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Page No 6.12:

Question 1:

Find the value of each of the following:
(i) 132
(ii) 73
(iii) 34

Answer:

We have
(i) 132 = 13 × 13 = 169
(ii) 73 = 7 × 7 × 7 = 343
(iii) 34 = 3 × 3 × 3 × 3 = 81

Page No 6.12:

Question 2:

Find the value of each of the following:
(i) (−7)2
(ii) (−3)4
(iii) (−5)5

Answer:

We know that if 'a' is natural number, then
(−a)even number = Positive number
(−a)odd number  = Negative number

We have
(i) (−7)2 = −7 × −7 = 49
(ii) (−3)4 = −3 × −3 × −3 × −3 = 81
(iii) (−5)5 = −5 × −5 × −5 × −5 × −5 = −3125

Page No 6.12:

Question 3:

Simplify:
(i) 3 × 102
(ii) 22 × 53
(ii) 33 × 52

Answer:

We have
(i) 3 × 102 = 3 × 100 = 300             [since 102 = 10 × 10 = 100]
(ii) 22 × 53 = 4 × 125 = 500            [since 22 = 2 × 2 = 4 and 53 = 5 × 5 × 5 = 125]
(iii) 33 × 52 = 27 × 25 = 675           [ since 33 = 3 × 3 × 3 = 27 and 52 = 5 × 5 = 25]

Page No 6.12:

Question 4:

Simplify:
(i) 32 × 104
(ii) 24 × 32
(ii) 52 × 34

Answer:

We have
(i) 32 × 104 = 9 × 10000 = 90000       [since 32 = 3 × 3 = 9 and 104 = 10 × 10 × 10 × 10 = 10000]
(ii) 24 × 32 = 16 × 9 = 144                  [since 24 = 2 × 2 × 2 × 2 = 16 and 32 = 3 × 3 = 9]
(iii) 52 × 34 = 25 × 81 = 2025              [since 52 = 5 × 5 = 25 and 34 = 3 × 3 × 3 × 3 = 81]

Page No 6.12:

Question 5:

Simplify:
(i) (−2) × (−3)3
(ii) (−3)2 × (−5)3
(iii) (−2)5 × (−10)2

Answer:

We know that if 'a' is natural number, then
(−a)even number = Positive number
(−a)odd number = Negative number

We have

(i) (−2) × (−3)3 = ( −2 )(−27) = 54               [since (−3)3 = −3 ×−3 × − 3 = −27]
(ii) (−3)2 × ( −5)3 = 9 (−125) =  −1125        [ since (−3)2 = −3 ×− 3 = 9 and (−5 )3 = −5 ×−5 × − 5 = −125]
(iii) ( −2)5 × (−10)2 = −32 × 100 = −3200    [ since (−2)5= −2 ×−2 × −2 ×−2 ×−2 = −32 and (−10)2 = −10 ×− 10 = 100]

Page No 6.12:

Question 6:

Simplify:
(i) 342
(ii) -234
(iii) -455

Answer:

We have

(i) 342 = 34×34 =916
(ii) -234 = -23×-23×-23×-23= 1681 
(iii) -455 = -45×-45×-45×-45×-45 =-10243125

Page No 6.12:

Question 7:

Identify the greater number in each of the following:
(i) 25 or 52
(ii) 34 or 43
(iii) 35 or 53

Answer:

We have
(i) 25 = 2 × 2 × 2 × 2 × 2 = 32  and 52 = 5 × 5 = 25
     Therefore, 32 > 25.
     Thus, 25 > 52.

(ii) 34 = 3 × 3 × 3 × 3 = 81 and 43= 4 × 4 × 4 = 64
      Therefore, 81 > 64.
      Thus, 34 > 43.

(iii) 35 = 3 × 3 × 3 × 3 × 3 = 243 and 53 = 5 × 5 × 5 = 125
       Therefore, 243 > 125.
       Thus, 35 > 53.

