Rs Aggarwal 2017 Solutions for Class 7 Math Chapter 5 Exponents are provided here with simple step-by-step explanations. These solutions for Exponents are extremely popular among Class 7 students for Math Exponents Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 7 Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

Page No 90:

Answer:

(i) 57×57×57×57 = 574

(ii) -43×-43×-43×-43×-43=-435

(iii) -16×-16×-16=-163

(iv) -8×-8×-8×-8×-8=-85



Page No 91:

Answer:

(i) 2536=5262                               [since 25 = 52 and 36 = 62]
          =562

(ii) -2764=-3343                     [since −27 = (−3)3 and 64 = 43]

              =-343

(iii) -32243=-2535                   [since −32 = (−2)5 and 243 = 35]
               =-235

(iv) -1128=-1727                    [since (−1)7 = −1 and 128 = 27]
              =-127

Page No 91:

Answer:

(i) 235=2535=2×2×2×2×23×3×3×3×3=32243

(ii) -853=-8353=-8×-8×-85×5×5=-512125

(iii) -13112=-132112=-13×-1311×11=169121

(iv) 163=1363=1×1×16×6×6=1216

(v) -125=-1525=-1×-1×-1×-1×-12×2×2×2×2=-132

(vi) -324=-3424=-3×-3×-3×-32×2×2×2=8116

(vii) -473=-4373=-4×-4×-47×7×7=-64343

(viii) -19=-1       [Since (-1) an odd natural number = -1]

Page No 91:

Answer:

(i) 4-1=41-1=141=14                                 [since ab-n=ban]

(ii) -6-1=-61-1=1-61=-16                    [since ab-n=ban]

(iii) 13-1=311=31                                           [since ab-n=ban]

(iv) -23-1=3-21=-32                                  [since ab-n=ban]
                     

Page No 91:

Answer:

We know that the reciprocal of abm is bam.

(i) Reciprocal of 384=834

(ii) Reciprocal of -5611=-6511

(iii) Reciprocal of 67 = Reciprocal of 617= 167

(iv) Reciprocal of (− 4)3 = Reciprocal of -413 = -143

Page No 91:

Answer:

(i) 80 = 1
(ii) (−3)0 = 1
(iii) 40 + 50 = 1 + 1 = 2
(iv) 60 × 70 = 1 × 1 = 1

Note: a0 = 1

Page No 91:

Answer:

(i) 324×152=3424×1252=81×116×25=81400

(ii) -235×-373=-2535×-3373
                                  = =-2573×-13335=-32×-1×33-5343=-32×-1×3-2343=-32×-1×1343×9=323087               since 3-2=19

                                
(iii) -125×23×342=-1525×23×3242
                                       =-1525×23×32(22)2
                                       =-1×23×3225×24
                                       =-1×23×3229 =-1×23-9×32  = -9×2-6=-926=-964                  since ab-1=ba1

(iv) 232×-353×722=2232×-3353×7222
                                             -1×33-2×7253=-1×31×7253=-1×3×49125=-147125

(v) -343--523×42=-3343--5323×42
                                             =-2764--1258×16
                                             =-2764+1258×16
                                             =-27+100064×16
                                              =97364×16=9734

Page No 91:

Answer:

(i) 496×49-4=496+-4                                             since an×am=an+m
                        =  4924292=4×49×9=1681

(ii) -78-3×-782=-78-3+2                                  since an×am=an+m
                             = -78-1            
                              =  8-71                                          since ab-1= ba1
                              = 8×-1-7×-1=-87

(iii) 43-3×43-2=43-3+-2                                    since an×am= an+m
                           = 43-5
                           = 345                                                since ab-1=ba1
                          = 3545=3×3×3×3×34×4×4×4×4=2431024

Page No 91:

Answer:

Note: ab-1=ba1


(i) 5−3 = 51-3=153=1353=1125


(ii) (−2)−5 = -21-5=1-25=15-25=1×-1-32×-1=-132


(iii) 14-4=414=4414=2561=256


(iv) -34-3=4-33 = 43-33=64-27=64×-1-27×-1=-6427


(v) -3-1×13-1=1-31×311=1×3-3×11=3-31=1-1=1×-1-1×-1=-11=-1


(vi) 57-1×74-1=751×471=7×45×71=45


(vii) 5-1-7-1-1=15-17-1=7-535-1
                                                     = 235-1=3521=352

(viii) 43-1-14-1-1 = 341-411-1=34-41-1
                                                               = 3-164-1=-134-1= 4-131=4×-1-13×-1= -413

(ix) 32-1÷-25-1=231÷5-21
                                 =23×-25=-415

(x) 23250=1          [since a0 = 1 for every integer a]

