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 Ã (â4)
(ii) 4 Ã (â3)
(iii)
(iv)
(v)
(vi) 3 Ã (â0.3)
(vii) 20 Ã (â0.4)
(viii) 9 Ã (â0.1)
Answer:
(i)
3 Ã (â4) = â(3 Ã 4)
= â12
(ii)
4 Ã (â3) = â(4 Ã 3)
= â12
(iii)
2 Ã = â
= â1
(iv)
6 Ã = â
= â2
(v)
20 Ã = â
= â4
(vi)
5 Ã (â0.3) = â(5 Ã 0.3)
=
=
= â1.5
(vii)
20 Ã (â0.4) = â(20 Ã 0.4)
=
= â8
(viii)
9 Ã (â0.1) = â(9 Ã 0.1)
=
= â0.9
Page No 49:
Question 2:
Some problems to do in your head.
(i)
(ii)
(iii)
(iv) 0.5 Ã (â0.3)
Answer:
(i)
(ii)
(iii)
à =
=
= â
(iv)
0.5 Ã (â0.3) = â(0.5 Ã 0.3)
= â0.15
Page No 50:
Question 1:
Using the rule, canât you do these in your head?
(i) (â3) Ã 4
(ii) (â4) Ã 7
(iii) (â1) Ã 9
(iv) (â5) Ã 4
(v) (â2) Ã 5
(vi)
(vii)
(viii)
(ix)
(x) (â0.5) Ã 0.2
(xi) (â0.4) Ã 0.3
Answer:
(i)
(â3) Ã 4 = â(3 Ã 4) {(âx) Ã y = â(xy), for all positive numbers x and y}
= â12
(ii)
(â4) Ã 7 = â(4 Ã 7) {(âx) Ã y = â(xy), for all positive numbers x and y}
= â28
(iii)
(â1) Ã 9 = â(1 Ã 9) {(âx) Ã y = â(xy), for all positive numbers x and y}
= â9
(iv)
(â5) Ã 4 = â(5 Ã 4) {(âx) Ã y = â(xy), for all positive numbers x and y}
= â20
(v)
(â2) Ã 5 = â(2 Ã 5) {(âx) Ã y = â(xy), for all positive numbers x and y}
= â10
(vi)
à = â {(âx) à y = â(xy), for all positive numbers x and y}
= â1
(vii)
à = {(âx) à y = â(xy), for all positive numbers x and y}
=
= â
(viii)
à = {(âx) à y = â(xy), for all positive numbers x and y}
=
= â
(ix)
à = {(âx) à y = â(xy), for all positive numbers x and y}
=
= â
(x)
(â0.5) Ã 0.2 = â(0.5 Ã 0.2) {(âx) Ã y = â(xy), for all positive numbers x and y}
= â0.1
(xi)
(â0.4) Ã 0.3 = â(0.4 Ã 0.3) {(âx) Ã y = â(xy), for all positive numbers x and y}
= â0.12
Page No 51:
Question 1:
Canât you do these in your head?