Page No 6.12:

Question 8:

Express each of the following in exponential form:
(i) (−5) × (−5) × (−5)
(ii) -57×-57×-57×-57
(iii) 43×43× 43 × 43× 43 

Answer:

We have
(i) (−5) × (−5) × (−5) = ( −5)3

(ii) -57×-57×-57×-57 = -574

(iii) 43×43×43×43×43 =435

Page No 6.12:

Question 9:

Express each of the following in exponential form:
(i) x × x × x × x × a × a × b × b × b
(ii) (−2) × (−2) × (−2) × (−2) × a × a × a
(iii) -23×-23×x×x×x

Answer:

We have
(i) x× x× x × x × a ×a ×b×b×b =x4a2b3
(ii) (-2) ×(-2) ×(-2) ×(-2) ×a ×a ×a =(-2)4×a3
(iii) -23×-23× x× x× x = -232×x3

Page No 6.12:

Question 10:

Express each of the following numbers in exponential form:
(i) 512
(ii) 625
(iii) 729

Answer:

We have
(i) Prime factorisation of 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 29

(ii) Prime factorisation of 625 = 5 x 5 x 5 x 5 = 54

(iii) Prime factorisation of 729 = 3 x 3 x 3 x 3 x 3 x 3 = 36

Page No 6.12:

Question 11:

Express each of the following numbers as a product of powers of their prime factors:
(i) 36
(ii) 675
(iii) 392

Answer:

We have
(i) Prime factorisation of 36 = 2 x 2 x 3 x 3 = 22 x 32

(ii) Prime factorisation of 675 = 3 x 3 x 3 x 5 x 5 = 33 x 52

(iii) Prime factorisation of 392 = 2 x 2 x 2 x 7 x 7 = 23 x 72



Page No 6.13:

Question 12:

Express each of the following numbers as a product of powers of their prime factors:
(i) 450
(ii) 2800
(iii) 24000

Answer:

We have
(i) Prime factorisation of 450 = 2 x 3 x 3 x 5 x 5 = 2 x 32 x 52

(ii) Prime factorisation of 2800 = 2 x 2 x 2 x 2 x 5 x 5 x 7 = 24 x 52 x 7

(iii) Prime factorisation of 24000 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 5 x 5 x 5 = 26 x 3 x 53

Page No 6.13:

Question 13:

Express each of the following as a rational number of the form pq:
(i) 372
(ii) 793
(iii) -234

Answer:

We have

(i) 372 =37×37=949
(ii) 793 =79×79×79= 343729

(iii) -234 = -23×-23×-23×-23=1681

Page No 6.13:

Question 14:

Express each of the following rational numbers in power notation:
(i) 4964
(ii) -64125
(iii) -1216

Answer:

We have
 (i) 4964=78×78=(7)2(8)2 =782
(ii) -64125 = -45×-45×-45=-(4)3(5)3=-453
(iii) -1216=-16×-16×-16=(-1)3(6)3=-163

Page No 6.13:

Question 15:

Find the value of each of the following:
(i) -122×23×342
(ii) -354×494×-15182

Answer:

We have

(i)
-122×23×342=14×8×916=14×92=98       Since -122 = 14, 23 = 8 and 342 =916

(ii)
    -354×494×-15182 =(-3)454×4494×-3×52×92=(-3)454×4492×92×-3×52×92=8154×4481×92×(-3)2×5222×92=154×4492×(-3)2×5222×92=154×4492×9×524×92=152×4392×19=1×4352×93 =6425×729=6418225        Since 43 =64, 52=25 and 93=729

Page No 6.13:

Question 16:

If a = 2 and b = 3, then find the values of each of the following:
(i) (a + b)a
(ii) (ab)b
(iii) bab
(iv) ab+baa

Answer:

We have a = 2 and b = 3.
Thus,
(i) (a + b)a = (2 + 3)2 = (5)2 = 25

(ii) (ab)b = (2 x 3 )3 = (6)3 = 216

(iii) bab=323 =32×32×32 =278

(iv) ab+baa=23+322 =4+962 =1362 =16936



Page No 6.28:

Question 1:

Using laws of exponents, simplify and write the answer in exponential form:
(i) 23× 24 × 25
(ii) 512 ​÷ 53
(iii) (72)3
(iv) (32)5 ​÷ 34
(v) 37 â€‹× 27
(vi) (521 ÷ 513) × 57

Answer:

