Mathematics Part i Solutions Solutions for Class 8 Math Chapter 3 Negative Numbers are provided here with simple step-by-step explanations. These solutions for Negative Numbers are extremely popular among class 8 students for Math Negative Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part i Solutions Book of class 8 Math Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part i Solutions Solutions. All Mathematics Part i Solutions Solutions for class 8 Math are prepared by experts and are 100% accurate.

#### Page No 49:

#### Question 1:

Can you work the multiplication out in your head?

(i) 3 × (−4)

(ii) 4 × (−3)

(iii)

(iv)

(v)

(vi) 3 × (−0.3)

(vii) 20 × (−0.4)

(viii) 9 × (−0.1)

#### Answer:

(i)

3 × (−4) = −(3 × 4)

= −12

(ii)

4 × (−3) = −(4 × 3)

= −12

(iii)

2 × = −

= −1

(iv)

6 × = −

= −2

(v)

20 × = −

= −4

(vi)

5 × (−0.3) = −(5 × 0.3)

=

=

= −1.5

(vii)

20 × (−0.4) = −(20 × 0.4)

=

= −8

(viii)

9 × (−0.1) = −(9 × 0.1)

=

= −0.9

#### Page No 49:

#### Question 2:

Some problems to do in your head.

(i)

(ii)

(iii)

(iv) 0.5 × (−0.3)

#### Answer:

(i)

(ii)

(iii)

× =

=

= −

(iv)

0.5 × (−0.3) = −(0.5 × 0.3)

= −0.15

#### Page No 50:

#### Question 1:

Using the rule, can’t you do these in your head?

(i) (−3) × 4

(ii) (−4) × 7

(iii) (−1) × 9

(iv) (−5) × 4

(v) (−2) × 5

(vi)

(vii)

(viii)

(ix)

(x) (−0.5) × 0.2

(xi) (−0.4) × 0.3

#### Answer:

(i)

(−3) × 4 = −(3 × 4) {(−*x*) × *y* = −(*xy*), for all positive numbers *x* and *y*}

= −12

(ii)

(−4) × 7 = −(4 × 7) {(−*x*) × *y* = −(*xy*), for all positive numbers *x* and *y*}

= −28

(iii)

(−1) × 9 = −(1 × 9) {(−*x*) × *y* = −(*xy*), for all positive numbers *x* and *y*}

= −9

(iv)

(−5) × 4 = −(5 × 4) {(−*x*) × *y* = −(*xy*), for all positive numbers *x* and *y*}

= −20

(v)

(−2) × 5 = −(2 × 5) {(−*x*) × *y* = −(*xy*), for all positive numbers *x* and *y*}

= −10

(vi)

× = − {(−*x*) × *y* = −(*xy*), for all positive numbers *x* and *y*}

= −1

(vii)

× = _{ }{(−*x*) × *y* = −(*xy*), for all positive numbers *x* and *y*}

=

= −

(viii)

× = _{ }{(−*x*) × *y* = −(*xy*), for all positive numbers *x* and *y*}

=

= −

(ix)

× = _{ }{(−*x*) × *y* = −(*xy*), for all positive numbers *x* and *y*}

=

= −

(x)

(−0.5) × 0.2 = −(0.5 × 0.2) {(−*x*) × *y* = −(*xy*), for all positive numbers *x* and *y*}

= −0.1

(xi)

(−0.4) × 0.3 = −(0.4 × 0.3) {(−*x*) × *y* = −(*xy*), for all positive numbers *x* and *y*}

= −0.12

#### Page No 51:

#### Question 1:

Can’t you do these in your head?

