surds and indices Model Questions & Answers, Practice Test for ibps po prelims 2023

Answer: (c)

$9^{3.5} × 3^{2.2} ÷ 9^(?) = 9^{1.4}$

⇒$3^{2 × 3.5} × 3^{2.2} ÷ 3^{2(?)} = 3^{2(1.4)}$

⇒$3^{7 + 2.2 – 2(?)} = 3^{2.8}$⇒9.2 – 2(?) = 2.8

⇒– 2(?) = – 6.4⇒? = 3.2

Answer: (c)

${(991)^3+(9)^3}/{(991)^2-991×9+(9)^2}$

As ${a^3+b^3}/{a^2-ab+b^2}={(a+b)(a^2-ab+b^2)}/{(a^2-ab+b^2)}$

=a+b

∴991+9=1000

Answer: (a)

Geven Exp. $=1/{1+a+b^{-1}}+1/{1+b+c^{-1}}+1/{1+c+a^{-1}}$

$=1/{1+a+b^{-1}}+b^{-1}/{1+b^{-1}c^{-1}+b^{-1}}+a/{a+ac+1}$

=$1/{1+a+b^{-1}}+b^{-1}/{1+b^{-1}+a}+a/{a+b^{-1}+1}$

$={1+a+b^{-1}}/{1+a+b^{-1}}=1$

$∵abc=1⇒(bc)^{-1}=a⇒b^{-1}c^{-1}$=a and ac=$b^{-1}$

Answer: (d)

$(3)^{3.5} × (3^2)^{2.2} ÷ 3^3 = (3)^{?}$

⇒$ 3^{3.5 + 4.4 - 3} = 3^{?} ⇒ 3^{4.9} = 3^{?}$ ⇒? = 4.9

Answer: (c)

$ {2^{(n+2)}-2(2^n)}/{2^(2n+2)}={2^n.2^2-2.2^n}/{2^2. 2^{2n}}={2.2^n(2-1)}/{2^2. 2^{2n}$

$=1/{2.2^n}=1/{2^(n+1)}$