Surds and Indices MCQ Quiz - Objective Question with Answer for Surds and Indices - Download Free PDF

Given:

27 × (81)2n+3 − 3m = 0

Concept used:

a^m × aⁿ = a^m+n

Calculation:

27 × (81)2n+3 − 3m = 0

⇒ 3³ × (3⁴)²ⁿ⁺³ - 3^m = 0

⇒  33 × (34)2n+3 = 3^m

⇒ 33 × 3⁸ⁿ⁺¹² = 3^m

⇒ 3^3+8n+12 = 3^m   [am × an = a^m+n]

⇒ 3⁸ⁿ⁺¹⁵ = 3^m

Now, both side as same base

⇒ 8n + 15 = m

⇒ m = 8n + 15

∴ The required value is 8n + 15.

(0.027) × (0.3)⁵ = (0.3)^? × (0.0081)²

⇒ (27/1000) × (3/10)⁵ = (3/10)^? × (81/10000)²

⇒ (3/10)³ × (3/10)⁵ = (3/10)^? × (3/10)⁸

By using the law of indices,

a^m × aⁿ = a^{{m + n}}

⇒ (0.3)⁸ = (0.3)^{? + 8}

⇒ (0.3)^{8 – 8} = (0.3)^?

⇒ (0.3)⁰ = (0.3)^?

∴ ? = 0

Given:

(27)m = (81)n

Formula Used:

If a^m = aⁿ

Then m = n 

(a^m)ⁿ = a^mn

Calculation: 

27 = 3 × 3 ×  3 = 3³

(27)m = 3^3m

81 = 3 × 3 × 3 × 3 = 3⁴

(81)n = 3⁴ⁿ

Now,

⇒ 33m = 34n

⇒ 3m = 4n

⇒ m/n = 4/3

Let assume that m = 4x then n = 3x

⇒ m² = (4x)² = 16x²

⇒ mn = (4x)(3x) = 12x²

m2 : mn = 16x2 ∶ 12x2

∴ m 2 : mn = 4 ∶ 3

To solve questions of this type, follow the laws of “Surds and indices’’ given below-

Laws of Indices:-

\(\begin{array}{l} 1 - :\;{a^m} \times {a^n} = {a^{\left\{ {m + n} \right\}}}\\ 2 - :\;{a^m} \div {a^n} = {a^{\left\{ {m - n} \right\}}}\\ 3 - :\;{\left( {{a^m}} \right)^n} = {a^{mn}}\\ 4 - :{\left( a \right)^{ - m}} = \frac{1}{{{a^m}}}\\ 5 - :{\left( a \right)^{\frac{m}{n}}} = \sqrt[n]{{{a^m}}}\\ 6 - :{\left( a \right)^0} = 1 \end{array}\)

Given,

\(\begin{array}{l} {\left( {\sqrt 5 } \right)^5} \times {25^2} = {5^x} \times 5\sqrt 5 \\ \Rightarrow {5^{\frac{5}{2}}} \times {5^4} = {5^x} \times 5 \times {5^{\frac{1}{2}}} \end{array}\)

⇒ 5^13/2 = 5^{x + 3/2}

⇒ 13/2 = x + 3/2

⇒ x = 5

Given:

(27)m = (81)n

Formula Used:

If a^m = aⁿ

Then m = n 

(a^m)ⁿ = a^mn

Calculation: 

27 = 3 × 3 ×  3 = 3³

(27)m = 3^3m

81 = 3 × 3 × 3 × 3 = 3⁴

(81)n = 3⁴ⁿ

Now,

⇒ 33m = 34n

⇒ 3m = 4n

⇒ m/n = 4/3

Let assume that m = 4x then n = 3x

⇒ m² = (4x)² = 16x²

⇒ mn = (4x)(3x) = 12x²

m2 : mn = 16x2 ∶ 12x2

∴ m 2 : mn = 4 ∶ 3

Formula used:

(a + b)² = a² + b² + 2ab

Calculation:

Given expression is:

\(\sqrt {8\; + \;2\sqrt {15} \;} \)

⇒  \(\sqrt {5\; + \;3\; + \;2\times \sqrt 5 \times \sqrt 3 \;} \)

⇒  \(\sqrt {{{(\sqrt 5 )}^2}\; + \;{{\left( {\sqrt 3 } \right)}^2}\; + \;2 \times \sqrt 5 \times \sqrt 3 \;} \)

⇒  \(\sqrt {{{\left( {\;\sqrt 5 \; + \;\sqrt 3 \;} \right)}^2}\;} \)

⇒  \(\sqrt 5 + \sqrt 3 \)

Concept:

We can find √x using the factorisation method.