 We have

(i)  23 x 24 x 25 = 2(3 + 4 + 5) = 212             [since am + an + ap = a(m+n+p)]

(ii) 512 ÷ 53 = 51253 = 512 - 3 = 59                [ since am ÷ an = am-n ]
(iii) (72)3 = 76                                                       [since (am)n = amn ]

(iv)(32)5 ÷ 34 = 310 ÷ 34                             [since (am)n = amn ]
                       = 3(10 - 4) = 36                      [since am ÷ an = am-n ]

(v) 37 â€‹× 27 = (3 x 2)7 = 67                          [since am x bm = (a x b)m ]

 (vi) (521 ÷  513) x 57 = 5(21 -13) x 57         [since am ÷ an = am-n ]
                                   = 58 x 57                  [since am x bn =a(m +n)]
                                   = 5(8+7)              
                                   = 515

Page No 6.28:

Question 2:

Simplify and express each of the following in exponential form:
(i) {(23)4×28}÷212
(ii) (82× 84) ÷ 83
(iii) 5752×53
(iv) 54×x10y554× x7y4

Answer:

We have
(i)  {(23)4 x 28} ÷ 212
    = {212 x 28} ÷ 212
    = 2(12 + 8) ÷ 212
 
   = 220 ÷ 212
   
= 2 (20 - 12) =  28

(ii) (82 x 84)  ÷ 83
    = 8(2 + 4) ÷ 83
    = 86 ÷ 83
   = 8(6-3) = 83 = (23)3 = 29

(iii) 5752 x 53 = 5(7-2) x 53
                        = 55 x 53
                        = 5(5 + 3 ) = 58

(iv) 54 ×x10y554×x7y4 = 5(4-4)×x(10-7)×y(5-4)
                         = 50×x3×y               [since 50 = 1]
                         = 1×x3y =x3y

Page No 6.28:

Question 3:

Simplify and express each of the following in exponential form:
(i) {(32)3×26}×56
(ii) xy12×y24×(23)4
(iii) 526×522
(iv) 235×355

Answer:

We have
(i)   {(32)3 x 26} x 56
    = {36 x 26} x 56             [since (am)n = amn]
    = 66 x 56                        [since am x bm= (a x b)m ]
    = 306

(ii)  
xy12×y24×(23)4
=x12y12×y24×212
=x12×y24y12×212
=x12×y24-12×212
=x12×y12×212
=(2xy)12 [since am×bm×cm=(a×b×c)m]
(iii)
526×522
=528 [since am×an=am+n]

(iv)
235×355
=23×355 [since am×bm=(a×b)m]
=255

Page No 6.28:

Question 4:

Write 9 × 9 × 9 × 9 × 9 in exponential form with base 3.

Answer:

We have
9 x 9 x 9 x 9 x 9  = (9)5 =(32)5 = 310

Page No 6.28:

Question 5:

Simplify and write each of the following in exponential form:
(i) (25)3÷ 53
(ii) (81)5 ÷ (32)5
(iii) 98×(x2)5(27)4× (x3)2
(iv) 32× 78× 136212× 913

Answer:

We have
(i) (25)3 ÷ 53
= (52)3÷ 53
= 56 ÷ 53
= 5653=56-3=53

(ii) (81)5 ÷ (32)5
= (34)5 ÷ (32)5
= (3)20 ÷ (3)10
= 320310=320-10=310

(iii)

98×(x2)5(27)4×(x3)2
=(32)8×(x2)5(33)4×(x3)2
=316×(x)10312×(x)6
=316-12× (x)10-6 = 34× x4= (3x)4

(iv)

32×78×136212×913
=32×72×76×136212×(13×7)3
=(21)2×76×136212×133×73
=76×136133×73
=916913=916-3=913

Page No 6.28:

Question 6:

Simplify:
(i) (35)11×(315)4-(35)18×(35)5
(ii) 16×2n+1-4×2n16×2n+2-2×2n+2
(iii) 10×5n+1+25×5n3×5n+2+10×5n+1
(iv) (16)7×(25)5× (81)3(15)7× (24)5 × (80)3

Answer:

