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ncert solutions class 8 maths chapter 12

 

ncert solutions class 8 maths chapter 12 deals with Exponents and Powers.

 

ncert solutions for class 8 maths chapter 12 exercise 12.1

class 8 maths chapter 12 exercise 12.1 (1)

1. Evaluate.

(i) (3)⁻²      (ii) (-4)⁻²       (iii) (½)⁻⁵

Solution to class 8 maths chapter 12 exercise 12.1 (1)

(i) (3)⁻²

= (¹/₃)²= ¹/₉     [∵ a⁻ᵐ= (¹/ₐ)ᵐ]

(ii) (-4)⁻²

= (- ¼)²= ¹/₁₆     [∵ a⁻ᵐ= (¹/ₐ)ᵐ]

(iii) (½)⁻⁵

= (²/₁)⁵= 32     [∵ a⁻ᵐ= (¹/ₐ)ᵐ]

class 8 maths chapter 12 exercise 12.1 (2)

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

(i) (-4)⁵ ÷ (-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 to class 8 maths chapter 12 exercise 12.1 (2)

(i) (-4)⁵ ÷ (-4)⁸ 

=(–4)5–8 [∵am÷an=am–n]

=(–4)–3=1(–4)3 [∵ a–m=1am]

(ii) 1232=12232 ∵ abm=ambm

=123×2=126

(iii)(–3)4×534=–34×5434

=–14×34×5434        [∵(ab)m=ambm]

=34–4×54        [∵am÷an=am–n]

=30×54=54           ∵a0=1

(iv) (3–7÷ 3–10)×3–5

=3–7–(–10)×3–5

=3–7+10×3–5=33×3–5

=33+(–5)           [∵am×an=am+n]

=3–2=132     ∵a–m=1am

(v) 2–3×–7–3=123×1–73   ∵a–m=1am

=12×–73=1–143   ∵abm=ambm

class 8 maths chapter 12 exercise 12.1 (3)

3  Find the value of 

(i) (3⁰+4⁻¹) ×2² (ii) (2⁻¹ ×4⁻¹)÷2⁻²

(iii) (½)⁻²+(⅓)⁻²+(¼)⁻² (iv) (3⁻¹+4⁻¹+5⁻¹)⁰

(v){(– ⅔)⁻²}²

Solution to class 8 maths chapter 12 exercise 12.1 (3)

(i) (3⁰+4⁻¹) ×2²=(1+¼) ×2²  (∵a⁻¹=¹/ₐ)

=(⁵/₄) ×4 =5

(ii) (2⁻¹ ×4⁻¹)÷2⁻²

=(½ ×¼) ÷ ¹/₍₂₎²

=⅛ ÷ ¼ =⅛ × 4=½

(iii) (½)⁻²+(⅓)⁻²+(¼)⁻²

=(2⁻¹)⁻² + (3⁻¹)⁻² +(4⁻¹)⁻²

=2⁻¹ˣ⁽⁻²⁾+3⁻¹ˣ⁽⁻²⁾+4⁻¹ˣ⁽⁻²⁾ {∵ (aᵐ)ⁿ =aᵐⁿ}

=2² + 3² + 4²

=4+9+16=29

(iv) (3⁻¹+4⁻¹+5⁻¹)⁰

=(⅓ + ¼ + ⅕)⁰

=[⁽²⁰⁺¹⁵⁺¹²⁾/₆₀]⁰

=(⁴⁷/₆₀)⁰

=1

(v){(– ⅔)⁻²}²

=(– ⅔)⁻⁴   {∵ (aᵐ)ⁿ =aᵐⁿ}

=( –  ³/₂)⁴      {∵a⁻ᵐ =¹/₍ₐ₎ᵐ}

= ⁸¹/₁₆

class 8 maths ch÷apter 12 exercise 12.1 (4)

4  Evaluate :

(i) 8–1×532–4          ii 5–1×2–1×6–1

Solution to class 8 maths chapter 12 exercise 12.1 (4)

(i) 8–1×532–4=23–1×532–4  ∵amn=am×n

=2–3––4×53

=2–3+4×53   ∵am÷an=am–n

=2×125=250

ii 5–1×2–1×6–1

=15×12×16  ∵a–m=1am

=110×16=160

class 8 maths ch÷apter 12 exercise 12.1 (5)

5. Find the value of m for which 5ᵐ÷5⁻³=5⁵

Solution to class 8 maths chapter 12 exercise 12.1 (5)

5ᵐ÷5⁻³=5⁵

⇒ 5ᵐ ⁻ ⁽⁻³⁾=5⁵

Comparing exponents both sides, we get 

⇒m+3=5

⇒m=5 –3 

⇒m=2

class 8 maths ch÷apter 12 exercise 12.1 (6)

6  Evaluate:

(i) 13–1–14–1–1           (ii) 58–7×85–4

Solution to class 8 maths chapter 12 exercise 12.1 (6)

(i)

13–1–14–1=311–411 

∵a–m=1am

=[3-4]=-1

(ii)

