Rs Aggarwal 2018 Solutions for Class 8 Math Chapter 2 Exponents are provided here with simple step-by-step explanations. These solutions for Exponents are extremely popular among Class 8 students for Math Exponents Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 8 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 33:

Question 1:

Evaluate:
(i) 4−3
(ii) 12-5
(iii) 43-3
(iv) -3-4
(v) -23-5

Answer:

(i) 4-3=143=164

(ii) 12-5=25=32

(iii) 43-3=343=3343=2764

(iv) -3-4=-134=(-1)434=181

(v) -23-5=-325=-3525=-24332

Page No 33:

Question 2:

Evaluate:
(i) 532×532
(ii) 566×56-4
(iii) 23-3×23-2
(iv) 98-3×982

Answer:

(i) 532×532= 534=5434=62581

(ii) 566×56-4=56(6+(-4))= 56(6-4)=562=5262=2536

(iii)23-3×23-2=23(-3-2)=23-5=325=3525=24332

(iv) 98-3×982=98(-3+2)=98-1=89

Page No 33:

Question 3:

Evaluate:
(i) 59-2×35-3×350
(ii) -35-4×-252
(iii) -23-3×-23-2

Answer:

(i)

59-2×35-3×350=59-2×35-3+0=59-2×35-3=952×533=9252×5333=32252×5333=3452×5333=3(4-3)×5(3-2)=3×5=15


(ii)

-35-4×-252=5-34×-252=54-34×-2252=5(4-2)×-22-34=52×-22-34=25×481=10081


(iii)

-23-3×-23-2=3-23×3-22=33-23×32-22=3(3+2)-2(3+2)=35-25=-24332

Page No 33:

Question 4:

Evaluate:
(i) -232-2
(ii) -132-2-1
(iii) 32-22

Answer:

(i) -232-2=-232×-2=-23-4=3-24=34(-2)4=3424=8116
(ii) -132-2-1=-132×-2-1=-13-4-1=-13-4×-1=-134=-1434=1434=181
(iii) 32-22=32-2×2=32-4=234=2434=1681

Page No 33:

Question 5:

Evaluate 13-3-12-3÷14-3.

Answer:

13-3-12-3÷14-3=33-23÷43=27-8÷64=1964

Page No 33:

Question 6:

Evaluate 43-1-14-1-1.

Answer:

43-1-14-1-1=341-411-1=34-41-1The L.C.M. of 4 and 1 is 4. 3×14×1-4×41×4-1= 34-164-1=3-164-1=-134-1=4-131=4-13=4×-1-13×-1=-413

Page No 33:

Question 7:

Evaluate 5-1×3-1-1÷6-1.

Answer:

5-1×3-1-1÷6-1=15×13- 1÷16=115- 1÷16=15×6=90

Page No 33:

Question 8:

Find the value of:
(i) (20 + 3−1) × 32
(ii) (2−1 × 3−1) ÷ 2−3
(iii) 12-2+13-2+14-2

Answer:

(i)
20+3-1×32=1+13×32  (because 20=1 and 3-1=13)=1×31×3+1×13×1×32=33+13×32=43×32=4×3(2-1)=4×3=12

(ii)

2-1×3-1÷2-3=12×13÷123 16÷13 23=16÷18=16×8=86=43


(iii)

12-2+13-2+14-2=212+312+412=22+32+42=4+9+16=29



Page No 34:

Question 9:

Find the value of x for which 53-4×53-5=533x.

Answer:

Consider the left side:
 53-4×53-5=53(-4+(-5))=53-9

Given:
53-9=533x
Comparing the powers:
-9=3x x=-3

Page No 34:

Question 10:

Find the value of x for which 494×49-7=492x-1.

Answer:

Given:
 494×49-7=492x-1

 49(4-7)=49-3=492x-1

 2x-1=-32x=-3+1=-2x=-1

Page No 34:

Question 11:

By what number should (−6)−1 be multiplied so that the product becomes 9−1?

Answer:

Let the required number be x.

 x×(-6)-1=9-1

x×1-6=19x-6=19or  x=-69
The greatest common divisor for the numerator and the denominator is 3.

 x=-69=(-6)÷39÷3=-23

Page No 34:

Question 12:

By what number should -23-3 be divided so that the quotient may be 427-2?

Answer:

Let the number be x.

-23-3÷x=427-2
3-23÷x=2742-323÷x=2742 -323×1x=2742-3323×1x=27242-278×1x=27242=27×274×4=27×274×2×2=27×278×2 1x=27×278×2-278 x=-27827×278×2=-278×8×227×27=-227

Page No 34:

Question 13:

If 52x + 1 ÷ 25 = 125, find the value of x.