We have
(i) (35)11× (315)4- (35)18×(35)5
   = 355 x 360 - 390 x 325
   = 3(55 + 60) - 3(90 + 25)
   = 3115 - 3115
   = 0

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

 (iii)

10×5n+1+25×5n3×5n+2+10×5n+1
=10×5n+1+(5)2×5n3×5n+2+2×5×5n+1
=10×5n+1+5×5n+13×5n+2+2×5×5n+1
=5n+1(10+5)3×5×5n+1+10×5n+1
=5n+1(15)5n+1(15+10)=5n+1×155n+1×25=1525=35


(iv)
(16)7×(25)5×(81)3(15)7×(24)5×(80)3
=(16)7×(52)5×(34)3(3×5)7×(3×8)5×16×53
=(16)7×(5)10×(3)1237×57×35×85×163×53
=(16)7×(5)10×(3)1237×35×57×53×85×163
=(16)7×(5)10×(3)12312×510×85×163
=(16)785×163
=(16)7-385=(16)485=(2×8)485=24×8485=248=168=2

Page No 6.28:

Question 7:

Find the values of n in each of the following:
(i) 52n×53=511
(ii) 9×3n=37
(iii) 8× 2n+2=32
(iv) 72n+1÷49=73
(v) 324×325=322n+1
(vi) 2310+3225=252n-2

Answer:

We have

(i) 52n x 53 = 511
= 52n+3 = 511
On equating the coefficients, we get
2n + 3 = 11
⇒2n = 11- 3
⇒2n = 8
⇒ n =82=4
(ii) 9 x 3n = 37
= (3)2 x 3n = 37
= (3)2+n = 37
On equating the coefficients, we get
2 + n = 7
⇒ n = 7 - 2  = 5

(iii) 8 x 2n+2 = 32
= (2)3 x 2n+2 = (2)5      [since 23 = 8 and 25 = 32]
= (2)3+n+2 = (2)5 
On equating the coefficients, we get
3 + n + 2 = 5
⇒ n + 5 = 5
⇒ n = 5 -5
⇒ n = 0

(iv) 72n+1 ÷ 49 = 73
= 72n+1 ÷ 72 = 73  [since 49 = 72]
=72n+172=73
=72n+1-2=73 [since aman=am-n]
= 72n-1=73
On equating the coefficients, we get
2n - 1 = 3
⇒ 2n = 3 + 1
⇒ 2n = 4
⇒ n = 42=2
(v) 324×325=322n+1
=32(4+5)=32(2n+1)
=329=322n+1
On equating the coefficients, we get
2n + 1 = 9
⇒ 2n = 9 - 1
⇒ 2n = 8
⇒ n =82=4
 (vi) 2310×3225=252n-2
=2310×3210=252n-2
=210×310310×210=252n-2
=1=252n-2
=250=252n-2 [since 250=1]
On equating the coefficients, we get
⇒ 0 = 2n - 2
⇒ 2n = 2
⇒ n = 22=1



Page No 6.29:

Question 8:

If 9n×32× 3n-(27)n(33)×23=127, find the value of n.

Answer:

We have

9n×32×3n-(27)n(33)5×23=127
=(32)n×32×3n-(33)n(3)15×23=127
=(3)2n+2+n-(3)3n(3)15×23=127
=(3)3n+2-(3)3n(3)15×23=127
=(3)3n×(3)2-(3)3n(3)15×23=127
=(3)3n(32-1)(3)15×23=127
=(3)3n ×8(3)15×23=127
=(3)3n ×23(3)15×23=127
=33n315=127
=33n-15=133
=33n-15=3-3
On equating the coefficients, we get
3n -15 = -3
⇒ 3n = -3 + 15
⇒ 3n = 12
⇒ n =123=4



Page No 6.30:

Question 1:

Express the following numbers in the standard form:
(i) 3908.78
(ii) 5,00,00,000
(iii) 3,18,65,00,000
(iv) 846 × 107
(v)723 × 109

Answer:

We have

(i) 3908.78 = 3.90878 x 103                                       [since the decimal point is moved 3 places to the left]

(ii) 5,00,00,000 = 5,00,00,000.00 = 5 x 107               [since the decimal point is moved 7 places to the left]

