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

Given:

Simplify the following expression:

\(\dfrac{(8.25)^3-3 \times(8.25)^2 \times 0.25+24.75 \times(0.25)^2-(0.25)^3}{4 \times 4 \times 4}\)

Formula used:

(a - b)³ = a³ - 3a²b + 3ab² - b³

Calculation:

\(\dfrac{(8.25)^3-3 \times(8.25)^2 \times 0.25+24.75 \times(0.25)^2-(0.25)^3}{4 \times 4 \times 4}\)

Let a = 8.25 and b = 0.25

⇒ a³ - 3a²b + 3ab² - b³ = (8.25 - 0.25)³

⇒ (8.25 - 0.25)3 = 8³

⇒ 8³ = 512

⇒ \(\dfrac{512}{4 \times 4 \times 4}\)

⇒ \(\dfrac{512}{64}\)

⇒ 8

∴ The correct answer is option (4).

Given:

(125)^x = 3125

Formula Used:

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Calculation:

We know that 125 = 5³, so we can rewrite the equation:

(5³)^x = 3125

Using the power of a power property, (a^m)ⁿ = a^m×n:

⇒ 5^3x = 3125

We also know that 3125 = 5⁵, so we can rewrite the equation:

⇒ 5^3x = 5⁵

Since the bases are the same, we can equate the exponents:

⇒ 3x = 5

⇒ x = 5/3

The correct answer is option 4.

Given:

(256)0.16 × (256)0.09 = ?

Calculation:

(256)0.16 × (256)0.09 = ?

Let the unknown number be x.

⇒ (256)0.16 × (256)0.09 = x

⇒ (256)⁽0.16 + 0.09) = x 

⇒ (256)^0.25 = x

⇒ x = (4⁴)^1/4 = 4

∴ (256)0.16 × (256)0.09 = 4

Given:

a^maⁿ = a^m+n

Calculation:

4(a3b6)16 × 16(a4b2)2

64 a³b⁶ × 32a⁴b²

64 × 32 × a³⁺⁴  b⁶⁺²

By using the above formula

∴ 4(a3b6)16 × 16(a4b2)2 = 2048 × a⁷b⁸

Formula Used:

a⁰ = 1

a^m × aⁿ = a^{m + n}

Calculation

\(\left(\frac{x}{y}\right)^{\mathrm{a}-\mathrm{b}}\left(\frac{x}{y}\right)^{\mathrm{b}-\mathrm{c}}\left(\frac{x}{y}\right)^{\mathrm{c}-\mathrm{a}} \)

\(⇒ \left(\frac{x}{y}\right)^{\mathrm{a}-\mathrm{b}+\mathrm{b}-\mathrm{c}+\mathrm{c}-\mathrm{a}} \)

\(⇒ (\frac{x}{y})^0 \)

⇒ 1

\(\therefore \left(\frac{x}{y}\right)^{\mathrm{a}-\mathrm{b}}\left(\frac{x}{y}\right)^{\mathrm{b}-\mathrm{c}}\left(\frac{x}{y}\right)^{\mathrm{c}-\mathrm{a}} \) = 1

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