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 Class 8 Math are prepared by experts and are 100% accurate.

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)

(i)

3 × (4) = (3 × 4)

= 12

(ii)

4 × (3) = (4 × 3)

= 12

(iii)

2 × = = 1

(iv)

6 × = =

(v)

20 × = =

(vi)

5 × (0.3) = (5 × 0.3)

= = = 1.5

(vii)

20 × (0.4) = (20 × 0.4)

= =

(viii)

9 × (0.1) = (9 × 0.1)

= = 0.9

Question 2:

Some problems to do in your head.

(i) (ii) (iii) (iv) 0.5 × (−0.3)

(i) (ii) (iii) × = = = (iv)

0.5 × (0.3) = (0.5 × 0.3)

= 0.15

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

(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}

=

(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

Question 1:

Can’t you do these in your head?

(i) (13 × 8) + (7 × 8)

(ii) (15 × 9) + (5 × 9)

(iii) (iv) (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

Question 1:

(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))

(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

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

(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

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

(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

Question 1:

Try to prove the following using the general principle.

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

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

(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.

Question 2:

Try simplifying these:

(i) (3x + 2y) − (x + y)

(ii) (3x + 2y) − (xy)

(iii) (x + y) − (xy)

(iv) (xy) − (x + y)

(v) (4x − 3y) − (2x − 5x)

(i)

(3x + 2y) (x + y

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

= (3x x + 2y)

= 2x + 2y y

= 2x + y

(ii)

(3x + 2y) (x y

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

= 3x x + 2y +

= 2x + 3y

(iii)

(x + y) (x y

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

= x x + y +

= 2y

(iv)

(x y) (x + y

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

= (x x y)

= y

= 2

(v)

(4x 3y) (2x 5x

= 4x 3y (3x

= 4x 3y + 3x

= 4x + 3x 3y

= 7x 3y

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