(iii) 3,18,65,00,000 = 3,18,65,00,000.00
                                = 3.1865 x 109                            [since the decimal point is moved 9 places to the left]

(iv) 846 × 107 = 8.46 x 102 x 107                               [since the decimal point is moved 2 places to the left]
                        = 8.46 x 109                                        [since am x an = am+n]

(v) 723 × 109 = 7.23 x 102 x 109                               [since the decimal point is moved 2 places to the left]
                      = 7.23 x 1011                                        [ since am x an = am+n]

Page No 6.30:

Question 2:

Write the following numbers in the usual form:
(i) 4.83 × 107
(ii) 3.21 × 105
(iii) 3.5 × 103

Answer:

We have
(i) 4.83 × 107 = 483 × 107-2                        [since the decimal point is moved two places to the right]
                      = 483 × 105 = 4,83,00,000

(ii) 3.21 × 105 = 321 x 105-2                       [since the decimal point is moved two places to the right]
                      = 321 x 103 = 3,21,000

(iii) 3.5 × 103 = 35 x 103-1                          [since the decimal point is moved one place to the right]
                       = 35 x 102 = 3,500

Page No 6.30:

Question 3:

Express the numbers appearing in the following statements in the standard form:
(i) The distance between the Earth and the Moon is 384,000,000 metres.
(ii) Diameter of the Earth is 1,27,56,000 metres.
(iii) Diameter of the Sun is 1,400,000,000 metres.
(iv) The universe is estimated to be about 12,000,000,000 years old.

Answer:

We have
(i) The distance between the Earth and the Moon is 3.84 x 108 metres.
     [Since the decimal point is moved 8 places to the left.]

(ii) The diameter of the Earth is 1.2756 x 107  metres.
      [Since the decimal point is moved 7 places to the left.]

(iii) The diameter of the Sun is 1.4 x 109 metres.
       [Since the decimal point is moved 9 places to the left.]

(iv) The universe is estimated to be about 1.2x 1010 years old.
       [Since the decimal point is moved 10 places to the left.]



Page No 6.31:

Question 1:

Write the following numbers in the expanded exponential forms:
(i) 20068
(ii) 420719
(iii) 7805192
(iv) 5004132
(v) 927303

Answer:

We have
(i) 20068 = 2 x 104 + 0 x 103 + 0 x 102 + 6 x 101 + 8 x 100
(ii) 420719 = 4 x 105 + 2 x 104 + 0 x 103 + 7 x 102 + 1 x 101 + 9 x 100
(iii) 7805192 = 7 x 106 + 8 x 105 + 0 x 104 + 5 x 103 + 1 x 102 + 9 x 101 + 2 x 100
(iv) 5004132 = 5 x 106 + 0 x 105 + 0 x 104  4 x 103 + 1 x 102 + 3 x 101 + 2 x 100
(v) 927303 = 9 x 105 + 2 x 104  + 7 x 103 + 3 x 102 +  0 x 101 + 3 x 100

Note: a0 = 1

Page No 6.31:

Question 2:

Find the number from each of the following expanded forms:
(i) 7 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
(ii) 5 × 105 + 4 × 104 + 2 × 103 + 3 × 100
(iii) 9 × 105 + 5 × 102 + 3 × 101
(iv) 3 × 104 + 4 × 102 + 5 × 100

Answer:

We have

(i) 7 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
    = 7 x 10000 + 6 x 1000 + 0 x 100 + 4 x 10 + 5 x 1 = 76045

(ii) 5 × 105 + 4 × 104 + 2 × 103 + 3 × 100
      = 
5 x 100000 + 4 x 10000 + 2 x 1000 + 3 x 1 = 542003

(iii) 9 × 105 + 5 × 102 + 3 × 101
       
= 9 x 100000 + 5 x 100 + 3 x 10 = 900530

(iv)
3 × 104 + 4 × 102 + 5 × 100
     = 3 x 10000 + 4 x 100 + 5 x 1 = 30405



Page No 6.32:

Question 1:

Mark the correct alternative in the following question:

6-1-8-1-1=a 124                           b 24                           c -24                           d -124

Answer:

6-1-8-1-1=16-18-1         As, x-1=1x=4-324-1=124-1=241                As, x-1=1x=24

Hence, the correct alternative is option (b).