Answer:

Given:
52x+1÷25=125
We know:25=5×5=52125=5×5×5=53 52x+152=535[(2x+1)-2]=53or 5[(2x+1)-2]=5[2x-1]=532x-1=32x=3+1=4x=42=2

x=2



Page No 36:

Question 1:

Write each of the following numbers in standard form:
(i) 57.36
(ii) 3500000
(iii) 273000
(iv) 168000000
(v) 4630000000000
(vi) 345 × 105

Answer:

(i) 57.36=5.736×10
(ii) 3500000=3.5×106
(iii) 273000=2.73×105
(iv) 168000000=1.68×108
(v) 4630000000000=4.63×1012
(vi) 345×105=3.45×107

Page No 36:

Question 2:

Write each of the following numbers in usual form:
(i) 3.74 × 105
(ii) 6.912 × 108
(iii) 4.1253 × 107
(iv) 2.5 × 104
(v) 5.17 × 106
(vi) 1.679 × 109

Answer:

(i) 3.74×105=374100×105=374×105102=374×10(5-2)=374×103=374000
(ii) 6.912×108=69121000×108=6912×108103=6912×10(8-3)=6912×105=691200000
(iii) 4.1253×107=4125310000×107=41253×107104=41253×10(7-4)=41253×103=41253000
(iv) 2.5×104=2510×104=25×10410=25×10(4-1)=25×103=25000
(v) 5.17×106=517100×106=517×106102=517×10(6-2)=517×104=5170000
(vi) 1.679×109=16791000×109=1679×109103=1679×10(9-3)=1679×106=1679000000

Page No 36:

Question 3:

(i) The height of Mount Everest is 8848 m. Write it in standard form.
(ii) The speed of light is 300000000 m/sec. Express it in standard form.
(iii) The distance from the earth to the sun is 149600000000 m. Write it in standard form.

Answer:

(i) The height of the Mount Everest is 8848 m.
In standard form, we have:
8848=8.848×1000 m=8.848×103m.
(ii) The speed of light is 300000000 m/s.
In standard form, we have: 
300000000=3×100000000 m/s=3×108 m/s
(iii) The Sun-Earth distance is 149600000000 m.
In standard form, we have: 149600000000=1496×100000000=1.496×1000×100000000=1.496×103×108=1.496×1011m

Page No 36:

Question 4:

Mass of earth is (5.97 × 1024) kg and mass of moon is (7.35 × 1022) kg. What is the total mass of the two?

Answer:

Mass of the Earth = 5.97×1024 kg
Now, 5.97×1024=5.97×10(2+22)=5.97×102×1022=597×1022
So, the mass of the Earth can also be written as 597×1022 kg.

Mass of the Moon = 7.35×1022 kg
Sum of the masses of the Earth and the Moon:=597×1022+7.35×1022=597+7.35×1022=604.35×1022 kg=6.0435×100×1022=6.0435×102×1022=6.0435×10(2+22)=6.0435×1024 kg

Page No 36:

Question 5:

Write each of the following numbers in standard form:
(i) 0.0006
(ii) 0.00000083
(iii) 0.0000000534
(iv) 0.0027
(v) 0.00000165
(vi) 0.00000000689

Answer:

(i) 0.0006=6104=6×10-4

(ii) 0.00000083=83108=8.3×10108=8.3×10(1-8)=8.3×10-7

(iii) 0.0000000534=5341010=5.34×1021010=5.34×10(2-10)=5.34×10-8

(iv) 0.0027=27104=2.7×10104=2.7×10(1-4)=2.7×10-3

(v) 0.00000165 = 165108 = 1.65×102108 = 1.65×10(2-8)  = 1.65×10-6

(vi) 0.00000000689 = 6891011 = 6.89×1021011 = 6.89×10(2-11)= 6.89×10-9

Page No 36:

Question 6:

(i) 1 micron = 11000000 m. Express it in standard form.
(ii) Size of a bacteria = 0.0000004 m. Express it in standard form.
(iii) Thickness of a paper = 0.03 mm. Express it in standard form.

Answer:

(i) 1 micron = 11000000 m=1×10-6 m

(ii) 0.0000004 m=4107m=4×10-7 m
(iii) Thickness of paper = 0.03 mm = 3102 mm=3×10-2 mm

Page No 36:

Question 7:

Write each of the following numbers in usual form:
(i) 2.06 × 10−5
(ii) 5 × 10−7
(iii) 6.82 × 10−6
(iv) 5.673 × 10−4
(v) 1.8 × 10−2
(vi) 4.129 × 10−3

Answer:

(i) 2.06×10-5=206100×1105=206102×105=206105+2=206107=20610000000=0.0000206

(ii) 5×10-7=5107=510000000=0.0000005

(iii) 6.82×10-6=682100×1106=682102×106=682102+6=682108=682100000000=0.00000682

(iv)5.673×10-4=56731000×1104=5673103×104=5673103+4=5673107=567310000000=0.0005673