(i) (13 × 8) + (7 × 8)

(ii) (15 × 9) + (5 × 9)

(iii)

(iv)

#### Answer:

(i)

(13 × 8) + (7 × 8) = (13 + 7) × 8

= 20 × 8

= 160

(ii)

(15 × 9) + (5 × 9) = (15 + 5) × 9

= 20 × 9

= 180

(iii)

+ =

=

(iv)

+ =

= 1

#### Page No 52:

#### Question 1:

Do these in your head:

(i) (12 × (−7)) + (8 × (−7))

(ii) (29 × (−2)) + (21 × (−2))

(iii)

(iv)

(v) (1.5 × (−9)) + (0.5 × (−9))

(vi) (3.4 × (−7)) + (0.6 × (−7))

(vii) (4 × (−0.19)) + (6 × (−0.19))

#### Answer:

(i)

(12 × (−7)) + (8 × (−7)) = (12 + 8) × (−7)

= 20 × (−7)

= −140

(ii)

(29 × (−2)) + (21 × (−2)) = (29 + 21) × (−2)

= 50 × (−2)

= −100

(iii)

+ =

_{= 1 × (−17)}

= −17

(iv)

+ =

= 2 × (−5)

= −10

(v)

(1.5 × (−9)) + (0.5 × (−9)) = (1.5 + 0.5) × (−9)

= 2 × (−9)

= −18

(vi)

(3.4 × (−7)) × (0.6 × (−7)) = (3.4 + 0.6) × (−7)

= 4 × (−7)

= −28

(vii)

(4 × (−0.19)) + (6 × (−0.19)) = (4 + 6) × (−0.19)

= 10 × (−0.19)

= −1.9

#### Page No 54:

#### Question 1:

Let’s find out

(i) (−2) × (−5)

(ii) (−5) × (−5)

(iii) (−1) × (−1)

(iv)

(v)

(vi) (−2) × (−5)

(vii) (−12) ÷ (−4)

(viii) (−12) ÷ (−12)

(ix)

(x)

(xi)

(xii) −2.5 × − 0.2

#### Answer:

(i)

(−2) × (−5) = 2 × 5 {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

= 10

(ii)

(−5) × (−5) = 5 × 5 {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

= 25

(iii)

(−1) × (−1) = 1 × 1 {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

= 1

(iv)

× (−2) = {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

=

(v)

× = {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

=

=

(vi)

(−2) × (−5) = 2 × 5 {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

= 10

(vii)

(−12) ÷ (−4) = (−12) ×

= {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

= 3

(viii)

(−12) ÷ (−12) = (−12) ×

= {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

= 1

(ix)

(−2) × = _{ } {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

=

(x)

(−3) ÷ = (−3) × (−2)

= 3 × 2 {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

= 6

(xi)

(−2) ÷ = (−2) × (−3)

= 2 × 3 {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

= 6

(xii)

(−2.5) × (−0.2) = 2.5 × 0.2 {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

= 0.5

#### Page No 54:

#### Question 2:

Simplify the following.

(i) (−1) × 2 × (−3)

(ii) 1 × (−2) × 3

(iii) (−2) × (−3) × (−4)

(iv) (−1) × (−1)

(v) (−1)^{3}

(vi) (−1)^{5}

(vii) (−1)^{99}

#### Answer:

(i)

(−1) × 2 × (−3) = (−1) × (2 × (−3))

= (−1) × (−6) {*x* × (−*y*) = −(*xy*), for all positive numbers *x* and *y*}

= 1 × 6 {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

= 6

(ii)

1 × (−2) × 3 = (1 × (−2)) × 3 {*x* × (−*y*) = −(*xy*), for all positive numbers *x* and *y*}

= (−2) × 3

= −(2 × 3) {(−*x*) × *y* = −(*xy*), for all positive numbers *x* and *y*}

= −6

(iii)

(−2) × (−3) × (−4) = ((−2) × (−3)) × (−4)

= 6 × (−4) {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

= −(6 × 4) {*x* × (−*y*) = −(*xy*), for all positive numbers *x* and *y*}

= −24

(iv)

(−1) × (−1) = 1 × 1 {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

= 1

(v)

(−1)^{3} = (−1) × (−1) × (−1)

= (1 × 1) × (−1) {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

= 1 × (−1)

= −(1 × 1) {*x* × (−*y*) = −(*xy*), for all positive numbers *x* and *y*}

= −1

(vi)

(−1)^{5} = (−1) × (−1) × (−1) × (−1) × (−1)

= (1 × 1) × (−1) × (−1) × (−1) {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

= 1 × (−1) × (−1) × (−1)

= −(1 × 1) × (−1) × (−1) {*x* × (−*y*) = −(*xy*), for all positive numbers *x* and *y*}

= (−1) × (−1) × (−1)

= (1 × 1) × (−1) {(−*x*) × (−*y*) = *xy*, for all positive numbers *x* and *y*}

= 1 × (−1)

= −(1 × 1) {*x* × (−*y*) = −(*xy*), for all positive numbers *x* and *y*}

= −1

(vii)

(−1)^{99} = (−1) × (−1) × (−1) × (−1) × (−1)….. 99 times

We have calculated that** **(−1)^{3} = −1

(−1)^{5} = −1

Thus, when (−1) is raised to any odd number, the result is −1.

