Class 7 > Olympiad Test: Surds And Indices

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Olympiad Test: Surds And Indices for Class 7 2023 is part of Class 7 preparation. The Olympiad Test: Surds And Indices questions and answers have been prepared
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Detailed Solution for Olympiad Test: Surds And Indices - Question 1

Here,

Detailed Solution for Olympiad Test: Surds And Indices - Question 2

Given Expression

= x^{(b – c)(b + c – a)}. x^{(c – a)(c + a – b)}

. x^{(a – b)(a + b – c) }

= x^{(b – c)(b + c) – a(b – c)} . x^{(c - a)(c + a) – b(c – a)}

. x^{(a - b)(a + b) - c(a - b)}

. x^{–a(b – c) – b(c – a) – c(a – b)}

= (x^{0} × x^{0})

= (1 × 1) = 1

Olympiad Test: Surds And Indices - Question 3

If m and n are whole numbers such that m^{n} = 121, the value of (m – 1)^{n + 1} is:

Detailed Solution for Olympiad Test: Surds And Indices - Question 3

We know that 11^{2} = 121 = mn (given)

Hence, putting m = 11 and n = 2, in

(m – 1)^{n + 1} = (11 – 1)^{(2 + 1)} = 10^{3} = 1000

Detailed Solution for Olympiad Test: Surds And Indices - Question 4

(256)^{0.16} × (256)^{0.09} = (256)^{(0.16 + 0.09)}

= (256)^{0.25}

= (256)^{(25/100)}

= (256)^{(1/4)}

= (4^{4})^{(1/4)}

= 4^{4(1/4)}

= 4^{1}

= 4

Detailed Solution for Olympiad Test: Surds And Indices - Question 5

(10)1^{50} ÷ (10)^{146}

= 10^{150 – 146}

= 10^{4}

= 10000

Olympiad Test: Surds And Indices - Question 6

If (25)^{7.5} × (5)^{2.5} ÷ (125)^{1.5} = 5^{x} then x = ?

Detailed Solution for Olympiad Test: Surds And Indices - Question 6

Given (25)^{7.5} × (5)^{2.5} ÷ (125)^{1.5} = 5^{x}

Then, (5^{2})^{7.5} × (5)^{2.5} ÷ (5^{3})^{1.5} = 5^{x}

⇒ 5^{(2 × 7.5)} × 5^{2.5} ÷ 5^{(3 × 1.5)} = 5^{x}

⇒ 5^{15} × 5^{2.5} ÷ 5^{4.5} = 5^{x}

⇒ 5^{x} = 5^{(15 + 2.5 – 4.5)}

⇒ 5^{x} = 5^{13}

⇒ x = 13

Detailed Solution for Olympiad Test: Surds And Indices - Question 7

(0.04)^{ – 1.5} = (4/100) ^{–1.5} = (1/25) ^{–3/2}

= (25)^{(3/2)}

= (5^{2})^{(3/2)}

= (5)^{2 × (3/2)}

= 5^{3}

= 125

Olympiad Test: Surds And Indices - Question 8

If 3^{(x – y)} = 27 and 3^{(x + y)} = 243, then x is equal to:

Detailed Solution for Olympiad Test: Surds And Indices - Question 8

3^{x – y} = 27 = 3^{3} ⇒ x – y = 3 ....(i)

3^{x + y} = 243 = 3^{5} ⇒ x + y = 5 .... (ii)

On solving (i) and (ii), we get x = 4.

Olympiad Test: Surds And Indices - Question 9

If 5^{a} = 3125, then the value of 5^{(a – 3)} is:

Detailed Solution for Olympiad Test: Surds And Indices - Question 9

5^{a} = 3125 ⇔ 5^{a} = 5^{5}

⇒ a = 5

∴ 5^{(a – 3)} = 5^{(5 – 3)} = 5^{2} = 25

Olympiad Test: Surds And Indices - Question 10

Given that 10^{0.48 }= x, 10^{0.70} = y and x^{z} = y^{2}, then the value of z is close to:

Detailed Solution for Olympiad Test: Surds And Indices - Question 10

x^{z} = y^{2}

⇒ 10^{(0.48z)} = 10^{(2 × 0.70)} = 10^{1.40}

⇒ 0.48z = 1.40

⇒ z =140/48 = 35/12 = 2.9 (approx.)

Detailed Solution for Olympiad Test: Surds And Indices - Question 11

Let (17)^{3.5} × (17)^{x} = 17^{8}

Then, (17)^{3.5 + x} = 17^{8}

∴ 3.5 + x = 8

⇒ x = (8 – 3.5)

⇒ x = 4.5

Detailed Solution for Olympiad Test: Surds And Indices - Question 12

= 10 × 1.732 = 17.32

Detailed Solution for Olympiad Test: Surds And Indices - Question 13

Detailed Solution for Olympiad Test: Surds And Indices - Question 14

Detailed Solution for Olympiad Test: Surds And Indices - Question 15

Detailed Solution for Olympiad Test: Surds And Indices - Question 16

(18a^{8}b^{6}) ÷ (3a^{2}b^{2}) = 18/3 × a^{8 – 2} × b^{6 – 2}

= 6a^{6}b^{4}

Olympiad Test: Surds And Indices - Question 17

Replace question mark with the suitable answer:

56 - 45- √? = √36

Detailed Solution for Olympiad Test: Surds And Indices - Question 17

56 - 45 -√? = √36

⇒ 11 - √36 =√?

⇒ 11 - 6 =√?

∴ ? = 25

Detailed Solution for Olympiad Test: Surds And Indices - Question 18

Detailed Solution for Olympiad Test: Surds And Indices - Question 19

Detailed Solution for Olympiad Test: Surds And Indices - Question 20

(0.003)^{3} = 0.003 × 0.003 × 0.003

= 0.000000027

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