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

Answer: (c)

$({x^5 - 1}/{x - 1}) + (x^3 + 2x^2 + x)$

${x^5 - 1}/{x - 1} + x^3 + 2x^2 + x$

= ${x^5 - 1 + x^4 + 2x^3 + x^2 - x^3 - 2x^2 - x}/{x - 1}$

= ${x^5 + x^4 + x^3 - x^2 - x - 1}/{x - 1}$

= ${x^3 (x^2 + x + 1) -1 (x^2 + x + 1)}/{(x - 1)}$

= ${(x^2 + x + 1)(x^3 - 1)}/{x - 1}$

= ${(x^2 + x + 1)(x - 1)(x^2 + x + 1)}/{(x - 1)}$

= $(x^2 + x + 1)^2$

∴ Square root of $[{x^5 - 1}/{x - 1} + x^3 + 2x^2 + x]$

= $√{(x^2 + x + 1)^2} = x^2 + x + 1$

Answer: (d)

According to the rule, the numbers having 2, 3, 7 and 8 at their unit's place are not perfect squares.

Therefore, here, we can see that options 1, 3 and 4 have 2, 3 and 7, respectively at their unit's place and so, consequently, these are not perfect squares.

Answer: (d)

${13^5 + 14^5 + 15^5 + 16^5}/{29} = {13^5 + 16^5}/{29} + {14^5 + 15^5}/{29}$

⇒${13^5 + 16^5}/{13 + 16} + {15^5 + 14^5}/{15 + 14}$

no remainder

as $x^n + a^n$ when n is odd is completely devisibel by x + a

Answer: (c)

$√{1 + 1/{1^2} + 1/{2^2}} + √{1 + 1/{2^2} + 1/{3^2}} +.....√{1 + 1/{2007^2} + 1/{2008^2}}$

⇒$√{({3^2}/2)} + √{(7/6)^2}.......√{{(2007 × 2008 + 1)^2}/{(2007)^2 (2008)^2}}$

$3/2 + 7/6 + {13}/{12}....{(2007 × 2008 + 1)}/{(2007)(2008)}$

⇒$1 + 1/{1 × 2} + 1 + 1/{2 × 3} + 1 + 1/{3 × 4}......1 + 1/{2007 × 2008}$

⇒2007 + 1 - $1/2 + 1/2 - 1/3 + 1/3 - 1/4.....1/{2007} - 1/{2008}$

⇒2008 - $1/{2008}$

Answer: (c)

Given that, $p^x = r^y = m and r^w = p^z = n$

Now, $p^x = r^y$

⇒$(p^x)^w = (r^y)^w⇒p^{xw} = r^{yw}$

⇒$p^{xw} = (r^w)^y$ ...(i)

Here, $r^w = p^z$ put in eq. (i)

then, $p^{xw} = (p^z)^y⇒p^{xw} = p^{zy}$ (Base is same)

∴ xw = zy