As 99 is an odd number, (−1)^{99 }= −1

#### Page No 60:

#### Question 1:

Try to prove the following using the general principle.

(i) (*x* − *y*) + *z* = *x* − (*y* − *z*) for all numbers *x*, *y*, *z*

(ii) (*x* + *y*) − *z* = *x* + (*y* − *z*) for all numbers *x*, *y*, *z*

#### Answer:

(i)

**To show: **(*x* − *y*) + *z *= *x* − (*y* − *z*) for all numbers *x*, *y*, *z*.

We know that (*x* + *y*) + *z *= *x* + (*y* + *z*) for all numbers *x*, *y*, *z*. ….(i)

Now, (*x* − *y*) + *z* = (*x* + (−*y*)) + *z*

= *x* + ((−*y*) + *z*) (Using (i))

= *x* + (−*y* + *z*)

= *x* − (*y* − *z*)

Thus, (*x* − *y*) + *z *= *x* − (*y* − *z*) for all numbers *x*, *y*, *z*.

(ii)

**To show: **(*x* + *y*) − *z *= *x* + (*y* − *z*) for all numbers *x*, *y*, *z*.

We know that (*x* + *y*) + *z *= *x* + (*y* + *z*) for all numbers *x*, *y*, *z*. ….(i)

Now, (*x* + *y*) − *z *= (*x* + *y*) + (−*z*)

= *x* + (*y* + (−*z*)) (Using (i))

= *x* + (*y* − *z*)

Thus, (*x* + *y*) − *z *= *x* + (*y* − *z*) for all numbers *x*, *y*, *z*.

#### Page No 60:

#### Question 2:

Try simplifying these:

(i) (3*x* + 2*y*) − (*x* + *y*)

(ii) (3*x* + 2*y*) − (*x* − *y*)

(iii) (*x* + *y*) − (*x* − *y*)

(iv) (*x* − *y*) − (*x* + *y*)

(v) (4*x* − 3*y*) − (2*x* − 5*x*)

#### Answer:

(i)

(3*x* + 2*y*) − (*x* + *y*)

= (3*x* + 2*y* − *x*) − *y *{*x* − (*y* + *z*) = (*x* − *y*) − *z* for all numbers *x*, *y*, *z*}

= (3*x* − *x* + 2*y*) − *y *

= 2*x* + 2*y* − *y*

= 2*x* + *y*

(ii)

(3*x* + 2*y*) − (*x* − *y*)

= 3*x* + 2*y* − *x* + *y *{*x* − (*y* − *z*) = *x* − *y* + *z* for all numbers *x*, *y*, *z*}

= 3*x *− *x *+ 2*y *+ *y *

= 2*x* + 3*y*

(iii)

(*x* + *y*) − (*x* − *y*)

= *x* + *y* − *x* + *y *{*x* − (*y* − *z*) = *x* − *y* + *z* for all numbers *x*, *y*, *z*}

= *x* − *x* + *y* + *y *

= 2*y*

(iv)

(*x* − *y*) − (*x* + *y*)

= (*x* − *y* − *x*) − *y *{*x* − (*y* + *z*) = (*x* − *y*) − *z* for all numbers *x*, *y*, *z*}

= (*x* − *x* − *y*) − *y *

= −*y *− *y *

= −2*y *

(v)

(4*x* − 3*y*) − (2*x* − 5*x*)

= 4*x* − 3*y* − (−3*x*)

= 4*x* − 3*y* + 3*x*

= 4*x* + 3*x* − 3*y*

= 7*x* − 3*y*

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