# NCERT Solutions for Class 8 Maths Chapter 10 Exponents and Powers (Updated Pattern)

An expression that represents repeated multiplication of the same factor is called a power. For example, in the case of 4^{2}, the number 4 is called the base, and the number 2 is the exponent. An exponent corresponds to the number of times the base is utilised as a factor in an expression.

**——————————Exercise 10.1——————————-**

**1. Evaluate:**

**(i) 3 ^{-2 }(ii) (-4)^{-2 }(iii) (1/2)^{-5} **

**Solution:**

(i) 3^{-2 }= (1/3)^{2}

= 1/9

(ii) **(-4) ^{-2 }= (1/-4)^{2}**

= 1/16

(iii) **(1/2) ^{-5} = (2/1)^{5}**

= 2^{5}

= 32

**2. Simplify and express the result in power notation with a positive exponent:**

**(i) (-4) ^{4 }÷(-4)^{8} **

**(ii) (1/2 ^{3})^{2} **

**(iii) -(3) ^{4}×(5/3)^{4} **

**(iv) (3 ^{-7}÷3^{-10})×3^{-5} **

**(v) 2 ^{-3}×(-7)^{-3}**

**Solution:**

(I)

= (-4)^{5}/(-4)^{8}

= (-4)^{5-8}

= 1/(-4)^{3}

(ii) **(1/2 ^{3})^{2} **

= 1^{2}/(2^{3})^{2}

= 1/2^{3×2} = 1/2^{6}

(iii) **-(3) ^{4}×(5/3)^{4} **

= (-1)^{4}×3^{4}×(5^{4}/3^{4} )

= 3^{(4-4)}×5^{4}

= 3^{0}×5^{4} = 5^{4}

**(iv) **

= (3^{-7}/3^{-10})× 3^{-5}

= 3^{-7 – (-10) }× 3^{-5}

= 3^{(-7+10)}×3^{-5}

= 3^{3}×3^{-5}

= 3^{(3+-5)}

= 3^{-2}

=1/3^{2}

**(v) 2 ^{-3}×(-7)^{ – 3}**

= (2×-7)^{-3}

(Because a^{m}×b^{m} = (ab)^{m})

= 1/(2×-7)^{3}

= 1/(-14)^{3}

**3. Find the value of:**

**(i) (3 ^{0}+4^{-1})×2^{2}**

**(ii) (2 ^{-1}×4^{-1})÷2^{ – 2}**

**(iii) (1/2) ^{-2}+(1/3)^{-2}+(1/4)^{-2}**

**(iv) (3 ^{-1}+4^{-1}+5^{-1})^{0}**

**(v) {(-2/3) ^{-2}}^{2}**

**Solution:**

**(i)(3 ^{0}+4^{– 1})×2^{2} = (1+(1/4))×2^{2}**

= ((4+1)/4 )×2^{2}

= (5/4)×2^{2}

= (5/2^{2})×2^{2}

= 5×2^{(2-2)}

= 5×2^{0}

= 5×1 = 5

**(ii)(2 ^{-1}×4^{-1})÷2^{-2}**

= [(1/2)×(1/4)] ÷(1/4)

= (1/2×1/2^{2} )÷ 1/4

= 1/2^{3}÷1/4

= (1/8)×(4)