Page No 91:

Answer:

(i)

-142-2-1=-142×-2-1                 since abmn=abmn
                          =-14-4-1
                           =-14-4×-1=-144=-1444=1256

(ii)

-2323=-232×3                               since abmn=abmn
                  = -236
                  = -2636=64729                      [since (−2)6 = 64 and (3)6 = 729]

(iii)

-323÷-326=-323-6                    since am÷an=am-n          
                             = -32-3 
                            =2-33                      since ab-1=ba1
                            =2×-1-3×-13=-233=-2333=-827

(iv)

-237÷-234=-237-4                   since am÷an=am-n
                            =-233
                             = -2333=-827

Page No 91:

Answer:

Let the required number be x.
(−5)-1 × x = (8)-1
1-5× x=18
x = 18×-5 = -58
Hence, the required number is -58.

Page No 91:

Answer:

Let the required number be x.
(3)-3 x x = 4
133×x=4
127×x=4
x = 4 x 27 = 108
Hence, the required number is 108.

Page No 91:

Answer:

Let the required number be x.
(-30)-1 ÷ x = 6-1
1-30×1x=16
1-30x=16
x = 6-30=1-5
                       =-15
Hence, the required number is -15.

Page No 91:

Answer:

353×35-6=352x-1
353+(-6)=352x-1      since am×an=am+n
35-3=352x-1
On equating the exponents:
−3 = 2x − 1
⇒ 2x = −3 + 1
⇒ 2x   = −2
x = -22=-1

Page No 91:

Answer:

35×105×2557×65=35×2×55×5257×2×35
                        =35×25×55×5257×25×35
                        =35×25×5735×25×57=35-5×25-5×57-7=30×20×50=1×1×1=1



Page No 92:

Answer:

    16×2n+1-4×2n16×2n+2-2×2n+2
24×2n+1-22×2n24×2n+2-2n+1×22
22×2n+3-2n22×2n+4-2n+1
2n×23-2n2n×24-2n×2
2n23-12n24-2=8-116-2=714=12

Page No 92:

Answer:

(i) 52n × 53 = 59
     52n+3 = 59          [since an × am = am+n]

     On equating the coefficients:
     2n + 3 = 9
     ⇒ 2n = 9 − 3
     ⇒ 2n = 6
     ∴ n =62=3

(ii) 8 × 2n+2 = 32
     ⇒ (2)3 × 2n+2 = (2)5      [since 23 = 8 and 25 = 32]
     ⇒ (2)3+ (n+2) = (2)5 
     On equating the coefficients:
      3 + n + 2 = 5
      ⇒ n + 5 = 5
      ⇒ n = 5 − 5
      ∴ n = 0

(iii) 62n+1 ÷ 36 = 63
       ⇒ 62n+1 ÷ 62 = 63       [since 36 = 62]
       ⇒ 62n+162=63
       ⇒ 62n+1-2=63          [since aman=am-n ]
       ⇒ 62n-1 = 63
        On equating the coefficients:
        2n - 1 = 3
        ⇒ 2n = 3 + 1
        ⇒ 2n = 4
         ∴ n = 42=2

Page No 92:

Answer:

    2n-7×5n-4=1250
2n27×5n54=2×54                       [since 1250 = 2 × 54]
2n×5n27×54=2×54
⇒  2n×5n=2×54×27×54           [using cross multiplication]
⇒  2n×5n=21+7×54+4               [since am × an = am+n ]
2n×5n=28×58
2×5n=2×58                      [since an × bn = (a × b)n ]
10n=108
n = 8

Page No 92:

Answer:

(i) 538 = 5.38 × 102                                        [since the decimal point is moved 2 places to the left]

(ii) 6428000 = 6.428 × 106                             [since the decimal point is moved 6 places to the left]

(iii) 82934000000 = 8.2934 × 1010                [since the decimal point is moved 10 places to the left]

(iv) 940000000000 = 9.4 × 1011                    [since the decimal point is moved 11 places to the left]

(v) 23000000 = 2.3 × 107                               [since the decimal point is moved 7 places to the left]

Page No 92:

Answer:

(i) Diameter of the Earth = 1.2756 × 107 m
     [since the decimal point is moved 7 places to the left]

(ii) Distance between the Earth and the Moon = 3.84 × 108 m
      [since the decimal point is moved 8 places to the left]

(iii) Population of India in March 2001 = 1.027 × 109
       [since the decimal point is moved 9 places to the left]

(iv) Number of stars in a galaxy = 1.0 × 1011
       [since the decimal point is moved 11 places to the left]

