ncert exemplar problems class 8 mathematics exponents and powers

NCERT Exemplar Problems Class 8 Mathematics  Chapter 8 Exponents and Powers

NCERT Exemplar SolutionsNCERT Solutions MathsRD Sharma Solutions

Multiple Choice Questions
Question. 1 In  2ⁿ, n is known as
(a) base (b) constant
(c) exponent (d) variable
Solution.
(c) We know that an is called the nth power of a; and is also read as a raised to the power n.
The rational number a is called the base and n is called the exponent (power or index). In the same way in 2ⁿ,n is known as exponent.

Question. 2 For a fixed base, if the exponent decreases by 1, the number becomes
(a) one-tenth of the previous number
(b) ten times of the previous number
(c) hundredth of the previous number
(d) hundred times of the previous number
Solution.
(a) For a fixed base, if the exponent decreases by 1, the number becomes one-tenth of the previous number.

Question. 3

Solution.

Question. 4 The value of 1/4^-2 is
Solution.

Question. 5 The value of 3⁵   ÷ 3^-6   is
(a) 3⁵     (b) 3^-6        (c) 3¹¹        (d) 3^-11
Solution.

Question. 6

Solution.

Question. 7

Solution.

Question. 8 The multiplicative inverse of 10^-100  is
(a) 10      (b) 100     (c) 10¹⁰⁰     (d)10^-100  
Solution.

Question.9  The value of (-2)^{2 x 3-1}  is
(a) 32      (b) 64       (c) -32    (d) -64
Solution.

Question.10

Solution.

Question. 11

Solution.

Question. 12 If x be any non-zero integer and w, n be negative integers, then x^m    x xⁿ  is equal to
(a) x^m         (b)x^(m+n)   (c) xⁿ               (d) x^(m-n)
Solution.

Question. 13 If y be any non-zero integer, then y⁰  is equal to
(a) 1                (b) 0                (c) –  1               (d) not defined
Solution.

Question.14 If x be any non-zero integer, then  _X^-1 is equal to
(a) x           (b) 1/x           (c) – x         (d) -1/x
Solution.

Question. 15

Solution.

Question. 16

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Question. 17

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Question. 18

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Question. 19

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Question. 20

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Question. 21

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Question. 22

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Question. 23

Solution.

Question.24 The standard form for 0.000064 is
(a) 64 x 10⁴   (b)  64 x 10^-4  (c) 6.4 x  10⁵  (d) 6.4 x 10^-5
Solution.
(d) Given, 0.000064 = 0.  64 x 10^-4 =6.4 x 10^-5
Hence, standard form of 0.000064 is 6.4 x 10^-5.

Question. 25 The standard form for 234000000 is
(a) 2.34 x 10⁸     (b) 0.234 x 10⁹    
(c) 2.34 x 10^-8     (d) 0.234 x 10^{– 9}    
Solution.
(a) Given, 234000000 = 234 x 10⁶ = 2.34 x 10⁺⁶  = 2.34 x 10⁸
Hence, standard form of 234000000 is 2.34 x 10⁸.

Question.26 The usual form for 2.03 x 10^{-5 is}      
(a) 0.203 (b) 0.00203 (c) 203000 (d) 0.0000203
Solution.

Question. 27

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Question. 28

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Question. 29

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Question. 30

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Question. 31

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Question. 32

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Question. 33

Solution.

Fill in the Blanks
In questions 34 to 65, fill in the blanks to make the statements true.

Question. 34 The multiplicative inverse of 10¹⁰  is_________
Solution.

Question.35  a³  x a^-10= _________
Solution.

Question.36 5⁰  = _________
Solution.

Question.37 5⁵  x 5^-5= _________
Solution.

Question.38

Solution.

Question. 39 The expression for 8^-2  as a power with the base 2 is_________
Solution.

Question. 40 Very small numbers can be expressed in standard form by using_________
exponents
Solution.
Very small numbers can be expressed in standard form by using negative exponents, i.e. 0.000023 = 2.3 x  10^-3

Question. 41 Very large numbers can be expressed in standard form by using
exponents.
Solution.
Very large numbers can be expressed in standard form by using positive exponents,
i.e. 23000 = 23 x 10³ =2.3 x  10³  x 10¹ =2.3 x 10⁴

Question. 42 By multiplying (10)⁵  by (10)^-10, we get
Solution.

