**RS Aggarwal Solutions Class 7 Maths Chapter 1 Ex 1.1: **Begin your Class 7 exam preparation with RS aggarwal Solutions Class 7. You can refer to RS Aggarwal Solutions Class 7 Maths Chapter 1 Ex 1.1 for all the accurate and well-explained solutions designed by the subject matter experts. They are designed as per the current CBSE Syllabus.

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## Download RS Aggarwal Solutions Class 7 Maths Chapter 1 Ex 1.1 PDF

RS Aggarwal Solutions Class 7 Maths Chapter 1 Ex 1.1

## RS Aggarwal Solutions Class 7 Maths Chapter 1 E 1.1

**Question 1.****Solution:**

(i) 15 + (-8) =15 – 8 = 7

(ii) (-16) +9 = -7

(iii) (-7) + (-23)= -7 – 23 = -30

(iv) (-32) + 47 = -32 + 47 = 15

(v) 53 + (-26) = 53 – 26 = 27

(vi) (-48) + (-36) = -48 – 36 = -84

**Question 2.****Solution:**

(i) 153 and -302 = 153 + (-302) = 153 – 302 = -149

(ii) 1005 and -277 = 1005 + (-277) = 1005 – 277 = 728

(iii) -2035 and 297 = -2035 + 297 = – 1738

(iv) -489 and -324 = -489 + (-324) = -489 – 324 = -813

(v) -1000 and438 = -1000 + 438 = -562

(vi) -238 and 500 = -238 + 500 = 262

**Question 3.****Solution:**

Additive inverse of

(i) -83 is – (-83) = 83

(ii) 256 is -256

(iii) 0 is 0

(iv) -2001 is – (-2001) = 2001

**Question 4.****Solution:**

(i) 28 from – 42 = -42 – (28) = -42 – 28 = -70

(ii) – 36 from 42 = 42 – (-36) = 42 + 36 = 78

(iii) -37 from -53 = -53 – (-37) = -53 + 37= -16

(iv) -66 from -34 = -34 – (-66) = -34 + 66 = 32

(v) 318 from 0 = 0 – (318) = -318

(vi) -153 from -240= -240 – (-153) = -240 + 153 = -87

(vii) -64 from 0 = 0 – (-64) = 0 + 64 = 64

(viii) – 56 from 144 = 144 – (-56) = 144 + 56 = 200

**Question 5.****Solution:**

– 34 – (-1032 + 878)

= -34 – (-154) = -34 + 154 = 120

**Question 6.****Solution:**

38 + (-87) – 134

= (38 – 87) – 134

= -49 – 134 = -183

**Question 7.****Solution:**

(i) {(-13) + 27} + (-41) = (-13) + {27 + (-41)} (By Associative Law of Addition)

(ii) (-26) + {(-49) + (-83)} = {(-26) + (-49)} +(-83) (By Associative Law of Addition)

(iii) 53 + (-37) = (-37) + (53) (By Commutative Law of Addition)

(iv) (-68) + (-76) = (-76) + (-68) (By Commutative Law of Addition)

(v) (-72) + (0) = -72 (Existence of Additive identity)

(vi) -(-83) = 83

(vii) (-60) – (………) = -59 => -60 – (-1) = -59

(viii) (-31) + (……….) = -40 => -31 + (-9) = -40

**Question 8.****Solution:**

{-13 – (-27)} + {-25 – (-40)}

= {-13 + 27} + {-25 + 40}

= 14 + 15 = 29

**Question 9.****Solution:**

36 – (- 64) = 36 + 64 = 100

(-64) – 36= -64 – 36 = -100

They are not equal

**Question 10.****Solution:**

(a + b) + c = {-8 + (-7)} + 6 = (-8 – 7) + 6 = -15 + 6 = -9

and a + (b + c) = -8 + (-7 + 6) = -8 + (-1) = -8 – 1 = -9 Hence proved

**Question 11.****Solution:**

LHS = (a -b) = -9 – (-6) = -9 + 6 = -3

RHS = (b – a) = -6 – (-9) = -6 + 9 = 3

LHS ≠ RHS.

Hence (a – b) ≠ (b – a)

**Question 12.****Solution:**

Sum of two integers = -16

One integer = 53

Second integer = -16 – (53) = -16 – 53 = (-69)

**Question 13.****Solution:**

Sum of two integers = 65

One integer = -31

Second integer = 65 – (-31) = 65 + 31 = 96

**Question 14.****Solution:**

Difference of a and (-6) = 4

a – (-6) = 4

⇒ a + 6 = 4

⇒ a = 4 – 6

⇒ a = -2

**Question 15.****Solution:**

(i) We can write any two integers having opposite signs

e.g. 5, -5

Sum = 5 + (-5) = 5 – 5 = 0

(ii) The sum is a negative integer

The greater integer must be negative and smaller integer be positive

e.g. -9, 6

Sum = -9 + 6 = -3

(iii) The sum is smaller than the both integers

Both integer will be negative -4, -6

Sum = -4 + (-6) = -4 – 6 = -10

(iv) The sum is greater than the both integers

Both integers will be positive

e.g. 6, 4

(v) The sum oftwo integers is smaller than one of these integers

The greater number will be positive and smaller be negative

e.g. 6, -4

Sum = 6 + (-4) = 2

**Question 16.****Solution:**

(i) False: Because, all negative integers are less than zero.

(ii) False: -10 is less than -7.

(iii) Tme: Every negative integer is less than zero.

(iv) True : Sum of two negative integers is negative.

(v) False: It is not always true.

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