(v) Present age of the universe = 1.2 × 1010 years
      [since the decimal point is moved 10 places to the left]



Page No 93:

Answer:

(i) 684502 = 6 x 105 + 8 x 104 + 4 x 103 + 5 x 102 + 0 x 101 + 2 x 100
(ii) 4007185 = 4 x 106 + 0 x 105 + 0 x 104 + 7 x 103 + 1 x 102 + 8 x 101 + 5 x 100
(iii) 5807294 = 5 x 106 + 8 x 105 + 0 x 104 + 7 x 103 + 2 x 102 + 9 x 101 + 4 x 100
(iv) 50074 = 5 x 104 + 0 x 103 + 0 x 102 + 7 x 101 + 4 x 100

Note: a0 = 1

Page No 93:

Answer:

(i) 6 × 104 + 3 × 103 + 0 × 102 + 7 × 101 + 8 × 100
     
= 6 x 10000 + 3 x 1000 + 0 x 100 + 7 x 10 + 8 x 1 = 63078
     
(ii) 9 × 106 + 7 × 105 + 0 × 104 + 3 × 103 + 4 × 102 + 6 × 101 + 2 × 100
      = 9 x 1000000 + 7 x 100000 + 0 x 10000 + 3 x 1000 + 4 x 100 + 6 x 10 + 2 x 1 = 9703462

(iii) 8 × 105 + 6 × 104 + 4 × 103 + 2 × 102 + 9 × 101 + 6 × 100
       = 8 x 100000 + 6 x 10000 + 4 x 1000 + 2 x 100 + 9 x 10 + 6 x 1 = 864296

Page No 93:

Answer:

(d) 24

6-1-8-1-1=16-18-1
                  = 4-324-1        [since L.C.M. of 6 and 8 is 24]
                  = 124-1
                  = 2411=24      since ab-1=ba1

Page No 93:

Answer:

(c) 15

We have:

5-1×3-1-1=15×13-1
                      = 115-1       
                      = 1511=15      since ab-1=ba1

Page No 93:

Answer:

(c) 116

We have:

2-1-4-12=12-142
                   = 2-142        [since L.C.M. of 2 and 4 is 4]
                   = 142
                   = 14×14=116     

Page No 93:

Answer:

(b) 29

We have:

12-2+13-2+14-2=212+312+412          sinceab-1=ba1
                                 = (22 + 32 + 42)
                                 = (4 + 9 + 16)
                                 = 29



Page No 94:

Answer:

(c) 65

We have:
6-1+32-1-1=16+23-1
                             = 1+46-1    [since L.C.M. of 3 and 6 is 6]
                             = 56-1         
                             = 651=65             sinceab-1=ba1

Page No 94:

Answer:

(b) 64
We have:
-12-6=2-16                            since ab-n=ban
                =-26=-2×-2×-2×-2×-2×-2 = 64

Page No 94:

Answer:

(b)  -38

34-1-14-1-1=43-41-1
                                  = 4-123-1     [ since L.C.M. of 1 and 3 is 3]
                                  = -83-1
                                  = 3-81                      since ab-1=ba1
                                  = 3×-1-8×-1=-38

Page No 94:

Answer:

(a) 116

-122-2-1=-122×-2-1                           sinceabmn=abmn
                          =-12-4-1
                           =-12-4×-1=-124=-1424=116

Page No 94:

Answer:

(c) 1

(a)0 = 1
560=1

Page No 94:

Answer:

(b) 24332

23-5=325                                                    since ab-n=ban
      
         = 3525=3×3×3×3×32×2×2×2×2=24332

Page No 94:

Answer:

(b) 138

1324=132×4=138                  sinceabmn=abmn

Page No 94:

Answer:

(b) -23
We have:
-32-1=2-31       since ab-n= ban
               = -23

Page No 94:

Answer:

(d) 1358

32-22×23-3=9-4×323         since ab-1=ba1
                             =5×33235×278 = 1358

Page No 94:

Answer:

(a) 1964
We have:
13-3-12-3÷14-3 = 313- 213 ÷ 413                               since ab-1=ba1
                                             = 33-23÷43
                                             = 27 - 8 ÷ 64
                                             = 19 ÷ 64
                                             = 19×164=1964



Page No 95:

Answer:

(c) (-5)5

We have:
-153÷-158=-153-8               since am÷an=am-n          
                             = -15-5 
                            =5-15                  Since ab-1=ba1
                            =5×-1-1×-15=-515=-55
                                

Page No 95:

Answer:

(a) 425

-257÷-255=-257-5            since am÷an=am-n
                            =-252
                             = -2252=425

Page No 95:

Answer:

(c) 49

-232=-23×-23=49

Page No 95:

Answer:

(b) -18
We have:
-123=-12×-12×-12=-18

Page No 95:

Answer:

(c) 34

53-5×5311=538x
53-5+11=538x        [ since am×an=am+n]
536=538x
 On equating the coefficients:
  6 = 8x
  ∴ x = 68=34

Page No 95:

Answer:

(c) -45
Let the required number be x.
(−8)-1 x x = (10)-1
⇒ 1-8×x=110
x = 110×(-8) = -45
Hence, the required number is -45.