Question.43

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Question.44

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Question.45

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Question.46

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Question.47

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Question.48

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Question.49

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Question.50

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Question.51

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Question.52

Solution.

Question.53 The value of 3 x 10^-7    is equal to_______
Solution
Given, 3 x 10^-7    = 3.0 x 10^-7    
Now, placing decimal seven place towards left of original position, we get 0.0000003. Hence, the value of 3 x 10^-7 is equal to 0.0000003.

Question.54 To add the numbers given in standard form, we first convert them into number with_______exponents.
Solution.
To add the numbers given in standard form, we first convert them into numbers with equal exponents.
e.g. 2.46 x 10⁶    + 24.6 x 105 = 2.46 x 10⁵ + 2.46 x 10⁶ = 4.92 x 10⁶

Question.55 The standard form for 32500000000 is_______.
Solution.
For standard form, 32500000000 = 3250 x 10² x 10² x 10³
= 3250 x 10⁷ = 3.250 x 10¹⁰ or 3.25 x 10¹⁰
Hence, the standard form for 32500000000 is 3.25 x 10¹⁰.

Question. 56 The standard form for 0.000000008 is_______.
Solution.
For standard form, 0.000000008 = 0.8 x  10^-8= 8 x  10^-9 =8.0 x  10^-9
Hence, the standard form for 0.000000008 is 8.0 x 10-9

Question.57 The usual form for 2.3 x  10^-10  is_______.
Solution. For usual form, 2.3 x  10^-10 = 0.23 x 10^-11
= 0.00000000023
Hence, the usual form for 2.3 x 10^-10 is 0.00000000023.

Question. 58 On dividing 8⁵ by_______. we get 8.
Solution.

Question. 59

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Question. 60

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Question.61

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Question.62

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Question.63

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Question.64

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Question.65

Solution.

True / False
In questions 66 to 90, state whether the given statements are True or False.

Question.66

Solution.

Question.67

Solution.

Question.68

Solution.

Question.69 24.58 = 2 x 10 + 4 x 1+5 x 10 + 8 x 100
Solution. False
R H S = 2 x 10+ 4 x 1+ 5 x 10+ 8 x 100=20+ 4 + 50+ 800=874 L H S ≠ R H S

Question.70 329.25 = 3 x  10²+ 2 x 10¹ + 9 x 10⁰  + 2 x 10^-1 + 5 x  10^-2
Solution.

Question.71

Solution.

Question.72

Solution.

Question.73

Solution.

Question. 74  5⁰ = 5
Solution.

Question. 75 (-2)⁰ = 2
Solution.

Question.76

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Question. 77 (-6) ° = – 1
Solution.

Question. 78

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Question. 79

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Question. 80

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Question. 81

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Question. 82

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Question. 83

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Question. 84

Solution.

Question.85 The standard form for 0.000037 is 3.7 x 10^-5
Solution. True
For standard form, 0.000037 = 0.37 x 10^-4= 3.7 x 10^-5

Question. 86 The standard form for 203000 is 2.03 x 105.
Solution. True
For standard form, 203000 = 203 x 10 x 10 x 10 = 203 x 10³
= 2.03 x 10² x 10³= 2.03 x  10⁵

Question. 87 The usual form for 2 x 10^-2 is not equal to 0.02.
Solution.

Question. 88 The value of 5^-2 is equal to 25.
Solution. False

Question. 89 Large numbers can be expressed in the standard form by using positive exponents.
Solution.True
e.g. 2360000 = 236 x 10 x 10 x 10 x 10= 236 x 10⁴
‘ = 2.36 x  10⁴ x 10²=2.36 x  10⁶

Question. 90

Solution.

Question. 91 Solve the following,

Solution.

Question. 92 Express 3^-5 x  3^-4 as a power of 3 with positive exponent.
Solution.

