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.
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Access Selina Solutions Concise Maths Class 7 Chapter 1: Integers Exercise 1C
Evaluate:
1. 18 – (20 – 15 ÷ 3)
Solution:
18 – (20 – 15 ÷ 3)
It can be written as
= 18 – (20 – 15/3)
By further calculation
= 18 – (20 – 5)
= 18 – 20 + 5
So we get
= 18 + 5 – 20
= 23 – 20
= 3
2. – 15 + 24 ÷ (15 – 13)
Solution:
– 15 + 24 ÷ (15 – 13)
It can be written as
= – 15 + 24 ÷ 2
So we get
= – 15 + 12
= – 3
3. 35 – {15 + 14 – (13 + 
)}
Solution:
35 – {15 + 14 – (13 +
)}
It can be written as
= 35 – [15 + 14 – (13 + 4)]
By further calculation
= 35 – [15 + 14 – 17]
Multiplying the negative sign
= 35 – 15 – 14 + 17
So we get
= 52 – 29
= 23
4. 27 – {13 + 4 – (8 + 4 – 
)}
Solution:
27 – {13 + 4 – (8 + 4 –
)}
It can be written as
= 27 – {13 + 4 – (8 + 4 – 4)}
By further calculation
= 27 – {13 + 4 – 8}
So we get
= 27 – {13 + (-4)}
= 27 – 9
= 18
5. 32 – [43 – {51 – (20 – 
)}]
Solution:
32 – [43 – {51 – (20 –
)}]
It can be written as
= 32 – [43 – {51 – (20 – 11)}]
By further calculation
= 32 – [43 – {51 – 9}]
So we get
= 32 – [43 – 42]
= 32 – 1
= 31
6. 46 – [26 – {14 – (15 – 4 ÷ 2 × 2)}]
Solution:
46 – [26 – {14 – (15 – 4 ÷ 2 × 2)}]
It can be written as
= 46 – [26 – {14 – (15 – 2 × 2)}]
By further calculation
= 46 – [26 – {14 – (15 – 4)}]
So we get
= 46 – [26 – {14 – 11}]
= 46 – [26 – 3]
Here
= 46 – 23
= 23
7. 45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]
Solution:
45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]
It can be written as
= 45 – [38 – {60 ÷ 3 – (6 – 3) ÷ 3}]
By further calculation
= 45 – [38 – {20 – 3 ÷ 3}]
So we get
= 45 – [38 – {20 – 1}]
By subtraction
= 45 – [38 – 19]
= 45 – 19
= 26
8. 17 – [17 – {17 – (17 – 
)}]
Solution:
17 – [17 – {17 – (17 –
)}]
It can be written as
= 17 – [17 – {17 – (17 – 0)}]
By further calculation
= 17 – [17 – {17 – 17}]
So we get
= 17 – [17 – 0]
= 17 – 17
= 0
9. 2550 – [510 – {270 – (90 – 
)}]
Solution:
2550 – [510 – {270 – (90 –
)}]
It can be written as
= 2550 – [510 – {270 – (90 – 87)}]
By further calculation
= 2550 – [510 – {270 – 3}]
So we get
= 2550 – [510 – 267]
= 2550 – 243
= 2307
10. 30 + [{-2 × (25 – 
)}]
Solution:
30 + [{-2 × (25 –
)}]
It can be written as
= 30 + [{-2 × (25 – 10)}]
By further calculation
= 30 + [{-2 × 15}]
So we get
= 30 + [-30]
= 30 – 30
= 0
11. 88 – {5 – (-48) ÷ (-16)}
Solution:
88 – {5 – (-48) ÷ (-16)}
It can be written as
= 88 – {5 – (- 48/ -16)}
By further calculation
= 88 – {5 – 3}
So we get
= 88 – 2
= 86
12. 9 × (8 – 
) – 2 (2 + 
)
Solution:
9 × (8 –
) – 2 (2 +
)
It can be written as
= 9 × (8 – 5) – 2 (2 + 6)
By further calculation
= 9 × 3 – 2 × 8
So we get
= 27 – 16
= 11
13. 2 – [3 – {6 – (5 – 
)}]
Solution:
2 – [3 – {6 – (5 –
)}]
It can be written as
= 2 – [3 – {6 – (5 – 1)}]
By further calculation
= 2 – [3 – {6 – 4}]
So we get
= 2 – [3 – 2]
= 2 – 1
= 1
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