Page No 6.32:

Question 2:

Mark the correct alternative in the following question:

232=a 64                           b 32                           c 256                           d 512

Answer:

Since, 232=29=512

Hence, the correct alternative is option (d).

Page No 6.32:

Question 3:

Mark the correct alternative in the following question:

3-1×5-1-1=a 115                           b -115                           c 15                           d -15

Answer:

3-1×5-1-1=13×15-1           As, x-1=1x=115-1=151                          As, x-1=1x=15

Hence, the correct alternative is option (c).

Page No 6.32:

Question 4:

Mark the correct alternative in the folowing question:

-35-1=a 35                             b 53                             c -53                             d -35

Answer:

Since, -35-1=-53             As, x-1=1x

Hence, the correct alternative is option (c).

Page No 6.32:

Question 5:

Mark the correct alternative in the folowing question:

-1301+-1302+-1303+...+-1400a 1                          b 101                          c 100                          d 0

Answer:

Since,

-1301+-1302+-1303+...+-1400=-1+1+-1+...+1        As, -1odd=-1 and -1even=1=-50+50             As, there are 50 -1's and 50 1's=0

Hence, the correct alternative is option is (d).

Page No 6.32:

Question 6:

Mark the correct alternative in the folowing question:

If a=25, then a250+a025=a 25                           b 26                           c 24                           d 0

Answer:

Since,

a250+a025=a1+a0            As, 250=1 and 025=0=a+1          As, a0=1=25+1=26

Hence, the correct alternative is option (b).

Page No 6.32:

Question 7:

Mark the correct alternative in the following question:

13-3-12-3÷14-3=a 1964                            b 6419                            c 2716                            d -1964

Answer:

Since,

13-3-12-3÷14-3=313-213÷413           As, x-1=1x=33-23÷43=27-8÷64=19÷64=1964

Hence, the correct alternative is option (a).

Page No 6.32:

Question 8:

Mark the correct alternative in the following question:

252-15232=a 4000                            b 8000                            c 3125                            d 1024

Answer:

Since,

252-15232=625-22532=40032=20232=202×32         As, xmn=xmn=203=8000

Hence, the correct alternative is option (b).

Page No 6.32:

Question 9:

Mark the correct alternative in the following question:

If 53-5×5311=538x, then x=?a -12                            b -34                            c 34                            d 43

Answer:

Since,

53-5×5311=538x53-5+11=538x         As, xm×n=xm+n536=538xComparing the exponent of both the sides, we get8x=6x=68 x=34

Hence, the correct alternative is option (c).

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Question 10:

Mark the correct alternative in the following question:

-132-2-1=a 181                            b 19                            c -181                            d -19

Answer:

-132-2-1=-132-2×-1                As, xmn=xmn=-132×2                                  As, xmn=xmn=-134=-1434                                          As, xym=xmym=181

Hence, the correct alternative is option (a).

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Question 11:

Mark the correct alternative in the following question:

14412+2561232-2=a 8                            b 4                            c -4                            d -8

Answer:

Since,

14412+2561232-2=12212+162129-2=122×12+162×127              As, xmn=xmn=12+167=287=4

Hence, the correct alternative is option (b).



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Question 12:

Mark the correct alternative in the following question:

1+3+5+7+9+1132=a 36                            b 216                            c 256                            d None of these

Answer:

Since,

1+3+5+7+9+1132=3632=6232=62×32               As, xmn=xmn=63=216

Hence, the correct alternative is option (b).

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Question 13:

Mark the correct alternative in the following question:

If abc=0, then xabcxbca=a 3                                    b 0                                    c -1                                    d 1

Answer:

xabcxbca=xabcxbac         As, xmn=xmn=xabcxabc             As, xmn=xmn=x0x0                As, abc=0=11                   As, x0=1=1

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Question 14:

Mark the correct alternative in the following question:

234=a 243                                    b 234                                    c 243                                    d None of these

Answer:

Since,

234=23×4                        As, xmn=xmn=24×3=243                        As, xmn=xmn

Hence, the correct alternative is option (c).