(i) (13 Ã 8) + (7 Ã 8)
(ii) (15 Ã 9) + (5 Ã 9)
(iii)
(iv)
Answer:
(i)
(13 Ã 8) + (7 Ã 8) = (13 + 7) Ã 8
= 20 Ã 8
= 160
(ii)
(15 Ã 9) + (5 Ã 9) = (15 + 5) Ã 9
= 20 Ã 9
= 180
(iii)
+ =
=
(iv)
+ =
= 1
Page No 52:
Question 1:
Do these in your head:
(i) (12 Ã (â7)) + (8 Ã (â7))
(ii) (29 Ã (â2)) + (21 Ã (â2))
(iii)
(iv)
(v) (1.5 Ã (â9)) + (0.5 Ã (â9))
(vi) (3.4 Ã (â7)) + (0.6 Ã (â7))
(vii) (4 Ã (â0.19)) + (6 Ã (â0.19))
Answer:
(i)
(12 Ã (â7)) + (8 Ã (â7)) = (12 + 8) Ã (â7)
= 20 Ã (â7)
= â140
(ii)
(29 Ã (â2)) + (21 Ã (â2)) = (29 + 21) Ã (â2)
= 50 Ã (â2)
= â100
(iii)
+ =
= 1 Ã (â17)
= â17
(iv)
+ =
= 2 Ã (â5)
= â10
(v)
(1.5 Ã (â9)) + (0.5 Ã (â9)) = (1.5 + 0.5) Ã (â9)
= 2 Ã (â9)
= â18
(vi)
(3.4 Ã (â7)) Ã (0.6 Ã (â7)) = (3.4 + 0.6) Ã (â7)
= 4 Ã (â7)
= â28
(vii)
(4 Ã (â0.19)) + (6 Ã (â0.19)) = (4 + 6) Ã (â0.19)
= 10 Ã (â0.19)
= â1.9
Page No 54:
Question 1:
Letâs find out
(i) (â2) Ã (â5)
(ii) (â5) Ã (â5)
(iii) (â1) Ã (â1)
(iv)
(v)
(vi) (â2) Ã (â5)
(vii) (â12) ÷ (â4)
(viii) (â12) ÷ (â12)
(ix)
(x)
(xi)
(xii) â2.5 Ã â 0.2
Answer:
(i)
(â2) Ã (â5) = 2 Ã 5 {(âx) Ã (ây) = xy, for all positive numbers x and y}
= 10
(ii)
(â5) Ã (â5) = 5 Ã 5 {(âx) Ã (ây) = xy, for all positive numbers x and y}
= 25
(iii)
(â1) Ã (â1) = 1 Ã 1 {(âx) Ã (ây) = xy, for all positive numbers x and y}
= 1
(iv)
à (â2) = {(âx) à (ây) = xy, for all positive numbers x and y}
=
(v)
à = {(âx) à (ây) = xy, for all positive numbers x and y}
=
=
(vi)
(â2) Ã (â5) = 2 Ã 5 {(âx) Ã (ây) = xy, for all positive numbers x and y}
= 10
(vii)
(â12) ÷ (â4) = (â12) Ã
= {(âx) Ã (ây) = xy, for all positive numbers x and y}
= 3
(viii)
(â12) ÷ (â12) = (â12) Ã
= {(âx) Ã (ây) = xy, for all positive numbers x and y}
= 1
(ix)
(â2) Ã = {(âx) Ã (ây) = xy, for all positive numbers x and y}
=
(x)
(â3) ÷ = (â3) à (â2)
= 3 Ã 2 {(âx) Ã (ây) = xy, for all positive numbers x and y}
= 6
(xi)
(â2) ÷ = (â2) à (â3)
= 2 Ã 3 {(âx) Ã (ây) = xy, for all positive numbers x and y}
= 6
(xii)
(â2.5) Ã (â0.2) = 2.5 Ã 0.2 {(âx) Ã (ây) = xy, for all positive numbers x and y}
= 0.5
Page No 54:
Question 2:
Simplify the following.
