Olympiad Test: Surds And Indices - Free MCQ with solutions Class 7 Mathematics

Here,

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)}
= (x⁰ × x⁰)
= (1 × 1) = 1

We know that 11² = 121 = mn (given)
Hence, putting m = 11 and n = 2, in
(m – 1)^{n + 1} = (11 – 1)^{(2 + 1)} = 10³ = 1000

(256)^0.16 × (256)^0.09 = (256)^{(0.16 + 0.09)}
= (256)^0.25
= (256)^(25/100)
= (256)^(1/4)
= (4⁴)^(1/4)
= 4^4(1/4)
= 4¹
= 4

(10)1⁵⁰ ÷ (10)¹⁴⁶ 
= 10^{150 – 146}
= 10⁴
= 10000

Given (25)^7.5 × (5)^2.5 ÷ (125)^1.5 = 5^x
Then, (5²)^7.5 × (5)^2.5 ÷ (5³)^1.5 = 5^x
⇒ 5^{(2 × 7.5)} × 5^2.5 ÷ 5^{(3 × 1.5)} = 5^x
⇒ 5¹⁵ × 5^2.5 ÷ 5^4.5 = 5^x
⇒ 5^x = 5^{(15 + 2.5 – 4.5)}
⇒ 5^x = 5¹³
⇒ x = 13

(0.04) ^{– 1.5} = (4/100) ^–1.5 = (1/25) ^–3/2
= (25)^(3/2)
= (5²)^(3/2)
= (5)^{2 × (3/2)}
= 5³
= 125

3^{x – y} = 27 = 3³ ⇒ x – y = 3 ....(i)
3^{x + y} = 243 = 3⁵ ⇒ x + y = 5 .... (ii)
On solving (i) and (ii), we get x = 4.

5^a = 3125 ⇔ 5^a = 5⁵
⇒ a = 5
∴ 5^{(a – 3)} = 5^{(5 – 3)} = 5² = 25

x^z = y²
⇒ 10^(0.48z) = 10^{(2 × 0.70)} = 10^1.40
⇒ 0.48z = 1.40
⇒ z =140/48 = 35/12 = 2.9 (approx.)

Let (17)^3.5 × (17)^x = 17⁸
Then, (17)^{3.5 + x} = 17⁸
∴ 3.5 + x = 8
⇒ x = (8 – 3.5)
⇒ x = 4.5

= 10 × 1.732 = 17.32

(18a⁸b⁶) ÷ (3a²b²) = 18/3 × a^{8 – 2} × b^{6 – 2}
= 6a⁶b⁴

56 - 45 -√? = √36
⇒ 11 - √36 =√?
⇒ 11 - 6 =√?
∴ ? = 25

(0.003)³ = 0.003 × 0.003 × 0.003
= 0.000000027