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Question 15:

Mark the correct alternative in the following question:

332-31257=a 64                                    b 16                                    c 32                                    d 4

Answer:

Since,

332-31257=1089-96157=12857=2757=27×57              As, xmn=xmn=25=32

Hence, the correct alternative is option (c).

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Question 16:

Mark the correct alternative in the following question:

If abc=0, then find the value of xabc.a 1                                    b a                                    c b                                    d c

Answer:

Since,

xabc=xabc                   As, xmn=xmn=xabc                      As, xmn=xmn=x0                          As, abc=0=1                            As, x0=1

Hence, the correct alternative is option (a).

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Question 17:

Mark the correct alternative in the following question:

If a=3-3-33 and b=33-3-3, then ab-ba=a 0                                    b 1                                    c -1                                    d 2

Answer:

Since, a=3-3-33=133-33        As, x-m=1xm=127-271=127-27×271×27=127-72927=1-72927=-72827Also, b=33-3-3=33-133           As, x-m=1xm=271-127=27×271×27-127=72927-127=729-127=72827Now,ab-ba=-7282772827-72827-72827=-1--1=-1+1=0

Hence, the correct alternative is option (a).

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Question 18:

Mark the correct alternative in the following question:

What should be multiplied to 6-2 so that the product may be equal to 216?a 64                                    b 65                                    c 63                                    d 6

Answer:

Let 6m should be multiplied to 6-2.A.T.Q.6-2×6m=2166-2+m=63Comparing th exponent of both the sides, we get-2+m=3m=3+2 m=5

So, 65 should be multiplied.

Hence, the correct alternative is option (b).

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Question 19:

Mark the correct alternative in the following question:

If xyz=0, then find the value of axyz+ayzx+azxy=a 3                                    b 2                                    c 1                                    d 0

Answer:

Since,

axyz+ayzx+azxy=axyz+ayzx+azxy                  As, xmn=xmn=axyz+axyz+axyz=a0+a0+a0=1+1+1                                 As, x0=1=3

Hence, the correct option is (a).

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Question 20:

Choose the correct alternative in the following question:

If 2n=4096, then 2n-5=a 128                              b 64                              c 256                              d 32

Answer:

As, 2n=40962n=212Comparing the exponent of both the sides, we getn=12Now, 2n-5=212-5=27=128

Hence, the correct alternative is option (a).

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Question 21:

Choose the correct alternative in the following question:

The number 4,70,394 in standard form is written as

(a) 4.70394 × 105                   (b) 4.70394 × 104                   (c) 47.0394 × 104                   (d) 4703.94 × 102

Answer:

Since, 4,70,394 = 4.70394 × 100000 = 4.70394 × 105.

So, the number 4,70,394 in standard form is written as 4.70394 × 105.

Hence, the correct alternative is option (a).

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Question 22:

Choose the correct alternative in the following question:

The number 2.35 × 104 in the usual form is written as

(a) 2.35 × 103                            (b) 23500                            (c) 2350000                            (d) 235 × 104

Answer:

Since, 2.35 × 104 =  2.35 × 10000 = 23500

So, the number 2.35 × 104 in the usual form is written as 23500.

Hence, the correct alternative is option (b).

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Question 23:

Choose the correct alternative in the following question:

If 3x=6561, then 3x-3=a 81                              b 243                              c 729                              d 27

Answer:

As, 3x=65613x=38Comparing the exponent of both the sides, we getx=8Now, 3x-3=38-3=35=243

Hence, the correct alternative is option (b).

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Question 24:

Choose the correct alternative in the following question:

If 2n=1024, then 2n2+2=a 64                                    b 128                                    c 256                                    d 512

Answer:

As, 2n=10242n=210Comparing the exponent of both the sides, we getn=10Now,2n2+2=2102+2=25+2=27=128

Hence, the correct alternative is option (b).

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Question 25:

Choose the correct alternative in the following question:

84+8212=a 84                                    b 877                                    c 72                                    d 865

Answer:

Since,

84+8212=82+2+8212=82×82+8212              As, xm+n=xm×xn=82×82+112              As, ab+ac=a×b+c=8212×82+112          As, abm=am×bm=82×12×64+112=8×6512=865

Hence, the correct alternative is option (d).



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