power indices and surds Model Questions & Answers, Practice Test for ssc mts tier 1 2024

Answer: (c)

$√2 = 2^{1/2} = (2^6)^{1/{12}} = (64)^{1/{12}}$

$^4$$√6 = 6^{1/4} = (6^3)^{1/{12}} = (216)^{1/{12}}$

$^3$$√4 = 4^{1/3} = (4^4)^{1/{12}} = (256)^{1/{12}}$

so, correct order is

⇒$(264)^{1/{12}} < (216)^{1/{12}} < (256)^{1/{12}}$

⇒$ √2 < ^4√6 < ^3√4 $

So, option (a) is correct.

Answer: (a)

($n^2 + 48) = x^2$

48 = $x^2 – n^2$

48 = (x – n) (x + n)

So the possible number of pairs of (x – n) and (x + n) are (1,48), (2,24), (3,16), (4,12), (6,8)

On solving the above pairs for (x – n) and (x + n), we get the integer values of n and x as

n = 1, x = 7

n = 4, x = 8

n = 11, x = 13

So, the total possible values of n are 3.

Answer: (a)

let two no. 8 be x and y

$x^2 + y^2$ = 208

$x^2$ + 18x = 208 ∴ $y^2$ = 18x

$x^2$ + 18x – 208 = 0

$x^2$ + 26x – 8x – 208 = 0

(x – 8) (x + 26) = 0

x = 8

$y^2$ = 18 × 8

$y^2$ = 144

y = 12

y – x = 12 – 8 = 4

Answer: (b)

Square of $2 + √2 = (2 + √2)^2$

= $(2)^2 + (√2)^2 + 2.2.√2 = 4 + 2 + 4√2$

= 6 + 4$√2$

6 + 4$√2$ is an irrational number.

Answer: (d)

Given, 27 × $(81)^{2n + 3} - 3^m$ = 0

⇒$3^3 × (3)^{8n + 12} = 3^m$

⇒$3^{8n + 15} = 3^m$⇒m = 8n + 15

(on comparing)