ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 1 Integers Ex 1.5
Question 1.
7 – 8 ÷ (-2) + 3 × (-4)
Solution:
7 – 8 ÷ (-2) + 3 × (-4)
= 7 + \(\frac { -8 }{ -2 }\) + 3 × (-4)
= 7 + 4 + 3 × (-4) (Use of BODMAS)
= 7 + 4 – 12
= 11 – 12
= -1
Question 2.
9 – {7 – 24 ÷ (8 + 6 × 2 – 16)}
Solution:
9 – {7 – 24 ÷ (8 + 6 × 2 – 16)}
= 9 – {7 – 24 ÷ (8 + 12 – 16)}
= 9 – {7 – 24 ÷ (4)}
= 9 – (7 – \(\frac { 24 }{ 4 }\))
= 9 – [7 – 6]
= 9 – 1
= 8
Question 3.
-11 – [-6 – {3 – 5(8 ÷ 4 – 1)}]
Solution:
-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 – [-6 + 2]
= -11 – (-4)
= -11 + 4
= -7
Question 4.
(-3) × (12) ÷ (-4) + 3 × 6
Solution:
(-3) × (12) ÷ (-4) + 3 × 6
= -3 × (\(\frac { -12 }{ -4 }\)) + 3 × 6
= -3 × 3 + 3 × 6
= -9 + 18
= 9
Question 5.
14 ÷ (3 of 2 – 3 + 4) – 9(5 – 3)
Solution:
14 ÷ (3 of 2 – 3 + 4) – 9(5 – 3)
= 14 ÷ (6 – 3 + 4) – 9(2)
= 14 ÷ 7 – 9 × 2
= 2 – 18
= -16