Exercise 1.4
Simplify each of the following
1. 3- (5-6÷3)
Solution:
Given 3- (5-6÷3)
According to removal of bracket rule firstly remove inner most bracket
We get 3 – (5 -6 ÷3) = 3- (5-2)
= 3-2
= 3
2. -25 + 14÷(5-3)
Solution:
Given -25 + 14÷(5-3)
According to removal of bracket rule firstly remove inner most bracket
We get -25 +14 ÷ (5-3) = -25 + 14 ÷ 2
= -25 + 7
= -18
Solution:
According to removal of bracket rule first we have to remove vinculum we get
= 25 – {5+4- (5-4)}
Now by removing the innermost bracket we get
= 25 – {5 + 4 -1 }
by removing the parentheses we get
= 25 – (8)
Now simplify we get
= 25 – 4
= 21
5. 36 – [18 – (14 – (15 – 4 ÷ 2 × 2))]
Solution:
36 – [18 – (14 – (15 – 4 ÷ 2 × 2))]
= 36 – [18 – (14 – (11 ÷ 2 × 2))]
= 36 – [18 – (14 – 11/2 × 2))]
= 36 – [18 – (14 – 11)]
= 36 – [18 – 3]
= 36 – 15
= 21
∴ 36 – [18 – (14 – (15 – 4 ÷ 2 × 2))] = 21
6. 45 – [38 – (60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3)]
Solution:
45 – [38 – (60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3)]
= 45 – [38 – (20 – (6 – 3) ÷ 3)]
= 45 – [38 – (20 – 3 ÷ 3)]
= 45 – [38 – (20 – 1)]
= 45 – [38 – 19]
= 45 – [19]
= 26
∴ 45 – [38 – (60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3)] = 26
7. 23 – [ 23 – (23- (23-))]
Solution:
23 – [ 23 – (23- (23-))]
= 23 – [23 – (23 – (23 – 0))]
= 23 – [23 – (23 – 23)]
= 23 – [23 – 0]
= 23 – 23
= 0
∴23 – [ 23 – (23- (23-))] = 0
8. 2550 – [ 510 – (270 – (90- ))]
Solution:
2550 – [ 510 – (270 – (90- ))]
= 2550 – [510 – (270 – (90 – 150))]
= 2550 – [510 – (270 – (–60))]
= 2550 – [510 – 330]
= 2550 – [180]
= 2550 – 180
= 2370
2550 – [ 510 – (270 – (90- ))] = 2370
12. 29 – ( – 2)(6 – (7 – 3))] ÷ [3 × (5 + ( – 3) × (-2))]
Solution:
[29 – (-2)(6 – (7 – 3))] ÷ [3 × (5 + (- 3) × (- 2))]
= [29 – (- 2)(6 – 4)] ÷ [3 × (5 + (3 × 2))]
= [29 – (-2)(2)] ÷ [3 × (5 + 6)]
= [29 + 4] ÷ [3 × 11]
= [33] ÷ [33]
= 1
∴ [29 – (-2) (6 -(7- 3))] ÷ [3 × (5 + (- 3) × (-2))] = 1
13. Using brackets, write a mathematical expression for each of the following:
(i) Nine multiplied by the sum of two and five.
(ii) Twelve divided by the sum of one and three.
(iii) Twenty divided by the difference of seven and two.
(iv) Eight subtracted from the product of two and three.
(v) Forty divided by one more than the sum of nine and ten.
(vi) Two multiplied by one less than the difference of nineteen and six.
Solution:
(i) 9 (2 + 5)
(ii) 12 ÷ (1 + 3)
(iii) 20 ÷ (7 – 2)
(iv) 2 × 3 – 8
(v) 40 ÷ [1 + (9 + 10)]
(vi) 2 × [(19 – 6) – 1]