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) (3x + 2y) − (x + y)
(ii) (3x + 2y) − (x − y)
(iii) (x + y) − (x − y)
(iv) (x − y) − (x + y)
(v) (4x − 3y) − (2x − 5x)
Answer:
(i)
(3x + 2y) − (x + y)
= (3x + 2y − x) − y {x − (y + z) = (x − y) − z for all numbers x, y, z}
= (3x − x + 2y) − y
= 2x + 2y − y
= 2x + y
(ii)
(3x + 2y) − (x − y)
= 3x + 2y − x + y {x − (y − z) = x − y + z for all numbers x, y, z}
= 3x − x + 2y + y
= 2x + 3y
(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
= 2y
(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
= −2y
(v)
(4x − 3y) − (2x − 5x)
= 4x − 3y − (−3x)
= 4x − 3y + 3x
= 4x + 3x − 3y
= 7x − 3y
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