# 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

Updated: March 6, 2019 — 3:12 pm

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