(i) (â1) Ã 2 Ã (â3)
(ii) 1 Ã (â2) Ã 3
(iii) (â2) Ã (â3) Ã (â4)
(iv) (â1) Ã (â1)
(v) (â1)3
(vi) (â1)5
(vii) (â1)99
Answer:
(i)
(â1) Ã 2 Ã (â3) = (â1) Ã (2 Ã (â3))
= (â1) à (â6) {x à (ây) = â(xy), for all positive numbers x and y}
= 1 Ã 6 {(âx) Ã (ây) = xy, for all positive numbers x and y}
= 6
(ii)
1 à (â2) à 3 = (1 à (â2)) à 3 {x à (ây) = â(xy), for all positive numbers x and y}
= (â2) Ã 3
= â(2 Ã 3) {(âx) Ã y = â(xy), for all positive numbers x and y}
= â6
(iii)
(â2) Ã (â3) Ã (â4) = ((â2) Ã (â3)) Ã (â4)
= 6 Ã (â4) {(âx) Ã (ây) = xy, for all positive numbers x and y}
= â(6 à 4) {x à (ây) = â(xy), for all positive numbers x and y}
= â24
(iv)
(â1) Ã (â1) = 1 Ã 1 {(âx) Ã (ây) = xy, for all positive numbers x and y}
= 1
(v)
(â1)3 = (â1) Ã (â1) Ã (â1)
= (1 Ã 1) Ã (â1) {(âx) Ã (ây) = xy, for all positive numbers x and y}
= 1 Ã (â1)
= â(1 à 1) {x à (ây) = â(xy), for all positive numbers x and y}
= â1
(vi)
(â1)5 = (â1) Ã (â1) Ã (â1) Ã (â1) Ã (â1)
= (1 Ã 1) Ã (â1) Ã (â1) Ã (â1) {(âx) Ã (ây) = xy, for all positive numbers x and y}
= 1 Ã (â1) Ã (â1) Ã (â1)
= â(1 à 1) à (â1) à (â1) {x à (ây) = â(xy), for all positive numbers x and y}
= (â1) Ã (â1) Ã (â1)
= (1 Ã 1) Ã (â1) {(âx) Ã (ây) = xy, for all positive numbers x and y}
= 1 Ã (â1)
= â(1 à 1) {x à (ây) = â(xy), for all positive numbers x and y}
= â1
(vii)
(â1)99 = (â1) à (â1) à (â1) à (â1) à (â1)â¦.. 99 times
We have calculated that (â1)3 = â1
(â1)5 = â1
Thus, when (â1) is raised to any odd number, the result is â1.
As 99 is an odd number, (â1)99 = â1
Page No 60:
Question 1:
Try to prove the following using the general principle.
(i) (x â y) + z = x â (y â z) for all numbers x, y, z
(ii) (x + y) â z = x + (y â z) for all numbers x, y, z
Answer:
(i)
To show: (x â y) + z = x â (y â z) for all numbers x, y, z.
We know that (x + y) + z = x + (y + z) for all numbers x, y, z. â¦.(i)
Now, (x â y) + z = (x + (ây)) + z
= x + ((ây) + z) (Using (i))
= x + (ây + z)
= x â (y â z)
Thus, (x â y) + z = x â (y â z) for all numbers x, y, z.
(ii)
To show: (x + y) â z = x + (y â z) for all numbers x, y, z.
We know that (x + y) + z = x + (y + z) for all numbers x, y, z. â¦.(i)
Now, (x + y) â z = (x + y) + (âz)
= x + (y + (âz)) (Using (i))
= x + (y â z)
Thus, (x + y) â z = x + (y â z) for all numbers x, y, z.
Page No 60:
Question 2:
Try simplifying these:
(i) (3x + 2y) â (x + y)
(ii) (3x + 2y) â (x â y)
(iii) (x + y) â (x â y)
(iv) (x â y) â (x + y)
(v) (4x â 3y) â (2x â 5x)
Answer:
(i)
(3x + 2y) â (x + y)
= (3x + 2y â x) â y {x â (y + z) = (x â y) â z for all numbers x, y, z}
= (3x â x + 2y) â y
= 2x + 2y â y
= 2x + y
(ii)
(3x + 2y) â (x â y)
= 3x + 2y â x + y {x â (y â z) = x â y + z for all numbers x, y, z}
= 3x â x + 2y + y
= 2x + 3y
(iii)
(x + y) â (x â y)
= x + y â x + y {x â (y â z) = x â y + z for all numbers x, y, z}
= x â x + y + y
= 2y
(iv)
(x â y) â (x + y)
= (x â y â x) â y {x â (y + z) = (x â y) â z for all numbers x, y, z}
= (x â x â y) â y
= ây â y
= â2y
(v)
(4x â 3y) â (2x â 5x)
= 4x â 3y â (â3x)
= 4x â 3y + 3x
= 4x + 3x â 3y
= 7x â 3y
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