Question. 93 Express 16^-2 as a power with the base 2.
Solution.

Question. 94

Solution.

Question. 95

Solution.

Question. 96 Express as a power of a rational number with negative exponent.

Solution.

Question. 97 Find the product of the cube of (-2) and the square of (+4).
Solution.

Question.98 Simplify

Solution.

Question. 99 Find the value of x, so that

Solution.

Question. 100 Divide 293 by 1000000 and express the result in standard form.
Solution.

Question. 101

Solution.

Q. 102 By what number should we multiply (-29) °, so that the product becomes (+29) °.
Solution.

Question. 103 By what number should (-15)^-1 be divided so that quotient may be equal to (-15)^-1  ?
Solution.

Question.104 Find the multiplicative inverse of (-7)²Ã· (90)^-1
Solution.

Question.105

Solution.

Question.106 Write 390000000 in the standard form.
Solution.

Question. 107 Write 0.000005678 in the standard form.
Solution.
For standard form, 0.000005678 = 0.5678 x 10^-5= 5.678 x 10^-5 x 10^-1= 5.678 x 10^-6  Hence, 5.678 x 10^-6 is the standard form of 0.000005678.

Question.108 Express the product of 3.2 x 10⁶ and 4.1 x 10¹ in the standard form.
Solution.

Question.109

Solution.

Question. 110 Some migratory birds travel as much as 15000 km to escape the extreme climatic conditions at home. Write the distance in metres using scientific notation.
Solution.

Question. 111 Pluto is 5913000000 m from the Sun. Express this in the standard form.
Solution.

Question. 112 Special balances can weigh something as 0.00000001 gram. Express this number in the standard form.
Solution.

Question. 113 A sugar factory has annual sales of 3 billion 720 million kilograms of sugar. Express this number in the standard form.
Solution.

Question. 114 The number of red blood cells per cubic millimetre of blood is approximately mm³)
Solution. The average body contain 5 L of blood.
Also, the number of red blood cells per cubic millimetre of blood is approximately 5.5 million.
Blood contained by body = 5 L = 5 x 100000 mm³
Red blood cells = 5 x 100000 mm³
Blood = 5.5 x 1000000 x 5 x 100000= 55 x 5 x  10^{5 + 5}
= 275 x  10¹⁰  = 2.75 x 10¹⁰ x 10² = 2.75 x 10¹²

Question. 115 Express each of the following in standard form:

Solution.

Question.116

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Question.117

Solution.

In questions 118 and 119, find the value of n.
Question.118

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Question.119

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Question.120

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Question.121

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Question.122

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Question.123 A new born bear weights 4 kg. How many kilograms might a five year old bear weight if its weight increases by the power of 2 in 5 yr?
Solution.
Weight of new born bear = 4 kg
Weight increases by the power of 2 in 5 yr.
Weight of bear in 5 yr = (4)² = 16 kg

Question.124 The cell of a bacteria doubles in every 30 min. A scientist begins with a single cell. How many cells will be thereafter (a) 12 h (b) 24 h ?
Solution.

Question.125 Planet A is at a distance of 9.35 x 10⁶ km from Earth and planet B is 6.27 x 107 km from Earth. Which planet is nearer to Earth?
Solution.
Distance between planet  A and Earth = 9.35 x 10⁶  km Distance between planet B and Earth = 6.27 x 10⁷ km
For finding difference between above two distances, we have to change both in same exponent of 10, i.e. 9.35 x.10⁶ = 0.935 x 10⁷, clearly 6.27 x 10⁷ is greater.
So, planet A is nearer to Earth.

Question.126 The cells of a bacteria double itself every hour. How many cells will be there after 8 h, if initially we start with 1 cell. Express the answer in powers.
Solution.

Question. 127 An insect is on the 0 point of a number line, hopping towards 1. She covers half the distance from her current location to 1 with each hop.
So, she will be at 1/2 after one hop, 3/4 after two hops and so on.

(a) Make a table showing the insect’s Location for the first 10 hops.
(b) Where will the insect be after n hops?
(c) Will the insect ever get to 1? Explain.
Solution.
(a) On the basis of given information in the question, we can arrange the following table which shows the insect’s location for the first 10 hops.

