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

Answer: (a)

2b = a + c (given)

⇒b + b = a + c

⇒a – b = b – c

$y^2$ = xz (given)

$x^{(b–c)} × y^{c–a} × z^{(a–b)} = x^{a–b} × y^{c–a} × z^{a–b}$ (∵ a – b = b – c)

$(xz)^{a–b} × y^{c–a} (because a^n × b^n = (ab)^n)$

$(y^2)^{a–b} × y^{c–a} (∵ y^2$ = xz)

$y^{2a – 2b + c – a} = y^{a+c–2b} = y^{2b–2b}$ (∵ a + c = 2b)

$y^0$ = 1

Answer: (c)

Given that, $(ab^{-1})^{2x - 1} = (ba^{-1})^{x - 2}$

⇒$(a/b)^{2x - 1} = (b/a)^{x - 2}$

⇒$(a/b)^{2x - 1} (a/b)^{x - 2} = 1$

⇒$(a/b)^{2x - 1 + x - 2 = (a/b)^0$

Base is same

So, 3x - 3 = 0⇒x = 1.

Answer: (a)

$1/{1 + x^{b - a} + x^{c - a}} + 1/{1 + x^{a - b} + x^{c - b}} + 1/{1 + x^{a - c} + x^{b - c}}$

= $1/{1 + {x^b}/{x^a} + {x^c}/{x^a}} + 1/{1 + {x^a}/{x^b} + {x^c}/{x^b}} + 1/{1 + {x^a}/{x^c} + {x^b}/{x^c}}$

= $1/{{x^a + x^b + x^c}/{x^a}} + 1/{{x^b + x^a + x^c}/{x^b}} + 1/{{x^c + x^a + x^b}/{x^c}}$

= ${x^a}/{x^a + x^b + x^c} + {x^b}/{x^a + x^b + x^c} + {x^c}/{x^a + x^b + x^c}$

= ${x^a + x^b + x^c}/{x^a + x^b + x^c} = 1$

Answer: (c)

Let, $(x)^{1/m} = (y)^{1/n} = (z)^{1/p} = k$ (say) ,

⇒$x = k^m, y = k^n and z = k^p$

Now multiply x, y and z

∴ xyz = $k^{m + n + p}$ (∵ xyz =1, given)

⇒1 = $k^{m + n + p} = k^0$

Here, base is same,

m + n + p = 0

Answer: (c)

Let number be x

According to the question

$x^2$ + x = 20

$x^2$ + x – 20 = 0

$x^2$ + 5x – 4x – 20 = 0

x(x + 5) –4 (x + 5) = 0

(x – 4) (x + 5) = 0

x – 4 = 0 ∴ x = 4

x + 5 = 0 ∴ x = –5