. 7 – 8 ÷ (-2) + 3 × (-4)

Answer:

7 – 8 ÷ (-2) + 3 × (-4)
= 7 + (-8/2)
+ 3 × (-4)
= 7 + 4 + 3 × (-4) ( use of BODMAS)
= 7 + 4 – 12
= 11 – 12
= – 1

Question 2. 9 – {7 – 24 ÷ (8 + 6 × 2 – 16)}

Answer:

9 – { 7 – 24 ÷ (8 + 6 × 2 – 16 ) }
= 9 – { 7 – 24 ÷ (8 + 12 – 16 ) }
= 9 – { 7 – 24 ÷ 4 }
= 9 – ( 7 – (24/4))
= 9 – [ 7 – 6 ]
= 9 – 1
= 8

Question 3. -11 – [-6 – {3 – 5(8 ÷ 4 – 1)}]

Answer:

-11 [ -6 –{ 3 – 5 ( 8 ÷ 4 – 1 ) } ]
= -11 [ -6 –{ 3 – 5 ( 2 – 1 ) } ]
= -11 [ -6 –{ 3 – 5 × 1} ]
= -11 [ -6 –{ 3 – 5 } ]
= -11 [ -6 – ( -2 ) ]
= -11 – (-4)
= -11 +4
= -7

Question 4. (-3) × (12) ÷ (-4) + 3 × 6

Answer:

(-3) × (12) ÷ (-4) + 3 × 6
= -3 × (-12/-4) + 3 × 6
= -9 + 18
= 9

Question 5. 14 ÷ (3 of 2 – 3 + 4) – 9(5 – 3)

Answer:

14 ÷ ( 6 – 3 + 4 ) – 9(2)
= 14 ÷ 7 – 9 × 2
= 2 – 18
= -16

—  : End of ML Aggarwal Integers Exe-1.5 Class 7 ICSE Maths Solutions :–

Return to –  ML Aggarwal Maths Solutions for ICSE Class -7

Thanks

 Share with your friends

Leave a Comment

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