Question. 128 Predicting the ones digit, copy and complete this table and answer the questions that follow.

Solution.
(a) On the basis of given pattern in 1^x  and 2^x , we can make more patterns for  3^x  4^x , 5^x ,6^x , 7^x , 8^x , 9^x , 10^x .
Thus, we have following table which shows all details about the patterns.

Question. 129 Astronomy The table shows the mass of the planets, the Sun and the Moon in our solar system.

Solution.

Question. 130 Investigating Solar System The table shows the average distance from each planet in our solar system to the Sun.

Solution.

Question. 131 This table shows the mass of one atom for five chemical elements.
Use it to answer the question given.

(a) Which is the heaviest element?
(b) Which element is lighter, Silver or Titanium?
(c) List all the five elements in order from lightest to heaviest.
Solution.

Question. 132 The planet Uranus is approximately 2,896,819,200,000 metres away from the Sun. What is this distance in standard form?
Solution.
Distance between the planet Uranus and the Sun is 2896819200000 m.
Standard form of 2896819200000 = 28968192 x 10 x 10 x 10 x 10 x 10
= 28968192 x 10⁵ = 2.8968192 x 10¹² m

Question. 133 An inch is approximately equal to 0.02543 metres. Write this distance in standard form.
Solution. Standard form of 0.02543 m = 0.2543 x  10^-1 m = 2.543 x 10^-2 m Hence,’ standard form of 0.025434s 2.543 x 10^-2 m.

Question.134 The volume of the Earth is approximately 7.67 x 10^-7 times the volume
of the Sun. Express this figure in usual form.
Solution.

Question.135 An electron’s mass is approximately 9.1093826 x 10^-31 kilograms. What is its mass in grams?
Solution.

Question. 136 At the end of the 20th century, the world population was approximately 6.1 x 10⁹ people. Express this population in usual form. How would you say this number in words?
Solution.
Given, at the end of the 20th century, the world population was 6,1 x 10⁹ (approx). People population in usual form = 6.1 x 10⁹ = 6100000000 Hence, population in usual form was six thousand one hundred million.

Question.137 While studying her family’s history, Shikha discovers records of ancestors 12 generations back. She wonders how many ancestors she had in the past 12 generations. She starts to make a diagram to help her figure this out. The diagram soon becomes very complex

Solution.
(a) On the basis of given diagram, we can make a table that shows the number of ancestors in each of the 12 generations.

Question. 138 About 230 billion litres of water flows through a river each day, how many litres of water flows through that river in a week? How many litres of water flows through the river in an year? Write your answer in standard notation.
Solution.

Question. 139 A half-life is the amount of time that it takes for a radioactive substance to decay one-half of its original quantity.
Suppose radioactive decay causes 300 grams of a substance to decrease 300 x 2^-3 grams after 3 half-lives. Evaluate 300 x  2^-3  to determine how many grams of the substance is left.
Explain why the expression 300 x 2^-n can be used to find the amount of the substance that remains after n half-lives.
Solution.

Question. 140 Consider a quantity of a radioactive substance. The fraction of this quantity that remains after t half-lives can be found by using the expression 3^-t.
(a) What fraction of the substance remains after 7 half-lives?
(b) After how many half-lives will the fraction be 1/243 of the original?
Solution.

Question. 141 One fermi is equal to  10^-15  metre. The radius of a proton is 1.3 fermi. Write the radius of a proton (in metres) in standard form.
Solution. The radius of a proton is 1.3 fermi.
One fermi is equal to  10^-15 m.
So, the radius of the proton is 1.3 x  10^-15 m.
Hence, standard form of radius of the proton is 1.3 x  10^-15  m.

Question. 142 The paper clip below has the indicated length. What is the length in Standard form.

Solution.

Length of the paper clip = 0.05 m
In standard form, 0.05 m = 0.5 x  10^-1  = 5.0 x 10^-2  m
Hence, the length of the paper clip in standard form is 5.0 x 10^-2  m

Question.143 Use the properties of exponents to verify that each statement is true.