# Selina Solutions Concise Maths Class 7 Chapter 1: Integers Exercise 1C

Selina Solutions Concise Maths Class 7 Chapter 1 Integers Exercise 1C provides students with a clear idea of the removal of brackets and performing various operations on the integers. Numerous solved examples are present to help students understand the method of solving problems in a shorter duration. Students can boost their exam preparation by solving the textbook problems, using the solutions created by experts. These solutions contain explanations in a simple language, keeping in mind the intelligent coefficient of students. Students can access Selina Solutions Concise Maths Class 7 Chapter 1 Integers Exercise 1C free PDF, from the links which are available below.

## Selina Solutions Concise Maths Class 7 Chapter 1: Integers Exercise 1C Download PDF

### Access other exercises of Selina Solutions Concise Maths Class 7 Chapter 1: Integers

Exercise 1A Solutions

Exercise 1B Solutions

Exercise 1D Solutions

### Access Selina Solutions Concise Maths Class 7 Chapter 1: Integers Exercise 1C

Evaluate:

1. 18 â€“ (20 â€“ 15 Ã· 3)

Solution:

18 â€“ (20 â€“ 15 Ã· 3)

It can be written as

= 18 â€“ (20 â€“ 15/3)

By further calculation

= 18 â€“ (20 â€“ 5)

= 18 â€“ 20 + 5

So we get

= 18 + 5 â€“ 20

= 23 â€“ 20

= 3

2. â€“ 15 + 24 Ã· (15 â€“ 13)

Solution:

â€“ 15 + 24 Ã· (15 â€“ 13)

It can be written as

= – 15 + 24 Ã· 2

So we get

= – 15 + 12

= – 3

3. 35 â€“ {15 + 14 â€“ (13 + )}

Solution:

35 â€“ {15 + 14 â€“ (13 +
)}

It can be written as

= 35 â€“ [15 + 14 â€“ (13 + 4)]

By further calculation

= 35 â€“ [15 + 14 â€“ 17]

Multiplying the negative sign

= 35 â€“ 15 â€“ 14 + 17

So we get

= 52 â€“ 29

= 23

4. 27 â€“ {13 + 4 â€“ (8 + 4 – )}

Solution:

27 â€“ {13 + 4 â€“ (8 + 4 –
)}

It can be written as

= 27 â€“ {13 + 4 â€“ (8 + 4 â€“ 4)}

By further calculation

= 27 â€“ {13 + 4 â€“ 8}

So we get

= 27 â€“ {13 + (-4)}

= 27 – 9

= 18

5. 32 â€“ [43 â€“ {51 â€“ (20 – )}]

Solution:

32 â€“ [43 â€“ {51 â€“ (20 –
)}]

It can be written as

= 32 â€“ [43 – {51 â€“ (20 â€“ 11)}]

By further calculation

= 32 â€“ [43 â€“ {51 â€“ 9}]

So we get

= 32 â€“ [43 â€“ 42]

= 32 â€“ 1

= 31

6. 46 â€“ [26 â€“ {14 â€“ (15 â€“ 4 Ã· 2 Ã— 2)}]

Solution:

46 â€“ [26 â€“ {14 â€“ (15 â€“ 4 Ã· 2 Ã— 2)}]

It can be written as

= 46 â€“ [26 â€“ {14 â€“ (15 â€“ 2 Ã— 2)}]

By further calculation

= 46 â€“ [26 â€“ {14 â€“ (15 â€“ 4)}]

So we get

= 46 â€“ [26 â€“ {14 â€“ 11}]

= 46 â€“ [26 â€“ 3]

Here

= 46 â€“ 23

= 23

7. 45 â€“ [38 â€“ {60 Ã· 3 â€“ (6 â€“ 9 Ã· 3) Ã· 3}]

Solution:

45 â€“ [38 â€“ {60 Ã· 3 â€“ (6 â€“ 9 Ã· 3) Ã· 3}]

It can be written as

= 45 â€“ [38 â€“ {60 Ã· 3 â€“ (6 â€“ 3) Ã· 3}]

By further calculation

= 45 â€“ [38 â€“ {20 â€“ 3 Ã· 3}]

So we get

= 45 â€“ [38 â€“ {20 â€“ 1}]

By subtraction

= 45 â€“ [38 â€“ 19]

= 45 â€“ 19

= 26

8. 17 â€“ [17 â€“ {17 â€“ (17 – )}]

Solution:

17 â€“ [17 â€“ {17 â€“ (17 –
)}]

It can be written as

= 17 â€“ [17 â€“ {17 â€“ (17 â€“ 0)}]

By further calculation

= 17 â€“ [17 â€“ {17 â€“ 17}]

So we get

= 17 â€“ [17 â€“ 0]

= 17 â€“ 17

= 0

9. 2550 â€“ [510 â€“ {270 â€“ (90 – )}]

Solution:

2550 â€“ [510 â€“ {270 â€“ (90 –
)}]

It can be written as

= 2550 â€“ [510 â€“ {270 â€“ (90 â€“ 87)}]

By further calculation

= 2550 â€“ [510 â€“ {270 â€“ 3}]

So we get

= 2550 â€“ [510 â€“ 267]

= 2550 â€“ 243

= 2307

10. 30 + [{-2 Ã— (25 – )}]

Solution:

30 + [{-2 Ã— (25 –
)}]

It can be written as

= 30 + [{-2 Ã— (25 â€“ 10)}]

By further calculation

= 30 + [{-2 Ã— 15}]

So we get

= 30 + [-30]

= 30 â€“ 30

= 0

11. 88 â€“ {5 â€“ (-48) Ã· (-16)}

Solution:

88 â€“ {5 â€“ (-48) Ã· (-16)}

It can be written as

= 88 â€“ {5 â€“ (- 48/ -16)}

By further calculation

= 88 â€“ {5 â€“ 3}

So we get

= 88 â€“ 2

= 86

12. 9 Ã— (8 – ) â€“ 2 (2 + )

Solution:

9 Ã— (8 –
) â€“ 2 (2 +
)

It can be written as

= 9 Ã— (8 â€“ 5) â€“ 2 (2 + 6)

By further calculation

= 9 Ã— 3 â€“ 2 Ã— 8

So we get

= 27 â€“ 16

= 11

13. 2 â€“ [3 â€“ {6 â€“ (5 – )}]

Solution:

2 â€“ [3 â€“ {6 â€“ (5 –
)}]

It can be written as

= 2 â€“ [3 â€“ {6 â€“ (5 â€“ 1)}]

By further calculation

= 2 â€“ [3 â€“ {6 â€“ 4}]

So we get

= 2 â€“ [3 â€“ 2]

= 2 â€“ 1

= 1