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**1. Express each of the following as a rational number of the form p/q, where p and q are integers and q ≠ 0:**

**(i) 2 ^{-3}**

**(ii) (-4) ^{-2}**

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

**(iv) (1/2) ^{-5}**

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

**Solution:**

**(i)** 2^{-3} = 1/2^{3} = 1/2×2×2 = 1/8 (we know that a^{-n} = 1/a^{n})

**(ii)** (-4)^{-2} = 1/-4^{2} = 1/-4×-4 = 1/16 (we know that a^{-n} = 1/a^{n})

**(iii)** 1/(3)^{-2} = 3^{2} = 3×3 = 9 (we know that 1/a^{-n} = a^{n})

**(iv)** (1/2)^{-5} = 2^{5 }/ 1^{5} = 2×2×2×2×2 = 32 (we know that a^{-n} = 1/a^{n})

**(v)** (2/3)^{-2} = 3^{2} / 2^{2} = 3×3 / 2×2 = 9/4 (we know that a^{-n} = 1/a^{n})

**2. Find the values of each of the following:**

**(i) 3 ^{-1} + 4^{-1}**

**(ii) (3 ^{0} + 4^{-1}) × 2^{2}**

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

**(iv) ((1/3) ^{-1} – (1/4)^{-1})^{-1}**

**Solution:**

**(i)** 3^{-1} + 4^{-1}

1/3 + 1/4 (we know that a^{-n} = 1/a^{n})

LCM of 3 and 4 is 12

(1×4 + 1×3)/12

(4+3)/12

7/12

**(ii)** (3^{0} + 4^{-1}) × 2^{2}

(1 + 1/4) × 4 (we know that a^{-n} = 1/a^{n}, a^{0} = 1)

LCM of 1 and 4 is 4

(1×4 + 1×1)/4 × 4

(4+1)/4 × 4

5/4 × 4

5

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

(We know that a^{0} = 1)

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

**(iv)** ((1/3)^{-1} – (1/4)^{-1})^{-1}

(3^{1} – 4^{1})^{-1} (we know that 1/a^{-n} = a^{n}, a^{-n} = 1/a^{n})

(3-4)^{-1}

(-1)^{-1}

1/-1 = -1

**3. Find the values of each of the following:**

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

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

**(iii) (2 ^{-1} × 4^{-1}) ÷ 2^{-2}**

**(iv) (5 ^{-1} × 2^{-1}) ÷ 6^{-1}**

**Solution:**

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

2^{1} + 3^{1} + 4^{1} (we know that 1/a^{-n} = a^{n})

2+3+4 = 9

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

2^{2} + 3^{2} + 4^{2} (we know that 1/a^{-n} = a^{n})

2×2 + 3×3 + 4×4

4+9+16 = 29

**(iii)** (2^{-1} × 4^{-1}) ÷ 2^{-2}

(1/2^{1} × 1/4^{1}) / (1/2^{2}) (we know that a^{-n} = 1/a^{n})

(1/2 × 1/4) × 4/1 (we know that 1/a ÷ 1/b = 1/a × b/1)

1/2

**(iv)** (5^{-1} × 2^{-1}) ÷ 6^{-1}

(1/5^{1} × 1/2^{1}) / (1/6^{1}) (we know that a^{-n} = 1/a^{n})

(1/5 × 1/2) × 6/1 (we know that 1/a ÷ 1/b = 1/a × b/1)

3/5

**4. Simplify:**

**(i) (4 ^{-1} × 3^{-1})^{2}**

**(ii) (5 ^{-1} ÷ 6^{-1})^{3}**

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

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

**Solution:**

**(i)** (4^{-1} × 3^{-1})^{2} (we know that a^{-n} = 1/a^{n})

(1/4 × 1/3)^{2}

(1/12)^{2}

(1×1 / 12×12)

1/144

**(ii)** (5^{-1} ÷ 6^{-1})^{3}

((1/5) / (1/6))^{3} (we know that a^{-n} = 1/a^{n})

((1/5) × 6)^{3} (we know that 1/a ÷ 1/b = 1/a × b/1)

(6/5)^{3}

6×6×6 / 5×5×5

216/125

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

(1/2 + 1/3)^{-1} (we know that a^{-n} = 1/a^{n})

LCM of 2 and 3 is 6

((1×3 + 1×2)/6)^{-1}

(5/6)^{-1}

6/5

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

(1/3 × 1/4)^{-1} × 1/5 (we know that a^{-n} = 1/a^{n})

(1/12)^{-1} × 1/5

12/5

**5. Simplify:**

**(i) (3 ^{2} + 2^{2}) × (1/2)^{3}**

**(ii) (3 ^{2} – 2^{2}) × (2/3)^{-3}**

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

**(iv) (2 ^{2} + 3^{2} – 4^{2}) ÷ (3/2)^{2}**

**Solution:**

**(i)** (3^{2} + 2^{2}) × (1/2)^{3}

(9 + 4) × 1/8 = 13/8

**(ii)** (3^{2} – 2^{2}) × (2/3)^{-3}

(9-4) × (3/2)^{3}

5 × (27/8)

135/8

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

(3^{3 }– 2^{3}) ÷ 4^{3} (we know that 1/a^{-n} = a^{n})

(27-8) ÷ 64

19 × 1/64 (we know that 1/a ÷ 1/b = 1/a × b/1)

19/64

**(iv)** (2^{2} + 3^{2} – 4^{2}) ÷ (3/2)^{2}

(4 + 9 – 16) ÷ (9/4)

(-3) × 4/9 (we know that 1/a ÷ 1/b = 1/a × b/1)

-4/3

**6. By what number should 5 ^{-1} be multiplied so that the product may be equal to (-7)^{-1}?**

**Solution:**

Let us consider a number x

So, 5^{-1} × x = (-7)^{-1}

1/5 × x = 1/-7

x = (-1/7) / (1/5)

= (-1/7) × (5/1)

= -5/7

**7. By what number should (1/2) ^{-1} be multiplied so that the product may be equal to (-4/7)^{-1}?**

**Solution:**

Let us consider a number x

So, (1/2)^{-1} × x = (-4/7)^{-1}

1/(1/2) × x = 1/(-4/7)

x = (-7/4) / (2/1)

= (-7/4) × (1/2)

= -7/8

**8. By what number should (-15) ^{-1} be divided so that the quotient may be equal to (-5)^{-1}?**

**Solution:**

Let us consider a number x

So, (-15)^{-1} ÷ x = (-5)^{-1}

1/-15 × 1/x = 1/-5

1/x = (1×-15)/-5

1/x = 3

x = 1/3

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