Calculation:

√[(10 + √25) (12 - √49)]

⇒ √[(10 + 5)(12 – 7)]

⇒ √(15 × 5)

⇒ √(3 × 5 × 5)

⇒ 5√3

Given,

2³ × 3⁴ × 1080 ÷ 15 = 6^x

⇒ 2³ × 3⁴ × 72 = 6^x

⇒ 2³ × 3⁴ × (2 × 6²) = 6^x

⇒ 2⁴ × 3⁴ × 6² = 6^x

⇒ (2 × 3)⁴ × 6² = 6^x           [∵ x^m × y^m = (xy)^m]

⇒ 6⁴ × 6² = 6^x

⇒ 6^{(4 + 2)} = 6^x

⇒ x = 6

Given:

√3n = 729

Formulas used:

(x^a)^b = x^ab

If x^a = x^b then a = b 

Calculation:

√3n = 729

⇒ √3n = (3²)³

⇒ (3n)^1/2 = (32)3

⇒ (3n)1/2 = 3⁶

⇒ n/2 = 6 

∴  n = 12 

Method I: (3 + 2√5)²

= (3² + (2√5)² + 2 × 3 × 2√5)

= 9 + 20 + 12√5 = 29 + 12√5

On comparing, 29 + 12√5 = 29 + K√5

we get,

K = 12

 Alternate Method 

29 + 12√5 = 29 + K√5

⇒ K√5 = 29 - 29 + 12√5

⇒ K√5 = 12√5

∴ K = 12

Statement I:

2√3 > 3√2
To Check either above given relation is correct or not, simplifying by squaring on both the sides.

⇒ (2√3)² > (3√2)²

⇒ 12 > 18 which is not true, as we know that 18 is greater than 12.

So, the given relation in statement I is not true.

Statement II:
Now, simplifying the values given in statement II

(Note: 2√8 = 2√(4 × 2) = 4√2)

4√2 > 2√8 on taking the square root from the right hand side.

⇒ 4√2 > 2 × 2√2

⇒ 4√2 > 4√2 which is not true, as the the value on left hand side is equal to the value on right hand side. 
So, the given relation in statement II is also not true.

Given:

(3/5)x = 81/625

Calculation:

We know,

3⁴ = 81 and 5⁴ = 625

⇒ (3/5)⁴ = 81/625

(3/5)x = 81/625

∴ On comparing both the equation, we get

x = 4

Now, 

 xx  = 4⁴ = 256

To solve questions of this type, follow the laws of “Surds and indices’’ given below:

Laws of Indices:

1. a^m × aⁿ = a^{{m + n}}

2. a^m ÷ aⁿ = a^{{m - n}}

3. (a^m)ⁿ = a^mn

4. (a)^-m = 1/a^m

5. (a)^m/n = ⁿ√a^m

6. (a)⁰ = 1

\({625^{0.17}} \times {625^{0.08}} = {25^?} \times {25^{- \frac{3}{2}}}\)

\(\Rightarrow {625^{0.17\; + \;0.08}} = {25^{? + (- \frac{3}{2})}}\)

\(\Rightarrow {625^{0.25}} = {25^{? - \frac{3}{2}}}\)

\(\Rightarrow {625^{\frac{1}{4}}} = {\left( {{5^2}} \right)^{? - \frac{3}{2}}}\)

\(\Rightarrow 5 = {5^{2 \times? - 3}}\)

⇒ 2 × ? - 3 = 1

⇒ ? = (1 + 3)/2

∴ ? = 2

Given:

2x = 4y = 8^z

\(\frac{1}{2x}+\frac{1}{4y}+\frac{1}{4z}=4\)

Calculation:

​\(\frac{1}{2x}+\frac{1}{4y}+\frac{1}{4z}=4\)---- (1)

2x = 4y = 8^z

⇒ 2x = 2²y = 2^3z

⇒ x = 2y = 3z

Converting y and z in x

2y = x, so 4y = 2x

3z = x, so 4z = 4x/3

Using the above value in equation (1)

⇒ \(\frac{1}{2x}+\frac{1}{2x}+\frac{3}{4x}=4 \)    

⇒ 7/4x = 4

∴ x = 7/16