58–7×85–4=5–78–7×8–45–4

∵abm=ambm

=5–7––4×8–4––7  ∵am÷an=am–n

=5–7+4×8–4+7

=5–3×83=8353  ∵a–m=1am

=512125

class 8 maths ch÷apter 12 exercise 12.1 (7)

7  Simplify:

(i) 25×t–45–3×10×t–8 t≠0  (ii)3–5×10–5×1255–7×6–5

Solution to class 8 maths chapter 12 exercise 12.1 (7)

(i)

25×t–45–3×10×t–8=52×t–45–3×5×2×t–8

=52––3–1×t–4––82 ∵am÷an=am–n

=52+3–1×t–4+82

=54×t42=6252 t4

(ii)

3–5×10–5×1255–7×6–5=3–5×2×5–5×535–7×2×3–5

=3–5×2–5×5–5×535–7×2–5×3–5  ∵abm=ambm

=3–5×2–5×5–5+35–7×2–5×3–5

=3–5×2–5×5–25–7×2–5×3–5  ∵ am×an=am+n

=3–5––5×2–5––5×5–2––7

∵am÷an=am–n

=3–5+5×2–5+5×5–2+7=30×20×55

=1×1×3125 

∵a0=1

=3125

ncert solutions for class 8 maths chapter 12 exercise 12.2

Scientific notation: 

Very large or very small numbers can be expressed in the simplest form by using scientific notation. In this notation any number can be written as the product of a number ‘x’  such that 1≤ x < 10 and a power of 10.

Example: 

 (i) 100.08 mm =1.0008 ×10² mm

(ii) 0.00000000000942=9.42 ×10⁻¹²

Note: 1) when we move the decimal point towards right side then multiply with negative power (power=number of places we moved) of 10.

Example: 0.00000000000942=9.42 ×10⁻¹²

2) when we move the decimal point towards left side then multiply with positive power (power=number of places we moved) of 10.

Example: 100.08 mm =1.0008 ×10² mm

class 8 maths chapter 12 exercise 12.2 (1)

1. Express the following numbers in standard form.

(i) 0.0000000000085 (ii) 0.00000000000942

(iii) 6020000000000000 (iv) 0.00000000837

(v) 31860000000

Solution to class 8 maths chapter 12 exercise 12.2 (1)

(i) 0.0000000000085 =8.5 ×10⁻¹²

(ii) 0.00000000000942=9.42 ×10⁻¹²

(iii) 6020000000000000=6.02 ×10¹⁵

(iv) 0.00000000837=8.37 ×10⁻⁹

(v) 31860000000 =3.186 ×10¹⁰

class 8 maths chapter 12 exercise 12.2 (2)

2. Express the following numbers in usual form.

(i) 3.06 ×10⁻⁶  (ii) 4.5 ×10⁴  (iii) 3 ×10⁻⁸

(iv) 1.0001 ×10⁹  (v) 5.8 ×10¹²

(vi) 3.61492 ×10⁶

Solution to class 8 maths chapter 12 exercise 12.2 (2)

(i) 3.06 ×10⁻⁶ = ⁽³·⁰⁶⁾/₍₁₀₀₀₀₀₀₎ =0.00000302

(ii) 4.5 ×10⁴ =4.5 ×10000 =45000

(iii) 3 ×10⁻⁸ =³/₍₁₀₀₀₀₀₀₀₀₎ =0.00000003

(iv) 1.0001 ×10⁹ =1.0001 ×1000000000

=1000100000

(v) 5.8 ×10¹² =5.8 ×1000000000000

=5800000000000

(vi) 3.61492 ×10⁶ =3.61492 ×1000000

=3614920

class 8 maths chapter 12 exercise 12.2 (3)

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

(i) 1 micron is equal to ¹/₍₁₀₀₀₀₀₀₎  m.

(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.

(iii) Size of a bacteria is 0.0000005 m.

(iv) Size of a plant cell is 0.00001275 m

(v) Thickness of a thick paper is 0.07 mm 

Solution to class 8 maths chapter 12 exercise 12.2 (3)

(i)1 micron is equal to ¹/₍₁₀₀₀₀₀₀₎ m.=1 ×10⁻⁶ m 

(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb

=1.6 ×10⁻¹⁹ coulomb.

(iii) Size of a bacteria is 0.0000005 m

=5.0 ×10⁻⁷ m 

(iv) Size of a plant cell is 0.00001275 m

= 1.275 ×10⁻⁵ m 

(v) Thickness of a thick paper is 0.07 mm 

= 7.0 ×10⁻² mm

class 8 maths chapter 12 exercise 12.2 (4)

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

Solution to class 8 maths chapter 12 exercise 12.2 (4)

Thickness of one book = 20 mm

Thickness of 5 books = 20 x 5 = 100 mm

Thickness of one paper = 0.016 mm

Thickness of 5 papers = 0.016 x 5 = 0.08 mm

Total thickness of a stack = 100 + 0.08 

=100.08 mm =1.0008 ×10² mm

 

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