= 1/2

(iii) **(1/2) ^{-2}+(1/3)^{-2}+(1/4)^{-2}**

= (2^{-1})^{-2}+(3^{-1})^{-2}+(4^{-1})^{-2}

= 2^{(-1×-2)}+3^{(-1×-2)}+4^{(-1×-2)}

= 2^{2}+3^{2}+4^{2}

= 4+9+16

=29

**(iv) (3 ^{-1}+4^{-1}+5^{-1})^{0}**

= 1

(v) **{(-2/3) ^{-2}}^{2} = (-2/3)^{-2×2}**

= (-2/3)^{-4}

= (-3/2)^{4}

= 81/16

**4. Evaluate:**

**(i) (8 ^{-1}×5^{3})/2^{-4}**

**(ii) (5 ^{-1}×2^{-2})×6^{-1} **

**Solution:**

**(i) (8 ^{-1}×5^{3})/2^{-4}**

=

= 2×125 = 250

**(ii)** **(5 ^{-1}×2^{-2})×6^{-1} **

= (1/10)×1/6

= 1/60

**NCERT Solutions for Class 8 Maths Chapter 9 Mensuration**

**5. Find the value of m for which 5^{m }÷ 5^{-3} = 5^{5}**

**Solution:**

5* ^{m}* ÷ 5

^{-3}= 5

^{5}

5^{(m-(-3) )} = 5^{5}

5^{m+3} =5^{5}

Comparing exponents on both sides, we get

m+3 = 5

m = 5-3

m = 2

**6. Evaluate:**

**(i) **

**(ii) **

**Solution:**

**(i)**

= 3-4

= -1

**(ii)**

=

=

=

= 512/125

**7. Simplify the following:**

**(i) **

**(ii) **

Solution:

**(i)**

=

=

**(ii)**

=

=

=

=

= 1×1×3125

= 3125

**——————————Exercise 10.2——————————-**

**1. Express the following numbers in standard form.**

**(i) 0.0000000000085**

**(ii) 0.00000000000942**

**(iii) 6020000000000000**

**(iv) 0.00000000837**

**(v) 31860000000**

**Solution:**

(i) 0.0000000000085 = 0.0000000000085×(10^{12}/10^{12}) = 8.5 ×10^{-12}

(ii) 0.00000000000942 = 0.00000000000942×(10^{12}/10^{12}) = 9.42×10^{-12}

(iii) 6020000000000000 = 6020000000000000×(10^{15}/10^{15}) = 6.02×10^{15}

(iv) 0.00000000837 = 0.00000000837×(10^{9}/10^{9}) = 8.37×10^{-9}

(v) 31860000000 = 31860000000×(10^{10}/10^{10}) = 3.186×10^{10}

**2. Express the following numbers in the usual form.**

**(i) 3.02×10 ^{-6}**

**(ii) 4.5×10 ^{4}**

**(iii) 3×10 ^{-8}**

**(iv) 1.0001×10 ^{9}**

**(v) 5.8×10 ^{12}**

**(vi) 3.61492×10 ^{6}**

Solution:

(i) 3.02**×**10^{-6} = 3.02/10^{6} = 0 .00000302

(ii) 4.5**×**10^{4} = 4.5**×**10000 = 45000

(iii) 3**×**10^{-8} = 3/10^{8} = 0.00000003

(iv) 1.0001**×**10^{9} = 1000100000

(v) 5.8**×**10^{12} = 5.8**×**1000000000000 = 5800000000000

(vi) 3.61492**×**10^{6 }= 3.61492**×**1000000 = 3614920

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**3. Express the number appearing in the following statements in standard form.**

**(i) 1 micron is equal to 1/1000000 m.(ii) The charge** of an electron is 0.000, 000, 000, 000, 000, 000, 16 coulomb.

(iii) Size of bacteria is 0.0000005 m

(iv) The size of a plant cell is 0.00001275 m

(v) The

**Thickness of a thick paper is 0.07 mm**

**Solution:**

**(i) 1 micron = 1/1000000**

= 1/10^{6}

= 1×10^{-6}

**(ii) Charge of an electron is 0.00000000000000000016 coulombs**

= 0.00000000000000000016**×**10^{19}/10^{19}

= 1.6**×**10^{-19 }coulomb

**(iii) Size of bacteria = 0.0000005**

= 5/10000000 = 5/10^{7} = 5×10^{-7} m

**(iv) Size of a plant cell is 0.00001275 m**

= 0.00001275×10^{5}/10^{5}

= 1.275×10^{-5}m

**(v) Thickness of a thick paper = 0.07 mm**

0.07 mm = 7/100 mm = 7/10^{2} = 7×10^{-2} mm

**4. In a stack, there are 5 books, each having a thickness of 20 mm and 5 paper sheets, each having a thickness of 0.016 mm. What is the total thickness of the stack?**

Solution:

The thickness of one book = 20 mm

Thickness of 5 books = 20×5 = 100 mm

The thickness of one paper = 0.016 mm

Thickness of 5 papers = 0.016×5 = 0.08 mm

Total thickness of a stack = 100+0.08 = 100.08 mm

= 100.08×10^{2}/10^{2} mm

mm