Surds and Indices

\text{1. } \left ( 100 + 10 + 1 \right )^0 = \,\, ?

 

A. 0

B. 111

C. 1

D. 10

Ans: C

Explanation:

\left ( 100 + 10 + 1 \right )^0 = \left ( 111 \right )^0 = 1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \left [ a^0 = 1 \right ]

 

 

\text{2. } 49 \times 49 \times 49 \times 49 = 7^?

 

A. 4

B. 6

C. 7

D. 8

Ans:  D

Explanation:

49 \times 49 \times 49 \times 49 = 7^?

 

\left ( 7^2 \times 7^2 \times 7^2 \times 7^2 \right ) = 7^?

 

7^{\left ( 2 + 2 + 2 + 2 \right )} = 7^?

 

7^8 = 7^?

 

\therefore \,\, ? = 8

 

 

\text{3. } \left ( 1000 \right )^7 \div 10^{18} = \,\, ?

 

A. 10

B. 100

C. 1000

D. 10000

Ans: C

Explanation:

\left ( 1000 \right )^7 \div 10^{18} = \frac{\left ( 1000 \right )^7}{10^{18}}

 

 

= \frac{\left ( 10^3 \right )^7}{10^{18}}

 

 

= \frac{10^{21}}{10^{18}}

 

 

= 10^{21} \times 10^{-18}

 

= 10^3

 

= 1000

 

 

\text{4. } \left ( 256 \right )^{0.16} \times \left ( 256 \right )^{0.09} = \,\, ?

 

A. 4

B. 16

C. 125

D. 625

Ans: A

Explanation:

\left ( 256 \right )^{0.16} \times \left ( 256 \right )^{0.09} = \left ( 256 \right )^{\left ( 0.16 \, + \, 0.09 \right )}

 

= \left ( 256 \right )^{0.25}

 

= \left ( 256 \right )^\frac{25}{100}

 

= \left ( 256 \right )^\frac{1}{4}

 

= \left ( 4^4 \right )^\frac{1}{4}

 

= 4

 

 

\text{5. The value of } \left ( 256 \right )^\frac{5}{4} \text \,\text{is}

 

A. 256

B. 512

C. 756

D. 1024

Ans: D

Explanation:

\left ( 256 \right )^\frac{5}{4} = \left ( 4^4 \right )^\frac{5}{4} = 4^{4 \times \frac{5}{4}} = 4^5 = 1024

 

 

\text{6. If }\sqrt{2^n} = 64, \text{then the value of n is}

 

A. 6

B. 8

C. 10

D. 12

Ans: D

Explanation:

\sqrt{2^n} = 64

 

\sqrt{2^n} = 2^6

 

\left ( 2^n \right )^\frac{1}{2} = 2^6 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \left [ \sqrt{a} = a^\frac{1}{2}\right ]

 

2^\frac{n}{2} = 2^6

 

\text{Since bases are equal exponents are equal.}

 

\frac{n}{2} = 6

 

n = 12

 

 

\text{7. If } \, 5^{\left ( x \, + \, 3 \right )} = \left ( 25 \right )^{\left ( 3x \, - \, 4 \right )} \text{then the value of} \, \, x \,\, \text{is}

 

\text{A. } \frac{5}{11}

 

\text{B. } \frac{11}{3}

 

\text{C. } \frac{5}{3}

 

\text{D. } \frac{11}{5}

 

 

Ans: D

Explanation:

5^{\left ( x \, + \, 3 \right )} = \left ( 25 \right )^{\left ( 3x \, - \, 4 \right )}

 

5^{\left ( x \, + \, 3 \right )} = \left ( 5^2 \right )^{\left ( 3x \, - \, 4 \right )}

 

5^{\left ( x \, + \, 3 \right )} = 5^{\left ( 6x \, - \, 8 \right )}

 

\text{Since bases are equal exponents are equal.}

 

x + 3 = 6x - 8

 

5x = 11

 

x = \frac{11}{5}

 

 

\text{8. If } 5^a = 3125 \, \, \text{then the value of } 5^{\left ( a \, - \, 3 \right )} \, \text{is }

 

A. 25

B. 625

C. 125

D. 5

Ans: A

Explanation:

5^a = 3125

 

5^a = 5^5

 

a = 5

 

\therefore 5^{\left ( a \, - \, 3 \right )} = 5^{\left ( 5 \, - \, 3 \right )} = 5^2 = 25

 

 

\text{9. If } \frac{9^n \times 3^5 \times \left ( 27 \right )^3}{3 \times \left ( 81 \right )^4} = 27 \,\, \text{then the value of n is}

 

 

A. 1

B. 2

C. 3

D. 4

Ans: C

Explanation:

\frac{9^n \times 3^5 \times \left ( 27 \right )^3}{3 \times \left ( 81 \right )^4} = 27

 

 

\frac{\left ( 3^2 \right )^n \times 3^5 \times \left ( 3^3 \right )^3}{3 \times \left ( 3^4 \right )^4} = 3^3

 

 

\frac{3^{2n} \times 3^5 \times 3^9}{3 \times 3^{16}} = 3^3

 

 

\frac{3^{2n} \times 3^{14}}{3^{17}} = 3^3

 

 

3^{2n} \times 3^{14} \times 3 ^{-17} = 3^3

 

3 \, ^{\left ( 2n \, + \, 14 \, - \, 17 \right )} = 3^3

 

3 \, ^{\left ( 2n \, - \, 3 \right )} =3^3

 

\text{Since bases are equal exponents are equal.}

 

2n - 3 = 3

 

2n = 6

 

n = 3

 

 

\text{10. The value of } \frac{1}{\left ( 216 \right )^{-\frac{2}{3}}} + \frac{1}{\left ( 256 \right )^{-\frac{3}{4}}} + \frac{1}{\left ( 32 \right )^{-\frac{1}{5}}} \,\, \text{is}

 

 

A. 92

B. 96

C. 98

D. 102

Ans: D

Explanation:

\frac{1}{\left ( 216 \right )^-\frac{2}{3}} + \frac{1}{\left ( 256 \right )^-\frac{3}{4}} + \frac{1}{\left ( 32 \right )^-\frac{1}{5}} = \frac{1}{\left ( 6^3 \right )^-\frac{2}{3}} + \frac{1}{\left ( 4^4 \right )^-\frac{3}{4}} + \frac{1}{\left ( 2^5 \right )^-\frac{1}{5}}

 

 

= \frac{1}{6\, ^{3 \, \times \, \left ( -\frac{2}{3} \right )}} + \frac{1}{4 \, ^{4 \, \times \, \left ( -\frac{3}{4} \right )}} + \frac{1}{2 \, ^{5 \, \times \, \left ( -\frac{1}{5} \right )}}

 

 

= \frac{1}{6^{-2}} + \frac{1}{4^{-3}} + \frac{1}{2^{-1}}

 

 

= 6^2 + 4^3 + 2^1

 

= 36 + 64 + 2

 

= 102