(v)1.8×10-2=1810×1102=1810×102=18101+2=18103=181000=0.018

(vi) 4.129×10-3=41291000×1103=4129103×103=4129103+3=4129106=41291000000=0.004129



Page No 37:

Question 1:

Tick (✓) the correct answer
The value of 25-3 is
(a) -8125
(b) 254
(c) 1258
(d) -25

Answer:

(c) 1258

25-3=523=5323=1258

Page No 37:

Question 2:

Tick (✓) the correct answer
The value of (−3)−4 is
(a) 12
(b) 81
(c) -112
(d) 181

Answer:

(d) 181

-3-4=1-34=1-14×34=134=181

Page No 37:

Question 3:

Tick (✓) the correct answer
The value of (−2)−5 is
(a) −32
(b) -132
(c) 32
(d) 132

Answer:

(b) -132

-2-5=1-25=1-32=1×-1-32×-1=-132

Page No 37:

Question 4:

Tick (✓) the correct answer
(2−5 ÷ 2−2) = ?
(a) 1128
(b) -1128
(c) -18
(d) 18

Answer:

(d) 18

2-5÷2-2=125÷122=132÷14=132×4=432=18

Page No 37:

Question 5:

Tick (✓) the correct answer
The value of (3−1 + 4−1)−1 ÷ 5−1 is
(a) 710
(b) 607
(c) 75
(d) 715

Answer:

(b) 607

3-1+4-1-1÷5-1=13+14-1÷15=4+312-1÷15=712-1÷15=127÷15=127×5=607

Page No 37:

Question 6:

Tick (✓) the correct answer
12-2+13-2+14-2=?
(a) 61144
(b) 14461
(c) 29
(d) 129

Answer:

(c) 29

12-2+13-2+14-2= 212+312+412=22+32+42=4+9+16=29

Page No 37:

Question 7:

Tick (✓) the correct answer
13-3-12-3÷14-3=?
(a) 1964
(b) 2716
(c) 6419
(d) 1625

Answer:

(a) 1964

13-3-12-3÷14-3=33-23÷43=27-8÷64=19÷64=1964

Page No 37:

Question 8:

Tick (✓) the correct answer
-122-2-1=?
(a) 116
(b) 16
(c) -116
(d) −16

Answer:

(a) 116

-122-2-1=-12-4-1=-12(-4×-1)=-124=116

Page No 37:

Question 9:

Tick (✓) the correct answer
The value of x for which 712-4×7123x=7125, is
(a) −1
(b) 1
(c) 2
(d) 3

Answer:

(d) 3

712-4×7123x=7125712-4+3x=71253x-4=53x=9or x=93=3

Page No 37:

Question 10:

Tick (✓) the correct answer
If (23x − 1 + 10) ÷ 7 = 6, then x is equal to
(a) −2
(b) 0
(c) 1
(d) 2

Answer:

(d) 2

23x-1+10÷7=623x-1+107=61On cross multiplying:23x-1+10×1=6×7=42

23x-1  = 42 -10
23x-1 = 32
23x-1  = 25
 3x-1 = 5
 3x = 6
Therefore,
x =  2

Page No 37:

Question 11:

Tick (✓) the correct answer
230=?
(a) 32
(b) 23
(c) 1
(d) 0

Answer:

(c) 1

Using the law of exponents ab0=1:
 230=1



Page No 38:

Question 12:

Tick (✓) the correct answer
-53-1=?
(a) 53
(b) 35
(c) -35
(d) none of these

Answer:

(c) -35

-53-1=3-51=3-5=3×-1-5×-1=-35

Page No 38:

Question 13:

Tick (✓) the correct answer
-123=?
(a) -16
(b) 16
(c) 18
(d) -18

Answer:

(d) -18

-123=-1323=-18

Page No 38:

Question 14:

Tick (✓) the correct answer
-342=?
(a) -916
(b) 916
(c) 169
(d) -169

Answer:

(b) 916

-342=-3242=916

Page No 38:

Question 15:

Tick (✓) the correct answer
3670000 in standard form is
(a) 367 × 104
(b) 36.7 × 105
(c) 3.67 × 106
(d) none of these

Answer:

(c) 3.67×106

3670000=367×104=3.67×100×104=3.67×102×104=3.67×102+4=3.67×106

Page No 38:

Question 16:

Tick (✓) the correct answer
0.0000463 in standard form is
(a) 463 × 10−7
(b) 4.63 × 10−5
(c) 4.63 × 10−9
(d) 46.3 × 10−6

Answer:

(b) 4.63×10-5

0.0000463=463107=4.63×102107=4.63×102-7=4.63×10-5

Page No 38:

Question 17:

Tick (✓) the correct answer
0.000367 × 104 in usual form is
(a) 3.67
(b) 36.7
(c) 0.367
(d) 0.0367

Answer:

(a) 3.67

0.000367×104=367106×104=367×104-6=367×10-2=367102=367100=3.67



Page No 39:

Question 1:

Evaluate:
(i) 3−4
(ii) (−4)3
(iii) 34-2
(iv) -23-5
(v) 570

Answer:

(i) 3-4=134=181

(ii) -43=-13×43=-1×64=-64

(iii) 34-2=432=4232=169

(iv) -23-5=3-25=35-25=243-32=243×-1-32×-1=-24332

(v) Using the property ab0=1, we have:
570 = 1

Page No 39:

Question 2:

Evaluate: -233-2.

