Integers

integers class 7 selina

Step by Step solutions of Concise Mathematics ICSE Class-7 Maths chapter 1- Integers by Selina is provided

Table of Contents

Exercise: 1-B

Q1: Divide

i. 117 ÷ 9 = 13

ii. (-117) ÷ 9 = -13

iii. 117 ÷ (-9) = -13

iv. (-117) ÷ (-9) = 13

v. 225 ÷ (-15) = -15

vi. (-552) ÷ 24 = -23

vii. (-798) ÷ (-21) = 38

viii. (-910) ÷ 26 = -35


Q2: Evaluate

i. (-234) ÷ 13 = -18

ii. 234 ÷ (-13) = -18

iii. (-234) ÷ (-13) = 18

iv. 374 ÷ (-17) = -22

v. (-374) ÷ 17 = -22

vi. (-374) ÷ (-17) = 22

vii. (-728) ÷ 14 = -52

viii. 272 ÷ (-17) = -16


Q3: Find the quotient in each of the following divisions:

i. 299 ÷ 23 = 13

ii. 299 ÷ (-23) = -13

iii. (-384) ÷ 16 = -24

iv. (-572) ÷ (-22) = 26

v. 408 ÷ (-17) = -24


Q4: Divide

i. 204 ÷ 17 = 12

ii. 152 ÷ (-19) = -8

iii. 0 ÷ 35 = 0

iv. 0 ÷ (-82) = 0

v. 5490 ÷ 10 = 549

vi. 762800 ÷ 100 = 7628


Q5: State, true or false :

i. 0 ÷ 32 = 0 → True

ii. 0 ÷ (-9) = 0 → True

iii. (-37) ÷ 0 = 0 → False

iv. 0 ÷ 0 = 0 → False


Q6: Evaluate

i. 42 ÷ 7 + 4
= 6 + 4 = 10

ii. 12 + 18 ÷ 3
= 12 + 6 = 18

iii. 19 – 20 ÷ 4
= 19 – 5 = 14

iv. 16 – 5 × 3 + 4
= 16 – 15 + 4 = 5

v. 6 – 8 – (-6) ÷ 2
= 6 – 8 + 3 = 1

vi. 13 – 12 ÷ 4 × 2
= 13 – 3 × 2 = 13 – 6 = 7

vii. 16 + 8 ÷ 4 – 2 × 3
= 16 + 2 – 6 = 12

viii. 16 ÷ 8 + 4 – 2 × 3
= 2 + 4 – 6 = 0

ix. 16 – 8 + 4 ÷ 2 × 3
= 8 + 2 × 3 = 8 + 6 = 14

x. (-4) + (-12) ÷ (-6)
= -4 + 2 = -2

xi. (-18) + 6 ÷ 3 + 5
= -18 + 2 + 5 = -11

xii. (-20) × (-1) + 14 ÷ 7
= 20 + 2 = 22



Q7: The product of two integers is -90. If one of them is 6, find the other integer.

Product = -90, one number = 6
⇒ other number = -90 ÷ 6 = -15



Q8: The product of two integers is 210. If one of them is -15, find the other.

Product = 210, one number = -15
⇒ other number = 210 ÷ (-15) = -14


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