Page No 95:

Answer:

(c) 2.156 × 106
A given number is said to be in standard form if it can be expressed as k x 10n, where k is a real number such that 1 ≤ k < 10 and n is a positive integer.
For example: 2.156 × 106



Page No 96:

Answer:

We know that the reciprocal of abm is bam.

(i) Reciprocal of 234= 324

(ii) Reciprocal of -3561=-5361

(iii) Reciprocal of 25 = Reciprocal of 215=125

(iv) Reciprocal of (−5)6 = Reciprocal of -516=-156

Page No 96:

Answer:

Let the required number be x.
(−6)-1 × x = (9)-1
⇒  1-6×x=19
x = 19×-6=-23
Hence, the required number is -23.

Page No 96:

Answer:

Let the required number be x.
(−20)-1 ÷ x = (−10)-1
⇒  1-20×1x=1-10
1-20x=1-10
x = -10-20=12=2-1
Hence, the required number is 2-1.

Page No 96:

Answer:


(i) 2000000 = 2.000000 × 106         [since the decimal point is moved 6 places to the left]
                    = 2 × 106     

(ii) 6.4 × 105 = 6.4 × 100000
                      = 640000

Page No 96:

Answer:

We have:
    16×2n+1-8×2n16×2n+2-4×2n+1

24×2n+1-23×2n24×2n+2-22×2n+1

23×2n+2-2n23×2n+3-2n

2n×22-2n2n×23-2n

2n22-12n23-1=4-18-1=37

Page No 96:

Answer:

We have:
    2n-7×5n-4=1250
2n27×5n54=2×54                      [since 1250 = 2 × 54]
2n×5n27×54=2×54
⇒  2n×5n=2×54×27×54           [using cross multiplication]
⇒  2n×5n=21+7×54+4               [since am × an = am+n ]
2n×5n=28×58
2×5n=2×58                      [since an × bn = (a × b)n ]
10n=108
n = 8

Page No 96:

Answer:

(c)  1
We know:
(a)0 = 1
340=1

Page No 96:

Answer:

(d) -6427

-34-3=4-33                           since ab-n=ban

           = 43-33

           = 4×4×4-3×-3×-3=64-27

           = 64×-1-27×-1=-6427
             
                                      

Page No 96:

Answer:

(b) -35

-53-1=3-51                                       since ab-n=ban
           = 3-5×-1-1=-35

Page No 96:

Answer:

(a) 19

13-3-12-3=313-213                                         since ab-1=ba1
                        = 33-23
                        = (27 − 8) = 19

Page No 96:

Answer:

(a) 49


-2310÷-238 = -2310-8            since am÷an=am-n
                              = -232
                              = -2232=49

Page No 96:

Answer:

(c) 3.263 x 105

A given number is said to be in standard form if it can be expressed as k x 10n, where k is a real number such that 1 ≤ k < 10 and n is a positive integer.
For example: 3.263 x 105

Page No 96:

Answer:

(i) If 9 × 3n = 36, then n = 4.
Explanation:
If 9 × 3n = 36
⇒ 32 × 3n = 36
⇒ 3(2 + n) = 36
Equating the powers:
⇒ ( 2 + n) = 6
n = (6 - 2) = 4

(ii) (8)0 = 1
Explanation:
By definition, we have a0 = 1 for every integer a.
(8)0 = 1

(iii) ab-16 = ba16
Explanation:
We know:
  ab-1=ba1

(iv) (−2)−5 = -132
Explanation:
(−2)−5 = -21-5=1-25         Since ab-1=ba1
= 15-25=1×1×1×1×1-2×-2×-2×-2×-2=1-32

Page No 96:

Answer:

(i) True
645 = 6.45 x 102                  [since the decimal point is moved 2 places to the left]

(ii) False
 27000 = 2.7 x 104               [since the decimal point is moved 4 places to the left]

(iii) False
(30 + 40 + 50) = 1               [since a0 = 1 for every integer a]

(iv) False
Reciprocal of 56 = Reciprocal of 516=156

(v) False
5−1 × x = 8−1 
15×x=18
x = 18×5=58
    



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