Question

Evaluate each of the following:
(a) (–30) ÷ 10
(b) 50 ÷ (–5)
(c) (–36) ÷ (–9)
(d) (– 49) ÷ (49)
(e) 13 ÷ [(–2) + 1]
(f) 0 ÷ (–12)
(g) (–31) ÷ [(–30) + (–1)]
(h) [(–36) ÷ 12] ÷ 3
(i) [(– 6) + 5)] ÷ [(–2) + 1]
NCERT Class 7th Mathematics Chapter 1 Integers

Answer 1

(a) (–30) ÷ 10 
= -3

(b) 50 ÷ (–5) 
= -10

(c) (–36) ÷ (–9)
= 4

(d) (– 49) ÷ (49) 
= -1

(e) 13 ÷ [(–2) + 1] 
=  13 ÷ [-1] 
= -13

(f) 0 ÷ (–12)
= 0

(g) (–31) ÷ [(–30) + (–1)]
= (–31) ÷ (–31)
= 1

(h) [(–36) ÷ 12] ÷ 3 
= (–3) ÷ 3 
= -1

(i) [(– 6) + 5)] ÷ [(–2) + 1]
= (-1) ÷ (-1)
= 1

Answer 2

Evaluation of each of the following are as follows:

(a) (–30) ÷ 10 = - 3

(b) 50 ÷ (–5) = - 10

(c) (–36) ÷ (–9) = 4

(d) (– 49) ÷ (49) = -1

(e) 13 ÷ [(–2) + 1] = -13

(f) 0 ÷ (–12) = 0

(g) (–31) ÷ [(–30) + (–1)] = 1

(h) [(–36) ÷ 12] ÷ 3= -1

(i) [(– 6) + 5)] ÷ [(–2) + 1] = 1

Concept used:

Simplification:

An expression involving integers (i.e., directed numbers) is simplified by the rule of BODMAS.

1. First, carry out the operations inside the brackets (B).

Brackets are removed in the order: line bracket (or vinculum), first (or common) brackets, second (or curly) brackets, third (or rectangular

or square) brackets.

2. Then change 'of' into 'x' and multiply (O).

3. Carry out the operations of division and multiplication, in that order (DM).

4. Finally, carry out the operations of addition and subtraction (AS).

Rules for the addition of integers:

• If we add the two integers with the same sign (+,+)(-,-), add the integers, and in the answer put the sign given with the numbers.

(+4) + (+2) = (+6)

(-8) + (-4) =( -12)

• If we add the two integers with different signs (+,-) (-,+), subtract the absolute value, and in the answer put the sign of the largest absolute value.

(+10) + (-5) = (+5)

(-8) + (+4) =( -4)

Rules for division of integers:

• If both the integers have the same sign (+,+)(-,-), then the sign in the answer is positive (+).
• If the integers have different signs (+,-) (-,+), then the sign in the answer is negative (-).

Detail solution:

(a) Given = (- 30) ÷ 10

= $\frac{-30} {10}$

(- 30) ÷ 10 = -3

(b) Given = 50 ÷ (–5)

= $\frac {50}{-5}$

50 ÷ (–5) = -10

(c) Given = (–36) ÷ (–9)

= $\frac {-36}{-9}$

(–36) ÷ (–9) = 4

(d) Given = (– 49) ÷ (49)

= $\frac {- 49}{49}$

(– 49) ÷ (49) = -1

(e) Given = 13 ÷ [(–2) + 1]

 = 13 ÷ [- 2 + 1]

 =  13 ÷ [-1]

= $\frac {13}{-1}$

13 ÷ [(–2) + 1] = -13

(f) Given = 0 ÷ (–12)

= $\frac {0}{-12}$

0 ÷ (–12) = 0

(g) Given = (–31) ÷ [(–30) + (–1)]

= (- 31) ÷ [ - 30 - 1]

= (- 31) ÷ (- 31)

= $\frac {- 31}{-31}$

(–31) ÷ [(–30) + (–1)] = 1

(h) Given =  [(–36) ÷ 12] ÷ 3

=  [- 3] ÷ 3

= $\frac {- 3}{3}$

[(–36) ÷ 12] ÷ 3 = -1

(i) Given = [(– 6) + 5)] ÷ [(–2) + 1]

= [ -  6 + 5] ÷ [ - 2 + 1]

= -1 ÷ -1

= $\frac {- 1}{- 1}$

[(– 6) + 5)] ÷ [(–2) + 1] = 1

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