## 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