Selina Concise Class 7 Math Chapter 1 Integer Exercise 1C Solution
EXERCISE 1C
Evaluate:
(1) 18 – (20 – 15 ÷ 3)
= 18 – (20 – 5)
= 18 – 15
= 3
(2) – 15 + 24 ÷ (15 – 13)
= – 15 + 24 ÷ 2
= – 15 + 12
= – 3
(3) 35 – [15 + 14 – {13 + (2 – 1 + 3)}]
= 35 – [15 + 14 – {13 + (1 + 3)}]
= 35 – [15 + 14 – (13 + 4)]
= 35 – [15 + 14 – 17]
= 35 – [15 – 3]
= 35 – 12
= 23
(4) 27 – [13 + 4 – {8 + 4 – (1 + 3)}]
= 27 – [13 + 4 – {8 + 4 – 4}]
= 27 – [13 + 4 – {12 – 4}]
= 27 – [13 + 4 – 8]
= 27 – (17 – 8)
= 27 – 9
= 18
(5) 32 – [43 – {51 – (20 – (18 – 7))}]
= 32 – [43 – {51 – (20 – 11)}]
= 32 – [43 – {51 – 9}]
= 32 – [43 – 42]
= 32 – 1
= 31
(6) 46 – [26 – {14 – (15 – 4 ÷ 2 × 2)}]
= 46 – [26 – {14 – (15 – 2 × 2)}]
= 46 – [26 – {14 – (15 – 4)}]
= 46 – [26 – {14 – 11}]
= 46 – [26 – 3]
= 46 – 23
= 23
(7) 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
(8) 17 – [17 – {17 – (17 – (17 – 17))}]
= 17 – [17 – {17 – (17 – 0)}]
= 17 – [17 – {17 – 17}]
= 17 – [17 – 0]
= 17 – 17
= 0
(9) 2550 – [510 – {270 – (90 – (80 + 7))}]
= 2550 – [510 – {270 – (90 – 87)}]
= 2550 – [510 – {270 – 3}]
= 2550 – [510 – 267]
=2550 – 243
= 2307
(10) 30 + [{- 2 × (25 – (13 – 3))}]
= 30 + [{- 2 × (25 – 10)}]
= 30 + [{- 2 × 15}]
= 30 – 30
= 0
(11) 88 – [5 – {(- 48) ÷ (- 16)}]
= 88 – [5 – 3]
= 88 – 2
= 86
(12) 9 × {(8 – (3 + 2)} – 2{2 + (3 + 3)}
= 9 × {8 – 5} – 2 × {2 + 6}
= 9 × 3 – 2 × 8
= 27 – 16
= 11
(13) 2 – [3 – {6 – (5 – (4 – 3))}]
= 2 – [3 – {6 – (5 – 1)}]
= 2 – [3 – {6 – 4}]
= 2 – [3 – 2]
= 2 – 1
= 1
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