Power Indices And Surds Pdf Model Questions & Answers, Practice Test for ibps po prelims 2024 2025

Answer: (c)

equ $a^x$ = b

Multiplying both sides by y in power

$(a^x)^y = b^y$

⇒$a^{x^y} = c (∵ b^y = c)$

Again multiplying both sides by z in power

$(a^{xy})^z = c^z$

⇒$a^{xyz} = c^z$

But xyz = 1 Given

So a = $c^z$

Answer: (a)

Given, $a^x = b^y = c^z = k$

⇒$a = k^{1/x}, b = k^{1/y} and c = k^{1/z}$

∴ abc = $k^{1/x + 1/y + 1/z}$

⇒ 1 = $k^{1/x + 1/y + 1/z}$ = k (∵ abc =1, given)

On comparing, we get

$1/x + 1/y + 1/z$ = 0

⇒xy + yz + zx = 0

Answer: (d)

Given that, $16 × 8^{n + 2} = 2^m$

⇒$(2)^4 × 2^{3(n + 2)} = 2^m$

⇒$(2)^{(4 + 3n + 6)} = 2^m$

⇒$2^{(3n + 10)} = 2^m$

Here base is same, so

3n + 10 = m

⇒m = 3n + 10

Answer: (c)

${6^2 + 7^2 + 8^2 + 9^2 + 10^2}/{√{7 + 4 √3} - √{4 + 2√3}}$

⇒${330}/{√{(2 + √3)^2 - √{(1 + √3)^2}}}$

⇒${300}/{2 + √3 - 1 - √3}$ = 330

Answer: (b)

$a^x = b^y = c^z$ = k (let)

a = $(k)^{1/x}, b = (k)^{1/y}, c = (k)^{1/z}$

Now, abc = 1⇒$(k)^{1/x + 1/y + 1/z} = (k)^0$

∴ $1/x + 1/y + 1/z$ = 0