Answer:

-233-2=-23-6=3-26=36-26=72964

Page No 39:

Question 3:

Simplify: (3-1+6-1)÷34-1.

Answer:

3-1+6-1÷34-1=13+16÷431=1×23×2+1×16×1÷43=2+16÷43=36÷43=12÷43=12×34=38

 3-1+6-1÷34-1=38

Page No 39:

Question 4:

By what number should -23-3 be divided so that the quotient is 49-2?

Answer:

Let the number be x.

 -23-3÷x=49-2

3-23÷x=942 3-23x=94233-23x=9242x= 33-23 9242=33-23 322222=33-23×222322=33-23×2434=33-23×2333×21311-1×2131=2-3=2×-1-3×-1=-23

Page No 39:

Question 5:

By what number should (−3)−1 be multiplied so that the product becomes 6−1?

Answer:

Let the number be x.

 -3-1×x=6-1
1-3 × x=16 1 × -1-3 × -1 × x=16 -x3 = 16On cross multiplying:-x × 6 = 1 × 3-6x = 36x = -3 x = -36 = -12

Page No 39:

Question 6:

Express each of the following in standard form:
(i) 345
(ii) 180000
(iii) 0.000003
(iv) 0.000027

Answer:

(i) 345=3.45×100=3.45×102

(ii) 180000=18×10000=18×104=1.8×10×104=1.8×101+4=1.8×105

(iii) 0.000003=31000000=3×10-6

(iv) 0.000027=27100000=27106=2.7×10106=2.7×101-6=2.7×10-5

Page No 39:

Question 7:

Mark (✓) against the correct answer
The value of (−3)−3 is
(a) −27
(b) 9
(c) -127
(d) 127

Answer:

(c) -127

-3-3=1-33=13-33=1-27=1×-1-27×-1=-127

Page No 39:

Question 8:

Mark (✓) against the correct answer
The value of 34-3 is
(a) -2764
(b) 6427
(c) -94
(d) 2764

Answer:

(b) 6427

34-3=433=4333=6427

Page No 39:

Question 9:

Mark (✓) against the correct answer
(3−6 ÷ 34) = ?
(a) 3−2
(b) 32
(c) 3−10
(d) 310

Answer:

(c) 3-10

3-6÷34=136÷34=136×134=136+4=1310=3-10

Page No 39:

Question 10:

Mark (✓) against the correct answer
If 512-4×5123x=5125, then x=?
(a) −1
(b) 1
(c) 2
(d) 3

Answer:

(d) 3

512-4×5123x=5125512-4+3x=5125 -4+3x=53x=5+4=9x=93=3

Page No 39:

Question 11:

Mark (✓) against the correct answer
350=?
(a) 53
(b) 35
(c) 1
(d) 0

Answer:

(c) 1

Using the law of exponents, which says ab0=1, we get:
 350=1

Page No 39:

Question 12:

Mark (✓) against the correct answer
-65-1=?
(a) 65
(b) -65
(c) 56
(d) -56

Answer:

(d) -56

-65-1=5-61=5-6=5×-1-6×-1=-56

Page No 39:

Question 13:

Mark (✓) against the correct answer
-133=?
(a) -19
(b) 19
(c) -127
(d) 127

Answer:

(c) -127

-133=-1333=-127



Page No 40:

Question 14:

Fill in the blanks.
(i) 360000 written in standard form is .........
(ii) 0.0000123 written in standard form is .........
(iii) -23-2=.........
(iv) 3 × 10−3 in usual form is .........
(v) 5.32 × 10−4 in usual form is .........

Answer:

(i) The standard form of 36000 is 3.6×105
360000=36×104=3.6×10×104=3.6×101+4=3.6×105

(ii) The standard form of 0.0000123 is 1.23×10-5.
0.0000123=12310000000=123107=1.23×100107=1.23×102107=1.23×102-7=1.23×10-5

(iii) -23-2=94
-23-2=3-22=32-22=94

(iv)The usual form of 3×10-3  is 0.003.
3×10-3=3103=31000=0.003

(v) The usual form of 5.32×10-4 is 0.000532.
5.32×10-4 =5.32104=5.3210000=0.000532



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